Online Instructor’s Manual for
Electronic Devices and Circuit Theory Eleventh Edition Robert L. Boylestad Louis Nashelsky
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10 9 8 7 6 5 4 3 2 1 ISBN10: 0-13-278373-8 ISBN13: 978-0-13-278373-6
Contents Solutions to Problems in Text
Solutions for Laboratory Manual
iii
1
209
iv
Chapter 1 1.
Copper has 20 orbiting electrons with only one electron in the outermost shell. The fact that the outermost shell with its 29th electron is incomplete (subshell can contain 2 electrons) and distant from the nucleus reveals that this electron is loosely bound to its parent atom. The application of an external electric field of the correct polarity can easily draw this loosely bound electron from its atomic structure for conduction. Both intrinsic silicon and germanium have complete outer shells due to the sharing (covalent bonding) of electrons between atoms. Electrons that are part of a complete shell structure require increased levels of applied attractive forces to be removed from their parent atom.
2.
Intrinsic material: an intrinsic semiconductor is one that has been refined to be as pure as physically possible. That is, one with the fewest possible number of impurities. Negative temperature coefficient: materials with negative temperature coefficients have decreasing resistance levels as the temperature increases. Covalent bonding: covalent bonding is the sharing of electrons between neighboring atoms to form complete outermost shells and a more stable lattice structure.
3.
4.
a.
W = QV = (12 µC)(6 V) = 72 μJ
b.
1 eV = 2.625 × 1014 eV 72 × 106 J = 19 1.6 10 J
5.
48 eV = 48(1.6 1019 J) = 76.8 1019 J W 76.8 1019 J Q= = = 2.40 1018 C 3.2 V V 6.4 1019 C is the charge associated with 4 electrons.
6.
GaP ZnS
7.
An n-type semiconductor material has an excess of electrons for conduction established by doping an intrinsic material with donor atoms having more valence electrons than needed to establish the covalent bonding. The majority carrier is the electron while the minority carrier is the hole.
Gallium Phosphide Zinc Sulfide
Eg = 2.24 eV Eg = 3.67 eV
A p-type semiconductor material is formed by doping an intrinsic material with acceptor atoms having an insufficient number of electrons in the valence shell to complete the covalent bonding thereby creating a hole in the covalent structure. The majority carrier is the hole while the minority carrier is the electron. 8.
A donor atom has five electrons in its outermost valence shell while an acceptor atom has only 3 electrons in the valence shell.
1
9.
Majority carriers are those carriers of a material that far exceed the number of any other carriers in the material. Minority carriers are those carriers of a material that are less in number than any other carrier of the material.
10.
Same basic appearance as Fig. 1.7 since arsenic also has 5 valence electrons (pentavalent).
11.
Same basic appearance as Fig. 1.9 since boron also has 3 valence electrons (trivalent).
12.
13.
14.
For forward bias, the positive potential is applied to the p-type material and the negative potential to the n-type material.
15.
a.
b.
kTK (1.38 1023 J/K)(20C 273C) VT q 1.6 1019 C 25.27 mV I D I s (eVD / nVT 1) 40 nA(e(0.5 V) / (2)(25.27mV) 1) 40 nA(e9.89 1) 0.789 mA
16.
k (TK ) (1.38 1023 J/K)(100C 273C) q 1.6 1019 32.17 mV
a.
VT
b.
I D I s (eVD / nVT 1) 40 nA(e(0.5 V) / (2)(32.17 mV) 1) 40 nA(e7.77 1) 11.84 mA
17.
a.
TK = 20 + 273 = 293 kT (1.38 1023 J/K)(293) VT K q 1.6 1019 C 25.27 mV
b.
I D I s (eVD / nVT 1)
0.1 A e 10/(2)(25.27 mV) 1 = 0.1 A(e197.86 1) 0.1 A
2
18.
kTK (1.38 1023 J/K)(25C 273C) VT q 1.6 1019 C =25.70 mV ID = I s (eVD / nVT 1)
8mA = I s (e(0.5V) / (1)(25.70 mV) 1) I s (28 108 ) 8 mA Is = 28.57 pA 2.8 108 19.
I D I s (eVD / nVT 1) 6 mA 1 nA(eVD /(1)(26 mV) 1) 6 106 eVD / 26 mV 1 eVD / 26 mV 6 106 1 6 106 log e eVD / 26 mV log e 6 106 VD = 15.61 26 mV VD = 15.61(26 mV) 0.41 V
20.
(a) x 0 1 2 3 4 5
y = ex 1 2.7182 7.389 20.086 54.6 148.4
(b) y = e0 = 1 (c) For x = 0, e0 = 1 and I = Is(1 1) = 0 mA 21.
T = 20C: T = 30C: T = 40C: T = 50C: T = 60C:
Is = 0.1 A Is = 2(0.1 A) = 0.2 A (Doubles every 10C rise in temperature) Is = 2(0.2 A) = 0.4 A Is = 2(0.4 A) = 0.8 A Is = 2(0.8 A) = 1.6 A
1.6 A: 0.1 A 16:1 increase due to rise in temperature of 40C. 22.
For most applications the silicon diode is the device of choice due to its higher temperature capability. Ge typically has a working limit of about 85 degrees centigrade while Si can be used at temperatures approaching 200 degrees centigrade. Silicon diodes also have a higher current handling capability. Germanium diodes are the better device for some RF small signal applications, where the smaller threshold voltage may prove advantageous.
3
23.
From 1.19: VF @ 10 mA Is
75C 1.1 V
25C 0.85 V
100C 1.0 V
200C 0.6 V
0.01 pA
1 pA
1 A
1.05 A
VF decreased with increase in temperature 1.7 V: 0.65 V 2.6:1 Is increased with increase in temperature 2 A: 0.1 A = 20:1 24.
An “ideal” device or system is one that has the characteristics we would prefer to have when using a device or system in a practical application. Usually, however, technology only permits a close replica of the desired characteristics. The “ideal” characteristics provide an excellent basis for comparison with the actual device characteristics permitting an estimate of how well the device or system will perform. On occasion, the “ideal” device or system can be assumed to obtain a good estimate of the overall response of the design. When assuming an “ideal” device or system there is no regard for component or manufacturing tolerances or any variation from device to device of a particular lot.
25.
In the forward-bias region the 0 V drop across the diode at any level of current results in a resistance level of zero ohms – the “on” state – conduction is established. In the reverse-bias region the zero current level at any reverse-bias voltage assures a very high resistance level the open circuit or “off” state conduction is interrupted.
26.
The most important difference between the characteristics of a diode and a simple switch is that the switch, being mechanical, is capable of conducting current in either direction while the diode only allows charge to flow through the element in one direction (specifically the direction defined by the arrow of the symbol using conventional current flow).
27.
VD 0.7 V, ID = 4 mA V 0.7 V RDC = D = 175 I D 4 mA
28.
At ID = 15 mA, VD = 0.82 V V 0.82 V RDC = D = 54.67 I D 15 mA As the forward diode current increases, the static resistance decreases.
4
29.
VD = 10 V, ID = Is = 0.1 A V 10 V RDC = D = 100 M I D 0.1 A VD = 30 V, ID = Is= 0.1 A V 30 V = 300 M RDC = D I D 0.1 A As the reverse voltage increases, the reverse resistance increases directly (since the diode leakage current remains constant).
30.
ID = 10 mA, VD = 0.76 V V 0.76 V RDC = D = 76 I D 10 mA Vd 0.79 V 0.76 V 0.03 V =3 15 mA 5 mA 10 mA I d RDC >> rd
rd =
31.
Vd 0.79 V 0.76 V 0.03 V =3 15 mA 5 mA 10 mA I d 26 mV 26 mV (b) rd = = 2.6 10 mA ID
(a) rd =
(c) quite close 32.
33.
ID = 1 mA, rd =
Vd 0.72 V 0.61 V = 55 2 mA 0 mA I d
ID = 15 mA, rd =
Vd 0.8 V 0.78 V =2 20 mA 10 mA I d
26 mV ID = 1 mA, rd = 2 = 2(26 ) = 52 vs 55 (#30) I D
ID = 15 mA, rd =
34.
35.
rav =
26 mV 26 mV = 1.73 vs 2 (#30) 15 mA ID
Vd 0.9 V 0.6 V = 24.4 I d 13.5 mA 1.2 mA
Vd 0.8 V 0.7 V 0.09 V = 22.5 I d 7 mA 3 mA 4 mA (relatively close to average value of 24.4 (#32))
rd =
5
Vd 0.9 V 0.7 V 0.2 V = 14.29 I d 14 mA 0 mA 14 mA
36.
rav =
37.
Using the best approximation to the curve beyond VD = 0.7 V: Vd 0.8 V 0.7 V 0.1 V =4 rav = 25 mA 0 mA 25 mA I d
38.
Germanium: 0.42 V 0.3 V rav 4 30 mA 0 mA
GaAa: rav
39.
1.32 V 1.2 V 4 30 mA 0 mA
(a) VR = 25 V: CT 0.75 pF VR = 10 V: CT 1.25 pF CT 1.25 pF 0.75 pF 0.5 pF = 0.033 pF/V 10 V 25 V 15 V VR
(b) VR = 10 V: CT 1.25 pF VR = 1 V: CT 3 pF CT 1.25 pF 3 pF 1.75 pF = 0.194 pF/V 10 V 1 V 9V VR
(c) 0.194 pF/V: 0.033 pF/V = 5.88:1 6:1 Increased sensitivity near VD = 0 V 40.
From Fig. 1.33 VD = 0 V, CD = 3.3 pF VD = 0.25 V, CD = 9 pF
6
41.
The transition capacitance is due to the depletion region acting like a dielectric in the reversebias region, while the diffusion capacitance is determined by the rate of charge injection into the region just outside the depletion boundaries of a forward-biased device. Both capacitances are present in both the reverse- and forward-bias directions, but the transition capacitance is the dominant effect for reverse-biased diodes and the diffusion capacitance is the dominant effect for forward-biased conditions.
42.
VD = 0.2 V, CD = 7.3 pF 1 1 = 3.64 k XC = 2 fC 2 (6 MHz)(7.3 pF) VD = 20 V, CT = 0.9 pF 1 1 = 29.47 k XC = 2 fC 2 (6 MHz)(0.9 pF)
43.
CT
C (0)
1 VR / VK
n
8 pF
1 5 V / 0.7 V
1/2
8pF 8 pF 8 pF 1/2 (1+7.14) 8.14 2.85 2.81 pF
44.
CT
C (0)
1 V /V R
4 pF =
k
10 pF
1 VR /0.7 V
1/3
(1 VR /0.7 V)1/3 2.5
1 VR /0.7 V (2.5)3 15.63 VR /0.7 V 15.63 1 14.63 VR (0.7)(14.63) 10.24 V
45.
10 V = 1 mA 10 k ts + tt = trr = 9 ns ts + 2ts = 9 ns ts = 3 ns tt = 2ts = 6 ns
If =
7
46.
47.
a.
As the magnitude of the reverse-bias potential increases, the capacitance drops rapidly from a level of about 5 pF with no bias. For reverse-bias potentials in excess of 10 V the capacitance levels off at about 1.5 pF.
b.
6 pF
c.
At VR 4 V, CT 2 pF C (0) CT n 1 VR / Vk 2 pF
6 pF
1 4V/0.7 V
n
1 4 V 0.7 V 3 n
(6.71) n 3 n log10 6.71 log10 3 n(0.827) 0.477 0.477 n 0.58 0.827
48.
At VD = 25 V, ID = 0.2 nA and at VD = 100 V, ID 0.45 nA. Although the change in IR is more than 100%, the level of IR and the resulting change is relatively small for most applications.
49.
TA = 25C, IR = 0.5 nA TA = 100C, IR = 60 nA The change is significant. 60 nA: 0.5 nA = 120:1 Yes, at 95C IR would increase to 64 nA starting with 0.5 nA (at 25C) (and double the level every 10C).
50.
IF = 0.1 mA: rd 700 IF = 1.5 mA: rd 70 IF = 20 mA: rd 6
Log scale:
The results support the fact that the dynamic or ac resistance decreases rapidly with increasing current levels.
8
51.
T = 25C: Pmax = 500 mW T = 100C: Pmax = 260 mW Pmax = VFIF P 500 mW IF = max = 714.29 mA 0.7 V VF Pmax 260 mW = 371.43 mA VF 0.7 V 714.29 mA: 371.43 mA = 1.92:1 2:1
IF =
52.
Using the bottom right graph of Fig. 1.37: IF = 500 mA @ T = 25C At IF = 250 mA, T 104C
53.
VZ 100% VZ (T1 T0 ) 0.75 V 0.072 = 100 10 V(T1 25) 7.5 0.072 = T1 25 7.5 = 104.17 T1 25 = 0.072 T1 = 104.17 + 25 = 129.17
54.
TC = +0.072% =
55.
TC =
56.
VZ 100% VZ (T1 T0 ) (5 V 4.8 V) = 100% = 0.053%/C 5 V(100 25)
(20 V 6.8 V) 100% = 77% (24 V 6.8 V) The 20 V Zener is therefore 77% of the distance between 6.8 V and 24 V measured from the 6.8 V characteristic.
9
At IZ = 0.1 mA, TC 0.06%/C (5 V 3.6 V) 100% = 44% (6.8 V 3.6 V) The 5 V Zener is therefore 44% of the distance between 3.6 V and 6.8 V measured from the 3.6 V characteristic. At IZ = 0.1 mA, TC 0.025%/C 57.
58.
24 V Zener: 0.2 mA: 400 1 mA: 95 10 mA: 13 The steeper the curve (higher dI/dV) the less the dynamic resistance.
59.
VK 2.0 V, which is considerably higher than germanium ( 0.3 V) or silicon ( 0.7 V). For germanium it is a 6.7:1 ratio, and for silicon a 2.86:1 ratio.
60.
1.6 1019 J 19 0.67 eV 1.072 10 J 1 eV Eg
hc
hc (6.626 1034 Js)(3 108 ) m/s Eg 1.072 1019 J
1850 nm Very low energy level.
61.
Fig. 1.53 (f) IF 13 mA Fig. 1.53 (e) VF 2.3 V
62.
(a) Relative efficiency @ 5 mA 0.82 @ 10 mA 1.02 1.02 0.82 100% = 24.4% increase 0.82 1.02 = 1.24 ratio: 0.82 (b) Relative efficiency @ 30 mA 1.38 @ 35 mA 1.42 1.42 1.38 100% = 2.9% increase 1.38 1.42 ratio: = 1.03 1.38
10
(c) For currents greater than about 30 mA the percent increase is significantly less than for increasing currents of lesser magnitude. 63.
(a)
0.75 = 0.25 3.0 From Fig. 1.53 (i) 75
(b) 0.5 = 40 64.
For the high-efficiency red unit of Fig. 1.53:
0.2 mA 20 mA C x 20 mA = 100C x= 0.2 mA/ C
11
Chapter 2 1.
The load line will intersect at ID = (a)
E 12 V = 16 mA and VD = 12 V. R 750
VDQ 0.85 V I DQ 15 mA
VR = E VDQ = 12 V 0.85 V = 11.15 V (b) VDQ 0.7 V I DQ 15 mA
VR = E VDQ = 12 V 0.7 V = 11.3 V (c)
VDQ 0 V I DQ 16 mA
VR = E VDQ = 12 V 0 V = 12 V For (a) and (b), levels of VDQ and I DQ are quite close. Levels of part (c) are reasonably close but as expected due to level of applied voltage E. 2.
E 6V = 30 mA R 0.2 k The load line extends from ID = 30 mA to VD = 6 V. VDQ 0.95 V, I DQ 25.3 mA
(a) ID =
E 6V = 12.77 mA R 0.47 k The load line extends from ID = 12.77 mA to VD = 6 V. VDQ 0.8 V, I DQ 11 mA
(b) ID =
E 6V = 8.82 mA R 0.68 k The load line extends from ID = 8.82 mA to VD = 6 V. VDQ 0.78 V, I DQ 78 mA
(c) ID =
The resulting values of VDQ are quite close, while I DQ extends from 7.8 mA to 25.3 mA. 3.
Load line through I DQ = 10 mA of characteristics and VD = 7 V will intersect ID axis as 11.3 mA. E 7V ID = 11.3 mA = R R 7V = 619.47 k 0.62 kΩ standard resistor with R = 11.3 mA
12
4.
E VD 30 V 0.7 V = 19.53 mA 1.5 k R VD = 0.7 V, VR = E VD = 30 V 0.7 V = 29.3 V
(a) ID = IR =
E VD 30 V 0 V = 20 mA 1.5 k R VD = 0 V, VR = 30 V
(b) ID =
Yes, since E VT the levels of ID and VR are quite close. 5.
(a) I = 0 mA; diode reverse-biased. (b) V20 = 20 V 0.7 V = 19.3 V (Kirchhoff’s voltage law) 19.3 V = 0.965 A I(20 Ω) = 20 V(10 Ω) = 20 V 0.7 V = 19.3 V 19.3 V I(10 Ω) = = 1.93 A 10 I = I(10 Ω) + I(20 Ω) = 2.895 A 10 V = 1 A; center branch open (c) I = 10
6.
(a) Diode forward-biased, Kirchhoff’s voltage law (CW): 5 V + 0.7 V Vo = 0 Vo = 4.3 V Vo 4.3 V IR = ID = = 1.955 mA 2.2 k R (b) Diode forward-biased, 8 V + 6 V 0.7 V = 2.25 mA ID = 1.2 k 4.7 k Vo = 8 V (2.25 mA)(1.2 kΩ) = 5.3 V
7.
10 k(12 V 0.7 V 0.3 V) = 9.17 V 2 k 10 k (b) Vo = 10 V
(a) Vo =
13
8.
(a) Determine the Thevenin equivalent circuit for the 10 mA source and 2.2 k resistor. ETh = IR = (10 mA)(2.2 k) = 22 V RTh = 2. 2k Diode forward-biased 22 V 0.7 V = 4.84 mA ID = 2.2 k 2.2 k Vo = ID(1.2 k) = (4.84 mA)(1.2 k) = 5.81 V (b) Diode forward-biased 20 V + 20 V 0.7 V = 5.78 mA ID = 6.8 k Kirchhoff’s voltage law (CW): +Vo 0.7 V + 20 V = 0 Vo = 19.3 V
9.
(a)
Vo1 = 12 V – 0.7 V = 11.3 V Vo2 = 1.2 V
(b) Vo1 = 0 V Vo2 = 0 V
10.
(a) Both diodes forward-biased Si diode turns on first and locks in 0.7 V drop. 12 V 0.7 V IR = 2.4 mA 4.7 k ID = IR = 2.4 mA Vo = 12 V 0.7 V = 11.3 V (b) Right diode forward-biased: 20 V + 4 V 0.7 V = 10.59 mA ID = 2.2 k Vo = 20 V 0.7 V = 19.3 V
11.
(a) Si diode “on” preventing GaAs diode from turning “on”: 1 V 0.7 V 0.3 V = 0.3 mA I= 1 k 1 k Vo = 1 V 0.7 V = 0.3 V 16 V 0.7 V 0.7 V + 4 V 18.6 V = 3.96 mA 4.7 k 4.7 k Vo = 16 V 0.7 V 0.7 V = 14.6 V
(b) I =
14
12.
Both diodes forward-biased: Vo1 = 0.7 V, Vo2 = 0.7 V 20 V 0.7 V 19.3 V = = 19.3 mA 1 k 1 k I0.47 k = 0 mA I = I1 kΩ I0.47 kΩ = 19.3 mA 0 mA = 19.3 mA
I1 k =
13.
1 k(9.3 V) 3.1 V 1 k 2 k 16 k(8.8 V) Vo2 (8.8 V) 2.93 V 1 k 2 k Vo = Vo1 Vo2 = 6.03 V
Superposition: Vo1 (9.3 V)
ID =
9.3 V 6.03 V = 1.635 mA 2 k
14.
Both diodes “off”. The threshold voltage of 0.7 V is unavailable for either diode. Vo = 0 V
15.
Both diodes “on”, Vo = 10 V 0.7 V = 9.3 V
16.
Both diodes “on”. Vo = 0.7 V
17.
Both diodes “off”, Vo = 10 V
18.
The Si diode with 5 V at the cathode is “on” while the other is “off”. The result is Vo = 5 V + 0.7 V = 4.3 V
19.
0 V at one terminal is “more positive” than 5 V at the other input terminal. Therefore assume lower diode “on” and upper diode “off”. The result: Vo = 0 V 0.7 V = 0.7 V The result supports the above assumptions.
20.
Since all the system terminals are at 10 V the required difference of 0.7 V across either diode cannot be established. Therefore, both diodes are “off” and Vo = +10 V as established by 10 V supply connected to 1 k resistor.
15
21.
The Si diode requires more terminal voltage than the Ge diode to turn “on”. Therefore, with 5 V at both input terminals, assume Si diode “off” and Ge diode “on”. The result: Vo = 5 V 0.3 V = 4.7 V The result supports the above assumptions.
22.
Vdc = 0.318 Vm Vm =
Im =
Vdc 2V = 6.28 V 0.318 0.318
Vm 6.28 V = 3.14 mA 2 k R
23.
Using Vdc 0.318(Vm VT) 2 V = 0.318(Vm 0.7 V) Solving: Vm = 6.98 V 10:1 for Vm:VT
24.
Vm =
Vdc 2V = 6.28 V 0.318 0.318
I Lmax =
6.28 V = 0.628 mA 10 k
16
Imax(2 k) = I Dmax I Lmax
6.28 V = 3.14 mA 2 k + Imax(2 k) = 0.678 mA + 3.14 mA = 3.77 mA
25.
Vm = 2 (120 V) = 169.68 V Vdc = 0.318Vm = 0.318(169.68 V) = 53.96 V
26.
Diode will conduct when vo = 0.7 V; that is, 1 k(vi ) vo = 0.7 V = 1 k 1 k Solving: vi = 1.4 V For vi 1.4 V Si diode is “on” and vo = 0.7 V. For vi < 1.4 V Si diode is open and level of vo is determined by voltage divider rule: 1 k(vi ) = 0.5 vi vo = 1 k 1 k For vi = 10 V: vo = 0.5(10 V) = 5 V When vo = 0.7 V, vRmax vimax 0.7 V I Rmax
Imax(reverse) =
= 10 V 0.7 V = 9.3 V 9.3 V = 9.3 mA 1 k
10 V = 0.5 mA 1 k 1 k
17
27.
(a) Pmax = 14 mW = (0.7 V)ID 14 mW = 20 mA ID = 0.7 V (b) Imax = 2 × 20 mA = 40 mA (c)
4.7 k 68 k = 4.4 kΩ VR = 160 V 0.7 V = 159.3 V 159.3 V Imax = = 36.2 mA 4.4 k I Id = max = 18.1 mA 2
(d) Total damage, 36.2 mA > 20 mA 28.
(a) Vm = 2 (120 V) = 169.7 V VLm = Vim 2VD = 169.7 V 2(0.7 V) = 169.7 V 1.4 V = 168.3 V Vdc = 0.636(168.3 V) = 107.04 V (b) PIV = Vm(load) + VD = 168.3 V + 0.7 V = 169 V (c) ID(max) =
VLm RL
168.3 V = 168.3 mA 1 k
(d) Pmax = VDID = (0.7 V)Imax = (0.7 V)(168.3 mA) = 117.81 mW 29.
Imax =
100 V = 45.45 mA 2.2 k
18
30.
Positive half-cycle of vi: Voltage-divider rule: 2.2 k(Vimax ) Vomax = 2.2 k 2.2 k 1 = (Vimax ) 2 1 = (100 V) 2 = 50 V Polarity of vo across the 2.2 k resistor acting as a load is the same. Voltage-divider rule: 2.2 k(Vimax ) Vomax = 2.2 k 2.2 k 1 = (Vimax ) 2 1 = (100 V) 2 = 50 V Vdc = 0.636Vm = 0.636 (50 V) = 31.8 V
31.
Positive pulse of vi: Top left diode “off”, bottom left diode “on” 2.2 k 2.2 k = 1.1 k 1.1 k(170 V) = 56.67 V Vopeak = 1.1 k 2.2 k Negative pulse of vi: Top left diode “on”, bottom left diode “off” 1.1 k(170 V) = 56.67 V Vopeak = 1.1 k 2.2 k Vdc = 0.636(56.67 V) = 36.04 V
32.
(a) Si diode open for positive pulse of vi and vo = 0 V For 20 V < vi 0.7 V diode “on” and vo = vi + 0.7 V. For vi = 20 V, vo = 20 V + 0.7 V = 19.3 V For vi = 0.7 V, vo = 0.7 V + 0.7 V = 0 V
19
(b) For vi 8 V the 8 V battery will ensure the diode is forward-biased and vo = vi 8 V. At vi = 8 V vo = 8 V 8 V = 0 V At vi = 20 V vo = 20 V 8 V = 28 V For vi > 8 V the diode is reverse-biased and vo = 0 V.
33.
(a) Positive pulse of vi: 1.8 k(12 V 0.7 V) = 5.09 V Vo = 1.8 k 2.2 k Negative pulse of vi: diode “open”, vo = 0 V (b) Positive pulse of vi: Vo = 12 V 0.7 V + 4 V = 15.3 V Negative pulse of vi: diode “open”, vo = 0 V
34.
(a) For vi = 20 V the diode is reverse-biased and vo = 0 V. For vi = 5 V, vi overpowers the 4 V battery and the diode is “on”. Applying Kirchhoff’s voltage law in the clockwise direction: 5 V + 4 V vo = 0 vo = 1 V
(b) For vi = 20 V the 20 V level overpowers the 5 V supply and the diode is “on”. Using the short-circuit equivalent for the diode we find vo = vi = 20 V. For vi = 5 V, both vi and the 5 V supply reverse-bias the diode and separate vi from vo. However, vo is connected directly through the 2.2 k resistor to the 5 V supply and vo = 5 V.
20
35.
(a) Diode “on” for vi 4.7 V For vi > 4.7 V, Vo = 4 V + 0.7 V = 4.7 V For vi < 4.7 V, diode “off” and vo = vi (b) Again, diode “on” for vi 3.7 V but vo now defined as the voltage across the diode For vi 3.7 V, vo = 0.7 V For vi < 3.7 V, diode “off”, ID = IR = 0 mA and V2.2 k = IR = (0 mA)R = 0 V Therefore, vo = vi 3 V At vi = 0 V, vo = 3 V vi = 8 V, vo = 8 V 3 V = 11 V
36.
For the positive region of vi: The right Si diode is reverse-biased. The left Si diode is “on” for levels of vi greater than 5.3 V + 0.7 V = 6 V. In fact, vo = 6 V for vi 6 V. For vi < 6 V both diodes are reverse-biased and vo = vi. For the negative region of vi: The left Si diode is reverse-biased. The right Si diode is “on” for levels of vi more negative than 7.3 V + 0.7 V = 8 V. In fact, vo = 8 V for vi 8 V. For vi > 8 V both diodes are reverse-biased and vo = vi.
iR: For 8 V < vi < 6 V there is no conduction through the 10 k resistor due to the lack of a complete circuit. Therefore, iR = 0 mA. For vi 6 V vR = vi vo = vi 6 V For vi = 10 V, vR = 10 V 6 V = 4 V 4V = 0.4 mA and iR = 10 k For vi 8 V vR = vi vo = vi + 8 V
21
For vi = 10 V vR = 10 V + 8 V = 2 V 2 V = 0.2 mA and iR = 10 k
37.
(a) Starting with vi = 20 V, the diode is in the “on” state and the capacitor quickly charges to 20 V+. During this interval of time vo is across the “on” diode (short-current equivalent) and vo = 0 V. When vi switches to the +20 V level the diode enters the “off” state (open-circuit equivalent) and vo = vi + vC = 20 V + 20 V = +40 V
(b) Starting with vi = 20 V, the diode is in the “on” state and the capacitor quickly charges up to 15 V+. Note that vi = +20 V and the 5 V supply are additive across the capacitor. During this time interval vo is across “on” diode and 5 V supply and vo = 5 V. When vi switches to the +20 V level the diode enters the “off” state and vo = vi + vC = 20 V + 15 V = 35 V.
22
38.
(a) For negative half cycle capacitor charges to peak value of 120 V = 120 V with polarity . The output vo is directly across the “on” diode resulting in vo = 0 V as a negative peak value. For next positive half cycle vo = vi + 120 V with peak value of vo = 120 V + 120 V = 240 V.
(b) For positive half cycle capacitor charges to peak value of 120 V 20 V = 100 V with polarity . The output vo = 20 V = 20 V For next negative half cycle vo = vi 100 V with negative peak value of vo = 120 V 100 V = 220 V.
39.
(a) = RC = (56 k)(0.1 F) = 5.6 ms 5 = 28 ms (b) 5 = 28 ms
T 1 ms = = 0.5 ms, 56:1 2 2
(c) Positive pulse of vi: Diode “on” and vo = 2 V + 0.7 V = 1.3 V Capacitor charges to 12 V + 2 V 0.7 V = 13.3 V Negative pulse of vi: Diode “off” and vo = 12 V 13.3 V = 25.3 V
40.
Solution is network of Fig. 2.181(b) using a 10 V supply in place of the 5 V source.
23
41.
Network of Fig. 2.178 with 2 V battery reversed.
42.
(a) In the absence of the Zener diode 180 (20 V) =9V VL = 180 220 VL = 9 V < VZ = 10 V and diode non-conducting 20 V = 50 mA 220 180 with IZ = 0 mA and VL = 9 V
Therefore, IL = IR =
(b) In the absence of the Zener diode 470 (20 V) = 13.62 V VL = 470 220 VL = 13.62 V > VZ = 10 V and Zener diode “on” Therefore, VL = 10 V and VRs = 10 V I Rs VRs / Rs 10 V/220 = 45.45 mA
and (c)
IL = VL/RL = 10 V/470 = 21.28 mA IZ = I Rs IL = 45.45 mA 21.28 mA = 24.17 mA
PZ max = 400 mW = VZIZ = (10 V)(IZ)
400 mW = 40 mA 10 V I Lmin = I Rs I Z max = 45.45 mA 40 mA = 5.45 mA
IZ =
RL =
VL 10 V = 1,834.86 I Lmin 5.45 mA
Large RL reduces IL and forces more of I Rs to pass through Zener diode. (d) In the absence of the Zener diode RL (20 V) VL = 10 V = RL 220 10RL + 2200 = 20RL 10RL = 2200 RL = 220
24
43.
(a) VZ = 12 V, RL =
VL 12 V = 60 I L 200 mA
RLVi 60 (16 V) RL RS 60 Rs 720 + 12Rs = 960 12Rs = 240 Rs = 20
VL = VZ = 12 V =
(b)
PZ max = VZ I Z
max
= (12 V)(200 mA) = 2.4 W 44.
Since IL =
VL VZ is fixed in magnitude the maximum value of I Rs will occur when IZ is a RL RL
maximum. The maximum level of I Rs will in turn determine the maximum permissible level of Vi. I Zmax
IL =
PZ max VZ
400 mW = 50 mA 8V
VL VZ 8V = 36.36 mA RL RL 220
I Rs = IZ + IL = 50 mA + 36.36 mA = 86.36 mA I Rs
Vi VZ Rs
or Vi = I Rs Rs + VZ = (86.36 mA)(91 ) + 8 V = 7.86 V + 8 V = 15.86 V Any value of vi that exceeds 15.86 V will result in a current IZ that will exceed the maximum value. 45.
At 30 V we have to be sure Zener diode is “on”. RLVi 1 k(30 V) VL = 20 V = RL Rs 1 k Rs Solving, Rs = 0.5 k At 50 V, I RS
50 V 20 V 20 V = 60 mA, IL = = 20 mA 0.5 k 1 k
IZM = I RS IL = 60 mA 20 mA = 40 mA 46.
For vi = +50 V: Z1 forward-biased at 0.7 V Z2 reverse-biased at the Zener potential and VZ2 = 10 V. Therefore, Vo = VZ1 VZ2 = 0.7 V + 10 V = 10.7 V
25
For vi = 50 V: Z1 reverse-biased at the Zener potential and VZ1 = 10 V. Z2 forward-biased at 0.7 V. Therefore, Vo = VZ1 VZ2 = 10.7 V
For a 5 V square wave neither Zener diode will reach its Zener potential. In fact, for either polarity of vi one Zener diode will be in an open-circuit state resulting in vo = vi.
47.
Vm = 1.414(120 V) = 169.68 V 2Vm = 2(169.68 V) = 339.36 V
48.
The PIV for each diode is 2Vm PIV = 2(1.414)(Vrms)
26
Chapter 3 1.
2.
A bipolar transistor utilizes holes and electrons in the injection or charge flow process, while unipolar devices utilize either electrons or holes, but not both, in the charge flow process.
3.
Forward- and reverse-biased.
4.
The leakage current ICO is the minority carrier current in the collector.
5.
6.
7.
8.
IE the largest IB the smallest IC IE
9.
1 I C IC = 100IB 100 IE = IC + IB = 100IB + IB = 101IB I 8 mA = 79.21 A IB = E 101 101 IC = 100IB = 100(79.21 A) = 7.921 mA
IB =
10.
11.
IE = 5 mA, VCB = 1 V: VBE = 800 mV VCB = 10 V: VBE = 770 mV VCB = 20 V: VBE = 750 mV The change in VCB is 20 V:1 V = 20:1 The resulting change in VBE is 800 mV:750 mV = 1.07:1 (very slight)
12.
13.
V 0.9 V 0.7 V = 25 I 8 mA 0 (b) Yes, since 25 is often negligible compared to the other resistance levels of the network.
(a) rav =
(a) IC IE = 3.5 mA (b) IC IE = 3.5 mA (c) negligible: change cannot be detected on this set of characteristics. (d) IC IE
27
14.
(a) Using Fig. 3.7 first, IE 2 mA Then Fig. 3.8 results in IC 2 mA (b) Using Fig. 3.8 first, IE 5 mA Then Fig. 3.7 results in VBE 0.77 V (c) Using Fig. 3.10(b) IE = 5 mA results in VBE 0.81 V (d) Using Fig. 3.10(c) IE = 5 mA results in VBE = 0.7 V (e) Yes, the difference in levels of VBE can be ignored for most applications if voltages of several volts are present in the network.
15.
(a) IC = IE = (0.998)(4 mA) = 3.992 mA (b) IE = IC + IB IC = IE IB = 2.8 mA 0.02 mA = 2.78 mA I 2.78 mA dc = C = 0.993 IE 2.8 mA 0.98 (c) IC = IB = IB (40 A) = 1.96 mA 1 1 0.98 I 1.96 mA IE = C = 2 mA 0.993
16.
17.
18.
(a) Fig. 3.13(b): IB 35A Fig. 3.13(a): IC 3.6 mA (b) Fig. 3.14(a): VCE 12 V Fig. 3.14(b): VBE 0.75 V
19.
20.
IC 2 mA = 111.11 I B 18 A 111.11 = 0.991 (b) = 1 111.11 1 (c) ICEO = 0.3 mA (d) ICBO = (1 )ICEO = (1 0.991)(0.3 mA) = 2.7 A
(a)
dc =
(a) Fig. 3.14(a): ICEO 0.3 mA (b) Fig. 3.14(a): IC 1.35 mA I 1.35 mA dc = C = 135 IB 10 A
28
(c) = ICBO
21.
135 = 0.9926 1 136 (1 )ICEO = (1 0.9926)(0.3 mA) = 2.2 A
(a) dc =
I C 5.25 mA = 87.5 IB 60 A
(b) dc =
I C 3.25 mA = 108.3 IB 30 A
(c) dc =
I C 1.35 mA = 135 IB 10 A
(d) Yes, highest at lowest levels of I C . 22.
(a) ac =
I C 5.9 mA 4.6 mA = = 65 70 A 50 A I B
(b) ac =
I C 4.1 mA 2.3 mA = = 90 40 A 20 A I B
(c) ac =
I C 2.4 mA 0.3 mA = = 105 20 A 0 A I B
(d) Yes, highest at lowest levels of IC. (e)
IC 5.25 mA 3.25 mA 1.35 mA
a. b. c.
dc
ac
87.5 108.3 135
65 90 105
VCE 4V 7V 10 V
dc/ac 1.35 1.20 1.29
dc > ac although ratio consistent over scope of characteristics. 23.
24.
I C 2.9 mA = 116 IB 25 A 116 = = 0.991 1 116 1 IE = IC/ = 2.9 mA/0.991 = 2.93 mA
dc =
(a) = (b) =
1
0.980 = 49 1 0.980
120 = 0.992 1 120 1
29
IC
2 mA = 16.66 A 120 IE = IC + IB = 2 mA + 16.66 A 2.017 mA
(c) IB =
25.
26.
Ve = Vi Vbe = 2 V 0.1 V = 1.9 V V 1.9 V = 0.95 1 Av = o Vi 2V Ie =
27.
VE 1.9 V = 1.9 mA (rms) RE 1 k
Output characteristics: Curves are essentially the same with new scales as shown.
Input characteristics: Common-emitter input characteristics may be used directly for common-collector calculations. 28.
PCmax = 35 mW = VCEIC PCmax
IC = I Cmax , VCE =
I Cmax
VCE = VCEmax , IC = VCE = 10 V, IC = IC = 5 mA, VCE = VCE = 15 V, IC =
=
PCmax VCEmax
PCmax VCE PCmax IC PCmax VCE
35 mW = 5.83 V 6 mA
=
35 mW = 2.33 mA 20 V
35 mW = 3.5 mA 10 V
35 mW = 7.00 V 5 mA
35 mW = 2.91 mA 12 V
30
29.
PCmax
IC = I Cmax , VCE =
I Cmax
VCB = VCBmax , IC =
VCB = 10 V, IC =
PCmax
VCBmax PCmax
IC = 4 mA, VCB =
42 mW =6V 7 mA
IC PCmax VCB
42 mW = 2.1 mA 20 V
42 mW = 10.5 V 4 mA
42 mW = 2.8 mA 15 V
30.
The operating temperature range is 55C TJ 150C 9 F = C + 32 5 9 = (55C) + 32 = 67F 5 9 F = (150C) + 32 = 302F 5 67F TJ 302F
31.
I Cmax = 200 mA, VCEmax = 30 V, PDmax = 625 mW
IC = I Cmax , VCE = VCE = VCEmax , IC =
PDmax I Cmax PDmax VCEmax
=
625 mW = 3.125 V 200 mA
625 mW = 20.83 mA 30 V
31
IC = 100 mA, VCE = VCE = 20 V, IC =
32.
PDmax IC
PDmax VCE
625 mW = 6.25 V 100 mA
625 mW = 31.25 mA 20 V
From Fig. 3.23 (a) ICBO = 50 nA max max avg = min 2 50 150 200 = = 2 2 = 100 ICEO ICBO = (100)(50 nA) = 5 A
33.
hFE (dc) with VCE = 1 V, T = 25C IC = 0.1 mA, hFE 0.43(100) = 43 IC = 10 mA, hFE 0.98(100) = 98 hfe(ac) with VCE = 10 V, T = 25C IC = 0.1 mA, hfe 72 IC = 10 mA, hfe 160 For both hFE and hfe the same increase in collector current resulted in a similar increase (relatively speaking) in the gain parameter. The levels are higher for hfe but note that VCE is higher also.
34.
As the reverse-bias potential increases in magnitude the input capacitance Cibo decreases (Fig. 3.23(b)). Increasing reverse-bias potentials causes the width of the depletion region to A increase, thereby reducing the capacitance C . d
32
35.
(a) At IC = 1 mA, hfe 120 At IC = 10 mA, hfe 160 (b) The results confirm the conclusions of problems 21 and 22 that beta tends to increase with increasing collector current.
36.
Tj = +125°C: 1.45 norm Tj = +25°C: 0.95 norm
Tj = 55°C: 0.5 norm
Yes, 34% drop between +125°C and 25°C. 65.5% drop between +125°C and 55°C. 37.
(a) ac =
I C 16 mA 12.2 mA 3.8 mA = = 190 80 A 60 A 20 A I B VCE 3 V
(b) dc =
IC 12 mA = 201.7 I B 59.5 A
(c) ac =
4 mA 2 mA 2 mA = 200 18 A 8 A 10 A
(d) dc =
I C 3 mA = 230.77 I B 13 A
(e) In both cases dc is slightly higher than ac ( 10%) (f)(g) In general dc and ac increase with increasing IC for fixed VCE and both decrease for decreasing levels of VCE for a fixed IE. However, if IC increases while VCE decreases when moving between two points on the characteristics, chances are the level of dc or ac may not change significantly. In other words, the expected increase due to an increase in collector current may be offset by a decrease in VCE. The above data reveals that this is a strong possibility since the levels of are relatively close.
33
Chapter 4 1.
VCC VBE 16 V 0.7 V 15.3 V = 30 A RB 510 k 510 k
(a)
I BQ
(b)
I CQ I BQ = (120)(30 A) = 3.6 mA
(c) VCEQ VCC I CQ RC = 16 V (3.6 mA)(1.8 k) = 9.52 V (d) VC = VCEQ = 6.48 V (e) VB = VBE = 0.7 V (f) 2.
VE = 0 V
(a) IC = IB = 80(40 A) = 3.2 mA (b) RC =
(c) RB =
VRC IC VRB IB
VCC VC 12 V 6 V 6V = 1.875 k IC 3.2 mA 3.2 mA
12 V 0.7 V 11.3 V = 282.5 k 40 A 40 A
(d) VCE = VC = 6 V 3.
(a) IC = IE IB = 4 mA 20 A = 3.98 mA 4 mA (b) VCC = VCE + ICRC = 7.2 V + (3.98 mA)(2.2 k) = 15.96 V 16 V (c) =
(d) RB =
I C 3.98 mA = 199 200 IB 20 A VRB IB
VCC VBE 15.96 V 0.7 V = 763 k IB 20 A
VCC 16 V = 5.93 mA RC 2.7 k
4.
I Csat =
5.
(a) Load line intersects vertical axis at IC =
21 V = 7 mA 3 k
and horizontal axis at VCE = 21 V. (b) IB = 25 A: RB = (c)
VCC VBE 21 V 0.7 V = 812 k IB 25 A
I CQ 3.4 mA, VCEQ 10.75 V
34
I C 3.4 mA = 136 IB 25 A
(d) =
136 136 = 0.992 1 136 1 137
(e) = (f)
I Csat =
VCC 21 V = 7 mA RC 3 k
(g) (h) PD = VCEQ I CQ = (10.75 V)(3.4 mA) = 36.55 mW
6.
(i)
Ps = VCC(IC + IB) = 21 V(3.4 mA + 25 A) = 71.92 mW
(j)
PR = Ps PD = 71.92 mW 36.55 mW = 35.37 mW
(b)
I BQ VCEQ
(c) =
7.
VCC VBE 16 V 0.7 V = 30 µA RB 510 k = 8.5 V, I CQ = 4.1 mA
I C 4.1 mA = 136.67 IB 30 A
16 V 0.7 V = 16.81 µA 910 k I BQ = (120)(16.81 µA) = 2.017 mA
I BQ I CQ
(from characteristics) VEQ = 11.5 V, I CQ = 2.4 mA 8.
VCC VBE 20 V 0.7 V 19.3 V = RB ( 1) RE 270 k (126)2.2 k 547.2 k = 35.27 A
(a)
I BQ
(b)
I CQ I BQ = (125)(35.27 A) = 4.41 mA
(c) VCEQ = VCC IC(RC + RE) = 20 V (4.41 mA)(470 k + 2.2 k) = 20 V 11.77 V = 8.23 V (d) VC = VCC ICRC = 20 V (4.41 mA)(470 k) = 20 V 2.07 V = 17.93 V (e) VB = VCC IBRB = 20 V (35.27 A)(270 k) = 20 V 9.52 V = 10.48 V (f)
VE = VC VCE = 17.93 V 8.23 V = 9.7 V
35
9.
(a)
I BQ = 35.27 µA using from problem 8. I sat =
20 V = 7.49 mA 2.2 k 470
(b)
I CQ = 4.7 mA, VCEQ = 7.5 V
(c)
IC 4.7 mA = 133.25 I B 35.27 A
(d) reasonably close (e) different and Q pt.
10.
(a) RC =
VCC VC 12 V 7.6 V 4.4 V = 2.2 k IC 2 mA 2 mA
(b) IE IC: RE =
(c) RB =
VRB IB
VE 2.4 V = 1.2 k I E 2 mA
VCC VBE VE 12 V 0.7 V 2.4 V 8.9 V = 356 k IB 2 mA/80 25 A
(d) VCE = VC VE = 7.6 V 2.4 V = 5.2 V (e) VB = VBE + VE = 0.7 V + 2.4 V = 3.1 V 11.
(a) IC IE =
VE 2.1 V = 3.09 mA RE 0.68 k
I C 3.09 mA = 154.5 IB 20 A
=
(b) VCC = VRC + VCE + VE = (3.09 mA)(2.7 k) + 7.3 V + 2.1 V = 8.34 V + 7.3 V + 2.1 V = 17.74 V VCC VBE VE 17.74 V 0.7 V 2.1 V IB IB 20 A 14.94 V = 747 k = 20 A
(c) RB =
12.
I Csot
VRB
VCC 20 V 20 V = 7.49 mA RC RE 470 2.2 k 2.67 k
36
13.
(a)
VCC 24 V RC RE RC 1.2 k 24 V = 3.529 k RC + 1.2 k = 6.8 mA RC = 2.33 k
I Csat = 6.8 mA =
IC 4 mA = 133.33 I B 30 A
(b) =
VCC VBE VE 24 V 0.7 V (4 mA)(1.2 k) IB IB 30 A 18.5 V = 616.67 k = 30 A
(c) RB =
VRB
(d) PD = VCEQ I CQ = (10 V)(4 mA) = 40 mW (e) P = I C2 RC = (4 mA)2(2.33 k) = 37.28 mW 14.
(a) Problem 1: I CQ = 3.6 mA, VCEQ = 6.48 V (b)
I BQ = 30 A (the same) I CQ I BQ = (180)(30 A) = 5.4 mA VCEQ VCC I CQ RC = 16 V (5.4 mA)(1.8 k) = 6.28 V
5.4 mA 3.6 mA 100% = 50% 3.6 mA 9.52 V 6.28 V 100% = 51.6% %VCE = 6.28 V About 50% change for each.
(c) %IC =
(d) Problem 8: I CQ = 4.41 mA, VCEQ = 8.23 V ( I BQ = 35.27 A) (e)
VCC VBE 20 V 0.7 V = 28.19 A RB ( 1) RE 270 k (187.5 1)(2.2 k) I BQ = (187.5)(28.19 A) = 5.29 mA
I BQ I CQ
= VCC IC(RC + RE) = 20 V (5.29 mA)(470 k + 2.2 k) = 5.88 V
37
(f)
5.29 mA 4.41 mA 100% = 20% 4.41 mA 5.88 V 8.23 V %VCE = 100% = 28.6% 8.23 V
%IC =
(g) For both IC and VCE the % change is less for the emitter-stabilized. ?
15.
RE 10R2
(80)(0.68 k) 10(9.1 k) 54.4 k 91 k (No!)
(a) Use exact approach: RTh = R1 R2 = 62 k 9.1 k = 7.94 k RV (9.1 k)(16 V) = 2.05 V ETh = 2 CC R2 R1 9.1 k 62 k ETh VBE 2.05 V 0.7 V RTh ( 1) RE 7.94 k (81)(0.68 k) = 21.42 A
I BQ
(b)
I CQ I BQ = (80)(21.42 A) = 1.71 mA
(c) VCEQ = VCC I CQ (RC + RE) = 16 V (1.71 mA)(3.9 k + 0.68 k) = 8.17 V (d) VC = VCC ICRC = 16 V (1.71 mA)(3.9 k) = 9.33 V (e) VE = IERE ICRE = (1.71 mA)(0.68 k) = 1.16 V
16.
(f)
VB = VE + VBE = 1.16 V + 0.7 V = 1.86 V
(a)
RTh 62 k 9.1 k = 7.94 k 9.1 k(16 V) ETh = 2.05 V 9.1 k + 62 k ETh VBE 2.05 V 0.7 V I BQ RTh ( 1) RE 7.94 k (140 1)(0.68 k) = 13 μA (vs. 21.42 μA) I CQ I BQ = (140)(13 A) = 1.82 mA (vs. 1.71 mA) VCEQ = VCC I CQ (RC + RE)
= 16 V (1.82 mA)(3.9 k + 0.68 k) = 7.66 V (vs. 8.17 V)
38
VC = VCC ICRC = 16 V (1.82 mA)(3.9 k) = 8.9 V (vs. 9.33 V) VE = IERE ICRE = (1.82 mA)(0.68 k) = 1.24 V (vs. 1.16 V) VB = VE + VBE = 1.24 V + 0.7 V = 1.94 V (vs. 1.86 V) (b)
17.
I BQ affected the most
(a) IC =
VCC VC 18 V 12 V = 1.28 mA RC 4.7 k
(b) VE = IERE ICRE = (1.28 mA)(1.2 k) = 1.54 V (c) VB = VBE + VE = 0.7 V + 1.54 V = 2.24 V (d) R1 =
VR1 I R1
VR1 = VCC VB = 18 V 2.24 V = 15.76 V
:
I R1 I R2
R1 =
18.
VR1 I R1
=
VB 2.24 V = 0.4 mA R2 5.6 k
15.76 V = 39.4 k 0.4 mA
(a) IC = IB = (100)(20 A) = 2 mA (b) IE = IC + IB = 2 mA + 20 A = 2.02 mA VE = IERE = (2.02 mA)(1.2 k) = 2.42 V (c) VCC = VC + ICRC = 10.6 V + (2 mA)(2.7 k) = 10.6 V + 5.4 V = 16 V (d) VCE = VC VE = 10.6 V 2.42 V = 8.18 V (e) VB = VE + VBE = 2.42 V + 0.7 V = 3.12 V (f)
I R1 I R2 I B
3.12 V + 20 A = 380.5 A + 20 A = 400.5 A 8.2 k V VB 16 V 3.12 V = 32.16 k R1 = CC I R1 400.5 A
=
39
VCC 16 V 16 V = = 3.49 mA RC RE 3.9 k 0.68 k 4.58 k
19.
I Csat =
20.
(a) Testing RE 10 Rz (140)(0.68 k ) 10(9.1 k ) 95.2 k 91 k satisfied (barely) Applying the approximation approved: RV (9.1 k)(16 V) = 2.05 V (vs. 1.95 V) VB = 2 CC R1 R2 62 k 9.1 k VE = VB VBE = 2.05 V 0.7 V = 1.35 V (vs. 1.24 V) V 1.35 V IC IE = E = 1.99 mA (vs. 1.82 mA) RE 0.68 k VC = VCC ICRC = 16 V (1.99 mA)(3.9 k) = 8.24 V (vs 8.9 V) VCE = VC VE = 8.24 V 1.35 V = 6.89 V (vs 7.66 V)
21.
(a) RE 10R2 (120)(1 k) 10(8.2 k) 120 k 82 k (checks) RV (8.2 k)(18 V) = 3.13 V VB = 2 CC R1 R2 39 k 8.2 k VE = VB VBE = 3.13 V 0.7 V = 2.43 V V 2.43 V IC IE = E = 2.43 mA 1 k RE (b) VCE = VCC IC(RC + RE) = 18 V (2.43 mA)(3.3 k + 1 k) = 7.55 V (c) IB =
IC
2.43 mA = 20.25 A 120
(d) VE = IERE ICRE = (2.43 mA)(1 k) = 2.43 V (e) VB = 3.13 V 22.
(a) RTh = R1 R2 = 39 k 8.2 k = 6.78 k R V 8.2 k(18 V) ETh = C CC = 3.13 V R1 R2 39 k 8.2 k ETh VBE 3.13 V 0.7 V RTh ( 1) RE 6.78 k (121)(1 k) 2.43 V = 19.02 A = 127.78 k IC = IB = (120)(19.02 A) = 2.28 mA (vs. 2.43 mA #16)
IB =
40
(b) VCE = VCC IC(RC + RE) = 18 V (2.28 mA)(3.3 k + 1 k) = 18 V 9.8 V = 8.2 V (vs. 7.55 V #16) (c) 19.02 A (vs. 20.25 A #16) (d) VE = IERE ICRE = (2.28 mA)(1 k) = 2.28 V (vs. 2.43 V #16) (e) VB = VBE + VE = 0.7 V + 2.28 V = 2.98 V (vs. 3.13 V #16) The results suggest that the approximate approach is valid if Eq. 4.33 is satisfied. 23.
(a) VB =
R2 9.1 k(16 V) = 2.05 V VCC 62 k 9.1 k R1 R2 VE = VB VBE = 2.05 V 0.7 V = 1.35 V V 1.35 V IE = E = 1.99 mA RE 0.68 k I CQ IE = 1.99 mA
VCEQ = VCC IC (RC + RE)
I BQ
= 16 V (1.99 mA)(3.9 k + 0.68 k) = 16 V 9.11 V = 6.89 V I CQ 1.99 mA = = 24.88 A 80
(b) From Problem 12: I CQ = 1.71 mA, VCEQ = 8.17 V, I BQ = 21.42 A (c) The differences of about 14% suggest that the exact approach should be employed when appropriate. 24.
(a)
VCC 24 V 24 V RC RE 3RE RE 4RE 24 V 24 V = 0.8 k RE = 4(7.5 mA) 30 mA
I Csat 7.5 mA
RC = 3RE = 3(0.8 k) = 2.4 k (b) VE = IERE ICRE = (5 mA)(0.8 k) = 4 V (c) VB = VE + VBE = 4 V + 0.7 V = 4.7 V RV R2 (24 V) (d) VB = 2 CC , 4.7 V = R2 R1 R2 24 k R2 = 5.84 k IC 5 mA = 129.8 (e) dc = I B 38.5 A
41
(f)
25.
RE 10R2
(129.8)(0.8 k) 10(5.84 k) 103.84 k 58.4 k (checks)
(a) From problem 12b, IC = 1.71 mA From problem 12c, VCE = 8.17 V (b) changed to 120: From problem 12a, ETh = 2.05 V, RTh = 7.94 k ETh VBE 2.05 V 0.7 V IB = RTh ( 1) RE 7.94 k + (121)(0.68 k) = 14.96 A IC = IB = (120)(14.96 A) = 1.8 mA VCE = VCC IC(RC + RE) = 16 V (1.8 mA)(3.9 k + 0.68 k) = 7.76 V (c)
% I C % VCE
(d) %IC %VCE
1.8 mA 1.71 mA 100% = 5.26% 1.71 mA 7.76 V 8.17 V 100% = 5.02% 8.17 V
11c 49.83% 48.70%
11f 34.59% 46.76%
20c 5.26% 5.02%
Fixed-bias
Emitter feedback
Voltagedivider
(e) Quite obviously, the voltage-divider configuration is the least sensitive to changes in . 26.
(a) Problem 21: Approximation approach: I CQ = 2.43 mA, VCEQ = 7.55 V Problem 22: Exact analysis: I CQ = 2.28 mA, VCEQ = 8.2 V The exact solution will be employed to demonstrate the effect of the change of . Using the approximate approach would result in %IC = 0% and %VCE = 0%. Problem 22: ETh = 3.13 V, RTh = 6.78 k ETH VBE 3.13 V 0.7 V 2.43 V IB = RTh ( 1) RE 6.78 k (180 1)1 k 187.78 k = 12.94 A IC = IB = (180)(12.94 A) = 2.33 mA VCE = VCC IC(RC + RE) = 18 V (2.33 mA)(3.3 k + 1 k) = 7.98 V
42
2.33 mA 2.28 mA 100% = 2.19% 2.28 mA 7.98 V 8.2 V %VCE = 100% = 2.68% 8.2 V
%IC =
For situations where RE > 10R2 the change in IC and/or VCE due to significant change in will be relatively small. %IC = 2.19% vs. 49.83% for problem 14. %VCE = 2.68% vs. 48.70% for problem 14. Voltage-divider configuration considerably less sensitive. (b) The resulting %IC and %VCE will be quite small. 27.
VCC VBE 16 V 0.7 V RB ( RC RE ) 270 k + (120)(3.6 k 1.2 k) = 18.09 A
(a) IB =
(b) IC = IB = (120)(18.09 A) = 2.17 mA (c) VC = VCC ICRC = 16 V (2.17 mA)(3.6 k) = 8.19 V 28.
VCC VBE 16 V 0.7 V = 3.19 mA RC RE 3.6 k 1.2 k
(a)
I CQ
(b)
I CQ = 3.19 mA vs 2.17 mA (not too close)
(c)
R RC RE 4.8 k , RF / 270 k /120 2.25 k R 2( RF / )
(d) Yes, I CQ I BQ
(VCC VBE ) VCC VBE RF ( RC RE ) RF / RC RE
For RC RE RF / , I CQ
(e)
I CQ
VCC VBE RC RE
16 V 0.7 V = 2.58 mA vs. 3.19 mA (much closer) 270 k 3.6 k 1.2 k 240
43
29.
(a) IB =
VCC VBE 30 V 0.7 V = 12.47 A RB ( RC RE ) 550 k 180(8.2 k 1.8 k)
IC = IB = (180)(12.47 A) = 2.24 mA (b) VC = VCC ICRC = 30 V (2.24 mA)(8.2 k) = 30 V 18.37 V = 11.63 V (c) VE = IERE ICRE = (2.24 mA)(1.8 k) = 4.03 V (d) VCE = VCC IC(RC + RE) = 30 V (2.24 mA)(8.2 k + 1.8 k) = 7.6 V 30.
(a)
R RC RE 8.2 k 1.8 k 10 k RF / 550 k/180 3.06 k R 3.27( RF / ) should be close
(b)
I CQ I CQ
31.
VCC VBE 30 V 0.7 V = 2.93 mA 8.2 k 1.8 k RC RE = 2.93 mA relatively close to 2.24 mA
VCC VBE 22 V 0.7 V RB ( RC RE ) 470 k (90)(9.1 k 9.1 k) = 10.09 A IC = IB = (90)(10.09 A) = 0.91 mA VCE = VCC IC(RC + RE) = 22 V (0.91 mA)(9.1 k + 9.1 k) = 5.44 V VCC VBE 22 V 0.7 V (b) = 135, IB = RB ( RC RE ) 470 k (135)(9.1 k 9.1 k) = 7.28 A
(a) IB =
IC = IB = (135)(7.28 A) = 0.983 mA VCE = VCC IC(RC + RE) = 22 V (0.983 mA)(9.1 k + 9.1 k) = 4.11 V (c)
% I C % VCE
0.983 mA 0.91 mA 100% = 8.02% 0.91 mA 4.11 V 5.44 V 100% = 24.45% 5.44 V
(d) The results for the collector feedback configuration are closer to the voltage-divider configuration than to the other two. However, the voltage-divider configuration continues to have the least sensitivities to change in .
44
32.
1 M = 0 , RB = 150 k VCC VBE 12 V 0.7 V IB = RB ( RC RE ) 150 k (180)(4.7 k 3.3 k) = 7.11 A IC = IB = (180)(7.11 A) = 1.28 mA VC = VCC ICRC = 12 V (1.28 mA)(4.7 k) = 5.98 V Full 1 M: RB = 1,000 k + 150 k = 1,150 k = 1.15 M VCC VBE 12 V 0.7 V IB = RB ( RC RE ) 1.15 M (180)(4.7 k 3.3 k) = 4.36 A IC = IB = (180)(4.36 A) = 0.785 mA VC = VCC ICRC = 12 V (0.785 mA)(4.7 k) = 8.31 V VC ranges from 5.98 V to 8.31 V
33.
(a) VE = VB VBE = 4 V 0.7 V = 3.3 V (b) IC IE =
VE 3.3 V = 2.75 mA RE 1.2 k
(c) VC = VCC ICRC = 18 V (2.75 mA)(2.2 k) = 11.95 V (d) VCE = VC VE = 11.95 V 3.3 V = 8.65 V (e) IB =
(f)
34.
=
VRB RB
VC VB 11.95 V 4 V = 24.09 A 330 k RB
IC 2.75 mA = 114.16 I B 24.09 A
VCC VEE VBE 6 V + 6 V 0.7 V 330 k (121)(1.2 k) RB ( 1) RE = 23.78 A IE = ( + 1)IB = (121)(23.78 A) = 2.88 mA VEE + IERE VE = 0 VE = VEE + IERE = 6 V + (2.88 mA)(1.2 k) = 2.54 V
(a) IB =
45
35.
82 k(12 V) = 9.46 V 82 k 22 k VE VB VBE = 9.46 V 0.7 V = 8.76 V V 8.76 V IE = E = 7.3 mA RE 1.2 k
(a) VB
IE 7.3 mA = 65.77 μA 1 111 I C I B (110)(65.77 A) = 7.23 mA
IB =
(b) VB 9.46 V, VC = 12 V, VE = 8.76 V (c) VBC = VB VC = 9.46 V 12 V = 2.54 V VCE = VC VE = 12 V 8.76 V = 3.24 V 36.
VEE VBE 12 V 0.7 V RB ( 1) RE 9.1 k (80 1)15 k = 9.26 A
(a) IB =
(b) IC = IB = (80)(9.26 A) = 0.741 mA (c) VCE = VCC + VEE IC(RC + RE) = 16 V + 12 V (0.741 mA)(27 k) =8V (d) VC = VCC ICRC = 16 V (0.741 mA)(12 k) = 7.11 V 37.
(a) IE =
8 V 0.7 V 7.3 V = 3.32 mA 2.2 k 2.2 k
(b) VC = 10 V (3.32 mA)(1.8 k) = 10 V 5.976 = 4.02 V (c) VCE = 10 V + 8 V (3.32 mA)(2.2 k + 1.8 k) = 18 V 13.28 V = 4.72 V 38.
(a) VE = 4 V 0.7 V = 3.3 V V V 3.3 V IE = E E = 3 mA RE RE 1.1 k IC IE = 3 mA VC = 8 V = VCC ICRC = 14 V (3 mA)RC RC = 2 k (b) IE = 3 mA, IB =
IE 3 mA = 32.97 μA 1 91
46
(c) VBC = VB VC = 4 V 8 V = 4 V VCE = VC VE = 8 V 3.3 V = 4.7 V 39.
(a) RE > 10R2 not satisfied Use exact approach: Network redrawn to determine the Thevenin equivalent:
510 k = 255 k 2 18 V + 18 V I= = 35.29 A 510 k 510 k ETh = 18 V + (35.29 A)(510 k) =0V
RTh =
18 V 0.7 V 255 k (130 1)(7.5 k) = 13.95 A
IB =
(b) IC = IB = (130)(13.95 A) = 1.81 mA (c) VE = 18 V + (1.81 mA)(7.5 k) = 18 V + 13.58 V = 4.42 V (d) VCE = 18 V + 18 V (1.81 mA)(9.1 k + 7.5 k) = 36 V 30.05 V = 5.95 V 40.
(a) IB =
VRB RB
VC VBE 8 V 0.7 V = 13.04 A 560 k RB
(b) IC =
VCC VC 18 V 8 V 10 V = = 2.56 mA 3.9 k 3.9 k RC
(c) =
IC 2.56 mA = 196.32 I B 13.04 A
(d) VCE = VC = 8 V
47
41.
IB = RB = RC =
IC
VRB IB VRC IC
2.5 mA = 31.25 A 80
VCC VBE 12 V 0.7 V = 361.6 k 31.25 A IB
VCC VC VCC VCEQ 12 V 6 V 6V 2.5 mA 2.5 mA IC I CQ
= 2.4 k Standard values: RB = 360 k RC = 2.4 k VCC = 10 mA RC RE 20 V 20 V 20 V = 10 mA = 10 mA 5RE = = 2 k 4RE RE 5 RE 10 mA 2 k = 400 RE = 5 RC = 4RE = 1.6 k I 5 mA IB = C = 41.67 A 120 20 V 0.7 V 5 mA(0.4 k) 19.3 2 V RB = VRB/IB = 41.67 A 41.67 A = 415.17 k Standard values: RE = 390 , RC = 1.6 k, RB = 430 k
42.
I Csat =
43.
RE =
VE VE 3V = 0.75 k I E I C 4 mA
VCC VC VCC (VCEQ VE ) IC IC IC 24 V (8 V + 3 V) 24 V 11 V 13 V = 3.25 k = 4 mA 4 mA 4 mA VB = VE + VBE = 3 V + 0.7 V = 3.7 V RV R (24 V) VB = 2 CC 3.7 V = 2 2 unknowns! R2 R1 R2 R1 use RE 10R2 for increased stability (110)(0.75 k) = 10R2 R2 = 8.25 k Choose R2 = 7.5 k
RC =
VRC
48
Substituting in the above equation: 7.5 k(24 V) 3.7 V = 7.5 k R1 R1 = 41.15 k Standard values: RE = 0.75 k, RC = 3.3 k, R2 = 7.5 k, R1 = 43 k 44.
1 1 VCC = (28 V) = 5.6 V 5 5 V 5.6 V = 1.12 k (use 1.1 k) RE = E I E 5 mA
VE =
VC = VRC
VCC 28 V + 5.6 V = 14 V + 5.6 V = 19.6 V VE 2 2 = VCC VC = 28 V 19.6 V = 8.4 V VRC
8.4 V = 1.68 k (use 1.6 k) 5 mA IC VB = VBE + VE = 0.7 V + 5.6 V = 6.3 V RV R (28 V) VB = 2 CC 6.3 V = 2 (2 unknowns) R2 R1 R2 R1
RC =
IC 5 mA = 135.14 I B 37 A RE = 10R2 (135.14)(1.12 k) = 10(R2) R2 = 15.14 k (use 15 k) (15.14 k)(28 V) Substituting: 6.3 V = 15.14 k R1 Solving, R1 = 52.15 k (use 51 k)
=
Standard values: RE = 1.1 k RC = 1.6 k R1 = 51 k R2 = 15 k 45.
4.7 k(20 V) = 4.14 V 4.7 k 18 k = 4.14 V 0.7 V = 3.44 V
(a) VB1 VE1
3.44 V = 3.44 mA 1 k VC1 20 V (3.44 mA)(2.2 k) = 12.43 V I C1 I E1
3.3 k(20 V) = 2.61 V 3.3 k 22 k = 2.61 V 0.7 V = 1.91 V
VB2 VE2
49
1.91 V = 1.59 mA 1.2 k 20 V (1.59 mA)(2.2 k) = 16.5 V
I E2 I C2 VC2
(b)
I B1 I B2
46.
I C1
I C2
3.44 mA = 21.5 μA, I C1 I E1 = 3.44 mA 160
1.59 mA = 17.67 μA, I C2 I E2 = 1.59 mA 90
(a)
D 12 = (50)(75) = 3750
(b)
I B1
RB ( D 1) RE = 4.19 μA I B2 (1 1) I B1 = (50 + 1)(4.19 µA) = 213.69 μA
(c)
I C1 1 I B1 = (50)(4.19 µA) = 0.21 mA
VCC VBE1 VBE2
18 V 0.7 V 0.7 V 2.2 M (3750+1)470
I C2 2 I B2 = (75)(213.69 µA) = 16.03 mA
(d) VC1 = 18 V, VC2 = 18 V VE2 I E RE I C2 RE (16.03 mA)(470 )
= 7.53 V VE1 VE2 + 0.7 V = 7.53 V + 0.7 V = 8.23 V 47.
3.3 k(22 V) = 4.48 V 3.3 k 4.7 k 8.2 k VE1 VB1 0.7 V = 3.78 V
(a) VB1
I E1 I E2 I C2 I C1 I B1 I B2
I C1 60 I C3 120
VE1 RE
3.78 V = 3.44 mA 1.1 k
= 57.33 μA, I C1 = 3.44 mA = 28.67 μA, I C2 = 3.44 mA
(b) VB1 = 4.48 V (4.7 k 3.3 k)(22 V) = 10.86 V 3.3 k 4.7 k 8.2 k = 3.78 V, VC1 VB2 0.7 V = 10.16 V
VB2 VE1
VE@ = 10.16 V, VC2 = 22 V (3.44 mA)(2 2 k) = 14.43 V
50
48.
(a)
I B1
VCC VEB1 RB 1 2 RC
12 V 0.7 V = 2.45 μA 1.8 M (80)(160)(220 )
I C1 1 I B1 (80)(2.45 A) = 196 μA I B2 I C1 = 196 μA I C2 2 I B2 (160)(196 A) = 31.36 mA
(b) VB1 I B1 RB (2.45 A)(1.8 M) = 4.41 V VE1 12 V I C (220 ) 12 V (31.36 mA)(220 ) = 5.1 V VB2 VBE2 = 0.7 V, VC2 VE1 = 5.1 V, VC1 VB2 = 0.7 V, VE2 = 0 V 18 V 0.7 V = 8.65 mA I 2 k
49.
I2 k =
50.
For current mirror: I(3 k) = I(2.4 k) = I = 2 mA
51.
6 V I B RB VBE I E RE 0 6 V I B 100 k 0.7 V ( 1) I B 1.2 k 0 6 V 0.7 V I B (100 k (120 1)(1.2 k)) 0 6 V 0.7 V IB = 21.61 µA 100 k 145 k I = I C I B (120)(21.61 A) = 2.59 mA
52.
53.
54.
4.3 k (18 V) = 9 V 4.3 k 4.3 k VE = 9 V 0.7 V = 9.7 V 18 V (9.7 V) = 4.6 mA = I IE = 1.8 k
VB
IE =
VZ VBE 5.1 V 0.7 V = 3.67 mA 1.2 k RE
VCC VBE 12 V 0.7 V 11.3 V = 22.16 A RB 510 k 510 k IC = IB = (100)(22.16 A) = 2.216 mA VC = VCC + ICRC = 12 V + (2.216 mA)(3.3 k) = 4.69 V VCE = VC = 4.69 V
IB =
51
55.
RE 10R2
(220)(0.75 k) 10(16 k) 165 k 160 k (checks) Use approximate approach: 16 k(22 V) = 3.59 V 16 k + 82 k VE = VB + 0.7 V = 3.59 V + 0.7 V = 2.89 V IC IE = VE/RE = 2.89/0.75 k = 3.85 mA I 3.85 mA IB = C = 17.5 A 220
VB
VC = VCC + ICRC = 22 V + (3.85 mA)(2.2 k) = 13.53 V V VBE 8 V 0.7 V 7.3 V = 2.212 mA 3.3 k 3.3 k RE VC = VCC + ICRC = 12 V + (2.212 mA)(3.9 k) = 3.37 V
56.
IE =
57.
I Csat
VCC 10 V = 4.167 mA RC 2.4 k
From characteristics I Bmax 31 A Vi VBE 10 V 0.7 V = 51.67 A 180 k RB 51.67 A 31 A, well saturated
IB =
Vo = 10 V (0.1 mA)(2.4 k) = 10 V 0.24 V = 9.76 V
52
58.
I Csat = 8 mA =
5V RC
5V = 0.625 k 8 mA IC 8 mA I Bmax = sat = 80 A 100 Use 1.2 (80 A) = 96 A 5 V 0.7 V = 44.79 k RB = 96 A
RC =
Standard values: RB = 43 k RC = 0.62 k 59.
(a) From Fig. 3.23c: IC = 2 mA: tf = 38 ns, tr = 48 ns, td = 120 ns, ts = 110 ns ton = tr + td = 48 ns + 120 ns = 168 ns toff = ts + tf = 110 ns + 38 ns = 148 ns (b) IC = 10 mA: tf = 12 ns, tr = 15 ns, td = 22 ns, ts = 120 ns ton = tr + td = 15 ns + 22 ns = 37 ns toff = ts + tf = 120 ns + 12 ns = 132 ns The turn-on time has dropped dramatically 168 ns:37 ns = 4.54:1 while the turn-off time is only slightly smaller 148 ns:132 ns = 1.12:1
60.
(a) Open-circuit in the base circuit Bad connection of emitter terminal Damaged transistor (b) Shorted base-emitter junction Open at collector terminal (c) Open-circuit in base circuit Open transistor
53
61.
(a) The base voltage of 9.4 V reveals that the 18 k resistor is not making contact with the base terminal of the transistor. If operating properly: VB
18 k(16 V) = 2.64 V vs. 9.4 V 18 k 91 k
As an emitter feedback bias circuit: VCC VBE 16 V 0.7 V R1 ( 1) RE 91 k (100 1)1.2 k = 72.1 A VB = VCC IB(R1) = 16 V (72.1 A)(91 k) = 9.4 V
IB =
(b) Since VE > VB the transistor should be “off” 18 k(16 V) = 2.64 V With IB = 0 A, VB = 18 k 91 k Assume base circuit “open” The 4 V at the emitter is the voltage that would exist if the transistor were shorted collector to emitter. 1.2 k(16 V) VE = =4V 1.2 k 3.6 k 62.
(a) RB, IB, IC, VC (b) , IC (c) Unchanged, I Csat not a function of (d) VCC, IB, IC (e) , IC, VRC , VRE , VCE
63.
(a) IB =
ETh VBE E VBE Th RTh ( 1) RE RTh RE
E VBE ETh VBE IC = IB = Th RTh RE RTh R
As ,
E
RTh
, IC, VRC VC = VCC VRC and VC
(b) R2 = open, IB, IC VCE = VCC IC(RC + RE) and VCE
54
(c) VCC, VB, VE, IE, IC (d) IB = 0 A, IC = ICEO and IC(RC + RE) negligible with VCE VCC = 20 V (e) Base-emitter junction = short IB but transistor action lost and IC = 0 mA with VCE = VCC = 20 V 64.
(a) RB open, IB = 0 A, IC = ICEO 0 mA and VC VCC = 18 V (b) , IC, VRC , VRE , VCE (c) RC, IB, IC, VE (d) Drop to a relatively low voltage 0.06 V (e) Open in the base circuit
65.
(a)
S ( I CO ) = 120
(b)
S (VBE ) /RB 120 / 510 k = 235 × 10-6 S
(c)
S ( ) I C1 /1 = 3.6 mA/120 = 30 × 10-6 A
(d)
I C S ( I CO )I CO S (VBE )VBE S ( ) (120)(10 A 0.2 A) (2.35 104 S )(0.5 V 0.7 V) (30 106 A)(150 120)
2.12 mA 66.
(a)
S ( I CO )
(1 RB /RE ) 125(1 270 k/2.2 k) RB /RE 125 270 k/2.2 k
= 62.44 (b)
S (VBE )
/ RE 125 / 2.2 k RB / RE 125 270 k / 2.2 k
= 229.3 × 106 S (c)
S ( )
I C1 (1 RB / RE )
1 (2 RB / RE )
4.41 mA(1 122.73) 125(156.25 122.73)
= 15.65 × 10-6 A
55
(d)
I C S ( I CO )I CO S (VBE )VBE S ( ) (62.44)(10 A 0.2 A) ( 229.36 106 S)(0.5 V 0.7 V) + 15.65 106 A(156.25 125) = 1.03 mA
67.
(1 RTh / RE ) (1 7.94 k / 0.68 k) (80) RTh / RE 80 7.94 k / 0.68 k
(a) S(ICO) =
= 11.06 / RE 80 / 0.68 k RTh / RE 80 7.94 k / 0.68 k = 1280 106S
(b) S(VBE) =
(c) S() =
I C1 (1 RTh / RE )
1 (2 RTh / RE )
1.71 mA(1 + 7.94 k / 0.68 k) 80(100 7.94 k / 0.68 k)
= 2.43 106 A
(d) IC = S(ICO)ICO + S(VBE) VBE + S() = (11.06)(10 A 0.2 A) + (1.28 103S)(0.5 V 0.7 V) + (2.43 106)(100 80) = 0.313 mA 68.
(a) S(ICO) = ( + 1)
(1 RB / RC ) 196.32(1 560 k / 3.9 k) RB / RC 196.32 560 k / 3.9 k
= 83.69 / RC 196.32 / 3.9 k RB / RC 196.32 560 k / 3.9 k = 1.48.04 106S
(b) S(VBE) =
(c) S() =
I C1 ( RB RC )
1 ( RB 2 RC )
2.56 mA(560 k 3.9 k) 196.32(560 k 245.4(3.9 k))
= 4.83 106A
(d) IC = S(ICO)ICO + S(VBE) VBE + S() = (83.69)(10 A 0.2 A) + (1.48.04 106S)(0.5 0.7 V) + (4.83 106A)(245.4 196.32) = 1.087 mA
56
69.
Type Fixed-bias Emitter-bias Voltage-divider Collector feedback S(ICO): S(VBE): S():
S(ICO) 120 62.44 11.06 83.69
S(VBE) 235 106S 229.36 106S 1280 106S 148.04 106S
S() 30 106 A 15.65 106 A 2.43 106A 4.83 106 A
Considerably less for the voltage-divider configuration compared to the other three. The voltage-divider configuration is the most sensitive . The voltage-divider configuration is the least sensitive with the fixed-bias configuration very sensitive.
In general, the voltage-divider configuration is the least sensitive with the fixed-bias the most sensitive. 70.
(a) Fixed-bias: S(ICO) = 120, IC = 1.176 mA S(VBE) = 235 106S, IC = 0.047 mA S() = 30 106A, IC = 0.90 mA (b) Voltage-divider bias: S(ICO) = 11.06, IC = 0.108 mA S(VBE) = 1280 106S, IC = 0.0256 mA S() = 2.43 106A, IC = 0.0048 × 106 A (c) For the fixed-bias configuration there is a strong sensitivity to changes in ICO and and much less to changes in VBE. For the voltage-divider configuration the opposite occurs with a high sensitivity to changes in VBE and less to changes in ICO and . In total the voltage-divider configuration is considerably more stable than the fixed-bias configuration.
57
Chapter 5 1.
(a) If the dc power supply is set to zero volts, the amplification will be zero. (b) Too low a dc level will result in a clipped output waveform. (c) Po = I2R = (5 mA)22.2 k = 55 mW Pi = VCCI = (18 V)(3.8 mA) = 68.4 mW P (ac) 55 mW o = 0.804 80.4% Pi (dc) 68.4 mW
2.
3.
No, because they intersect to define the Q point.
4.
1 1 = 15.92 2 fC 2 (1 kHz)(10 F) f = 100 kHz: XC = 0.159 Yes, better at 100 kHz
XC =
5.
6.
(a)
ro
(b)
ro
7.
VA VCEQ I CQ
100 V 8 V = 27 k 4 mA
VCE VCE 6 V 6V I C = 0.222 mA 27 k I C ro ro
Vi 10 mV I i 0.5 mA = 20 (=re)
(a) Zi =
(b) Vo = IcRL = IcRL = (0.98)(0.5 mA)(1.2 k) = 0.588 V Vo 0.588 V 10 mV Vi = 58.8
(c) Av =
(d) Zo =
58
(e) Ai = (f)
8.
Io Ie = = = 0.98 Ii Ie
Ib = Ie Ic = 0.5 mA 0.49 mA = 10 A 26 mV 26 mV = 13 2 mA I E (dc) Zi = re = (80)(13 ) = 1.04 k
(a) re =
(b) Ib = =
IC
Ie I Ie e 1 1
2 mA = 24.69 A 81
(c) Ai =
Io I L Ii Ib
IL =
ro ( I b ) ro RL
ro Ib ro RL ro Ai = Ib ro RL 40 k = (80) 40 k 1.2 k = 77.67
(d) Av =
RL ro
re 1.165 k = 13 = 89.6
9.
1.2 k 40 k 13
(a) Zi = re = (140)re = 1200 1200 = 8.57 re = 140 V 30 mV = 25 A (b) Ib = i Z i 1.2 k (c) Ic = Ib = (140)(25 A) = 3.5 mA
59
(d) IL = Ai =
ro I c (50 k)(3.5 mA) = 3.321 mA 50 k 2.7 k ro RL I L 3.321 mA = 132.84 25 A Ii
Vo Ai RL (2.7 k) = (132.84) 1.2 k Vi Zi = 298.89 Vi 48 mV (a) re = = 15 I i 3.2 mA
(e) Av =
10.
(b) Zi = re = 15 (c) IC = Ie = (0.99)(3.2 mA) = 3.168 mA (d) Vo = ICRL = (3.168 mA)(2.2 k) = 6.97 V (e) Av = (f)
11.
Vo 6.97 V = 145.21 Vi 48 mV
Ib = (1 )Ie = (1 0.99)Ie = (0.01)(3.2 mA) = 32 A VCC VBE 12 V 0.7 V = 51.36 A 220 k RB IE = ( + 1)IB = (60 + 1)(51.36 A) = 3.13 mA 26 mV 26 mV = 8.31 re = IE 3.13 mA Zi = RB re = 220 k (60)(8.31 ) = 220 k 498.6 = 497.47 ro 10RC Zo = RC = 2.2 k
(a) re:
IB =
(b) Av =
RC 2.2 k = 264.74 8.31 re
(c) Zi = 497.47 (the same) Zo = ro RC = 20 k 2.2 k = 1.98 k RC ro 1.98 k = 238.27 Av = re 8.31 12.
Av =
RC R 4.7 k re = C = 23.5 re (200) Av
60
re =
26 mV 26 mV 26 mV IE = = 1.106 mA IE re 23.5
IE 1.106 mA = 12.15 A 1 91 V VBE IB = CC VCC = IBRB + VBE RB = (12.15 A)(1 M) + 0.7 V = 12.15 V + 0.7 V = 12.85 V VCC VBE 8 V 0.7 V = 18.72 A (a) IB = RB 390 k IE = ( + 1)IB = (101)(18.72 A) = 1.89 mA 26 mV 26 mV = 13.76 re = IE 1.89 mA IC = IB = (100)(18.72 A) = 1.87 mA
IB =
13.
(b) Zi = RB re = 390 k (100)(13.76 ) = 390 k 1.38 k = 1.38 k ro 10RC Zo = RC = 5.6 k (c) Av =
(d) Av =
14.
Av =
RC 5.6 k = 406.98 13.76 re RC ro re
(5.6 k) (30 k) 13.76
4.72 k = 343.03 13.76
RC 406.98 = 203.49 2 re RC (203.49)(13.76 ) = 2.8 k
15.
(a) Test RE 10R2 ? (100)(1.2 k) 10(4.7 k) 120 k > 47 k (satisfied) Use approximate approach: RV 4.7 k(16 V) VB = 2 CC = 1.721 V R1 R2 39 k 4.7 k VE = VB VBE = 1.721 V 0.7 V = 1.021 V V 1.021 V IE = E = 0.8507 mA RE 1.2 k 26 mV 26 mV re = = 30.56 IE 0.8507 mA
61
(b) Zi = R1 R2 re = 4.7 k 39 k (100)(30.56 ) = 1.77 k ro 10RC Zo RC = 3.9 k (c) Av =
RC 3.9 k = 127.6 30.56 re
(d) ro = 25 k (b) Zi(unchanged) = 1.77 k Zo = RC ro = 3.9 k 25 k = 3.37 k (c) Av =
( RC ro )
(3.9 k) (25 k)
re 30.56 = 110.28 (vs. 127.6)
3.37 k 30.56
?
16.
RE 10R2
(100)(1 k) 10(5.6 k) 100 k > 56 k (checks!) & ro 10RC 50 k 10(3.3k) 33 k (checks) Use approximate approach: R R 3.3 k = 20.625 Av = C re C 160 re Av 26 mV 26 mV 26 mV re = = 1.261 mA IE 20.625 IE re VE VE = IERE = (1.261 mA)(1 k) = 1.261 V RE VB = VBE + VE = 0.7 V + 1.261 V = 1.961 V 5.6 k VCC = 1.961 V VB = 5.6 k 82 k 5.6 k VCC = (1.961 V)(87.6 k) VCC = 30.68 V
IE =
17.
Test RE 10R2 ? (180)(2.2 k) 10(56 k) 396 k < 560 k (not satisfied) Use exact analysis: (a) RTh = 56 k 220 k = 44.64 k 56 k(20 V) = 4.058 V ETh = 220 k 56 k ETh VBE 4.058 V 0.7 V IB = RTh ( 1) RE 44.64 k (181)(2.2 k) = 7.58 A 62
IE = ( + 1)IB = (181)(7.58 A) = 1.372 mA 26 mV 26 mV = 18.95 re = IE 1.372 mA (b) VE = IERE = (1.372 mA)(2.2 k) = 3.02 V VB = VE + VBE = 3.02 V + 0.7 V = 3.72 V VC = VCC ICRC = 20 V IBRC = 20 V (180)(7.58 A)(4.7 k) = 13.59 V (c) Zi = R1 R2 re = 56 k 220 k (180)(18.95 k) = 44.64 k 3.41 k = 3.17 k ro 10 RC 33.33 k 10(4.7 k) 47 k (checks) Av =
RC ro re
=
(6.8 k) (50 k)
18.95 = 298.15
18.
(a) Test: RE 10 R2 (70)(2.2 k) 10(68 k) 154 k 680 k No! Exact method: RTh 27 k 68 k 19.32 k 68 k(12 V) ETh = 8.59 V 68 k 27 k ETh VBE 9.59 V 0.7 V IB RTh ( 1) RE 19.32 k (70 1)(2.2 k) 44.95 A I E ( 1) I B (70 1)(44.95 A) = 3.19 mA 26 mV 26 mV = 8.15 re 3.19 mA IE
(b) VB VBE I E RE 0.7 V + (3.19 mA)(2.2 k) 7.72 V
63
VC 24 V I C RC 24 V ( I B )(3.3 k) 24 V (70)(44.95 A)(3.3 k) 13.61 V VCB VC VB 13.61 V 7.72 V = 5.89 V VE VB 0.7 7.72 V 0.7 V = 7.02 V VCE VC VE 13.61 V 7.02 V = 6.59 V
(c)
Z i 68 k 27 k RE 19.3 k (70)(2.2 k) 19.32 k 154 k 17.17 k
(d)
19.
A
RC 3.3 k = 1.5 RE 2.2 k
VCC VBE 20 V 0.7 V RB ( 1) RE 390 k (141)(1.2 k) 19.3 V = = 34.51 A 559.2 k IE = ( + 1)IB = (140 + 1)(34.51 A) = 4.866 mA 26 mV 26 mV = 5.34 re = 4.866 mA IE
(a) IB =
(b) Zb = re + ( + 1)RE = (140)(5.34 k) + (140 + 1)(1.2 k) = 747.6 + 169.9 k = 169.95 k Zi = RB Zb = 390 k 169.95 k = 118.37 k Zo = RC = 2.2 k (c) Av =
RC Zb
=
(140)(2.2 k) = 1.81 169.95 k
( 1) RC / ro (d) Zb = re + RE 1 ( RC RE ) / ro (141) 2.2 k / 20 k = 747.6 1.2 k 1 (3.4 k) / 20 k = 747.6 + 144.72 k = 145.47 k
Zi = RB Zb = 390 k 145.47 k = 105.95 k Zo = RC = 2.2 k (any level of ro)
64
Av =
Vo Vi
RC
re RC 1 Z b ro ro R 1 C ro
(140)(2.2 k) 5.34 2.2 k 1 145.47 k 20 k 20 k = 2.2 k 1 20 k 2.117 0.11 = = 1.81 1.11
20.
(a) dc analysis the same re = 5.34 (as in #19) (b) Zi = RB Zb = RB re = 390 k (140)(5.34 ) = 746.17 vs. 118.37 k in #32 Zo = RC = 2.2 k (as in #32) (c) Av =
RC 2.2 k = 411.99 vs 1.81 in #19 5.34 re
(d) Zi = 746.17 vs. 105.95 k for #19 Zo = RC ro = 2.2 k 20 k = 1.98 k vs. 2.2 k in #19 RC ro 1.98 k Av = = 370.79 vs. 1.81 in #19 5.34 re Significant difference in the results for Av. 21.
ro 10 RC 100 k 10(8.2 k) 82 k Therefore:
Av =
RC Zb
RC R C = 10 RE RE
RC 8.2 k = 0.82 k 10 10 26 mV 26 mV = 6.842 mA IE = 3.8 re VE = IERE = (6.842 mA)(0.82 k) = 5.61 V VB = VE + VBE = 5.61 V + 0.7 V = 6.31 V IE 6.842 mA IB = = 56.55 A ( 1) 121 VR V VB 20 V 6.31 V = 242.09 k and RB = B CC 56.55 A IB IB
RE =
65
22.
VCC VBE RB ( 1) RE 22 V 0.7 V 21.3 V = 330 k (81)(1.2 k 0.47 k) 465.27 k = 45.78 A IE = ( + 1)IB = (81)(45.78 A) = 3.71 mA 26 mV 26 mV =7 re = 3.71 mA IE
(a) IB =
(b) ro < 10(RC + RE) ( 1) RC / ro Zb = re + RE 1 ( RC RE ) / ro (81) 5.6 k / 40 k = (80)(7 ) + 1.2 k 1 6.8 k / 40 k 81 0.14 1.2 k = 560 + 1 0.17
(note that ( + 1) = 81 RC/ro = 0.14) = 560 + [81.14 /1.17]1.2 k = 560 + 83.22 k = 83.78 k Zi = RB Zb = 330 k 83.78 k = 66.82 k
Av =
RC Zb
re RC 1 r r o o RC 1 ro
7 5.6 k (80)(5.6 k) 1 83.78 k 40 k 40 k = 1 5.6 k/40 k (5.35) 0.14 = 1 0.14 = 4.57
23.
(a) Test: RE 10 R2 (200)(1.2 k) 10(120 k) 240 k 1.2 M No!
66
Exact method: RTh 430 k 120 k 93.82 k 120 k(16 V) ETh = 3.49 V 120 k 430 k ETh VBE 3.49 V 0.7 V IB RTh ( 1) RE 93.82 k (200 1)(1.2 k) 8.33 A I E ( 1) I B (200 1)(8.33 A) = 1.674 mA 26 mV 26 mV = 15.53 re 1.674 mA IE
(b) VB VE 0.7 V ( I E )(RE ) 0.7 V + (1.674 mA)(1.2 k) 0.7 V 2.71V I C I B (200)(8.33 A) = 1.67 mA VC VCC I C RC 16 V (1.67 mA)(4.7 k) 16 V 7.85 V 8.15 V VE I E RE (1.674 mA)(1.2 k) = 2.01 V VCE VC VE 8.15 V 2.01 V = 6.14 V VCB VC VB 8.15 V 2.71 V = 5.44 V
(c)
Z i 120 k 430 k RE 93.82 k (200)(1.2 k)
93.82 k 240 k 67.45 k Z o = 4.7 k
24.
RC 4.7 k = 3.92 1.2 k RE
(d)
A
(e)
Ai A
Z i (3.92)(67.45 k) (4.7 k) RC = 56.26
VCC VBE 16 V 0.7 V 15.3 V RB ( 1) RE 270 k (111)(2.7 k) 569.7 k = 26.86 A IE = ( + 1)IB = (110 + 1)(26.86 A) = 2.98 mA 26 mV 26 mV re = = 8.72 2.98 mA IE re = (110)(8.72 ) = 959.2
(a) IB =
67
(b) Zb = re + ( + 1)RE = 959.2 + (111)(2.7 k) = 300.66 k Zi = RB Zb = 270 k 300.66 k = 142.25 k Zo = RE re = 2.7 k 8.72 = 8.69 (c) Av =
25.
RE 2.7 k 0.997 RE re 2.7 k 8.69
VCE VBE 8 V 0.7 V = 6.84 A RB ( 1) RE 390 k (121)5.6 k IE = ( + 1)IB = (121)(6.84 A) = 0.828 mA 26 mV 26 mV = 31.4 re = 0.828 mA IE ro < 10RE: ( 1) RE Zb = re + 1 RE / ro (121)(5.6 k) = (120)(31.4 ) + 1 5.6 k/40 k = 3.77 k + 594.39 k = 598.16 k Zi = RB Zb = 390 k 598.16 k = 236.1 k Zo RE re = 5.6 k 31.4 = 31.2
(a) IB =
( 1) RE / Z b 1 RE / ro (121)(5.6 k) / 598.16 k = 1 5.6 k / 40 k = 0.994
(b) Av =
V0 = 0.994 Vi Vo = AvVi = (0.994)(1 mV) = 0.994 mV
(c) Av =
68
26.
(a)
? Test RE 10R2 (200)(2 k) 10(8.2 k) 400 k 82 k (checks)!
Use approximate approach: 8.2 k(20 V) = 2.5545 V VB = 8.2 k 56 k VE = VB VBE = 2.5545 V 0.7 V 1.855 V V 1.855 V IE = E = 0.927 mA RE 2 k IE 0.927 mA = 4.61 A ( 1) (200 + 1) IC = IB = (200)(4.61 A) = 0.922 mA
IB =
(b) re =
26 mV 26 mV = 28.05 0.927 mA IE
(c) Zb = re + ( + 1)RE = (200)(28.05 ) + (200 + 1)2 k = 5.61 k + 402 k = 407.61 k Zi = 56 k 8.2 k 407.61 k = 7.15 k 407.61 k = 7.03 k Zo = RE re = 2 k 28.05 = 27.66 (d) Av =
27.
RE 2 k = 0.986 RE re 2 k 28.05
VEE VBE 6 V 0.7 V = 0.779 mA RE 6.8 k 26 mV 26 mV re = = 33.38 0.779 mA IE
(a) IE =
(b) Zi = RE re = 6.8 k 33.38 = 33.22 Zo = RC = 4.7 k (c) Av =
RC
re = 140.52
(0.998)(4.7 k) 33.38
69
28.
75 = 0.9868 1 76 V VBE 5 V 0.7 V 4.3 V IE = EE = 1.1 mA RE 3.9 k 3.9 k 26 mV 26 mV re = = 23.58 1.1 mA IE
=
Av =
29.
RC (0.9868)(3.9 k) = 163.2 23.58 re
VCC VBE 12 V 0.7 V RF RC 220 k 120(3.9 k) = 16.42 A IE = ( + 1)IB = (120 + 1)(16.42 A) = 1.987 mA
(a) IB =
re =
26 mV 26 mV = 13.08 1.987 mA IE
(b) Zi = re
RF Av
Need Av! RC 3.9 k = 298 Av = 13.08 re 220 k Zi = (120)(13.08 ) 298 = 1.5696 k 738 = 501.98 Zo = RC RF = 3.9 k 220 k = 3.83 k (c) From above, Av = 298 30.
RC = 160 re RC = 160(re) = 160(10 ) = 1.6 k
Av =
Ai =
RF 200 RF = 19 19 = RF 200(1.6 k) RF RC 19RF + 3800RC = 200RF 3800 RC 3800(1.6 k) = 181 181 = 33.59 k
RF =
70
VCC VBE RF RC IB(RF + RC) = VCC VBE and VCC = VBE + IB(RF + RC) 26 mV 26 mV = 2.6 mA with IE = 10 re
IB =
IE 2.6 mA = 12.94 A 1 200 1 VCC = VBE + IB(RF + RC) = 0.7 V + (12.94 A)(33.59 k + (200)(1.6 k)) = 5.28 V
IB =
31.
(c)
(a) Av: Vi = Ibre + ( + 1)IbRE Io + I = IC = Ib but Ii = I + Ib and I = Ii Ib Substituting, Io + (Ii Ib) = Ib and Io = ( + 1)Ib Ii Assuming ( + 1)Ib Ii Io ( + 1)Ib and Vo = IoRC = ( + 1)IbRC
Therefore,
Vo ( 1) I b RC Vi I b re ( 1) I b RE
and Av =
I b RC I b re I b RE
Vo RC R C Vi re RE RE
71
(b) Vi Ib(re + RE) For re RE Vi IbRE Now Ii = I + Ib V Vo Ib = i RF Since Vo Vi V Ii = o I b RF or Ib = Ii +
Vo RF
and Vi = IbRE Vi = REIi + but Vo = AvVi and Vi = REIi + or
so
Vi
Vo RE RF
AvVi RE RF
Av RE Vi = REIi RF
A R Vi 1 v E RE I i RF V RE RE RF Zi = i A R Ii RF ( Av ) RE 1 v E RF
Zi = with Zi = Zi
Vi = x y Ii
where x = RE and y = RF/Av
x y ( RE )( RF / Av ) x y RE RF / Av
RE RF RE Av RF
Zo: Set Vi = 0
72
Vi = Ibre + ( + 1)IbRE Vi Ib(re + RE) = 0 since , re + RE 0
Ib = 0 and Ib = 0
1 1 RC RF V R R 1 C F = RC RF and Zo = o 1 1 Io RC RF RC RF
Io =
(c) Av Zi
Vo V o Vo RC RF
RC 2.2 k = 1.83 1.2 k RE
RE RF
RE Av RF = 40.8 k Zo RC RF = 2.2 k 120 k = 2.16 k
32.
(90)(1.2 k)(120 k) (90)(1.2 k)(1.83) 120 k
VCC VBE 9 V 0.7 V RF RC (39 k 22 k) (80)(1.8 k) 8.3 V 8.3 V = 40.49 A = 61 k 144 k 205 k IE = ( + 1)IB = (80 + 1)(40.49 A) = 3.28 mA 26 mV 26 mV = 7.93 re = 3.28 mA IE
(a) IB =
Zi = RF1 re = 39 k (80)(7.93 ) = 39 k 634.4 = 0.62 k Zo = RC RF2 = 1.8 k 22 k = 1.66 k
73
1.8 k 22 k R RC RF2 re re 7.93 1.664 k = 209.82 = 7.93
(b) Av =
33.
(a)
RE
Zi
0.68 k 1 (0.68 k 1.8 k) 80 61 k
1 ( RE RC ) RF 0.68 k 0.68 k 3 3 12.5 10 4.066 10 53.16 103 12.79 k Z o RC RF 1.8 k 61 k = 1.75 k
(b)
34.
A
RC 1.8 k = 2.65 0.68 k RE
VCC VBE 18 V 0.7 V = 25.44 A RB 680 k IE = ( + 1)IB = (100 + 1)(25.44 A) = 2.57 mA 26 mV = 10.116 re = 2.57 mA R 3.3 k = 326.22 AvNL C 10.116 re Zi = RB re = 680 k (100)(10.116 ) = 680 k 1,011.6 = 1.01 k Zo = RC = 3.3 k
(a) IB =
(b) RL 4.7 k (326.22) AvNL 4.7 k 3.3 k RL Ro = 191.65
(c)
AvL =
(d)
AiL = Av
Zi (1.01 k) (191.65) 4.7 k RL = 41.18
74
35.
(a)
AvNL = 326.22 RL Av RL Ro NL
AvL =
4.7 k (326.22) = 191.65 4.7 k 3.3 k 2.2 k RL = 2.2 k: AvL = (326.22) = 130.49 2.2 k 3.3 k 0.5 k (326.22) = 42.92 RL = 0.5 k: AvL = 0.5 k 2.3 k As RL, AvL
RL = 4.7 k: AvL =
(b) No change for Zi, Zo, and AvNL ! 36.
VCC VBE 12 V 0.7 V = 11.3 A RB 1 M IE = ( + 1)IB = (181)(11.3 A) = 2.045 mA 26 mV 26 mV = 12.71 re = IE 2.045 mA
(a) IB =
AVNL =
RC 3 k = 236 re 12.71
Zi = RB re = 1 M (180)(12.71 ) = 1 M 2.288 k = 2.283 k Zo = RC = 3 k (b) (c) No-load: Av = AvNL = 236 (d)
Zi 2.283 k(236) AvNL 2.283 k 0.6 k Z i Rs = 186.9
Avs
(e) No change! (f)
Avs
Zi 2.283 k( 236) ( AvNL ) = 164.1 2.283 k 1 k Z i Rs
Rs, Avs (g) No change! (h)
Ai A
Zi (2.283 k) (236) 3 k RC
179.6
75
37.
VCC VBE 24 V 0.7 V = 41.61 A RB 500 k IE = ( + 1)IB = (80 + 1)(41.61 A) = 3.37 mA 26 mV 26 mV = 7.715 re = IE 3.37 mA
(a) IB =
RL 4.3 k = 557.36 re 7.715 Zi = RB re = 560 k (80)(7.715 ) = 560 k 617.2 = 616.52 Zo = RC = 4.3 k AvNL =
(b) (c)
Vo RL 2.7 k(557.36) AvNL Vi RL Ro 2.7 k 4.3 k = 214.98 V V V Avs o o i Vs Vi Vs AvL =
Vi =
Z iVs 616.52 Vs = 0.381 Vs Z i Rs 616.52 1 k
Avs (214.98)(0.381)
= 81.91 (d)
R Zi Ais Avs s RL
(e)
AvL =
1 k 616.52 (81.91) 2.7 k = 49.04
Vo RL 5.6 k(557.36) = 315.27 AvNL Vi RL Ro 5.6 k 4.3 k
Vi the same = 0.381 Vs Avs
Vo Vi = (315.27)(0.381) = 120.12 Vi Vs
As RL, Avs (f)
AvL the same = 214.98 Vi Zi 616.52 = 0.552 Vs Z i Rs 616.52 0.5 k Avs
Vo Vi = (214.98)(0.552) = 118.67 Vi Vs
As Rs, Avs (g) No change!
76
38.
(a) Exact analysis: R2 16 k(16 V) ETh = = 3.048 V VCC R1 R2 68 k 16 k RTh = R1 R2 = 68 k 16 k = 12.95 k ETh VBE 3.048 V 0.7 V RTh ( 1) RE 12.95 k (101)(0.75 k) = 26.47 A IE = ( + 1)IB = (101)(26.47 A) = 2.673 mA 26 mV 26 mV = 9.726 re = IE 2.673 mA
IB =
RC 2.2 k = 226.2 re 9.726 Zi = 68 k 16 k re = 12.95 k (100)(9.726 ) = 12.95 k 972.6 = 904.66 Zo = RC = 2.2 k AvNL
(b) RL 5.6 k(226.2) ( AvNL ) = 162.4 RL Z o 5.6 k 2.2 k
(c)
AvL =
(d)
AiL = AvL
Zi RL
= (162.4)
(904.66 ) 5.6 k
= 26.24 (e)
AvL =
RC Re re
2.2 k 5.6 k 9.726
= 162.4 Zi = 68 k 16 k 972.6
re = 904.66 Z AiL = AvL i RL (162.4)(904.66 ) = 5.6 k = 26.24 Zo = RC = 2.2 k Same results!
77
39.
(a)
AvL =
RL Av RL Z o NL
4.7 k (226.4) = 154.2 4.7 k 2.2 k 2.2 k = (226.4) = 113.2 2.2 k 2.2 k 0.5 k = (226.4) = 41.93 0.5 k 2.2 k
RL = 4.7 k: AvL = RL = 2.2 k: AvL RL = 0.5 k: AvL RL, AvL (b) Unaffected! 40.
VCC VBE 18 V 0.7 V RB ( 1) RE 680 k (111)(0.82 k) = 22.44 A IE = ( + 1)IB = (110 + 1)(22.44 A) = 2.49 mA 26 mV 26 mV = 10.44 re = IE 2.49 mA
(a) IB =
RC 3 k re RE 10.44 0.82 k = 3.61 Zi RB Zb = 680 k (re + ( + 1)RE) = 680 k (610)(10.44 ) + (110 + 1)(0.82 k) = 680 k 92.17 k = 81.17 k Zo RC = 3 k AvNL
(b) (c)
Vo RL 4.7 k(3.61) AvNL Vi RL Ro 4.7 k 3 k = 2.2 V V V Avs o o i Vs Vi Vs AvL =
Vi =
Z iVs 81.17 k (Vs ) = 0.992 Vs Z i Rs 81.17 k 0.6 k
Avs = (2.2)(0.992)
= 2.18 (d) None!
78
(e)
AvL – none! Vi Zi 81.17 k = 0.988 Vs Z i Rs 81.17 k 1 k Avs = (2.2)(0.988)
= 2.17 Rs, Avs , (but only slightly for moderate changes in Rs since Zi is typically much larger than Rs) (f)
Ai A L
Zi (81.17 k) (2.2) RL 4.7 k
37.99
41.
Using the exact approach: ETh VBE RTh ( 1) RE 2.33 V 0.7 V = 10.6 k (121)(1.2 k) = 10.46 A
IB =
R2 VCC R1 R2 12 k = (20 V) = 2.33 V 91 k 12 k RTh = R1 R2 = 91 k 12 k = 10.6 k
ETh =
IE = ( + 1)IB = (121)(10.46 A) = 1.266 mA 26 mV 26 mV = 20.54 re = IE 1.266 mA (a)
RE 1.2 k = 0.983 re RE 20.54 1.2 k Zi = R1 R2 (re + ( +1)RE) = 91 k 12 k ((120)(20.54 ) + (120 + 1)(1.2 k)) = 10.6 k (2.46 k + 145.2 k) = 10.6 k 147.66 k = 9.89 k AvNL
Zo = RE re = 1.2 k 20.54 = 20.19 (b) (c)
RL 2.7 k(0.983) AvNL RL Z o 2.7 k 20.19 = 0.976 Zi 9.89 k(0.976) Avs AvL Z i Rs 9.89 k 0.6 k = 0.92 AvL =
79
(d)
AvL = 0.976 (unaffected by change in Rs) Zi 9.89 k(0.976) AvL Z i Rs 9.89 k 1 k = 0.886 (vs. 0.92 with Rs = 0.6 k) As Rs, Avs Avs
(e) Changing Rs will have no effect on AvNL , Zi, or Zo. (f)
(g)
RL 5.6 k(0.983) AvNL RL Z o 5.6 k 20.19 = 0.979 (vs. 0.976 with RL = 2.7 k) Zi 9.89 k(0.979) ( AvL ) Avs Z i Rs 9.89 k 0.6 k = 0.923 (vs. 0.92 with RL = 2.7 k) As RL, Av, Avs
AvL =
Ai A L
Zi (9.89 k) (0.979) RL 2.7 k
3.59
42.
VEE VBE 6 V 0.7 V RE 2.2 k = 2.41 mA 26 mV 26 mV re = = 10.79 IE 2.41 mA
(a) IE =
RC 4.7 k = 435.59 re 10.79 Zi = RE re = 2.2 k 10.79 = 10.74 Zo = RC = 4.7 k AvNL =
(b) (c)
AvL =
Vi =
RL 5.6 k(435.59) = 236.83 AvNL RL Ro 5.6 k 4.7 k
Zi 10.74 (Vs ) = 0.097 Vs Vs 10.74 100 Z i Rs
Vo Vi = (236.83)(0.097) Vi Vs = 22.97 (d) Vi = Ie re Vo = IoRL 4.7 k( I e ) Io = = 0.4563Ie 4.7 k 5.6 k Avs
80
AvL =
Vo (0.4563 I e ) RL 0.4563(5.6 k) = Vi 10.79 I e re = 236.82 (vs. 236.83 for part c)
Avs : 2.2 k 10.79 = 10.74
Vi =
Zi 10.74 (Vs ) Vs = 0.097 Vs 10.74 100 Z i Rs
Vo Vi = (236.82)(0.097) Vi Vs = 22.97 (same results)
Avs
(e)
RL 2.2 k (435.59) AvNL RL Ro 2.2 k 4.7 k = 138.88 V V Vi Zi 10.74 = 0.021 Avs 0 i , Vi Vs Vs Z i Rs 10.74 500 AvL =
Avs = (138.88)(0.021) = 2.92 Avs very sensitive to increase in Rs due to relatively small Zi; Rs, Avs AvL sensitive to RL; RL, Av
(f)
Zo = RC = 4.7 k unaffected by value of Rs!
(g) Zi = RE re = 10.74 unaffected by value of RL! (h)
Z i (236.82)(10.74 k) RL 5.6 k
Ai A L
= 0.454
43.
(a)
Av1 Av2
(b)
RL AvNL RL Ro RL AvNL RL Ro
1 k(420) = 97.67 1 k 3.3 k
2.7 k(420) = 189 2.7 k 3.3 k
AvL = Av1 Av2 = (97.67)(189) = 18.46 103 Avs
Vo Vo Vo1 Vi1 Vs Vi2 Vi1 Vs
= Av2 Av1 Vi =
Vi Vs
Z iVs 1 k(Vs ) = 0.625 Z i Rs 1 k 0.6 k
Avs = (189)(97.67)(0.625)
= 11.54 103
81
(c)
Ai1 Ai2
(d)
Av Z i (97.67)(1 k) = 97.67 RL 1 k
Av Z i (189)(1 k) = 70 RL 2.7 k
AiL Ai1 Ai2 = (97.67)(70) = 6.84 103
(e) No effect! (f)
No effect!
(g) In phase 44.
(a)
Av1
Z i2 Z i2 Z o1
Av1NL
1.2 k (1) 1.2 k 20
= 0.984 RL 2.2 k (640) Av2 = Av2 NL RL Z o2 2.2 k 4.6 k = 207.06 (b)
AvL Av1 Av2 = (0.984)(207.06)
= 203.74 Zi Avs Av Z i Rs L 50 k (203.74) = 50 k 1 k = 199.75 (c)
Ai1 Av1
Z i1 Z i2
= (0.984) = 41 Ai2 Av2
(50 k) 1.2 k
Z i2 RL
= (207.06)
(1.2 k) 2.2 k
= 112.94 (d)
AiL AvL
Z i1 RL
= (203.74) = 4.63 103
(50 k) 2.2 k
82
(e) A load on an emitter-follower configuration will contribute to the emitter resistance (in fact, lower the value) and therefore affect Zi (reduce its magnitude). (f)
The fact that the second stage is a CE amplifier will isolate Zo from the first stage and Rs.
(g) The emitter-follower has zero phase shift while the common-emitter amplifier has a 180 phase shift. The system, therefore, has a total phase shift of 180 as noted by the negative sign in front of the gain for AvT in part b. 45.
For each stage: 6.2 k (15 V) = 3.08 V VB = 24 k 6.2 k VE = VB 0.7 V = 3.08 V 0.7 V = 2.38 V V 2.38 V IE IC = E = 1.59 mA RE 1.5 k VC = VCC ICRC = 15 V (1.59 mA)(5.1 k) = 6.89 V
46.
(a) re =
26 mV 26 mV = 16.35 IE 1.59 mA
Ri1 Ri2 = R1 R2 re = 6.2 k 24 k (150)(16.35 )
= 1.64 k RC Ri2 5.1 k 1.64 k = 75.8 Av1 16.35 re Av2
RC 5.1 k = 311.9 re 16.35
Av = Av1 Av2 = (75.8)(311.9) = 23,642 (b)
AiT
Io Z AT i Ii RL
(23,642)(1.64 k) 5.1 k 7602.5
47.
3.9 k (20 V) = 4.4 V 3.9 k 6.2 k 7.5 k 6.2 k 3.9 k VB2 (20 V) = 11.48 V 3.9 k 6.2 k 7.5 k VE1 VB1 0.7 V = 4.4 V 0.7 V = 3.7 V VB1
I C1 I E1
VE1 RE
3.7 V = 3.7 mA I E2 I C2 1 k
VC2 = VCC ICRC = 20 V (3.7 mA)(1.5 k) VC1 VB2 VBE2
= 14.45 V 11.48 V 0.7 V = 10.78 V
83
48.
re =
26 mV 26 mV =7 IE 3.7 mA
Av1 Av2
re = 1 re
RC 1.5 k 214 7 re
AvT Av1 Av2 = (1)(214) = 214
Vo = AvT Vi = (214)(10 mV) = 2.14 V 49.
(Vo (from problem #48) = 2.14 V) Ro = RD = 1.5 k RL 10 k Vo(load) = (Vo ) (2.14 V) Ro RL 10 k 1.5 k = 1.86 V
50.
(a)
D 12 (50)(120) 6000 IB
VCC VBE1 VBE2 RB D RE
16 V 1.4 V 2.4 M (6000)(510 )
2.67 A VB1 16 V ( I B )(2.4 M) 16 V (2.67 A)(2.4 M) 9.59 V VC1 = 16 V
VE2 I E2 RE D I B RE (6000)(2.67 A)(510 ) 8.17 V VCB1 VC1 VB1 16 V 9.59 V = 6.41 V VCE2 16 V VE2 16 V 8.17 V = 7.83 V
(b)
I B1 = 2.67 μA, I B2 I C1 1 I B1 = (50)(2.67 µA) = 133.5 μA I E2 VE2 / RE 8.17 V / 510 = 16.02 mA
(c)
Z i RB D RE 2.4 M (6000)(510 ) 2.4 M 3.06 M = 1.35 M re Z o 1 re2 with I E1 ( 1) I B1 (51)(2.67 µA) = 136.17 µA
2
and re1 re2
26 mW 26 mW = 190.94 136.17 A I E1 26 mW 26 mW = 1.62 16.02 mA I E2
190.94 1.62 1.59 1.62 120 = 3.21
Zo
84
(d)
A 1
D RB (6000)(2.4 M) 14.4 103 M RB D RE 2.4 M (6000)(510 ) 5.46 M 3 = 2.64 × 10
Ai
51.
(a) and (b) the same as problem 50 (a) VB1 = 9.59 V, VC1 = 16 V, VE2 = 8.17 V, VCB1 = 6.41 V, VCE2 = 7.83 V (b)
I B1 = 2.67 μA, I B2 = 133.5 μA, I E2 = 16.02 mA
(c)
Z i RB ( RE RL ) 2.4 M (6000)(510 1.2 k) 2.4 M 2.15 M = 1.13 M Zo
(d)
re1
2
re2 not affected by RL , Z o = 3.21 from problem 50
A 1
D RB (6000)(2.4 M) RB D ( RE RL ) 2.4 M (6000)(510 0.2 k) = 3.16 × 103
Ai
52.
From Problem 51, Z i = 1.35 M, and A NL 1 1.35 M(Vs ) 0.999 Vs 1.35 M 1.2 k A Vo / Vs 1 Vi
53.
(a)
IB
VCC VBET RB D RE
16 V 1.6 V 3.24 A 2.4 M (4000)(510 )
VB1 VCC I B RB 16 V (3.24 A)(2.4 M) = 8.22 V VE2 I E2 RE D I B RE (4000)(3.24 A)(510 ) = 6.61 V VC2 VCC I C2 RC 16 V I E2 RC 16 V D I B RC 16 V (4000)(3.24 A)(470 ) = 9.91 V VCE2 VC2 VE2 9.91 V 6.61 V = 3.3 V VCB1 VC1 VB1 VC2 VB1 9.91 V 8.22 V = 1.69 V
85
(c)
I E2 D I B (4000)(3.24 A) = 12.96 mA re2
26 mV 26 mW =2 12.96 mA I E2
Z i 2.4 M D re2 2.4 M (4000)(2 ) 2.4 M 8 k
8 k Z o RC = 470
54.
(d)
A
D RC (4000)(470 ) = 235 Zi 8 k
(e)
Ai
A Z i (235)(8 k) = 4 × 103 RC 470
(a)
I B1
VCC VBE1 RB 1 2 RC
16 V 0.7 V 4.16 A 1.5 M (160)(200)(68 )
VB1 I B1 RB (4.16 A)(1.5 M) = 6.24 V VE2 = 0 V VC2 VE1 16 V I C 68 16 V (12 I B1 )68 16 V (160)(200)(4.16 A)(68 ) 6.95 V VB2 VBE2 = 0.7 V VC1 VB2 = 0.7 V VE1 VC2 = 6.95 V
(b)
I B1 = 4.16 μA I C1 1 I B1 (160)(4.16 µA) = 0.666 mA I B2 I C1 = 0.666mA, I C2 I C 12 I B1 = 133.12 mA I E2 I C2 = 133.12 mA
(c)
Z i RB 12 RC 1.5 M (160)(200)(68 ) = 2.17 M Z o RC ro1 68
(d)
A = 1
(e)
Ai
1 2 RB (160)(200)(1.5 M) 13.06 × 103 RB 12 RC 1.5 M (160)(200)(68 )
86
55.
(a)
I B1
VCC VBE1 RB 1 2 RC
16 V 0.7 V 4.16 A 1.5 M (160)(200)(68 )
VB1 I B1 RB (4.16 A)(1.5 M) = 6.24 V VE2 I E2 RB I C2 RE (12 I B1 ) RE (160)(200)(4.16 A)(22 ) = 2.93 V VC2 16 V ( I C )(68 ) 16 V (12 I B1 )(68 ) 16 V (160)(200)(4.16 A)(68 ) 6.95 V VB2 VBE2 VE2 = 0.7 V + 2.93 V = 3.63V VC1 VB2 = 3.63 V VE1 VC2 = 6.95 V
(b)
I B1 = 4.16 μA I C1 1 I B1 (160)(4.16 µA) = 0.666 mA I B2 I C1 = 0.666mA, I C2 I C 12 I B1 = 133.12 mA I E2 I C2 = 133.12 mA
(c)
Z i RB 12 RC 1.5 M (160)(200)(68 ) = 0.887 M Z o RC ro1 68
56.
(d)
A 1
(e)
Ai
1 2 RB (160)(200)(1.5 M) 13.06 × 103 RB 12 RC 1.5 M (160)(200)(68 )
(a) dc same as 54 (b) dc same as 54 (c)
Z i RB 12 ( RC RL ) 1.5 (160)(200)(68 1.2 k) =1.5 (160)(200)(64.35 ) 1.5 M 2.06 M) = 0.867 M Z o 68
87
(d)
A 1 Ai
12 RB (160)(200)(1.5 M) RB 12 ( RC RL ) 1.5 M (160)(200)(64.55 )
= 13.49 × 103 57.
re =
26 mV 26 mV = 21.67 I E (dc) 1.2 mA
re = (120)(21.67 ) = 2.6 k 58.
59.
60.
61.
62.
(a) Av =
Vo = 160 Vi Vo = 160 Vi
(b) Ib = =
Vi hreVo Vi hre AvVi Vi (1 hre Av ) hie hie hie
Vi 1 (2 104 )(160)
1 k Ib = 9.68 104Vi
(c) Ib =
Vi = 1 103Vi 1 k
(d) % Difference =
1 103 Vi 9.68 104 Vi 1 103 Vi
100%
= 3.2 % (e) Valid first approximation 63.
% difference in total load =
RL RL 1/ hoe RL
100%
2.2 k (2.2 k 50 k)
100% 2.2 k 2.2 k 2.1073 k = 100% 2.2 k = 4.2 % =
In this case the effect of 1/hoe can be ignored.
88
64.
(hie = 4 k, hre = 4.05 104)
(a) Vo = 180Vi
Vi (4.05 104 )(180Vi ) 4 k = 2.32 104Vi
(b) Ib =
(c) Ib =
Vi V i = 2.5 104Vi hie 4 k
(d) % Difference =
2.5 104 Vi 2.32 104 Vi 2.5 104 Vi
100% = 7.2%
(e) Yes, less than 10% 65.
From Fig. 5.18 min max hoe: 1 S
30 S
Avg =
(1 30) S = 15.5 S 2
1/ hoe 64.53 k 3.3 k (3.3 k 64.52 k) 100% 4.88% 3.3 k Ignore effects
66.
(a) hfe = = 120 hie re = (120)(4.5 ) = 540 1 1 = 25 S hoe = ro 40 k 1 k = 11.11 90 = hfe = 90 1 1 ro = hoe 20 S = 50 k (a) re = 8.31 (from problem 26)
(b) re
67.
(b)
hie
h fe = = 60 hie = re = (60)(8.31 ) = 498.6
(c) Zi = RB hie = 220 k 498.6 = 497.47 Zo = RC = 2.2 k (d) Av =
h fe RC
hie Ai hfe = 60
(60)(2.2 k) = 264.74 498.6
89
(e) Zi = 497.47 (the same) 1 = 40 k Zo = ro RC, ro = 25 S = 40 k 2.2 k = 2.09 k (f)
68.
h fe (ro RC )
(60)(2.085 k) = 250.90 hie 498.6 Ai = AvZi/RC = (250.90)(497.47 )/2.2 k = 56.73
Av =
(a) 68 k 12 k = 10.2 k Zi = 10.2 k hie = 10.2 k 2.75 k = 2.166 k Zo = RC ro = 2.2 k 40 k = 2.085 k h fe RC
RC = RC ro = 2.085 k hie (180)(2.085 k) = 136.5 = 2.75 k
(b) Av =
Ai =
I o I o I i I i I i I i
h fe 10.2 k = 1 hoe RL 10.2 k 2.68 k 180 = (0.792) 1 (25 S)(2.2 k) = 135.13
69.
(a) Zi = RE hib = 1.2 k 9.45 = 9.38 1 1 = 2.7 k = 2.7 k 1 M 2.7 k Zo = RC hob 1 106 A/V (b) Av =
h fb ( RC 1/ hob )
hib
(0.992)( 2.7 k) 9.45
= 284.43 Ai 1
90
(c) = hfb = (0.992) = 0.992 0.992 = = 124 1 1 0.992 re = hib = 9.45 1 1 = 1 M ro = hob 1 A/V 70.
(a)
Z i hie
h fe hre RL
2.75 k
(180)(2 104 )(2.2 k) (1 25S)(2.2 k)
1 hoe RL 2.68 k Zi = 10.2 k Z i = 2.12 k 1 1 Z o hoe (h fe hre / hie ) 25 S (180)(2 104 ) / 2.75 k
83.75 k Zo = 2.2 k 83.75 k = 2.14 k
(b)
h fe RL
Av
hie (hie hoe ) h fe hre ) RL
(180)(2.2 k)
140.3
(c)
Ai
Ai =
h fe 1 hoe RL
(180) 170.62 1 (25S)(2.2 k)
Io Io Ii 10.2 k (170.62) 10.2 k 2.68 k Ii I Ii i 135.13
71.
(a)
Z i = hie
h fe hre RL 1 hoe RL
(140)(1.5 104 )(2.2 k) 1 (25 S)(2.2 k) = 0.86 k 43.79 = 816.21 Z i = RB Z i = 470 k 816.21 = 814.8
= 0.86 k
(b) Av =
2.75 k (2.75 k)(25S) (180) * 2 104 ) 2.2 k
h fe RL hie (hie hoe h fe hre ) RL
(140)(2.2 k) 0.86 k ((0.86 k)(25 S) (140)(1.5 104 ))2.2 k = 357.68
=
91
h fe Io 140 I i 1 hoe RL 1 (25 S)(2.2 k) = 132.70 I I I 470 k I i Ai o o i Ii = 470 k 0.816 k I i I i I i
(c) Ai =
Ii
= (132.70)(0.998)
I i
= 0.998
= 132.43 (d) Zo =
1
hoe h fe hre /(hie Rs )
=
=
72.
(a)
(b)
Z i hib
h fb hrb RL
1
25 S (140)(1.5 104 ) /(0.86 k 1 k)
1 72.9 k 13.71 S
9.45 k
(0.997)(1 104 )(2.2 k) 1 (0.5 A/V)(2.2 k)
1 hob RL 9.67 Zi = 1.2 k Z i = 1.2 k 9.67 = 9.59 h fb
Ai Ai
1 hob RL
0.997 0.996 1 (0.5 A/V)(2.2 k)
Ii 1.2 k (0.996) 1.2 k 9.67 k Ii Ii Io
0.988
(c)
Av
h fL RL hib (hib hob h fb h fb ) RL
(0.997)(2.2 k)
226.21
(d)
Z o
1 hob h fb hrb / hib 1 0.5 A/V (0.997)(1 104 ) / 9.45
90.5 k
Zo = 2.2 k Z o = 2.15 k 73.
9.45 (9.45 )(0.5 A/V) (0.997)(1 104 ) (2.2 k)
92
74.
(a) hfe (0.2 mA) 0.6 (normalized) hfe (1 mA) = 1.0 % change =
h fe (0.2 mA) h fe (1 mA) h fe (0.2 mA)
100%
0.6 1 100% 0.6 = 66.7%
=
(b) hfe(1 mA) = 1.0 hfe(5 mA) 1.5 h fe (1 mA) h fe (5 mA) % change = 100% h fe (1 mA) 1 1.5 100% 1 = 50%
=
75.
Log-log scale! (a) Ic = 0.2 mA, hie = 4 (normalized) Ic = 1 mA, hie = 1(normalized) 4 1 % change = 100% = 75% 4 (b) Ie = 5 mA, hie = 0.3 (normalized) 1 0.3 % change = 100% = 70% 1
76.
(a) hoe = 20 S @ 1 mA Ic = 0.2 mA, hoe = 0.2(hoe @ 1 mA) = 0.2(20 S) = 4 S 1 1 = 250 k 6.8 k hoe 4 S Ignore 1/hoe
(b) ro =
77.
(a) Ic = 10 mA, hoe = 10(20 S) = 200 S 1 1 = 5 k vs. 8.6 k hoe 200S Not a good approximation
(b) ro =
93
78.
(a) hre(0.1 mA) = 4(hre(1 mA)) = 4(2 104) = 8 104 (b) hreVce = hreAv Vi = (8 104)(210)Vi = 0.168 Vi In this case hreVce is too large a factor to be ignored.
79.
(a) hfe (b) hoe (c) hoe 30 µS (normalized) to hoe 0.1 µS (normalized) at low levels of Ic (d) mid-region
80.
(a) hie is the most temperature-sensitive parameter of Fig. 5.33. (b) hoe exhibited the smallest change. (c) Normalized: hfe(max) = 1.5, hfe(min) = 0.5 For hfe = 100 the range would extend from 50 to 150—certainly significant. (d) On a normalized basis re increased from 0.3 at 65C to 3 at 200C—a significant change. (e) The parameters show the least change in the region 0 100C.
81.
(a) Test:
RE 10R2
70(1.5 k) 10(39 k) ? 105 k 390 k No! RTh = 39 k 150 k = 30.95 k 39 k(14 V) = 2.89 V ETh = 39 k 150 k ETh VBE 2.89 V 0.7 V IB = RTh ( 1) RE 30.95 k (71)(1.5 k) = 15.93 A VB = ETh IBRTh = 2.89 V (15.93 A)(30.95 k) = 2.397 V
94
VE = 2.397 V 0.7 V = 1.697 V V 1.697 V and IE = E = 1.13 mA RE 1.5 k VCE = VCC IC(RC + RE) = 14 V 1.13 mA(2.2 k + 1.5 k) = 9.819 V Biasing OK (b) R2 not connected at base: VCC 0 14 V 0.7 V IB = = 51.85 A RB ( 1) RE 150 k + (71)(1.5 k) VB = VCC IBRB = 14 V (51.85 A)(150 k) = 6.22 V as noted in Fig. 5.194.
95
Chapter 6 1.
2.
From Fig. 6.11: VGS = 0 V, ID = 8 mA VGS = 1 V, ID = 4.5 mA VGS = 1.5 V, ID = 3.25 mA VGS = 1.8 V, ID = 2.5 mA VGS = 4 V, ID = 0 mA VGS = 6 V, ID = 0 mA
3.
4.
(a) VDS 1.4 V (b) rd =
V 1.4 V = 233.33 I 6 mA
(c) VDS 1.6 V (d) rd =
V 1.6 V = 533.33 I 3 mA
(e) VDS 1.4 V (f)
rd =
V 1.4 V = 933.33 I 1.5 mA
(g) ro = 233.33 ro 233.33 233.33 rd = 2 2 1 VGS VP 1 (1 V) ( 4 V) 0.5625 = 414.81 (h) rd =
(i)
5.
233.33
1 (2 V)
( 4 V)
533.33 vs. 414.81 933.33 vs 933.2
2
233.33 = 933.2 0.25
Eq. (6.1) is valid!
(a) VGS = 0 V, ID = 8 mA (for VDS > VP) VGS = 1 V, ID = 4.5 mA ID = 3.5 mA (b) VGS = 1 V, ID = 4.5 mA VGS = 2 V, ID = 2 mA ID = 2.5 mA
96
(c) VGS = 2 V, ID = 2 mA VGS = 3 V, ID = 0.5 mA ID = 1.5 mA (d) VGS = 3 V, ID = 0.5 mA VGS = 4 V, ID = 0 mA ID = 0.5 mA (e) As VGS becomes more negative, the change in ID gets progressively smaller for the same change in VGS. (f)
Non-linear. Even though the change in VGS is fixed at 1 V, the change in ID drops from a maximum of 3.5 mA to a minimum of 0.5 mA—a 7:1 change in ID.
6.
The collector characteristics of a BJT transistor are a plot of output current versus the output voltage for different levels of input current. The drain characteristics of a JFET transistor are a plot of the output current versus input voltage. For the BJT transistor increasing levels of input current result in increasing levels of output current. For JFETs, increasing magnitudes of input voltage result in lower levels of output current. The spacing between curves for a BJT are sufficiently similar to permit the use of a single beta (on an approximate basis) to represent the device for the dc and ac analysis. For JFETs, however, the spacing between the curves changes quite dramatically with increasing levels of input voltage requiring the use of Shockley’s equation to define the relationship between ID and VGS. VCsat and VP define the region of nonlinearity for each device.
7.
(a) The input current IG for a JFET is effectively zero since the JFET gate-source junction is reverse-biased for linear operation, and a reverse-biased junction has a very high resistance. (b) The input impedance of the JFET is high due to the reverse-biased junction between the gate and source. (c) The terminology is appropriate since it is the electric field established by the applied gate to source voltage that controls the level of drain current. The term “field” is appropriate due to the absence of a conductive path between gate and source (or drain).
8.
VGS = 0 V, ID = IDSS = 12 mA VGS = VP = 6 V, ID = 0 mA Shockley’s equation: VGS = 1 V, ID = 8.33 mA; VGS = 2 V, ID = 5.33 mA; VGS = 3 V, ID = 3 mA; VGS = 4 V, ID = 1.33 mA; VGS = 5 V, ID = 0.333 mA.
9.
For a p-channel JFET, all the voltage polarities in the network are reversed as compared to an n-channel device. In addition, the drain current has reversed direction.
10.
(b) IDSS = 10 mA, VP = 6 V
97
11.
VGS = 0 V, ID = IDSS = 12 mA VGS = VP = 4 V, ID = 0 mA I V VGS = P = 2 V, ID = DSS = 3 mA 2 4 VGS = 0.3VP = 1.2 V, ID = 6 mA VGS = 3 V, ID = 0.75 mA (Shockley’s equation)
12.
(a) ID = IDSS = 9 mA (b) ID = IDSS(1 VGS/VP)2 = 9 mA(1 (2)/(4))2 = 2.25 mA (c) VGS = VP = 4 V, ID = 0 mA (d)
VGS VP , ID = 0 mA
13.
VGS = 0 V, ID = 16 mA VGS = 0.3VP = 0.3(5 V) = 1.5 V, ID = IDSS/2 = 8 mA VGS = 0.5VP = 0.5(5 V) = 2.5 V, ID = IDSS/4 = 4 mA VGS = VP = 5 V, ID = 0 mA
14.
ID = IDSS(1 VGS/VP)2 4 mA = 12 mA(1 (3)/VP)2 ( 3) 0.333 1 VP 0.333 1
( 3) VP
0.577 1
( 3) VP
0.423 VP
15.
2
3 VP 3 7.1 V 0.423
(a) ID = IDSS(1 VGS/VP)2 = 6 mA(1 (2 V)/(4.5 V))2 = 1.852 mA ID = IDSS(1 VGS/VP)2 = 6 mA(1 (3.6 V)/(4.5 V))2 = 0.24 mA ID 3 mA ( 4.5 V) 1 (b) VGS = VP 1 6 mA I DSS = 1.318 V VGS = VP 1
ID I DSS
( 4.5 V) 1
= 0.192 V
98
5.5 mA 6 mA
16.
ID = IDSS(1 VGS/VP)2 3 mA = IDSS(1 (3 V)/(6 V))2 3 mA = IDSS(0.25) IDSS = 12 mA
17.
VGS = 0 V, ID = IDSS = 7.5 mA VGS = 0.3VP = (0.3)(4 V) = 1.2 V, ID = IDSS/2 = 7.5 mA/2 = 3.75 mA VGS = 0.5VP = (0.5)(4 V) = 2 V, ID = IDSS/4 = 7.5 mA/4 = 1.875 mA VGS = VP = 4 V, ID = 0 mA
18.
From Fig. 6.22: 0.5 V < VP < 6 V 1 mA < IDSS < 5 mA For IDSS = 5 mA and VP = 6 V: VGS = 0 V, ID = 5 mA VGS = 0.3VP = 1.8 V, ID = 2.5 mA VGS = VP/2 = 3 V, ID = 1.25 mA VGS = VP = 6 V, ID = 0 mA For IDSS = 1 mA and VP = 0.5 V: VGS = 0 V, ID = 1 mA VGS = 0.3VP = 0.15 V, ID = 0.5 mA VGS = VP/2 = 0.25 V, ID = 0.25 mA VGS = VP = 0.5 V, ID = 0 mA
19.
At 25°C, PD = 625 mW 45°C 25°C = 20°C 5 mW 20 C 100 mW C PD 625 mW 100 mW = 525 mW
99
20.
VDS = VDSmax = 25 V, ID = ID = IDSS = 10 mA, VDS =
PDmax VDSmax PDmax
I DSS VD 20 V , I D 6.67 mA
100 mW = 4 mA 25 V
100 mW = 10 V 10 mA
VD 15 V , I D 5 mA
21.
VGS = 0.5 V, ID = 6.5 mA 2.5 mA VGS = 1 V, ID = 4 mA Determine ID above 4 mA line: 2.5 mA x x = 1.5 mA 0.5 V 0.3 V ID = 4 mA + 1.5 mA = 5.5 mA corresponding with values determined from a purely graphical approach.
22.
Yes, all knees of VGS curves at or below VP = 3 V.
23.
From Fig 6.25, IDSS 9 mA At VGS = 1 V, ID = 4 mA ID = IDSS(1 VGS/VP)2 ID I DSS
= 1 VGS/VP
VGS =1 VP
ID I DSS
VGS
VP = 1
ID I DSS
1 V 4 mA 1 9 mA
= 3 V (an exact match) 24.
ID = IDSS(1 VGS/VP)2 = 9 mA(1 (1 V)/(3 V))2 = 4 mA, which compares very well with the level obtained using Fig. 6.25. 100
25.
(a) VDS 0.7 V @ ID = 4 mA (for VGS = 0 V) VDS 0.7 V 0 V = 175 r= I D 4 mA 0 mA (b) For VGS = 0.5 V, @ ID = 3 mA, VDS = 0.7 V 0.7 V = 233 r= 3 mA ro 175 (c) rd = 2 (1 VGS / VP ) (1 ( 0.5 V)/( 3 V)2 = 252 vs. 233 from part (b)
26.
27.
The construction of a depletion-type MOSFET and an enchancement-type MOSFET are identical except for the doping in the channel region. In the depletion MOSFET the channel is established by the doping process and exists with no gate-to-source voltage applied. As the gate-to-source voltage increases in magnitude the channel decreases in size until pinch-off occurs. The enhancement MOSFET does not have a channel established by the doping sequence but relies on the gate-to-source voltage to create a channel. The larger the magnitude of the applied gate-to-source voltage, the larger the available channel.
28.
At VGS = 0 V, ID = 6 mA At VGS = 1 V, ID = 6 mA(1 (1 V)/(3 V))2 = 2.66 mA At VGS = +1 V, ID = 6 mA(1 (+1 V)/(3 V))2 = 6 mA(1.333)2 = 10.667 mA At VGS = +2 V, ID = 6 mA(1 (+2 V)/(3 V))2 = 6 mA(1.667)2 = 16.67 mA I D 4.67 mA VGS ID 2.66 mA 1 V I D 3.34 mA 0 6.0 mA I D 4.67 mA +1 V 10.67 mA
29.
+2 V
16.67 mA
I D 6 mA
From 1 V to 0 V, ID = 3.34 mA while from +1 V to +2 V, ID = 6 mA almost a 2:1 margin. In fact, as VGS becomes more and more positive, ID will increase at a faster and faster rate due to the squared term in Shockley’s equation. 30.
VGS = 0 V, ID = IDSS = 12 mA; VGS = 8 V, ID = 0 mA; VGS = VGS = 0.3VP = 2.4 V, ID = 6 mA; VGS = 6 V, ID = 0.75 mA
101
VP = 4 V, ID = 3 mA; 2
31.
From problem 20: VGS 1 V 1 V 1 V VP = ID 14 mA 1 1.473 1 1.21395 1 1 9.5 mA I DSS =
1 4.67 V 0.21395
32.
ID = IDSS(1 VGS/VP)2 ID 4 mA = 11.11 mA IDSS = 2 1 VGS VP (1 ( 2 V) ( 5 V))2
33.
From problem 14(b): VGS = VP 1
ID I DSS
( 5 V) 1
20 mA 2.9 mA
= (5 V)(1 2.626) = (5 V)(1.626) = 8.13 V 34.
From Fig. 6.34, PDmax = 200 mW, ID = 8 mA P = VDSID P 200 mW = 25 V and VDS = max ID 8 mA
35.
(a) In a depletion-type MOSFET the channel exists in the device and the applied voltage VGS controls the size of the channel. In an enhancement-type MOSFET the channel is not established by the construction pattern but induced by the applied control voltage VGS. (b) (c) Briefly, an applied gate-to-source voltage greater than VT will establish a channel between drain and source for the flow of charge in the output circuit.
36.
(a) ID = k(VGS VT)2 = 0.4 103(VGS 3.5)2 VGS 3.5 V 4V 5V 6V 7V 8V
ID 0 0.1 mA 0.9 mA 2.5 mA 4.9 mA 8.1 mA
102
(b) ID = 0.8 103(VGS 3.5)2 ID 0 0.2 mA 1.8 mA 5.0 mA 9.8 mA 16.2 mA
VGS 3.5 V 4V 5V 6V 7V 8V
37.
(a) k =
V
I D (on)
GS (on)
VT
2
For same levels of VGS, ID attains twice the current level as part (a). Transfer curve has steeper slope. For both curves, ID = 0 mA for VGS < 3.5 V.
4 mA = 1 mA/V2 2 (6 V 4 V)
ID = k(VGS VT) = 1 103(VGS 4 V)2 2
38.
(b)
VGS 4V 5V 6V 7V 8V
ID 0 mA 1 mA 4 mA 9 mA 16 mA
For VGS < VT = 4 V, ID = 0 mA
(c)
VGS 2V 5V 10 V
ID 0 mA 1 mA 36 mA
(VGS < VT)
From Fig. 6.58, VT = 2.0 V At ID = 6.5 mA, VGS = 5.5 V: ID = 5.31 104(VGS 2)2
39.
ID = k VGS (on) VT
and VGS (on) VT
2
VGS (on) VT
ID = k(VGS VT)2 6.5 mA = k(5.5 V 2 V)2 k = 5.31 104
2
ID k ID k
ID k 3 mA =4V =4V 0.4 10 3 = 4 V 2.739 V = 1.261 V
VT = VGS (on)
7.5 V
103
40.
ID = k(VGS VT)2 ID = (VGS VT)2 k ID = VGS VT k ID =5V+ k = 27.36 V
30 mA 0.06 10 3
VGS = VT +
41.
Enhancement-type MOSFET: ID = k (VGS VT ) 2 dI D d 2k (VGS VT ) (VGS VT ) dVGS dVGS 1 dI D = 2k(VGS VT) dVGS Depletion-type MOSFET: ID = IDSS(1 VGS/VP)2
V dI D d = I DSS 1 GS dVGS dVGS VP I DSS 2 1
2
VGS V d 0 GS VP dVGS VP
= 2 I DSS 1
VGS VP
=
2 I DSS V 1 GS VP VP
=
2 I DSS VP VP VP
1
1 VP
1 VP
VGS VP
dI D 2 I DSS = (VGS VP ) dVGS VP2 dI D = k1(VGS K2) dVGS revealing that the drain current of each will increase at about the same rate.
For both devices
42.
ID = k(VGS VT)2 = 0.45 103(VGS (5 V))2 = 0.45 103(VGS + 5 V)2 VGS = 5 V, ID = 0 mA; VGS = 6 V, ID = 0.45 mA; VGS = 7 V, ID = 1.8 mA; VGS = 8 V, ID = 4.05 mA; VGS = 9 V, ID = 7.2 mA; VGS = 10 V, ID = 11.25 mA
104
43.
44.
45.
46.
V 0.1 V = 25 ohms I 4 mA V 4.9 V For the “off” transistor: R = = 9.8 M I 0.5 A Absolutely, the high resistance of the “off” resistance will ensure Vo is very close to 5 V.
(b) For the “on” transistor: R =
47.
105
Chapter 7 1.
(a) VGS = 0 V, ID = IDSS = 12 mA VGS = VP = 4 V, ID = 0 mA VGS = VP/2 = 2 V, ID = IDSS/4 = 3 mA VGS = 0.3VP = 1.2 V, ID = IDSS/2 = 6 mA (b)
(c)
I DQ 4.7 mA VDSQ = VDD I DQ RD = 14 V (4.7 mA)(1.8 k)
= 5.54 V (d)
I DQ = IDSS(1 VGS/VP)2 = 12 mA(1 (1.5 V)/(4 V))2
= 4.69 mA VDSQ = VDD I DQ RD = 14 V (4.69 mA)(1.8 k)
= 5.56 V excellent comparison 2.
(a)
I DQ = IDSS (1 VGS / VP )2
I DQ
= 10 mA 1 ( 3 V) ( 4.5 V) = 10 mA(0.333)2 = 1.11 mA
2
(b) VGSQ = 3 V (c) VDS = VDD ID(RD + RS) = 16 V (1.11 mA)(2.2 k) = 16 V 2.444 V = 13.56 V VD = VDS = 13.56 V VG = VGSQ = 3 V VS = 0 V
106
3.
(a)
I DQ
VDD VD 12 V 6 V = 2.727 mA 2.2 k RD
(b) VDS = VD VS = 6 V 0 V = 6 V (c) ID = IDSS(1 VGS/VP)2 VGS = VP 1 VGS = (4 V) 1
ID I DSS
2.727 mA 8 mA
= 1.66 V VGG = 1.66 V 4.
VGSQ = 0 V, ID = IDSS = 5 mA
VD = VDD IDRD = 20 V (5 mA)(2.2 k) = 20 V 11 V =9V 5.
VGS = VP = 4 V I DQ = 0 mA and VD = VDD I DQ RD = 18 V (0)(2 k) = 18 V
6.
(a)(b)VGS = 0 V, ID = 10 mA VGS = VP = 4 V, ID = 0 mA V VGS = P = 2 V, ID = 2.5 mA 2 VGS = 0.3VP = 1.2 V, ID = 5 mA VGS = IDRS ID = 5 mA: VGS = (5 mA)(0.75 k) = 3.75 V (c)
I DQ 2.7 mA VGSQ 1.9 V
(d) VDS = VDD ID(RD + RS) = 18 V (2.7 mA)(1.5 k + 0.75 k) = 11.93 V VD = VDD IDRD = 18 V (2.7 mA)(1.5 k) = 13.95 V VG = 0 V VS = ISRS = IDRS = (2.7 mA)(0.75 k) = 2.03 V
107
7.
ID = IDSS(1 VGS/VP)2 = I DSS 1
2 I D RS I D2 RS2 2 VP VP
I DSS RS2 2 2 I DSS RS 1 I D I DSS = 0 ID 2 VP VP
Substituting: 351.56 I D2 4.75ID + 10 mA = 0 ID = I DQ
b b 2 4ac = 10.91 mA, 2.60 mA 2a = 2.6 mA (exact match #6)
VGS = IDRS = (2.60 mA)(0.75 k) = 1.95 V vs. 2 V (#6) 8.
VGS = 0 V, ID = IDSS = 6 mA VGS = VP = 6 V, ID = 0 mA V VGS = P = 3 V, ID = 1.5 mA 2 VGS = 0.3VP = 1.8 V, ID = 3 mA VGS = IDRS ID = 2 mA: VGS = (2 mA)(1.6 k) = 3.2 V (a)
I DQ = 1.7 mA VGSQ = 2.8 V
(b) VDS = VDD ID(RD + RS) = 12 V (1.7 mA)(2.2 k + 1.6 k) = 5.54 V VD = VDD IDRD = 12 V (1.7 mA)(2.2 k) = 8.26 V VG = 0 V VS = ISRS = IDRS = (1.7 mA)(1.6 k) = 2.72 V (vs. 2.8 V from VS = (VGSQ ) ) 9.
(a)
I DQ = IS =
VS 1.7 V = 3.33 mA RS 0.51 k
(b) VGSQ I DQ RS = (3.33 mA)(0.51 k) 1.7 V
108
(c) ID = IDSS(1 VGS/VP)2 3.33 mA = IDSS(1 (1.7 V)/(4 V))2 3.33 mA = IDSS(0.331) IDSS = 10.06 mA (d) VD = VDD I DQ RD = 18 V (3.33 mA)(2 k) = 18 V 6.66 V = 11.34 V (e) VDS = VD VS = 11.34 V 1.7 V = 9.64 V 10.
(a) VGS = 0 V ID = IDSS = 4.5 mA (b) VDS = VDD ID(RD + RS) = 20 V + 4 V (4.5 mA)(2.2 k + 0.68 k) = 24 V 12.96 = 11.04 V (c) VD = VDD IDRD = 20 V (4.5 mA)(2.2 k) = 10.1 V (d) 4 V + ISRS VS = 0 VS = ISRS 4 V = IDRS 4 V = (4.5 mA)(0.68 kΩ) 4 V = 3.06 V 4 V = 0.94 V
11.
Network redrawn: VGS = 0 V, ID = IDSS = 6 mA VGS = VP = 6 V, ID = 0 mA V VGS = P = 3 V, ID = 1.5 mA 2 VGS = 0.3VP = 1.8 V, ID = 3 mA VGS = IDRS = ID(0.39 k) For ID = 5 mA, VGS = 1.95 V
109
From graph I DQ 3.55 mA VGSQ 1.4 V
VS = VGSQ = (1.4 V) = +1.4 V
12.
R2 110 k(20 V) VDD = R1 R2 910 k 110 k = 2.16 V VGS = 0 V, ID = IDSS = 10 mA VGS = VP = 3.5 V, ID = 0 mA V VGS = P = 1.75 V, ID = 2.5 mA 2
(a) VG =
VGS = 0.3VP = 1.05 V, ID = 5 mA VGSQ = VG IDRS VGSQ = 2.16 ID(1.1 k)
ID = 0: VGSQ = VG = 2.16 V VGSQ = 0 V, ID =
(b)
2.16 V = 1.96 mA 1.1 k
I DQ 3.3 mA VGSQ 1.5 V
(c) VD = VDD I DQ RD = 20 V (3.3 mA)(2.2 k) = 12.74 V VS = ISRS = IDRS = (3.3 mA)(1.1 k) = 3.63 V (d) VDSQ = VDD I DQ ( RD RS ) = 20 V (3.3 mA)(2.2 k + 1.1 k) = 20 V 10.89 V = 9.11 V
110
13.
(a) ID = IDSS = 10 mA, VP = 3.5 V VGS = 0 V, ID = IDSS = 10 mA VGS = VP = 3.5 V, ID = 0 mA
110 k(20 V) 110 k 910 k 2.16 V
VG
VP 3.5 V = 1.75 V, ID = 2.5 mA 2 2 VGS = 0.3VP = 1.05 V, ID = 5 mA
VGS =
I DQ 5.8 mA vs. 3.3 mA (#12) VGSQ 0.85 V vs. 1.5 V (#12) VD 20 V I D (2.2 k) 20 V (5.8 mA)(2.2 k) 20 V 12.76 V = 7.24 V
VS I D RS (5.8 mA)(1.1 k) 6.38 V
VDSQ VD VS 7.24 V 6.38 V
= 0.86 V
(b) As RS decreases, the intersection on the vertical axis increases. The maximum occurs at ID = IDSS = 10 mA. V 2.16 V = 216 RSmin G I DSS 10 mA 14.
(a)
ID
VRD RD
VDD VD 18 V 12 V 6 V = 3 mA RD 2 k 2 k
(b) VS I S RS I D RS (3 mA)(0.68 k) = 2.04 V VDS VDD I D ( RD RS ) 18 V (3 mA)(2 k 0.68 k) 18 V 8.04 V = 9.96 V 110 k(12 V) = 1.67 V 110 k 680 k VG I S RS VG I D RS VG VS 1.67 V 2.04 V
(c) VG VGS
0.37 V VGS
(d) VP 1
ID I DSS
0.37 V 0.37 V 0.37 V 0.388 3 mA 1 0.612 1 8 mA
0.954 V
111
15.
16.
16 V 10 V 6 V 3mA 2 k 2 k 12 k(16 V) VG 4V 12 k 36 k 8V V 4 V + 4 V = 2.67 k RS 3 mA 3 mA I I DQ
(a) VGS VGS VGS VGS
= 0 V, I D I DSS = 6 mA VP = 6 V, I D = 0 mA VP / 2 = 3 V, I D = 1.5 mA 0.3VP = 1.8 V, I D = 3 mA
VGS VSS I D RS VGS = 4 V I D (2.2 k) 4V VGS = 0 V, I D = 1.818 mA 2.2 k I D = 0 mA, VGS = 4 V I DQ 2.7 mA, VGSQ 2 V
(b) VDS VDD VSS I D ( RD RS ) 16 V + 4 V (2.7 mA)(4.4 k) 8.12 V VS VSS I D RS 4 V (2.7 mA)(2.2 k) 1.94 V
or VS (VGSQ ) (2 V) 2 V
112
17.
(a) ID =
VDD VSS VDS 20 V 2 V 4 V 14 V = 3.3 mA 3 k 1.2 k 4.2 k RD RS
(b) VD = VDD IDRD = 20 V (3.33 mA)(3 k) = 10 V VS = ISRS + VSS = IDRS + VSS = (3.33 mA)(1.2 k) + (2 V) =4V+ 2V =6V (c) VGS = VG VS =0V6V = 6 V 18.
I DQ = 4 mA
(a)
(b) VDQ 18 V 4 mA(1.8 k) 18 V 7.2 V = 10.8 V VDSQ VD VS 10.8 V I D RS 10.8 V (4 mA)(1.2 k) 10.8 V 4.8 V = 6 V
(c) Ps = (18 V)(4 mA) = 72 mW Pd = (6 V)(4 mA) = 24 mW 19.
(1.2 k)(16 V) =4V 1.2 k 3.6 k 4V IS ID = 4 mA 1 k VD 16 V I D (1.8 k) = 16 V (4 mA)(1.8 k) VG
8.8 V VGS VG VS 4 V 4 V = 0 V (requiring I D I DSS = 4 mA)
20.
VGS = 0 V, ID = IDSS = 6 mA VGS = VP = 4 V, ID = 0 mA VGS = VP/2 = 2 V, ID = IDSS/4 = 1.5 mA VGS = 0.3VP = 1.2 V, ID = IDSS/2 = 3 mA
VGS = IDRS = ID(0.43 k) ID = 4 mA, VGS = 1.72 V
(b) VDS = VDD ID(RD + RS) = 14 V 2.9 mA(1.2 k + 0.43 k) = 9.27 V VD = VDD IDRD = 14 V (2.9 mA)(1.2 k) = 10.52 V
113
21.
(a) VGS = 0 V, ID = IDSS = 8 mA VGS = VP = 8 V, ID = 0 mA V VGS = P = 4 V, ID = 2 mA 2 VGS = 0.3VP = 2.4 V, ID = 4 mA VGS = +1 V, ID = 10.125 mA VGS = +2 V, ID = 12.5 mA VGS = VSS IDRS = (4 V) ID(0.39 k) VGS = 4 ID(0.39 k) ID = 0: VGS = +4 V 4 VGS = 0: ID = = 10.26 mA 0.39 k I DQ 9 mA VGSQ +0.5 V
(b) VDS = VDD ID(RD + RS) + VSS = 18 V 9 mA(1.2 k + 0.39 k) + 4 V = 22 V 14.31 V = 7.69 V
VS = VGSQ = 0.5 V
114
22.
ID = k(VGS VT)2 I D (on) k= VGS (on) VTh
2
VDS = VDD ID(RD + RS) VDD VDS = 0 V; ID = RD RS 22 V = 1.2 k 0.51 k = 12.87 mA ID = 0 mA, VDS = VDD = 22 V
=
5 mA 5 mA 2 (7 V 4 V) 9 V2
Ie = 0.556 103 A/V2 and ID = 0.556 103(VGS 4 V)2
(c) VD = VDD IDRD = 22 V (8.25 mA)(1.2 k) = 12.1 V VS = ISRS = IDRS = (8.25 mA)(0.51 k) = 4.21 V (d) VDS = VD VS = 12.1 V 4.21 V = 7.89 V vs. 7.9 V obtained graphically
115
23.
(a) VG =
R2 6.8 M VDD (24 V) = 9.71 V 10 M 6.8 M R1 R2
VGS = VG IDRS VGS = 9.71 ID(0.75 k) At ID = 0 mA, VGS = 9.71 V 9.71 V = 12.95 mA At VGS = 0 V, ID = 0.75 k k=
V
I D (on)
GS (on)
VGS (Th ) 3
2
5 mA 5 mA 2 (6 V 3 V) (3 V) 2
= 0.556 10 A/V2 ID = 0.556 103(VGS 3 V)2 VGS 3V 4V 5V 6V 7V
ID 0 mA 0.556 mA 2.22 mA 5 mA 8.9 mA
I DQ 5 mA VGSQ 6 V
116
(b) VD = VDD IDRD = 24 V (5 mA)(2.2 k) = 13 V VS = ISRS = IDRS = (5 mA)(0.75 k) = 3.75 V 24.
R2 18 k VCC (20 V) R1 R2 91 k 18 k = 3.3 V
(a) VG =
(b) VGS = 0 V, ID = IDSS = 6 mA VGS = VP = 6 V, ID = 0 mA V VGS = P = 3 V, ID = 1.5 mA 2 VGS = VP = 1.8 V, ID = 3 mA I DQ 3.75 mA VGSQ 1.25 V
(c) IE = ID = 3.75 mA (d) IB =
IC
3.75 mA = 23.44 A 160
(e) VD = VE = VB VBE = VCC IBRB VBE = 20 V (23.44 A)(330 k) 0.7 V = 11.56 V (f)
25.
VC = VCC ICRC = 20 V (3.75 mA)(1.1 k) = 15.88 V
Testing: RE 10R2 (100)(1.2 k) 10(10 k) 120 k > 100 k (satisfied) (a) VB = VG =
R2VDD 10 k(16 V) R1 R2 40 k 10 k
= 3.2 V (b) VE = VB VBE = 3.2 V 0.7 V = 2.5 V VE 2.5 V = 2.08 mA RE 1.2 k IC IE = 2.08 mA ID = IC = 2.08 mA
(c) IE =
117
(d) IB =
IC
2.08 mA = 20.8 A 100
(e) VC = VG VGS VGS = VP 1
ID I DSS
= ( 6 V) 1
2.08 mA 6 mA
= 2.47 V VC = 3.2 (2.47 V) = 5.67 V VS = VC = 5.67 V VD = VDD IDRD = 16 V (2.08 mA)(2.2 k) = 11.42 V (f)
VCE = VC VE = 5.67 V 2.5 V = 3.17 V
(g) VDS = VD VS = 11.42 V 5.67 V = 5.75 V 26.
VGS = VP 1
ID I DSS
= (6 V) 1
4 mA 8 mA
= 1.75 V
VGS ( 1.75 V) = 0.44 k 4 mA ID RD = 3RS = 3(0.44 k) = 1.32 k Standard values: RS = 0.43 k RD = 1.3 k
VGS = IDRS: RS =
27.
VGS = VP 1
ID I DSS
= (4 V) 1
2.5 mA 10 mA
= 2 V VGS = VG VS and VS = VG VGS = 4 V (2 V) =6V VS 6V = 2.4 k (a standard value) RS = I D 2.5 mA RD = 2.5RS = 2.5(2.4 k) = 6 k use 6.2 k RV R2 (24 V) 88 M + 4R2 = 24R2 VG = 2 DD 4 V = R1 R2 22 M R2 20R2 = 88 M R2 = 4.4 M Use R2 = 4.3 M
118
28.
ID = k(VGS VT)2 ID = (VGS VT)2 k ID = VGS VT k ID 6 mA =4V+ = 7.46 V k 0.5 10 3 A/V 2 VR V VDS VDD VGS 16 V 7.46 V 8.54 V RD = D DD 6 mA 6 mA ID ID ID = 1.42 k Standard value: RD = 0.75 k RG = 10 M
and VGS = VT +
29.
VS 4V = 4 mA RS 1 k VDS = VDD ID(RD + RS) = 12 V (4 mA)(2 k + 1 k) = 12 V (4 mA)(3 k) = 12 V 12 V =0V JFET in saturation!
(a) ID = IS =
(b) VS = 0 V reveals that the JFET is nonconducting and the JFET is either defective or an open-circuit exists in the output circuit. VS is at the same potential as the grounded side of the 1 k resistor. (c) Typically, the voltage across the 1 M resistor is 0 V. The fact that the voltage across the 1 M resistor is equal to VDD suggests that there is a short-circuit connection from gate to drain with ID = 0 mA. Either the JFET is defective or an improper circuit connection was made. 30.
75 k(20 V) = 3.7 V (seems correct!) 75 k 330 k VGS = 3.7 V 6.25 V = 2.55 V (possibly okay) ID = IDSS(1 VGS/VP)2 = 10 mA(1 (2.55 V)/(6 V))2 = 3.3 mA (reasonable)
VG =
However, IS =
VS 6.25 V = 6.25 mA 3.3 mA 1 k RS
VRD = IDRD = ISRD = (6.25 mA)(2.2 k)
= 13.75 V and VRS VRD = 6.25 V + 13.75 V = 20 V = VDD VDS = 0 V 1. 2.
Possible short-circuit from D-S. Actual IDSS and/or VP may be larger in magnitude than specified.
119
31.
VS 6.25 V = 6.25 mA 1 k RS VDS = VDD ID(RD + RS) = 20 V (6.25 mA)(2.2 k + 1 k) = 20 V 20 V = 0 V (saturation condition)
ID = IS =
R2VDD 75 k(20 V) = 3.7 V (as it should be) R1 R2 330 k 75 k VGS = VG VS = 3.7 V 6.25 V = 2.55 V
VG =
ID = I DSS
V 1 GS VP
2
= 10 mA(1 (2.55 V)/(6 V))2
= 3.3 mA 6.25 mA In all probability, an open-circuit exists between the voltage divider network and the gate terminal of the JFET with the transistor exhibiting saturation conditions. 32.
(a) VGS = 0 V, ID = IDSS = 8 mA VGS = VP = +4 V, ID = 0 mA V VGS = P = +2 V, ID = 2 mA 2 VGS = 0.3VP = 1.2 V, ID = 4 mA VGS = IDRS ID = 4 mA; VGS = (4 mA)(0.51 k) = 2.04 V I DQ = 3 mA, VGSQ = 1.55 V (b) VDS = VDD + ID(RD + RS) = 18 V + (3 mA)(2.71 k) = 9.87 V (c) VD = VDD IDRD = 18 V (3 mA)(2.2 k) = 11.4 V
120
33.
k=
I D (on)
V
GS (on)
VGS (Th )
2
4 mA
7 V (3 V)
2
4 mA ( 4 V) 2
= 0.25 103 A/V2 ID = 0.25 103(VGS + 3 V)2 VGS 3 V 4 V 5 V 6 V 7 V 8 V
ID 0 mA 0.25 mA 1 mA 2.25 mA 4 mA 6.25 mA
VGS = VDS = VDD + IDRD At ID = 0 mA, VGS = VDD = 16 V V 16 V = 8 mA At VGS = 0 V, ID = DD RD 2 k
(b) VDS = VGS = 7.25 V (c) VD = VDS = 7.25 V or VDS = VDD + IDRD = 16 V + (4.4 mA)(2 k) = 16 V + 8.8 V VDS = 7.2 V = VD 34.
VGS 1.5 V = 0.375 4V VP
Find 0.375 on the horizontal axis. Then move vertically to the ID = IDSS(1 VGS/VP)2 curve. Finally, move horizontally from the intersection with the curve to the left to the ID/IDSS axis. ID
= 0.39 I DSS and ID = 0.39(12 mA) = 4.68 mA vs. 4.7 mA (#1) VDSQ = VDD IDRD = 14 V (4.7 mA)(1.8 k) = 5.54 V vs. 5.54 V (#1)
121
35.
VP
m=
4V (10 mA)(0.75 k)
I DSS RS = 0.533 V 0.533(0) M = m GG 4V VP =0 Draw a straight line from M = 0 through m = 0.533 until it crosses the normalized curve of ID
= I DSS 1
VGS VP
2
. At the intersection with the curve drop a line down to determine
VGS = 0.49 VP
so that VGSQ = 0.49VP = 0.49(4 V) = 1.96 V (vs. 1.9 V #6) If a horizontal line is drawn from the intersection to the left vertical axis we find ID = 0.27 I DSS and ID = 0.27(IDSS) = 0.27(10 mA) = 2.7 mA (vs. 2.7 mA from #6) (a) VGSQ = 1.96 V, I DQ = 2.7 mA (b) (c) (d) VDS = VDD ID(RD + RS) = 11.93 V (like #6) VD = VDD IDRD = 13.95 V (like #6) VG = 0 V, VS = IDRS = 2.03 V (like #6) 36.
VGG = m=
R2VDD 110 k(20 V) = 2.16 V R1 R2 110 k 910 k VP
I DSS RS V M = m GG VP
3.5 V = 0.318 (10 mA)(1.1 k) (2.16 V) = 0.196 0.318 3.5
Find 0.196 on the vertical axis labeled M and mark the location. Move horizontally to the vertical axis labeled m and then add m = 0.318 to the vertical height ( 1.318 in total)—mark the spot. Draw a straight line through the two points located above, as shown below.
122
Continue the line until it intersects the ID = IDSS(1 VGS/VP)2 curve. At the intersection move horizontally to obtain the ID/IDSS ratio and move down vertically to obtain the VGS / V p ratio. ID I DSS
= 0.33 and I DQ = 0.33(10 mA) = 3.3 mA vs. 3.3 mA (#12)
VGS = 0.425 and VGSQ = 0.425(3.5 V) VP = 1.49 V vs. 1.5 V (#12)
37.
m=
VP
6V (6 mA)(2.2 k)
I DSS RS = 0.4545 V (4 V) M = m GG 0.4545 (6 V) VP = 0.303
Find 0.303 on the vertical M axis. Draw a horizontal line from M = 0.303 to the vertical m axis. Add 0.4545 to the vertical location on the m axis defined by the horizontal line. Draw a straight line between M = 0.303 and the point on the m axis resulting from the addition of m = 0.4545. Continue the straight line as shown below until it crosses the normalized ID = IDSS(1 VGS/VP)2 curve:
123
At the intersection drop a vertical line to determine VGS = 0.34 VP and VGSQ = 0.34(6 V) = 2.04 V (vs. 2 V from problem 15) At the intersection draw a horizontal line to the ID/IDSS axis to determine ID = 0.46 I DSS and I DQ = 0.46(6 mA) = 2.76 mA (vs. 2.7 mA from problem 15) (a)
I DQ = 2.76 mA, VGSQ = 2.04 V
(b) VDS = VDD + VSS ID(RD + RS) = 16 V + 4 V (2.76 mA)(4.4 k) = 7.86 V (vs. 8.12 V from problem 15) VS = VSS + IDRS = 4 V + (2.76 mA)(2.2 k) = 4 V + 6.07 V = 2.07 V (vs. 1.94 V from problem 15)
124
Chapter 8 1.
gm0 =
2 I DSS 2(12 mA) = 6 mS 4 V VP
2.
gm0 =
2 I DSS VP
VP
2 I DSS 2(12 mA) = 2.4 V 10 mS g m0
VP = 2.4 V ( g )( V ) 5 mS(4 V) 2 I DSS IDSS = mO P = 10 mA VP 2 2
3.
gm0 =
4.
gm = g m 0 1
5.
gm =
VGSQ
VP
2(12 mA) 0.5 V 1 3 V 3 V
= 6.67 mS
VGSQ 2 I DSS 1 VP VP
2 I DSS 1 V 1 2.5 V 2.5 V IDSS = 12.5 mA
6 mS =
6.
gm = g m 0 =
7.
gm0 =
ID I DSS
20 mA 1 5V 2
2 I DSS VP
I DSS / 4 2(10 mA) 1 I DSS 5V 4
= 2 mS
2 I DSS 2(8 mA) = 3.2 mS VP 5V
gm = g m 0 1
VGSQ VP
3.2 mS 1
VP / 4 VP
= 3.2 mS 1
= 2.4 mS 8.
(a) gm = yfs = 4.5 mS (b) rd =
9.
1 1 = 40 k yos 25S
gm = gfs = 4.5 mS 1 1 = 40 k rd = g os 25S Zo = rd = 40 k Av(FET) = gmrd = (4.5 mS)(40 k) = 180
125
1 3 3.2 mS 4 4
10.
Av = gmrd gm =
11.
(a) gm0 =
(b) gm =
Av ( 200) = 2 mS rd (100 k)
2 I DSS 2(10 mA) = 4 mS VP 5V
I D 9 mA 7 mA = 3.64 mS VGS 0.85 V 0.3 V
(c) Eq. 8.6: gm = g m 0 1
(d) gm =
(a) rd =
VP
4 mS 1
0.5 V 5 V
= 3.6 mS
I D 8 mA 5 mA = 3 mS VGS 1.5 V 0.5 V
(e) gm = g m 0 1
12.
VGSQ
VDS I D
VGSQ
4 mS 1
VP
VGS constant
1 V 5 V
= 3.2 mS
(15 V 5 V) 10 V = 33.33 k (9.1 mA 8.8 mA) 0.3 mA
(b) At VDS = 10 V, ID = 9 mA on VGS = 0 V curve 2I 2(9 mA) gm0 = DSS = 4.5 mS VP 4V 13.
From 2N4220 data: (a) gm = yfs = 750 S = 0.75 mS 1 1 = 100 k (b) rd = yos 10 S
14.
(a) gm (@ VGS = 6 V) = 0, gm (@ VGS = 0 V) = gm0 =
2 I DSS 2(12 mA) = 4 mS VP 6V
(b) gm (@ ID = 0 mA) = 0, gm (@ ID = IDSS = 12 mA) = gm0 =
15.
gm = yfs = 5.6 mS, rd =
1 1 = 66.67 k yos 15 S
126
2 I DSS 2(12 mA) = 4 mS VP 6V
16.
17.
2 V 2(10 mA) 1 4V 4 V
gm =
VGSQ 2 I DSS 1 VP VP
rd =
1 1 = 40 k yos 25 S
= 2.5 mS
Graphically, VGSQ = 1.5 V gm =
VGSQ 2 I DSS 1 VP VP
1.5 V 2(10 mA) 1 6V 6 V
= 2.5 mS
Zi = RG = 1 M Zo = RD rd = 1.8 k 40 k = 1.72 k Av = gm(RD rd) = (2.5 mS)(1.72 k) = 4.3 18.
(a) VGSQ = 1.5 V, Graphically, I D = 1.25 mA Q
gm =
VGSQ 2 I DSS 1 VP VP
1.5 V 2(5 mA) 1 3V 3 V
= 1.67 mS
Zi = RG = 1 M Zo = RD rd = 1.8 k 40 k = 1.72 k Av = gm(RD rd) = (1.67 mS)(1.72 k) = 2.87 (b) 19.
Z i and Z o the same, 33% drop in Av .
(a) VGSQ = 2.5 V gm =
2 I DSS V 2(10 mA) 2.5 V 1 GS 1 VP VP 4V 4 V
= 5 mS[1 0.625] 5 mS[0.375] 1.875 mS Z i = 2 M Zo = 4.7 k 20 k = 3.81 k Av = gm(rd RD) = 1.875 mS(3.81 k) = 7.14 (b)
Z i = 2 M (the same), Zo = 4.7 k 40 k = 4.21 k (increased) Av = gm(RD rd) = 1.875 mS[4.21 k] = 7.89 (Av increased)
127
20.
gm = yfs = 3000 S = 3 mS 1 1 = 20 k rd = yos 50 S Zi = RG = 10 M Zo = rd RD = 20 k 3.3 k = 2.83 k Av = gm(rd RD) = (3 mS)(2.83 k) = 8.49
21.
gm = 3 mS, rd = 20 k Zi = 10 M RD 3.3 k Zo = R RS 3.3 k 1.1 k 1 (3 mS)(1.1 k) 1 g m RS D 20 k rd 3.3 k 3.3 k = = 730 1 3.3 0.22 4.52 g m RD (3 mS)(3.3 k) R RS 3.3 k 1.1 k 1 (3 mS)(1.1 k) 1 g m RS D 20 k rd 9.9 9.9 = = 2.19 1 3.3 0.22 4.52
Av =
22.
gm = yfs = 3000 S = 3 mS 1 1 = 100 k rd = yos 10 S Zi = RG = 10 M (the same) Zo = rd RD = 100 k 3.3 k = 3.195 k (higher) Av = gm(rd RD) = (3 mS)(3.195 k) = 9.59 (higher)
23.
(a) Av = gmRD 2 = gm ( 2.7 k) 2 gm = 0.74 mS 2.7 k 2I V gm = DSS 1 GS VP VP V 2(9 mA) 1 GS 8V 8 V V V 0.74 mS = 2.25 mS 1 GS = 2.25 mS + 2.25 mS GS 8 V 8V V 1.51 mS = 2.25 mS GS 8V
0.74 mS =
128
VGS , VGSQ = (0.671)(8 V) = 5.37 V 8V = I D RD
0.671 = VGSQ RS
VGSQ ID
5.37 V (from Q pt. on graph) = 3.83 k 1.4 mA
(b) Av = gmRD 2 = gm (2.7 k 30 k) = gm(2.48 kΩ) 2 gm = 0.81 mS 2.48 k 2I V V 2(9 mA) 1 GS gm = DSS 1 GS = 8V 8 V VP VP 0.81 mS = 2.25 mS + 2.25 mS 1.44 mS = 2.25 mS
= 2.25 mS + 2.25 mS
VGS 8V
VGS 8V
VGS 8V
VGS , VGSQ = (0.640)(8 V) = 5.12 V 8V = I D RD
0.640 VGSQ
RS
VGSQ I DQ
5.12 V (from Q pt. on graph) = 3.41 k 1.5 mA
Repeat with new value of gm. RS decreased to 3.41 kΩ from 3.83 kΩ. 24.
VGSQ = 0 V, gm = gm0 =
2 I DSS 2(6 mA) 1 1 = 2 mS, rd = = 25 k VP 6V yos 40 S
Zi = 1 M Zo = rd RD = 25 k 2 k = 1.852 k Av = gm(rd RD) = (2 mS)(1.852 k) 3.7 25.
VGSQ = 0.95 V
gm =
VGSQ 2 I DSS 1 VP VP
2(12 mA) 0.95 V 1 3V 3 V = 5.47 mS
=
Zi = 82 M 11 M = 9.7 M Zo = rd RD = 50 k 2 k = 1.92 k Av = gm(rd RD) = (5.47 mS)(1.92 k) = 10.5 Vo = AvVi = (10.5)(20 mV) = 210 mV
129
26.
VGSQ = 0.95 V (as before), gm = 5.47 mS (as before)
Zi = 9.7 M as before RD Zo = R RS 1 g m RS D rd but rd 10(RD + RS) RD 2 k 2 k 2 k Zo = 1 g m RS 1 (5.47 mS)(0.61 k) 1 3.337 4.337 = 461.1 g m RD since rd 10(RD + RS) Av = 1 g m RS (5.47 mS)(2 k) 10.94 = 2.52 (a big reduction) 4.337 (from above) 4.337 Vo = AvVi = (2.52)(20 mV) = 50.40 mV (compared to 214.4 mV earlier)
=
27.
VGSQ = 0.95 V, gm (problem 23) = 5.47 mS
Zi (the same) = 9.7 M Zo (reduced) = rd RD = 20 k 2 k = 1.82 k Av (reduced) = gm(rd RD) = (5.47 mS)(1.82 k) = 9.94 Vo (reduced) = AvVi = (9.94)(20 mV) = 198.8 mV 28.
VGSQ = 0.95 V (as before), gm = 5.47 mS (as before)
Zi = 9.7 M as before RD since rd < 10(RD + RS) Zo = RD RS 1 g m RS rd 2 k = 2 k 0.61 k 1 (5.47 mS)(0.61 k) 20 k 2 k 2 k = 1 3.33 0.13 4.46 = 448.4 (slightly less than 461.1 obtained in problem 24) g m RD R RS 1 g m RS D rd (5.47 mS)(2 k) = 2 k 0.61 k 1 (5.47 mS)(0.61 k) 20 k 10.94 10.94 = = 2.45 slightly less than 2.52 obtained in problem 24) 1 3.33 0.13 4.46
Av =
130
29.
VGSQ = 1.75 V, gm = 2.14 mS
rd 10RD, Zi RS 1/gm = 1.5 k 1/2.14 mS = 1.5 k 467.29 = 356.3 rd 10RD, Zo RD = 3.3 k rd 10RD, Av gmRD = (2.14 mS)(3.3 k) = 7.06 Vo = AvVi = (7.06)(4 mV) = 28.24 mV 30.
VGSQ = 1.75 V, gm =
Zi = RS
VGSQ 2 I DSS 1 VP VP
1.75 V 2(8 mA) 1 2.8 V 2.8 V
= 2.14 mS
rd RD 25 k 3.3 k 28.3 k 1.5 k 1.5 k 1 g m rd 1 (2.14 mS)(25 k) 54.5
= 1.5 k 0.52 k = 386.1 Zo = RD rd = 3.3 k 25 k = 2.92 k g R RD / rd (2.14 mS)(3.3 k) 3.3 k / 25 k Av = m D 1 RD / rd 1 3.3 k / 25 k 7.062 0.132 7.194 = 6.36 = 1 0.132 1.132 Vo = AvVi = (6.36)(4 mV) = 25.44 mV 31.
VGSQ 1.2 V, gm = 2.63 mS
rd 10RD,
32.
Zi RS 1/gm = 1 k 1/2.63 mS = 1 k 380.2 = 275.5 Zo RD = 2.2 k Av gmRD = (2.63 mS)(2.2 k) = 5.79
VGSQ = 2.85 V, gm =
VGSQ 2 I DSS 1 VP VP
2.85 V 2(9 mA) 1 4.5 V 4.5 V
= 1.47 mS
Zi = RG = 10 M Zo = rd RS 1/gm = 40 k 2.2 k 1/1.47 mS = 512.9
Av =
1 g m (rd = 0.754
33.
680.27 (1.47 mS)( 40 k 2.2 k) 3.065 RS ) 1 (1.47 mS)( 40 k 2.2 k) 1 3.065
g m (rd RS )
VGSQ = 2.85 V, gm = 1.47 mS
Zi = 10 M (as in problem 32) Zo = rd RS 1/gm = 20 k 2.2 k 680.27 = 506.4 < 512.9 (#27) 1.982 k 1.47 mS( 20 k 2.2 k) 2.914 Av = 1 g m (rd RS ) 1 1.47 mS( 20 k 2.2 k) 1 2.914 = 0.745 < 0.754 (#32) g m (rd RS )
131
34.
VGSQ = 3.8 V
gm =
VGSQ 2 I DSS 1 VP VP
3.8 V 2(6 mA) 1 6 V 6V
= 0.733 mS
The network now has the format examined in the text and Zi = RG = 10 M
rd = rd + RD = 30 k + 3.3 k = 33.3 k g r (0.733 mS)(30 k) 21.99 = 0.66 mS Zo = rd RS 1/ g m = g m m d 30 k 3.3 k 33.3 k rd RD = 33.3 k 3.3 k 1/0.66 mS = 3 k 1.52 k 1 k
Av =
g m (rd RS )
1 g m (rd RS ) = 0.66
35.
36.
0.66 mS(3 k) 1.98 1.98 1 0.66 mS(3 k) 1 1.98 2.98
1 1 = 50 k, VGSQ = 0 V yos 20 S 2I 2(8 mA) = 5.33 mS gm = gm0 = DSS VP 3 Av = gmRD = (5.33 mS)(1.1 k) = 5.863 Vo = AvVi = (5.863)(2 mV) = 11.73 mV
rd =
VGSQ = 0.75 V, gm = 5.4 mS
Zi = 10 M ro 10RD, ro 10RD,
Zo RD = 1.8 k Av gmRD = (5.4 mS)(1.8 k) = 9.72
132
37.
Zi = 10 M Zo = rd RD = 25 k 1.8 k = 1.68 k Av = gm(rd RD) VGSQ 2I 0.75 V 2(12 mA) gm = DSS 1 1 VP VP 3.5 V 3.5 V
= 5.4 mS
Av = (5.4 mS)(1.68 k) = 9.07 38.
gm = yfs = 6000 S = 6 mS 1 1 = 28.57 k rd = yos 35 S rd 10RD, Av = gm(rd RD) = (6 mS)(28.57 k 6.8 k) = 32.94 Vo = AvVi = (32.94)(1.8 mV) = 59.29 mV
39.
5.49 k
Zi = 10 M 91 M 9 M VGSQ 2I 1.45 V 2(12 mA) gm = DSS 1 1 VP VP 3 V 3V
= 4.13 mS
Zo = rd RS 1/gm = 45 k 1.1 k 1/4.13 mS = 1.074 k 242.1 = 197.6 g m (rd RS ) (4.13 mS)( 45 k 1.1 k) Av = 1 g m (rd RS ) 1 (4.13 mS)( 45 k 1.1 k) (4.13 mS)(1.074 k) 4.436 1 (4.13 mS)(1.074 k) 1 4.436 0.816
40.
gm = 2k VGSQ VGS (Th ) 3
= 2(0.3 10 )(8 V 3 V) = 3 mS
133
41.
VGSQ = 6.7 V
gm = 2k VGSQ VT = 2(0.3 103)(6.7 V 3 V) = 2.22 mS Zi = =
RF rd RD 1 g m (rd RD )
10 M 100 k 2.2 k 1 (2.22 mS)(100 k 2.2 k)
10 M 2.15 k 1.73 M 1 2.22 mS(2.15 k)
Zo = RF rd RD = 10 M 100 k 2.2 k = 2.15 k Av = gm(RF rd RD) = 2.22 mS(2.15 k) = 4.77 42.
gm = 2k VGSQ VT = 2(0.2 103)(6.7 V 3 V) = 1.48 mS RF rd RD 10 M 100 k 2.2 k Zi = 1 g m (rd RD ) 1 (1.48 mS)(100 k 2.2 k) =
10 M 2.15 k = 2.39 M > 1.73 M (#39) 1 (1.48 mS)(2.15 k)
Zo = RF rd RD = 2.15 k = 2.15 k (#39) Av = gm(RF rd RD) = (1.48 mS)(2.15 k) = 3.182 < 4.77 (#39) 43.
VGSQ = 5.7 V, gm = 2k VGSQ VT = 2(0.3 103)(5.7 V 3.5 V)
= 1.32 mS 1 = 33.33 k 30 S Av = gm(RF rd RD) = 1.32 mS(22 M 33.33 k 10 k) = 10.15 Vo = AvVi = (10.15)(20 mV) = 203 mV
rd =
44.
ID = k(VGS VT)2 I D (on) 4 mA k = = = 0.444 103 2 2 (VGS (on) VT ) (7 V 4 V) gm = 2k (VGSQ VGS (Th ) ) = 2(0.444 103)(7 V 4 V) = 2.66 mS Av = gm(RF rd RD) = (2.66 mS)(22 M 50 k 10 k) = 22.16 8.33 k 8.33 k Vo = AvVi = (22.16)(4 mV) = 88.64 mV
134
45.
VGSQ = 4.8 V, gm = 2k VGSQ VGS(Th ) = 2(0.4 103)(4.8 V 3 V) = 1.44 mS
Av = gm(rd RD) = (1.44 mS)(40 k 3.3 k) = 4.39 Vo = AvVi = (4.39)(0.8 mV) = 3.51 mV 46.
rd = VGSQ
1 1 = 50 k yos 20 S 2I 2(8 mA) = 0 V, gm = gm0 = DSS = 6.4 mS VP 2.5 V
Av = gm(rd RD)
8 = (6.4 mS)(50 k RD) 50 k RD 8 = 1.25 k = 50 k RD 6.4 mS and RD = 1.28 k Use RD = 1.3 k 47.
1 1 VGSQ VP ( 3 V) = 1 V 3 3 2 VGSQ 1 V 12 mA 1 I DQ I DSS 1 3 V VP
2
= 5.33 mA
RS =
VS 1V = 187.62 Use RS = 180 I DQ 5.33 mA
gm =
VGSQ 2 I DSS 1 VP VP
2(12 mA) 1 V 1 3V 3 V
= 5.33 mS
Av = gm(RD rd) = 10 10 = 1.876 k or RD 40 k = 5.33 mS = 1.876 k 40 kRD = 1.876 kRD + 75.04 k2 38.124RD = 75.04 k RD = 1.97 k RD = 2 k 48.
Z i RG = 1 M Z o rd RD RD = 2.7 k From graph VGSQ 3.4 V
gm
2( I DSS ) V 1 GS VP VP 3.4 V 2[10 mA] 1 6 V 6V
3.33 mS[1 0.567]
3.33 mS[0.433] 1.44 mS AvNL g m RD ( 1.44 mS)(2.7 k) 3.89
135
(b) (c)
AvL
RL 4.7 k AvNL ( 3.89) RL Ro 4.7 k 2.7 k
2.47 Avs
Ri 1 M ( 2.47) AvL 1 M 0.6 k Ri Rsig
2.47
(d)
AvL the same 1 M ( 2.47) 1 M 10 k 2.45 Slight drop in AVs Avs
(e) remain the same 49.
(a)
Z i RG = 2 M From graph VGSQ 3.5 V 2( I DSS ) V 2[6 mA] 3.5 V 1 GS 1 VP VP 6V 6 V
gm
.834 mS Z o RS 1/ g m 3.3 k 1/ 0.834 mS = 0.72 k g m RS (0.834)(3.3 k) AvNL 1 g m RS 1 (0.834)(3.3 k) 0.733
(c)
AvL
RL 2.2 k AvNL 0.552 RL Ro 2.2 k 0.72 k
Avs
Ri 2 M AvL (0.552) Ri Rsig 2 M 0.5 k
0.552
(d)
AvL
RL 4.7 k AvNL (0.733) RL Ro 4.7 k 0.72 k
0.670 slight drop in gain AvS the same
136
(e)
AvL the same Avs
Ri 2 M AvL (0.552) Ri Rsig 2 M 20 k
0.546 slight drop in Avs
(e) 50.
Z i and Z o the same.
(a) From graph VGSQ 1.75 V gm
2 I DSS V 2[5 mA] 1.75 V 1 GS 1 VP VP 4V 4V
1.41 mS Z i RS 1/ g m 1.2 k 1/1.41 mS = 1.2 k 709.22 446 Z o RD = 3.3 k AvNL g m RD (1.41 mS)(3.3 k) 4.65
(b) (c)
(d)
RL 4.7 k AvNL (4.65) 2.73 RL RO 4.7 k 3.3 k Ri 446 Avs AvL (2.73) 1.29 Ri Rsig 446 500
AvL
RL 2.2 k AvNL (4.65) 1.86 RL Ro 2.2 k 3.3 k Ri 446 Avs AvL (1.86) 0.877 Ri Rsig 446 500 AvL
severe drop in AvL + Avs (e)
(f)
RL 4.7 k AvNL (4.65) 2.73 (no change) RL Ro 4.7 k 3.3 k Ri 446 Avs AvL (2.73) 2.23 (improved gain) Ri Rsig 446 100 AvL
RL 2.2 k(4.65) AvNL 1.86 (as in part (d)) RL Ro 2.2 k 3.3 k Ri 446 Avs AvL (1.86) 1.52 Ri Rsig 446 100 AvL
(g) Increasing RL and decreasing Rsig will improve the gain.
137
51.
From the graph: VGSQ 1.45 V, I DQ = 3.7 mA VD VDD I D RD 18 V (3.7 mA)(2.2 k) = 9.86 V VS I D RS (3.7 mA)(390 ) = 1.44 V VDS VD VS 9.86 V 1.44 V = 8.42 V VG = 0 V
52.
From Problem 51, VGSQ = 1.45 V gm
2 I DSS V 2[8 mA] 1 1.45 V 1 GS 1 VP VP 4.5 V 4.5 V
2.41 mS Av1 Av2 g m RD (2.41 mS)(2.2 k) = 5.30 AvT Av1 Av2 ( 5.30)( 5.30) = 28.1
53.
From graph VGSQ 1.4 V , I DQ 3.6 mA VD VDD I D RD 18 V (3.6 mA)(2.2 k) 18 V 7.92 V = 10.08 V VS I D RS (3.6 mA)(390 ) = 1.4 V VDS VD VS 10.08 V 1.4 V = 8.68 V VG = 0 V
54.
From problem 53: VGSQ = 1.4 V gm
2 I DSS V 2[12 mA] 1.4 V 1 GS 1 VP VP 3 3V
8 mS[1 0.467] 0.533(8 mS) 4.26 mS Av1 Av2 g m ( RD rd ) (4.26 mS)(2.2 k 40 k) 8.9 AvT Av1 Av2 ( 8.9)( 8.9) = 79.21
Increase in AvT compared to problem 52. 55.
gos 25S , rd 1/ 25 S 40 k Z i = 10 M Z o = 2.7 k
138
56.
From graph VGSQ = 0.95 V, I DQ = 2.9 mA VD 10 V I D (1.8 k) 10 V (2.9 mA)(1.8 k) 10 V 5.22 V = 4.78 V VS I D RS (2.9 mA)(330 ) = 0.957 V VDS VD VS 4.78 V 0.957 V = 3.82 V VG = 0 V I D I S = 2.9 mA IG = 0 A BRE 10 R2 (150)(2.2 k) 10(8.2 k) 330 k 82 k checks 8.2 k(10 V) = 2.55 V 8.2 k 24 k VE VB VBE 2.55 V 0.7 V = 1.85 V V 1.85 V = 0.84 mA I E IC E RE 2.2 k VC 10 V I C 2.7 k 10 V (0.84 mA)(2.7 k VB
10 V 2.27 V = 7.73 V VCE VC VE 7.73 V 1.85 V = 5.88 V I 0.84 mA IB C = 5.6 μA 150
57.
Z i2 8.2 k 24 k (2.2 k) = 6.11 k 330 k 6 k Av1 g m ( RD Z i2 ) gm
2 I DSS V 2(6 mA) 0.95 V 1 GS 1 VP VP 3 3V
2.73 mS
139
Av1 (2.73 mS)(1.8 k 6 k) 3.77 R Av2 C re 26 mV 26 mV = 30.95 re 0.84 mA IE 2.7 k Av2 = 87.2 30.95 AvT Av1 Av2 ( 3.77)( 87.2) = 328.74
58.
Z i = 10 M Z o RC = 2.7 k
140
Chapter 9 1.
(a) 3, 1.699, 1.151 (b) 6.908, 3.912, 0.347 (c) results differ by magnitude of 2.3
2.
(a) log10 0.24 106 = 5.38 (b) loge (0.24 106) = 2.3 log10(0.24 106) = 12.37 (c) loge (0.24 106) = 12.39
3.
(a) same 22.92 (b) same 23.98 (c) same 0.903
4.
(a) dB = 10 log10
Po 100 W = 10 log10 = 10 log10 20 = 10(1.301) 5W Pi = 13.01 dB
100 mW = 10 log1020 = 10(1.301) 5 mW = 13.01 dB
(b) dB = 10 log10
(c) dB = 10 log10
100 W = 10 log105 = 10(0.6987) 20 W
= 6.99 dB P2 25 W 10 log10 1 mW 600 1 mW 600 = 43.98 dBm
5.
GdBm = 10 log10
6.
GdB = 20 log10
V2 220 V = 20 log10 = 20 log102 = 20(0.301) V1 110 V
= 6.02 dB V2 25 V = 20 log10 2500 20 log10 10 mV V1 = 20(3.398) = 67.96 dB
7.
GdB = 20 log10
8.
(a) Gain of stage 1 = A dB Gain of stage 2 = 2 A dB Gain of stage 3 = 2.7 A dB A + 2A +2.7A = 120 A = 21.05 dB
141
(b) Stage 1: Av1 = 21.05 dB = 20 log10
Vo1 Vi1
V 21.05 = 1.0526 = log10 o1 20 Vi 1
101.0526 = and
Vo1
Vo1 Vi1
= 11.288
Vi1
Vo2
Stage 2: Av2 = 42.1 dB = 20 log10 2.105 = log10 102.105 = and
Vo2 Vi2
Vo2 Vi2
Vo2 Vi2
= 127.35
Stage 3: : Av3 = 56.835 dB = 20 log10 2.8418 = log10 102.8418 = and
Vo3 Vi3
Vi2
Vo3 Vi3
Vo3 Vi3
Vo3 Vi3
= 694.624
AvT Av1 Av2 AV3 = (11.288)(127.35)(694.624) = 99,8541.1 ? AT = 120 dB 20 log1099,8541.1 120 dB 119.99 dB (difference due to level of accuracy carried through calculations)
9.
(a) GdB = 20 log10
P2 48 W = 69.83 dB 10 log10 P1 5 W
Vo 20 log10 Vi = 82.83 dB
(b) Gv = 20 log10
Po Ro Vi
20 log10 (48 W)(40 k) 100 mV
142
Vi 2 (100 mV)2 (c) Ri = = 2 k P 5 W
(d) Po = 10.
Vo2 Vo = Ro
Po Ro (48 W)(40 k) = 1385.64 V
(a) Same shape except Av = 190 is now level of 1. In fact, all levels of Av are divided by 190 to obtain normalized plot. 0.707(190) = 134.33 defining cutoff frequencies at low end f1 230 Hz (remember this is a log scale) at high end f2 160 kHz (b)
11.
Av
(a)
Vo 1 Vi 1 ( f1 / f ) 2
f1 =
1 1 2 RC 2 (1.2 k)(0.068 F)
= 1950.43 Hz 1
Av
1950.43 Hz 1 f
(b)
2
AVdB
100 Hz: Av = 0.051
25.8
1 kHz:
Av = 0.456
6.81
2 kHz:
Av = 0.716
2.90
5 kHz:
Av = 0.932
0.615
10 kHz: Av = 0.982
0.162
(c) f1 1950 Hz (d)(e)
143
12.
1 = 1.95 kHz 2 RC f1 1.95 kHz = tan1 f = tan1 f
(a) f1 =
(b) f
= tan1
100 Hz 1 kHz 2 kHz 5 kHz 10 kHz
(c) f1 =
1.95 kHz f
87.06 62.85 44.27 21.3 11.03
1 = 1.95 kHz 2 RC
(d) First find = 45 at f1 = 1.95 kHz. Then sketch an approach to 90 at low frequencies and 0 at high frequencies. Use an expected shape for the curve noting that the greatest change in occurs near f1. The resulting curve should be quite close to that plotted above. 13.
(a) 10 kHz (b) 1 kHz (c) 20 kHz 10 kHz 5 kHz (d) 1 kHz 10 kHz 100 kHz
144
14.
From example 9.9, re = 15.76 RC RL ro 4 k 2.2 k 40 k Av = re 15.76 = 86.97 (vs. 90 for Ex. 9.9) f Ls : ro does not affect Ri f Ls =
1 the same 6.86 Hz 2 ( Rs Ri )CS
1 1 2 ( Ro RL )CC 2 ( RC ro RL )CC RC ro = 4 k 40 k = 5.636 k 1 f LC 2 (5.636 k 2 k)(1 F) = 28.23 Hz (vs. 25.68 Hz for Ex. 9.9)
f LC =
1 327 Hz is the same. 2 Re CE In total, the effect of ro on the frequency response was to slightly reduce the mid-band gain.
f LE : Re not affected by ro, therefore, f LE
15.
(a) RE 10R2 (120)(1.2 k) 10(10 k) 144 k 100 k (checks!) 10 k(14 V) = 1.795 V VB = 10 k 68 k VE = VB VBE = 1.795 V 0.7 V = 1.095 V V 1.095 V = 0.913 mA IE = E RE 1.2 k 26 mV 26 mV re = = 28.48 0.913 mA IE (b)
AVmid
( RL RC )
re = 72.91
(3.3 k 5.6 k) 28.48
(c) Zi = R1 R2 re = 68 k 10 k (120)(28.48 ) 3.418 k = 2.455 k
145
1 1 = 137.93 Hz 2 Ri CS 2 (2.455 k)(0.47 F) 1 1 = = 38.05 Hz 2 ( Ro RL )CC 2 (5.6 k 3.3 k)(0.47 F)
f Ls
(d)
f LC
f LE =
R R 1 with Re = RE 1 2 re 2 Re CE
R1 R2 68 k 10 k = 8.72 k R1 R2 8.72 k = 72.67 120 Re 1.2 k 72.67 28.48 1.2 k 101.15 = 93.29 1 1 f LE = = 85.30 Hz 2 RE CE 2 (93.29 )(20 F)
(e) (f)
f1 f Ls = 137.93 Hz
(g) 16.
VCC VBE 20 V 0.7 V 19.3 V RB ( 1) RE 470 k (111)(0.91 k) 470 k 101.01 k = 33.8 A IE = ( + 1)IB = (111)(33.8 A) = 3.752 mA 26 mV = 6.93 re = 3.752 mA
(a) IB =
(b)
Vo ( RC RL ) (3 k 4.7 k) 1.831 k 6.93 6.93 Vi re = 264.24
Avmid
(c) Zi = RB re = 470 k (110)(6.93 ) = 470 k 762.3 = 761.07 (d)
1 1 2 Z i CS 2 (761.07 )(1 F) = 209.12 Hz
f LS =
1 1 2 ( Ro RL )CC 2 (3 k 4.7 k)(1 F) = 20.67 Hz
f LC =
146
f LE =
1 2 Re CE
Re RE
RB
0.91 k
470 k 6.93 100
0.91 k 4.71 k 762.65
1 = 2 (762.65 )(6.8 F) = 30.69 Hz
(e) f1 f Ls 209.12 Hz (f)
(g) 17.
(a) RE 10R2 (100)(2.2 k) 10(30 k) 220 k 300 k (No!) RTh = R1 R2 = 120 k 30 k = 24 k 30 k(14 V) = 2.8 V ETh = 30 k 120 k ETh VBE 2.8 V 0.7 V IB = RTh ( 1) RE 24 k 222.2 k = 8.53 A IE = ( + 1)IB = (101)(8.53 A) = 0.86 mA 26 mV 6 mV = 30.23 re = 0.86 mA IE (b)
Avmid =
RE RL re RE RL
2.2 k 8.2 k 30.23 2.2 k 8.2 k = 0.983
=
RE = RE RL = 2.2 k 8.2 k = 1.735 k (c) Zi = R1 R2 (re + RE ) = 120 k 30 k (100)(30.23 + 1.735 k) = 21.13 k
147
(d)
f Ls
1 2 Ri Cs
1 2 (21.13 k)(0.1 F) = 75.32 Hz
=
f LC
1 2 ( Ro RL )CC
Ro = RE
Rs
Rs R1 R2 120 k 30 k 24 k
re
24 k 30.23 100 = 2.2 k 270.23 240.69
= 2.2 k
1 2 (240.69 8.2 k)(0.1 F) = 188.57 Hz
f LC
(e) (f)
f1low f LC 188.57 Hz
(g) 18.
VEE VEB 4 V 0.7 V = 2.75 mA RE 1.2 k 26 mV 26 mV re = = 9.45 IE 2.75 mA
(a) IE =
(b)
RC RL 3.3 k 4.7 k re 9.45 = 205.1
Avmid =
(c) Zi = RE re = 1.2 k 9.45 = 9.38 (d)
f LC
(e)
1 1 2 Z i Cs 2 (9.38 )(10 F) 1.7 kHz 1 1 = 2 ( Ro RL )CE 2 (3.3 k 4.7 k)(10 F) = 1.989 Hz
f Ls =
f1 f Ls 1.7 kHz
148
(f)
(g) 19.
From problem 15 (a) re 28.48 (b)
AVmid = 72.91
(c) Zi = 2.455 k f Ls
(d)
1 1 2 ( Ri RS )CS 2 (2.455 k 0.82 k)(0.47 F)
f Ls = 103.4 Hz f LC = 38.05 Hz from problem 15 f LE =
1 2 Re CE
with Re = RE
RS
re
and RS RS R1 R2
RS 0.82 k 68 k 10 k 810.2 k 10 k 749.48 RS 749.48 re 28.48 6.25 28.48 34.73 120 1 f LE = = 235.79 Hz 2 (33.75 )(20 F)
(e) (f)
f1 f LE = 235.79 Hz
(g) 20.
From problem 16 (a) re 6.93 (b)
Avmid = 264.24
(c) Zi = 761.07
149
(d)
1 1 2 ( Rs Z i )Cs 2 (600 761.07 )(1 F) = 116.93 Hz
f LS =
1 1 2 ( Ro RL )CC 2 (3 k 4.7 k)(1 F) = 20.67 Hz
f LC =
f LE =
1 2 Re CE
Re RE
1 2 (12.21 )(6.8 F) = 1.917 Hz
Rs
re 0.91 k
Rs RB
0.6 k 470 k 6.93 100 0.91 k 12.38
=
0.91 k 12.21
(e) f1 = f LE 1.917 kHz (f)
(g) 21.
From problem 17 (a) re = 30.23 (b)
Avmid = 0.983
(c) Zi = 21.13 k (d)
1 1 = = 71.92 Hz 2 (1 k 21.13 k)(0.1 F) 2 ( Rs Ri )Cs 1 R Ro RE S re 2 ( Ro RL )CC
f Ls f LC
1 2 (39.12 8.2 k)(0.1 F) 193.16 Hz
Rs Rs R1 R2 1 k 120 k 30 k 0.96 k Ro (2.2 k) 39.12
(e) (f)
re
f1low f LC 193.16 Hz
(g)
150
(0.96 k 30.23 100
22.
From problem 18 (a) re = 9.45 (b)
Avmid = 205.1
(c) Zi = 9.38 (d)
f LC
(e) (f)
1 1 = 145.5 kHz 2 ( Rs Z i )Cs 2 (100 9.38 )(10 F) 1 1 = = 1.989 Hz 2 ( Ro RL )CE 2 (3.3 k 4.7 k)(10 F)
f Ls =
f1 f Ls = 145.5 Hz
(g)
23.
(a) VGS = IDRS ID = I DSS 1
(b) gm0 =
VGS VP
VGSQ 2.45 V
2
I DQ 2.1 mA
2 I DSS 2(6 mA) = 2 mS 6V VP
gm = g m 0 1
VGSQ VP
2 mS 1
( 2.45 V) ( 6 V)
= 1.18 mS (c)
Avmid = gm(RD RL)
= 1.18 mS(3 k 3.9 k) = 1.18 mS(1.6956 k) = 2 (d) Zi = RG = 1 M (e)
Avs = Av = 2
151
(f)
1 1 2 ( Rsig Ri )CG 2 (1 k 1 M)(0.1 F) = 1.59 Hz
f LG
1 2 ( Ro RL )CC 1 = 2 (3 k 3.9 k)(4.7 F) = 4.91 Hz
f LC
f LS
1 2 Req CS
Req RS
1 2 (496.69 )(10 F) = 32.04 Hz
=
1 1 = 1.2 k 847.46 1.2 k 1.18 mS gm
= 496.69
(g) f1 f LS 32 Hz (h, i)
24.
(a) same as problem 19 VGSQ 2.45 V, I DQ 2.1 mA (b) gm0 = 2 mS, gm = 1.18 mS (rd has no effect!) (c)
= gm (RD RL rd)
Avmid
= 1.18 mS(3 k 3.9 k 100 k) = 1.18 mS (1.67 k) = 1.971 (vs. 2 for problem 19) (d) Zi = RG = 1 M (the same) (e)
Avs ( mid) =
Zi 1 M ( Avmid ) (1.971) Z i Rsig 1 M 1 k
= 1.969 vs. 2 for problem 19
152
(f)
f LG = 1.59 Hz (no effect) f LC : Ro = RD rd = 3 k 100 k = 2.91 k 1 1 2 ( Ro RL )CC 2 (2.91 k 3.9 k)(4.7 F) = 4.97 Hz vs. 4.91 Hz for problem 19 RS f LS : Req = 1 RS (1 g m rd ) /(rd ( RD RL )) 1.2 k = 1 (1.2 k)(1 (1.18 mS)(100 k)) /(100 k 3 k 3.9 k) 1.2 k = 1 1.404 499.2 1 1 f LS : = 2 Req CS 2 (499.2 )(10 F) = 31.88 Hz vs. 32.04 for problem 19. Effect of rd = 100 k insignificant! f LC =
25.
68 k(20 V) = 4.72 V 68 k 220 k VGS = VG IDRS VGSQ 2.55 V VGS = 4.72 V ID(2.2 k) 2 ID = IDSS(1 VGS/VP) I DQ 3.3 mA
(a) VG =
(b) gm0 =
2 I DSS 2(10 mA) = 3.33 mS 6V VP
gm = g m 0 1 (c)
VGS ( 2.55 V) 3.33 mS 1 VP 6 V
= 1.91 mS Avmid = gm(RD RL) = (1.91 mS)(3.9 k 5.6 k) = 4.39
(d) Zi = 68 k 220 k = 51.94 k (e)
Vo Vi Vi Vs Vi Zi 51.94 k = = 0.972 Vs Z i Rs 51.94 k 1.5 k
Avs ( mid) =
Avs ( mid) = (4.39)(0.972) = 4.27
153
(f)
1 1 2 ( Rsig Ri )CG 2 (1.5 k 51.94 k)(1 F) = 2.98 Hz
f LG
1 1 2 ( Ro RL )CC 2 (3.9 k 5.6 k)(6.8 F) = 2.46 Hz
f LC
f LS
1 2 Req CS
Req = RS
1 2 (388.1 )(10 F) = 41 Hz
=
(g) f1 = f LS 41 Hz (h, i)
26.
(a)
CM i (1 Av )C f
1 ( 120) (10 pF)
= 1210 pF
(b) CM O (1
1 1 )C f 1 10 pF Av ( 120)
= 10.08 pF
(c) absolutely
154
1 1 = 1.5 k 1.91 mS gm = 1.5 k 523.56 = 388.1
27.
(a)
RTh1 Rs R1 R2 Ri
1 2 RTh1 Ci
f Hi =
0.82 k 68 k 10 k 3.418 k
1 2 (614.56 )(931.92 pF) = 277.89 kHz
=
=
= 5 pF + 40 pF + 12 pF(1 (72.91)) = 931.92 pF Prob. 15 RTh2 = RC RL = 5.6 k 3.3 k = 2.08 k Co = CWo Cce CM o
1 2 (2.08 k)(28 pF) = 2.73 MHz
=
= 8 pF + 8 pF + 12 pF = 28 pF
1 1 = 2 (120)(28.48 )(40 pF + 12 pF) 2mid re (Cbe Cbc ) = 895.56 kHz Prob. 15 fT = f = (120)(895.56 kHz) = 107.47 MHz
(b) f
(c)
(d)
GBP ( Avmid )(BW) (72.91)(250 kHz) = 18.23 MHz
28.
(a)
f Hi
2.547 k
= 614.56 Ci = CW1 Cbe Cbc (1 Av )
1 2 RTh2 Co
f Ho =
0.81 k
1 2 RTh1 Ci
RTh1 = Rs RB Ri VCC VBE 20 V 0.7 V RB ( 1) RE 470 k (111)(0.91 k) = 33.8 A
Ri: IB =
IE = ( + 1)IB = (110 + 1)(33.8 A) = 3.75 mA
155
26 mV 26 mV = 6.93 IE 3.75 mA Ri = re = (110)(6.93 ) = 762.3 RTh1 = Rs RB Ri = 0.6 k 470 k 762.3
re =
= 335.50 1 f Hi = 2 (335.50 )(Ci ) Ci: Ci = CWi Cbe + (1 Av)Cbc ( RL RC ) (4.7 k 3 k) re 6.93 = 264.2 Ci = 7 pF + 20 pF + (1 (264.2)6 pF = 1.62 nF 1 f Hi = 2 (335.50 )(1.62 nF)
Av: Avmid =
293 kHz f Ho =
1 2 RTh2 Co
RTh2 = RC RL = 3 k 4.7 k = 1.831 k
Co = CWo Cce CM o Cf = Cbc = 11 pF + 10 pF + 6 pF = 27 pF 1 2 (1.831 k)(27 pF) = 3.22 MHz
f Ho =
1 2mid re (Cbe Cbc ) 1 = 2 (110)(6.93 )(20 pF + 6 pF) = 8.03 MHz
(b) f =
fT = midf = (110)(8.03 MHz) = 883.3 MHz
156
(c)
(d)
GBP ( Avmid )(BW) (264.24)(293 kHz) = 77.42 MHz
29.
(a)
f Hi =
1 2 RTh1 Ci
1 2 (955 )(58 pF) = 2.87 MHz
=
RTh1 Rs R1 R2 Z b
Zb = re + ( + 1)(RE RL) = (100)(30.23 ) + (101)(2.2 k 8.2 k) = 3.023 k + 175.2 k = 178.2 k RTh1 = 1 k 120 k 30 k 178.2 k = 955 Ci = CWi Cbe Cbc (No Miller effect) = 8 pF + 30 pF + 20 pF = 58 pF 24 k
f Ho =
1 2 RTh2 Co
1 2 (38.94 )(32 pF) = 127.72 MHz
RTh2 = RE RL re
R1 R2 R3
= 2.2 k 8.2 k 30.23
=
= 1.735 k (30.23 + 9.6 ) = 1.735 k 39.83 = 38.94 Co = CWo Cce = 10 pF + 12 pF = 32 pF
1 2mid re (Cbe Cbc ) 1 = 2 (100)(30.23 )(30 pF + 20 pF) = 1.05 MHz fT = mid f = 100(1.05 MHz) = 105 MHz
(b) f =
157
24 k 1 k 100
(c)
(d)
GBP ( Avmid )(BW) (0.983)(800 kHz) = 786.4 MHz
30.
(a)
f Hi =
1 2 RTh1 Ci
RTh1 = Rs RE Ri VEE VBE 4 V 0.7 V = 2.75 mA RE 1.2 k 26 mV 26 mV re = = 9.45 IE 2.75 mA Ri = RE re = 1.2 k 9.45 = 9.38
Ri: IE =
Ci: Ci = CWi Cbe (no Miller cap-noninverting!) = 8 pF + 24 pF = 32 pF Ri = 0.1 k 1.2 k 9.38 = 8.52 1 584 MHz 2 (8.52 )(32 pF) 1 = RTh2 = RC RL = 3.3 k 4.7 k = 1.94 k 2 RTh2 Co
f Hi = f Ho
Co = CWo Cbc (no Miller) = 10 pF + 18 pF = 28 pF 1 2 (1.94 k)(28 pF) = 2.93 MHz
f Ho =
158
1 2mid re (Cbe Cbc ) 1 = 2 (80)(9.45 )(24 pF + 18 pF) = 5.01 MHz fT = midf = (80)(5.01 MHz) = 400.8 MHz
(b) f =
(c)
(d)
GBP ( Avmid )(BW) (205.1)(2 MHz) = 410.2 MHz
31.
(a) From problem 23 gm0 = 2 mS, gm = 1.18 mS (b) From problem 23 (c)
f Hi
Avmid Avs (mid) = 2 RTh1 = Rsig RG
1 2 RTh1 Ci
= 1 k 1 M = 999 Ci = CWi C gs CM i
1 2 (999 )(21 pF) = 7.59 MHz
f Hi
f Ho =
CM i = (1 Av)Cgd
= (1 (2)4 pF = 12 pF Ci = 3 pF + 6 pF + 12 pF = 21 pF RTh2 = RD RL
1 2 RTh2 Co
= 3 k 3.9 k = 1.696 k Co = CWo Cds CM o
1 2 (1.696 k)(12 pF) = 7.82 MHz
=
1 4 pF 2 = (1.5)(4 pF) = 6 pF Co = 5 pF + 1 pF + 6 pF = 12 pF CM o = 1
159
(d)
(e)
GBP ( Avmid )(BW) (2)(6 MHz) = 12 MHz
32.
2 I DSS 2(10 mA) = 3.33 mS VP 6V
(a) gm0 =
From problem #21 VGSQ 2.55 V, I DQ 3.3 mA gm = g m 0 1 (b)
VGSQ VP
= 3.33 mS 1
2.55 V 6 V
= 1.91 mS
Avmid = gm(RD RL)
= (1.91 mS)(3.9 k 5.6 k) = 4.39 Zi = 68 k 220 k = 51.94 k Vi Zi 51.94 k = 0.972 Vs Z i Rsig 51.94 k 1.5 k Avs (mid) = (4.39)(0.972)
= 4.27 (c)
f Hi =
1 2 RTh1 Ci
RTh1 = Rsig R1 R2
= 1.5 k 51.94 k = 1.46 k
Ci = CWi C gs (1 Av )C gd = 4 pF + 12 pF + (1 (4.39))8 pF = 59.12 pF 1 2 (1.46 k)(59.12 pF) = 1.84 MHz
f Hi =
f Ho =
1 2 RTh2 Co
RTh2 = RD RL = 3.9 k 5.6 k = 2.3 k
160
Co = CWo Cds 1
1 C gd Av
= 6 pF + 3 pF + 1
1 8 pF ( 4.39)
= 18.82 pF 1 2 (2.3 k)(18.82 pF) = 3.68 MHz
f Ho =
(d)
GBP ( Avmid )(BW)
(e)
(4.27)(1.3 MHz) = 5.55 MHz
33.
AvT = Av1 Av2 Av3 Av4
= Av4 = (20)4 = 16 104
34.
f 2 =
21/ n 1 f 2
=
21/ 4 1 (2.5 MHz)
1.18 = 0.435(2.5 MHz) = 1.09 MHz 35.
f1
f1
21/ n 1 40 Hz = 0.435 = 91.96 Hz
40 Hz 21/4 1
161
36.
4
Vm sin 2 f s t 1 sin 2 (3 f s )t 1 sin 2 (5 f s )t 3 5 1 1 sin 2 (7 f s )t sin 2 (9 f s )t 7 9 1 3 3 = 12.73 103 (sin 2 (100 10 )t 3 sin 2 (300 10 )t
(a) v =
1 1 1 sin 2 (500 103 )t sin 2 (700 103 )t sin 2 (900 103 )t ) 5 7 9 0.35 tr 0.35 0.7 s 500 kHz
(b) BW
(c) P = f Lo
At 90% or 81 mV, t 0.75s At 10% or 9 mV, t 0.05s tr 0.75s 0.05s = 0.7s
V V 90 mV 80 mV = 0.111 V 90 mV P (0.111)(100 kHz) fs 3.53 kHz
162
Chapter 10 1.
Vo =
2.
Av =
3.
Vo =
RF 250 k V1 (1.5 V) = 18.75 V R1 20 k
Vo R F Vi R1 For R1 = 10 k: 500 k = 50 Av = 10 k For R1 = 20 k: 500 k = 25 Av = 20 k Rf R1
V1 =
4.
Vo =
V1
1 M V1 = 2 V 20 k
2V = 40 mV 50
RF 200 k V1 V1 = 10 V1 20 k R1
For V1 = 0.1 V: Vo = 10(0.1 V) = 1 V For V1 = 0.5 V: Vo = 10(0.5 V) = 5 V 5.
Vo = 1
Vo ranges from 1 V to 5 V
RF 360 k ( 0.3 V) V1 1 12 k R1
= 31(0.3 V) = 9.3 V 6.
Vo = 1 V1 =
7.
Vo = 1
RF 360 k V1 1 V1 = 2.4 V 12 k R1 2.4 V = 77.42 mV 31 RF V1 R1
For R1 = 10 k: 200 k Vo = 1 (0.5 V) = 21(0.5 V) = 10.5 V 10 k For R1 = 20 k:
163
200 k (0.5 V) = 11(0.5 V) = 5.5 V 20 k Vo ranges from 5.5 V to 10.5 V.
Vo = 1
8.
Vo =
Rf R1
V1
Rf R2
V2
Rf R3
V3
330 k 330 k 330 k (0.2 V) + ( 0.5 V)+ (0.8 V) 33 k 22 k 12 k = [10(0.2 V) + 15(0.5 V) + 27.5(0.8 V)] = [2 V + (7.5 V) + 2.2 V] = [24 V 7.5 V] = 16.5 V
=
9.
Vo =
RF R R V1 F V2 F V3 R1 R2 R3
68 k 68 k 68 k (0.2 V)+ ( 0.5 V) + ( 0.8 V) 33 k 22 k 12 k = [0.41 V 1.55 V + 4.53 V] = 3.39 V
=
1 v1 (t ) dt RC 1 = 1.5 dt (200 k)(0.1 F) = 50(1.5t) = 75t
10.
vo(t) =
11.
Vo = V1 = +0.5 V
12.
Vo =
13.
V2 =
100 k (1.5 V) 20 k R1 = 5(1.5 V) = 7.5 V Rf
V1
200 k (0.2 V) = 2 V 20 k
V3 = 1
200 k (0.2 V) = +4.2 V 10 k
164
400 k 100 k 100 k (0.1 V) (0.1 V) 20 k 20 k 10 k = (2.1 V)(5) + (10)(0.1 V) = 10.5 V 1 V = 11.5 V
14.
Vo = 1
15.
Vo =
600 k 600 k (25 mV) + ( 20 mV) 15 k 30 k
300 k 30 k
300 k ( 20 mV) 15 k = [40(25 mV) + (20)(20 mV)](10) + (20)(20 mV) = [1 V 0.4 V](10) + 0.4 V = 6 V + 0.4 V = 6.4 V
16.
Vo = 1
Rf R1
VIo I Io R f
200 k (6 mV) + (120 nA)(200 k) 2 k = 101(6 mV) + 24 mV = 606 mV + 24 mV = 630 mV
= 1
17.
I Io 4 nA 20 nA + = 22 nA 2 2 I 4 nA Io 20 nA = 18 nA 2 2
I IB I IB I IB I IB
18.
f1 = 800 kHz f 800 kHz fc = 1 = 5.3 Hz Av2 150 103
19.
ACL =
20.
ACL =
SR 2.4 V/s = 80 Vi / t 0.3 V/10 s
200 k = 100 2 k R1 K = ACL Vi = 100(50 mV) = 5 V SR 0.4 V / s = 80 103 rad/s ws K 5V w 80 103 fs = s = 12.73 kHz 2 2 Rf
165
21.
VIo = 1 mV, typical IIo = 20 nA, typical Vo(offset) = 1
Rf R1
VIo + IIoRf
200 k (1 mV) + (200 k)(20 nA) 20 k = 101(1 mV) + 4000 106 = 101 mV + 4 mV = 105 mV
= 1
22.
Typical characteristics for 741 Ro = 25 , A = 200 K 200 k = 100 R1 2 k (b) Zi = R1 = 2 k Ro 25 (c) Zo = 1 1 A 1 (200,000) 100 25 = = 0.0125 2001
(a) ACL =
23.
Rf
Vo 120 mV = 120 Vd 1 mV V 20 V = 20 103 Ac = o Vc 1 mV A 120 Gain (dB) = 20 log d 20log Ac 20 10 3 = 20 log(6 103) = 75.56 dB
Ad =
166
24.
Vd = Vi1 Vi2 = 200 V 140 V = 60 V V V (200 V 140 V) = 170 V Vc = i1 i 2 2 2 Ad = 200 Ac A 6000 = 30 Ac = d 200 200 A (b) CMRR = d 105 Ac A 6000 Ac = d5 = 0.06 = 60 103 10 105
(a) CMRR =
Using Vo = AdVd 1
1 Vc CMRR Vd
(a) Vo = 6000(60 V) 1 (b)
1 170 V = 365.1 mV 200 60 V
Vo = 6000(60 V) 1
1 170 V = 360.01 mV 105 60 V
167
Chapter 11 RF 180 k V1 (3.5 mV) = 175 mV R1 3.6 k
1.
Vo =
2.
Vo = 1
RF 750 k (150 mV, rms) V1 1 R1 36 k
= 3.275 V, rms 0 3.
510 k 680 k (20 V) 18 k 22 k = (29.33)(30.91)(22.73)(20 V) = 412 mV
Vo = 1
750 k 33 k
4.
1
420 k R1
420 k 14 R1 = 71.4 k
R1 =
= +15
420 k = 22 R2
420 k 22 R2 = 19.1 k
420 k = 30 R2 420 k 30 R3 = 14 k
R2 =
R3 =
Vo = (+15)(22)(30)V1 = 9000(80 V) = 792 mV = 0.792 V 5. Vo1 =
Vo2 =
168
RF 1 150 k V1 V1 R1 R1 Vo1 150 k Av1 15 V1 R1 150 k = 10 k R1 = 15 RF 2 150 k V1 V1 R2 R2 Vo 2 150 k Av2 30 V1 R2 150 k = 5 k R2 = 30
6.
Vo =
RF R 470 k 470 k V1 F V2 (40 mV) + (20 mV) R1 R2 47 k 12 k
= [400 mV + 783.3 mV] = 1.18 V 10 k 150 k 300 k 300 k V1 V2 10 k 10 k 150 k 150 k = 0.5(3)(1 V) 2(2 V) = 1.5 V 4 V = 2.5 V
7.
Vo =
8.
Vo =
330 k 470 k 470 k (12 mV) (18 mV) 33 k 47 k 47 k = [(120 mV)(10) + 180 mV] = [1.2 V + 0.18 V] = +1.02 V
9.
10.
V1 12 V = 6 mA R1 2 k
11.
IL =
12.
Vo = I1R1 = (2.5 mA)(10 k) = 25 V
13.
I o RF 1 V1 R1 Rs Io
14.
100 k 1 (10 mV) = 0.5 mA 200 k 10
Vo = 1 = 1
2R [V2 V1] Rp 2(5000) [1 V 3 V] = 22 V 1000
169
15.
fOH =
16.
fOL =
17.
1 1 2 R1C1 2 (2.2 k)(0.05 F) = 1.45 kHz 1 1 2 R1C1 2 (20 k)(0.02 F) = 397.9 Hz
1 1 = 318.3 Hz 2 R1C1 2 (10 k)(0.05 F) 1 1 fOH = 2 R2C2 2 (20 k)(0.02 F) = 397.9 Hz
fOL =
170
Chapter 12 1.
2.5 mA
VCC VBE 18 V 0.7 V = 14.42 mA RB 1.2 k I BQ = 40(14.42 mA) = 576.67 mA
I BQ
I CQ
Pi = VCCIdc VCC I CQ = (18 V)(576.67 mA) 10.4 W IC(rms) = IB(rms) = 40(5 mA) = 200 mA 2 Po = I C (rms)RC = (200 mA)2(16 ) = 640 mW 2.
VCC VBE 18 V 0.7 V = 11.5 mA RB 1.5 k = I BQ = 40(11.5 mA) = 460 mA
I BQ = I CQ
Pi(dc) = VCCIdc = VCC I CQ I BQ
= 18 V(460 mA + 11.5 mA) = 8.5 W Pi VCC I CQ = 18 V(460 mA) = 8.3 W
3.
From problem 2: I CQ = 460 mA, Pi = 8.3 W. For maximum efficiency of 25%: Po P 100% = 25% % = 100% o = Pi 8.3 W Po = 0.25(8.3 W) = 21 W [If dc bias condition also is considered: VC = VCC I CQ RC = 18 V (460 mA)(16 ) = 10.64 V collector may vary 7.36 V about Q-point, resulting in maximum output power: 2 VCE ( P ) (7.36 V)2 Po = 1.69 W 2 RC 2(16)
171
4.
Assuming maximum efficiency of 25% with Po(max) = 1.5 W P % = o 100% Pi 1.5 W =6W Pi = 0.25 Assuming dc bias at mid-point, VC = 9 V V VC 18 V 9 V = 0.5625 A I CQ CC RC 16 Pi(dc) = VCC I CQ = (18 V)(0.5625 A) = 10.38 W at this input: P 1.5 W 100% = 14.45% % = o 100% = 10.38 W Pi N1 N2
2
25 Rs = 1
2
5.
Rp =
6.
R2 = a2R1 R 8 k = 1000 a2 = 2 R1 8
(4 ) = 2.5 k
a = 1000 = 31.6 7.
R2 = a2R1 8 k = a2(4 ) 8 k = 2000 a2 = 4 a = 2000 = 44.7
8.
(a) Ppri = PL = 2 W V2 (b) PL = L RL VL =
PL RL (2 W)(16 )
= 32 = 5.66 V (c) R2 = a R1 = (3.87)2(16 ) = 239.6 2 V pri =2W Ppri = R pri 2
2 V pri = (2 W)(239.6 )
Vpri = 479.2 = 21.89 V [or, Vpri = aVL = (3.87)(5.66 V) = 21.9 V]
172
(d) PL = I L2 RL IL =
PL 2W = 353.55 mA 16 RL
2 2 R pri = (239.6 ) I pri Ppri = 2 W = I pri
2W = 91.36 mA 239.6 I 353.55 mA = 91.36 mA or, Ipri = L 3.87 a
Ipri =
9.
Idc = I CQ = 150 mA Pi = VCC I CQ = (36 V)(150 mA) = 5.4 W % =
Po 2W 100% 100% = 37% 5.4 W Pi
10.
11.
173
12.
(a) Pi = VCCIdc = (25 V)(1.75 A) = 43.77 W 2 2 V p 2 22 V = 1.75 A Where, Idc = I p RL 8 (b) Po =
V p2 2 RL
(c) % = 13.
(22 V)2 = 30.25 W 2(8 )
Po 30.75 W 100% = 100% = 69% 43.77 W Pi
(a) max Pi = VCCIdc 2 VCC 2 25 V (25 V) RL 8 = 49.74 W V2 (25 V)2 = 39.06 W (b) max Po = CC 2 RL 2(8 )
= VCC
max Po 39.06 W 100% 100% max Pi 49.74 W = 78.5%
(c) max % =
14.
(a) VL(peak) = 20 V Pi = VCCIdc = VCC
2 VL RL
= (22 V)
2 20 V = 70 W 4
VL2 (20 V)2 = 50 W 2 RL 2(4 ) P 50 W 100% = 71.4% % = o 100% = Pi 70 W
Po =
(b) Pi = (22 V)
2 4V = 14 W 4
(4)2 =2W 2(4) P 2W 100% = 14.3% % = o 100% = 14 W Pi
Po =
174
15.
16.
(a) max Po(ac) for VLpeak = 30 V: max Po(ac) =
VL2 (30 V) 2 = 56.25 W 2 RL 2(8 )
(b) max Pi(dc) = VCCIdc = VCC
(c) max % =
(d) max PZQ
17.
max Po 56.25 W 100% = 100% 71.62 W max Pi = 78.54% 2 2 VCC 2 (30) 2 = 22.8 W 2 RL 2 8
(a) Pi(dc) = VCCIdc = VCC
2 Vo RL
= 30 V
(b) Po(ac) =
(c) % =
2 Vo 2 30 V VCC = 71.62 W 8
RL
2
2 8 = 27 W 8
VL2 (rms) (8 V) 2 =8W RL 8
Po 8W 100% = 29.6% 100% 27 W Pi
(d) P2Q = Pi Po = 27 W 8 W = 19 W
175
18.
(a) Po(ac) =
VL2 (rms) (18 V) 2 = 40.5 W RL 8
(b) Pi(dc) = VCCIdc = VCC = (40 V)
(c) % =
20.
2 18 2 V = 81 W 8
Po 40.5 W 100% = 100% = 50% 81 W Pi
P2Q = Pi Po = 81 W 40.5 W = 40.5 W
(d)
19.
2 VLpeak RL
%D2 =
A2 0.3 V 100% = 100% 14.3% 2.1 V A1
%D3 =
A3 0.1 V 100% = 100% 4.8% 2.1 V A1
%D4 =
A4 0.05 V 100% = 100% 2.4% 2.1 V A1
%THD =
D22 D32 D42 100%
= (0.143) 2 (0.048) 2 (0.024) 2 100% = 15.3%
21.
1 VCEmax VCEmin D2 = 2 VCEmax VCEmin
100%
1 (20 V + 2.4 V) 10 V = 2 100% 20 V 2.4 V
=
22.
1.2 V 100% = 6.8% 17.6 V
THD =
D22 D32 D42 (0.15) 2 (0.01)2 (0.05) 2
0.16 I 2R (3.3 A) 2 (4 ) P1 = 1 C = 21.8 W 2 2 P = (1 + THD2)P1 = [1 + (0.16)2)]21.8 W = 22.36 W
176
23.
PD (150C) = PD(25C) (T150 T25) (Derating Factor) = 100 W (150C 25C)(0.6 W/C) = 100 W 125(0.6) = 100 75 = 25 W
24.
PD =
25.
PD =
TJ TA 200C 80C JC CS SA 0.5 C/W + 0.8 C/W + 1.5 C/W 120C = 42.9 W = 2.8 C/W TJ TA
JA
200C 80C 120C (40C/W) 40C/W =3W
=
177
Chapter 13 1.
2.
3.
4.
5.
178
6.
7.
Circuit operates as a window detector. 9.1 k (12 V) = 7.1 V 9.1 k + 6.2 k 1 k (12 V) = 1.7 V Output goes low for input below 1 k + 6.2 k Output is high for input between +1.7 V and +7.1 V.
Output goes low for input above
8.
9.
11010 26 (16 V) = (16 V) = 13 V 5 32 2
10.
Resolution =
11.
See section 13.3.
12.
Maximum number of count steps = 212 = 4096
13.
212 = 4096 steps at T =
VREF 10 V 10 V 12 = 2.4 mV/count 4096 2n 2
1 1 = 50 ns/count f 20 MHz ns Period = 4096 counts 50 = 204.8 s count
179
14. f=
1.44 = 350 kHz ( RA 2 RB )C
1.44 7.5 k + 2(7.5 k)(350 kHz) 183 pF
C=
15. T = 1.1 RAC 20 s = 1.1(7.5 k)C 20 10 6 1.1(7.5 103 ) = 2.4 109 = 2400 1012 = 2400 pF
C=
16.
T=
1 1 = 100 s f 10 kHz
T = 1.1 RAC = 1.1(5.1 k)(5 nF) = 28 s
180
17.
2 V VC R1C1 V V+ = 12 V R3 11 k VC = (V ) ( 12 V) = 10.3 V R2 R3 1.8 k + 11 k
fo =
2 12 V 10.3 V (4.7 k)(0.001 F) 12 V 3 = 60.3 10 60 kHz
fo =
18.
With potentiometer set at top: R3 R4 5 k 18 k VC = V (12 V) = 11.74 V R2 R3 R4 510 5 k 18 k resulting in a lower cutoff frequency of 2 V VC 2 12 V 11.74 V fo = 3 12 V R1C1 (10 10 )(0.001 F) V = 4.3 kHz With potentiometer set at bottom: R4 18 k VC = V (12 V) R2 R3 R4 510 5 k 18 k = 9.19 V resulting in a higher cutoff frequency of 2 V VC 2 12 V 9.19 V fo = (10 k)(0.001 F) 12 V R1C1 V = 61.2 kHz
19.
V+ = 12 V R3 10 k VC = V (12 V) = 10.4 V R2 R3 1.5 k + 10 k fo =
2 V VC 2 12 V 10.4 V 10 k(C1 ) 12 V R1C1 V = 200 kHz
2 (0.133) 10 k(200 kHz) = 133 1012 = 133 pF
C1 =
20.
fo =
21.
C1 =
0.3 0.3 R1C1 (4.7 k)(0.001 F) = 63.8 kHz 0.3 0.3 = 300 pF R1 f (10 k)(100 kHz)
181
22.
23.
8 fo V 8(63.8 103 ) = 6V = 85.1 kHz
fL =
fo
0.3 0.3 R1C1 4.7 k(0.001 F) = 63.8 kHz
For current loop:
mark = 20 mA space = 0 mA
For RS 232 C:
mark = 12 V space = +12 V
24.
A line (or lines) onto which data bits are connected.
25.
Open-collector is active-LOW only. Tri-state is active-HIGH or active-LOW.
182
Chapter 14 1.
2.
3.
4.
5.
Af =
dA f Af
A 1 A
1 dA
A A
2000 2000 = 9.95 1 201 1 ( 2000) 10
1 1 20
1000
(10%) = 0.2%
300 300 = 14.3 1 21 1 ( 300) 15 Rif = (1 + A)Ri = 21(1.5 k) = 31.5 k Ro 50 k = 2.4 k Rof = 1 A 21
Af =
A 1 A
Ro RD = 40 k 8 k = 6.7 k Ro RD A = gmRL = (5000 106)(6.7 103) = 33.5 R2 200 k = = 0.2 R1 R2 200 k 800 k A 33.5 33.5 Af = 1 A 1 ( 0.2)( 33.5) 7.7 = 4.4
RL =
DC bias: IB =
VCC VBE 16 V 0.7 V RB ( 1) RE 600 k 76(1.2 k) 15.3 V = 22.1 A = 691.2 k
IE = (1 + )IB = 76(22.1 A) = 1.68 mA [VCE = VCC IC(RC + RE) = 16 V 1.68 mA(4.7 k + 1.2 k) 6.1 V] 26 mV 26 mV 15.5 I E (mA) 1.68 mA hie = (1 + )re = 76(15.5 ) = 1.18 k = Zi Zo = RC = 4.7 k
re =
183
h fe
75 = 31.5 103 hie RE 1.18 k 1.2 k = RE = 1.2 103 (1 + A) = 1 + (1.2 103)(31.5 103) = 38.8
Av =
Af = Av f
Av 31.5 10 3 = 811.86 106 1 Av 38.8 = Af RC = (811.86 106)(4.7 103) = 3.82
Z i f = (1 + Av)Zi = (38.8)(1.18 k) = 45.8 k Z o f = (1 + Av)Zo = (38.8)(4.7 k) = 182.4 k
without feedback (RE bypassed): RC 4.7 k = 303.2 Av = 15.5 re 6.
C=
7.
fo =
1
1
2 Rf 6 2 (10 10 )(2.5 103 ) 6 = 2.6 109 = 2600 pF = 0.0026 F
=
1 2 RC
3
1 64
Rc R
1 1 12 2 (6 10 )(1500 10 ) 6 4(18 103 / 6 103 ) 3
= 4.17 kHz 4.2 kHz 1 1 3 2 RC 2 (10 10 )(2400 1012 ) = 6.6 kHz
8.
fo =
9.
Ceq =
C1C2 (750 pF)(2000 pF) = 577 pF C1 C2 750 pF + 2000 pF 1 1 fo = 2 LCeq 2 40 106 )(577 10 12 )
= 1.05 MHz 10.
fo =
1 , 2 LCeq
where Ceq =
C1C2 C1 C2
(0.005 F)(0.01 F) 0.005 F + 0.01 F = 3300 pF
1 2 (100 H)(3300 pF) = 277 kHz
=
=
184
11.
fo =
Leq = L1 + L2 + 2 M = 1.5 mH + 1.5 mH + 2(0.5 mH) = 4 mH
1 , 2 Leq C 1
=
2 (4 10 )(250 10 12 ) = 159.2 kHz
12.
fo =
3
where Leq = L1 + L2 + 2 M = 750 H + 750 H + 2(150 H) = 1800 H
1 , 2 LCeq
1 2 (1800 H)(150 pF) = 306.3 kHz
=
13.
See Fig. 14.33a and Fig. 14.34.
14.
fo =
1 RT CT ln(1/(1 )) for = 0.5: 1.5 fo RT CT (a) Using RT = 1 k 1.5 1.5 = 1.5 F CT = RT f o (1 k)(1 kHz)
(b) Using RT = 10 k 1.5 1.5 = 1000 pF CT = RT f o (10 k)(150 kHz)
185
Chapter 15 1. 2.
Vr (rms) 2 V / 2 = 0.028 Vdc 50 V V VFL 28 V 25 V %VR = NL 100% = 100% = 12% VFL 25 V
ripple factor =
3.
Vdc = 0.318Vm V 20 V = 62.89 V Vm = dc 0.318 0.318 Vr = 0.385Vm = 0.385(62.89 V) = 24.2 V
4.
Vdc = 0.636Vm V 8V = 12.6 V Vm = dc 0.636 0.636 Vr = 0.308Vm = 0.308(12.6 V) = 3.88 V
5.
6.
Vr (rms) 100% Vdc 8.5 14.5 V = 1.2 V Vr(rms) = rVdc = 100 VNL = Vm = 18 V VFL = 17 V V VFL 18 V 17 V 100% = 100% %VR = NL VFL 17 V = 5.88%
%r =
7.
Vm = 18 V C = 400 F IL = 100 mA
8.
Vr =
9.
C = 100 F Vdc = 12 V RL = 2.4 k
2.4 I dc 2.4(100) = 0.6 V, rms C 400 4.17I dc Vdc = Vm C 4.17(100) = 18 V = 16.96 V 400 17 V
Vr =
2.4 I dc 2.4(120) = 1.44 V C 200
I dc
Vdc 12 V 5 mA RL 2.4 k
186
2.4 I dc 2.4(5) = 0.12 V C 100
Vr(rms) =
2.4 I dc 2.4(150) = 100 F rVdc (0.15)(24)
10.
C=
11.
C = 500 F Idc = 200 mA R = 8% = 0.08 Using
2.4I dc CVdc 2.4 I dc 2.4(200) Vdc = = 12 V rC 0.08(500)
r=
Vm = Vdc + 2.4 I dc Vr
4.17I dc (200)(4.17) = 12 V + C 500 = 12 V + 1.7 V = 13.7 V 2.4(200) = 6857 F (0.07)
12.
C=
13.
C = 120 F Idc = 80 mA Vm = 25 V 4.17I dc 4.17(80) = 25 V Vdc = Vm C 120 = 22.2 V %r =
2.4I dc 2.4(80) 100% = 100% (120)(22.2) CVdc = 7.2% r Vdc 2(80) = 1.6 V, rms 100 100
14.
Vr
15.
Vr = 2 V Vdc = 24 V R = 33 , C = 120 F 1.3 1.3 XC = = 10.8 C 120
%r =
XC 10.8 Vr (2 V) = 0.65 V R 33 Vdc = Vdc IdcR = 24 V 33 (100 mA) = 20.7 V
Vr
187
Vr 2V 100% = 100% Vdc 24 V = 8.3%
%r =
Vr 0.65 V 100% = 100% = 3.1% Vdc 20.7 V
16.
RL Vdc R RL 500 (40 V) = 50 500 = 36.4 V V 36.4 V = 72.8 mA Idc = dc RL 500 Vdc
17.
1.3 1.3 = 13 C 100 X 13 Vr C Vr (2.5 V) R 100 = 0.325 V, rms
XC =
18.
VNL = 60 V RL 1 k VFL = (50 V) = 45.46 V Vdc R RL 100 1 k V VFL 50 V 45.46 V 100% = 100% %VR = NL VFL 45.46 V = 10 %
19.
Vo = VZ VBE = 8.3 V 0.7 V = 7.6 V VCE = Vi Vo = 15 V 7.6 V = 7.4 V V VZ 15 V 8.3 V = 3.7 mA IR = i R 1.8 k V 7.6 V = 3.8 mA IL = o RL 2 k I 3.8 mA IB = C = 38 A 100
188
IZ = IR IB = 3.7 mA 38 A = 3.66 mA
189
R1 R2 (Vz VBE2 ) R2 33 k 22 k (10 V + 0.7 V) = 22 k = 26.75 V
20.
Vo =
21.
Vo = 1
22.
Vo = VL = 10 V + 0.7 V = 10.7 V
R1 12 k VZ 1 10 V R2 8.2 k = 24.6 V
23.
24.
IL = 250 mA Vm = Vr(rms) Vrpeak =
2 =
2 (20 V) = 28.3 V 2.4 I dc 3 Vr(rms) = 3 C 2.4(250) = 2.1 V 500 = 28.3 V 2.1 V = 26.2 V
=
Vdc = Vm Vrpeak
3
Vi(low) = Vdc Vrpeak = 26.2 V 2.1 V = 24.1 V 25.
To maintain Vi(min) 7.3 V (see Table 15.1) Vrpeak Vm Vi(min) = 12 V 7.3 V = 4.7 V so that Vrpeak
4.7 V = 2.7 V 1.73 3 The maximum value of load current is then
Vr(rms) =
Idc =
Vr (rms)C (2.7 V)(200) = 225 mA 2.4 2.4
190
26.
Vo = Vref 1
R2 R1
+ IadjRL
1.8 k + 100 A(2.4 k) 240 = 1.25 V(8.5) + 0.24 V = 10.87 V
= 1.25 V 1
27.
Vo = Vref 1
R2 R1
= 1.25 V 1
+ Iadj R2 1.5 k 220
+ 100 A(1.5 k)
= 9.9 V
191
Chapter 16 1.
(a) The Schottky Barrier diode is constructed using an n-type semiconductor material and a metal contact to form the diode junction, while the conventional p-n junction diode uses both p- and n-type semiconductor materials to form the junction. (b)
2.
(a) In the forward-biased region the dynamic resistance is about the same as that for a p-n junction diode. Note that the slope of the curves in the forward-biased region is about the same at different levels of diode current. (b)
3.
In the reverse-biased region the reverse saturation current is larger in magnitude than for a p-n junction diode, and the Zener breakdown voltage is lower for the Schottky diode than for the conventional p-n junction diode.
I R 100 A 0.5 A 1.33 A/ C 75 C C I R (1.33 A/ C)C (1.33 A/ C)(25C) 33.25 A I R 0.5 A+33.25 A 33.75 A
4.
XC
1 1 22.7 kΩ 2 fC 2 (1 MHz)(7 pF)
RDC
VF 400 mV = 40 IF 10 mA
5.
6.
(a) VF 0.7 V (b) VF 0.5 V (c)
7.
As T , VF
CD 6.2 pF, X C
1 1 = 25.67 k 2 fC 2 (1 MHz)(6.2 pF)
191
8.
(a) CT(VR) =
=
C (0)
1 VR
VT
n
80 pF
1
4.2 V 0.7 V
1/ 3
80 pF = 41.85 pF 1.912
(b) k = CT(VT + VR)n = 41.85 pF(0.7 V + 4.2 V)1/3 12
1.698
71 10 9.
(a) At 3 V, C = 40 pF At 12 V, C = 20 pF C = 40 pF 20 pF = 20 pF C 40 pF = 2 pF/V VR 20 V C 60 pF = 6.67 pF/V At 2 V, 9V VR C increases at less negative values of VR. VR
(b) At 8 V,
10.
Ratio =
Ct ( 1 V) 92 pF = 16.73 Ct ( 8 V) 5.5 pF Ct ( 1.25 V) = 13 Ct ( 7 V)
11.
Ct 15 pF Q
12.
1 1 2 fRs Ct 2 (10 MHz)(3 )(15 pF)
354.61 vs 350 on chart C C 100% 100% T1 = + To TCC = Co (T1 T0 ) TCC (Co ) (0.11 pF)(100) + 25 (0.02)(22 pF) = 50C
=
13.
VR from 2 V to 8 V Ct(2 V) = 60 pF, Ct(8 V) = 6 pF Ratio =
Ct ( 2 V) 60 pF = 10 Ct ( 8 V) 6 pF
192
14.
Q(1 V) = 82, Q(10 V) = 5000 Ratio =
Q ( 10 V) 5000 = 60.98 Q( 1 V) 82
BW =
f o 10 106 Hz = 121.95 kHz Q 82
BW =
f o 10 106 Hz = 2 kHz 5000 Q
Ct CC (55 pF)(40 pF) = 23.16 pF Ct CC 55 pF + 40 pF 1 1 fr 739.5 kHz 2 LC 2 (2mH)(23.16 pF)
15.
Ct 55 pF, CT
16.
% =
Pmax 100% ( Acm2 )(100 mW/cm 2 )
Pmax 100% (2 cm )(100 mW/cm 2 ) Pmax = 18 mW
9% =
2
17.
The greatest rate of increase in power will occur at low illumination levels. At higher illumination levels, the change in VOC drops to nearly zero, while the current continues to rise linearly. At low illumination levels the voltage increases logarithmically with the linear increase in current.
18.
(a)
I SC 100 mA 100 mA = 5 mA/fc f c 2(20fc ) 20 f c 20f c
(b) At f c1 20 f c , x I SC1 = 100 mA I SC (5 mA/f c )( f c ) (5 mA/fc )(8 f c ) = 40 mA I SC I SC1 + 40 mA = 100 mA + 40 mA = 140 mA
19.
20.
(a)
VOC = 0.57 V 0.54 V = 0.03 V VOC 0.03 V = 0.375 mV/fc f c 80fc
(b)
VOC (0.375 mV/f c )( f c ) (0.375 mV/f c )(20 f c ) = 7.5 mV VOC = 0.54 V + 7.5 mV = 547.5 mV
(a) At knee, I SC 90 mA, VOC 0.5 V At VOC = 0 V, I SC = 100 mA At I SC = 0 A, VOC = 0.56 V
193
(b) Plotting P VI we find Pmax VI (0.5 V)(90 mA) = 45 mW
21.
(c)
Pmax = 45 mW vs. P VI (0.5 V)(180 mA) = 90 mW for f c2
(a)
W
(b) W
h
h
(6.624 10 34 Js)(3 108 m/s) = 422.81 × 1021 J 470 nm(blue)
(6.624 10 3 Js)(3 108 m/s) = 305.72 × 1021 J 650 nm(red)
(c) yes (d) Yes, higher energy level. (e) Low wavelength of emitted light has high energy content. 22.
4 10 9 W/m 2 = 2,486 fc 1.609 10 12 From the intersection of VA = 30 V and 2,486 fc we find I 440 A
23.
Note that V is given and not V. At the intersection of V = 25 V and 3000fc we find I 500 A and VR = IR = (500 106 A)(100 103 ) = 50 V
24.
Drawing a straight line approximation to the curve through the intersection of the axis will result in m
y 200 A 0.2 A/f c x 1000fc
y mx b I (0.2 A/f c )( f c ) 0 I (0.2 A/f c )( f c )
25.
(a) Extending the curve: 0.1 k 1000fc, 1 k 25fc R (1 0.1) 103 = 0.92 /fc 0.9 /fc f c (1000 25) f c (b) 1 k 25fc, 10 k 1.3fc R (10 1) 103 = 379.75 /fc 380 /fc f c (25 1.3) fc
194
(c) 10 k 1.3fc, 100 k 0.15fc R (100 10) 103 = 78,260.87 /fc 78 103 /fc f c (1.3 0.15) f c The greatest rate of change in resistance occurs in the low illumination region. 26.
The “dark current” of a photodiode is the diode current level when no light is striking the diode. It is essentially the reverse saturation leakage current of the diode, comprised mainly of minority carriers.
27.
10fc R 2 k Vo = 6 V =
(2 103 )Vi 2 103 5 103
Vi = 21 V 28.
29.
Except for low illumination levels (0.01fc) the % conductance curves appear above the 100% level for the range of temperature. In addition, it is interesting to note that for other than the low illumination levels the % conductance is higher above and below room temperature (25C). In general, the % conductance level is not adversely affected by temperature for the illumination levels examined.
(a)
(b)
(c) Increased levels of illumination result in reduced rise and decay times.
195
30.
The highest % sensitivity occurs between 5250A and 5750A. Fig 16.20 reveals that the CdS unit would be most sensitive to yellow. The % sensitivity of the CdS unit of Fig. 16.30 is at the 30% level for the range 4800A 7000A. This range includes green, yellow, and orange in Fig. 16.20.
31.
(a) 5 mW radiant flux (b) 3.5 mW
32.
3.5 mW = 2.34 1010 lms 13 1.496 10 W/lm
(a) Relative radiant intensity 0.8. (b)
Relative radiant vs degrees off vertical >30 and relative radiant intensity essentially zeroDrops off very sharply after 25!
33.
At IF = 60 mA, 4.4 mW At 5, relative radiant intensity = 0.8 (0.8)(4.4 mW) = 3.52 mW
34.
6, 7, 8
35.
36.
The LED generates a light source in response to the application of an electric voltage. The LCD depends on ambient light to utilize the change in either reflectivity or transmissivity caused by the application of an electric voltage.
37.
The LCD display has the advantage of using approximately 1000 times less power than the LED for the same display, since much of the power in the LED is used to produce the light, while the LCD utilizes ambient light to see the display. The LCD is usually more visible in daylight than the LED since the sun’s brightness makes the LCD easier to see. The LCD, however, requires a light source, either internal or external, and the temperature range of the LCD is limited to temperatures above freezing.
196
38.
Since log scales are present, the differentials must be as small as possible. R (7000 1000) 6000 = 150 /C T (40 0) 40
R (3 1) 2 = 0.05 /C 40 40 T
From the above 150 /C: 0.05 /C = 3000:1 Therefore, the highest rate of change occurs at lower temperatures such as 20C. 39.
No. 1 Fenwall Electronics Thermistor material. Specific resistance 104 = 10,000 cm 2x R = 2 (10,000 ) = 20 k R= A twice
40.
(a) 105 A = 10 A (b) Power 0.1 mW, R 107 = 10 M (c)
41.
42.
Log scale 0.3 mW V = IR + IRunk + Vm V = I(R + Runk) + 0 V V Runk = R I 0.2 V = 10 2 mA = 100 10 = 90
The primary difference between the standard p-n junction diode and the tunnel diode is that the tunnel diode is doped at a level from 100 to several thousand times the doping level of a p-n junction diode, thus producing a diode with a “negative resistance” region in its characteristic curve.
197
43.
1 1 6 2 fC 2 (1 10 Hz)(5 10 12 F) = 31.83 k
At 1 MHz: XC =
1 2 (100 10 Hz)(5 10 12 F) = 318.3
At 100 MHz: XC =
6
At 1 MHz: X LS = 2fL = 2(1 106 Hz)(6 109 H) = 0.0337 At 100 MHz: X LS = 2(100 106 Hz)(6 109 H) = 3.769 Ls effect is negligible! R and C in parallel: f = 1 MHz
(152 180)(31.83 k 90) 152 j 31.83 k = 152.05 0.27 152 0
ZT =
f = 100 MHz (152 180)(318.3 90) ZT = 152 j318.3 = 137.16 25.52 152 0 At very high frequencies XC has some impact! 44.
The heavy doping greatly reduces the width of the depletion region resulting in lower levels of Zener voltage. Consequently, small levels of reverse voltage can result in a significant current levels.
45.
At VT = 0.1 V, IF 5.5 mA At VT = 0.3 V IF 2.3 mA V 0.3 V 0.1 V I 2.3 mA 5.5 mA 0.2 V = 62.5 = 3.2 mA
R=
46.
E 2V 5.13 mA R 0.39 k From graph: Stable operating points: IT 5 mA, VT 60 mV IT 2.8 mA, VT = 900 mV
Isat =
198
47.
48.
E 0.5 V = 9.8 mA R 51 Draw load line on characteristics.
Isat =
fs = =
1 2 LC
1
Rl2C L
1 2 (5 10 3 H)(1 10 6 F)
1
(10 ) 2 (1 10 6 F) 5 10 3 H
= (2250.79 Hz)(0.9899) 2228 Hz
199
Chapter 17 1.
2.
3.
4.
(a) p-n junction diode (b) The SCR will not fire once the gate current is reduced to a level that will cause the forward blocking region to extend beyond the chosen anode-to-cathode voltage. In general, as IG decreases, the blocking voltage required for conduction increases. (c) The SCR will fire once the anode-to-cathode voltage is less than the forward blocking region determined by the gate current chosen. (d) The holding current increases with decreasing levels of gate current.
5.
(a) Yes (b) No (c) No. As noted in Fig. 17.8b the minimum gate voltage required to trigger all units is 3 V. (d) VG = 6 V, IG = 800 mA is a good choice (center of preferred firing area). VG = 4 V, IG = 1.6 A is less preferable due to higher power dissipation in the gate. Not in preferred firing area.
6.
In the conduction state, the SCR has characteristics very similar to those of a p-n junction diode (where VT = 0.7 V).
7.
The smaller the level of R1, the higher the peak value of the gate current. The higher the peak value of the gate current the sooner the triggering level will be reached and conduction initiated.
8.
(a) VP =
Vsec (rms) 2 2 117 V = 2 = 82.78 V 2 VDC = 0.636(82.78 V) = 52.65 V
(b) VAK = VDC VBatt = 52.65 V 11 V = 41.65 V
200
(c) VR = VZ + VGK = 11 V + 3 V = 14 V At 14 V, SCR2 conducts and stops the charging process. (d) At least 3 V to turn on SCR2. (e) V2 9.
1 1 VP = (82.78 V) = 41.39 V 2 2
(a) Full-wave rectified waveform with peak value of 168.28 V (b)
P VI 100 W (0.707)(168.28 V)(I rms ) I rms 840.5 mA, I peak = 1.19 A
(c) 1.19 A (d)
1 1 1 tr T 4 4 60 Hz
(e)
RC (510 k)(0.1 F) = 51 ms
(f)
open
= 4.17 ms
(g) dc level established = 0.636 (Vpeak) = 0.636(168.28 V) 107 V C 107 V(1 e t / 51 ms ) = 40 V t = 23.86 ms (h) turn on (g) forced commutation 10.
(a) Vpeak = (1.414)(6.3 V) = 8.91 V Vpeak across 6 V lamp = 8.91 V 0.7 V = 8.21 V 150 (6 V) = 0.783 V 150 1 k VC1 (peak) 8.21 V 0.783 V = 7.43 V
(b) VR2
(c) Vpeak = 8.21 V 0.7 V 5 V = 2.51 V (d) Vlamp = 6 V (e) P = VI = 2 W = (6 V)(I) 2W I = 0.333 A 6V
201
11.
12.
Negative pulse at the cathode terminal.
13.
(a)
VGK 12 V VR , VR
VGK 12 V
R (24 V) R RS
R (24 V) R RS
(b)
RS R , VGK = 0 V, scs off
(c)
VGK 2 V= 12 V+
10 k(24 V) 10 k RS
RS = 14 k
14.
(d)
I
12 V = 60 mA 200
(e)
I
12 V = 0.12 mA 100 k
(f)
Inductive element in alarm could establish spikes. Use protective capacitive element.
(a) Charge toward 200 V but will be limited by the development of a negative voltage
VGK VZ VC that will eventually turn the GTO off. 1 (b) = R3C1 = (20 k)(0.1 F) = 2 ms 5 = 10 ms (c)
15.
1 (5 ) = 5 ms = 5RGTO C1 2 5 ms 5 ms 1 RGTO = = 10 k (20 k above) 6 2 5C1 5(0.1 10 F)
5 =
(a) 0.7 mW/cm2 (b) 0C 0.82 mW/cm2 100C 0.16 mW/cm2 0.82 0.16 100% 80.5% 0.82
202
16.
VC = VBR + VGK = 6 V + 3 V = 9 V VC = 40(1 et/RC) = 9 40 40et/RC = 9 40et/RC = 31 t/RC = 31/40 = 0.775 e RC = (10 103 )(0.2 106 F) = 2 103 s loge(et/RC) = loge 0.775 t/RC = t/2 103 = 0.255 and t = 0.255(2 103) = 0.51 ms
17.
18.
VBR1 VBR2 10% VBR2
= 6.4 V 0.64 V 5.76 V 7.04 V 19.
(1 M jX Cb )Vi 0
VG =
10 M 1 M jX Cb
(1 M jX Cb )Vi 0 11 M jX Cb
For 45° shift in denominator X Cb 11 M . In the numerator the result is 1 M j11 M j11 M 11 M 90 . (11 M 90)(Vi 0) VG 0.707Vi 45 15.56 M 45
A 45° shift is established with X Cb 11 M X CB
20.
1 2 fCb
Cb
1 1 241 pF 2 fX C 2 (60 Hz)(11 M)
( X C 90)(170 V0) 12 V R jX C ( X C )((170 V)( 90) 12 V R 2 X C2
VXC
( X C )(170 V)
R 2 X C2
12 V
1 1 2.65 k 2 fC 2 (60 Hz)(1 F) (2.65 k)(170 V) 12 V R 2 (2.65 k)2
XC
R = 25.25 k
203
21.
V VP > R1 IP 40 V [0.6(40 V) + 0.7 V] = 1.53 M > R1 10 10 6 V VV 40 V 1 V < R1 = 4.875 k < R1 8 mA IV
1.53 M > R1 > 4.875 k 22.
(a) =
RB1 RB1 RB2
0.65 = IE 0
(b) RBB = RB1 RB2 (c)
IE 0
2 k 2 k RB2
RB2 = 1.08 k
= 2 k + 1.08 k = 3.08 k
VRB = VBB = 0.65(20 V) = 13 V 1
(d) VP = VBB + VD = 13 V + 0.7 V = 13.7 V 23.
(a) =
RB1 RBB
0.55 =
IE 0
RB1
10 k RB1 = 5.5 k
RBB = RB1 RB2 10 k = 5.5 k + RB2 RB2 = 4.5 k
(b) VP = VBB + VD = (0.55)(20 V) + 0.7 V = 11.7 V (c) R1 <
V VP 20 V 11.7 V = 166 k 50 A Ip
ok: 68 k < 166 k V VV 18.8 = (68 103)(0.1 106) loge = 5.56 ms 8.3 V VP V 11.7 t2 = ( RB1 R2 ) C loge P = (0.2 k + 2.2 k)(0.1 106) loge VV 1.2 = 0.546 ms T = t1 + t2 = 6.106 ms 1 1 = 163.77 Hz f= T 6.106 ms
(d) t1 = R1C loge
204
(e)
(f) R2V 2.2 k(20 V) R2 RBB 2.2 k 10 k = 3.61 V R (V 0.7 V) VR2 2 P R2 RB1 VR2
2.2 k(11.7 V 0.7 V) 2.2 k 0.2 k = 10.08 V
=
1 1 = 184.16 Hz R1C log e (1/(1 )) (6.8 k)(0.1 F)log e 2.22 difference in frequency levels is partly due to the fact that t2 10% of t1.
(g) f
24.
IB = 25 A IC = h fe I B = (40)(25 A) = 1 mA
25.
205
26.
I T 0.95 0 0.95 = = 1.26%/C 25 ( 50) 75
(a) DF =
(b) Yes, curve flattens after 25C. 27.
(a) At 25C, ICEO 2 nA At 50C, ICEO 30 nA I CEO (30 2) 109 A 28 nA = 1.12 nA/C T (50 25)C 25C ICEO (35C) = ICEO(25C) + (1.12 nA/C)(35C 25C) = 2 nA + 11.2 nA = 13.2 nA From Fig. 17.55 ICEO (35C) 4 nA Derating factors, therefore, cannot be defined for large regions of non-linear curves. Although the curve of Fig. 17.55 appears to be linear, the fact that the vertical axis is a log scale reveals that ICEO and T( C) have a non-linear relationship.
28.
29.
Io IC 20 mA = 0.44 I i I F 45 mA Yes, relatively efficient.
(a) PD = VCEIC = 200 mW PD 200 mW IC = = 6.67 mA @ VCE = 30 V 30 V VCEmax PD 200 mW = 20 V @ IC = 10 mA 10 mA IC P 200 mW IC = D = 8.0 mA @ VCE = 25 V 25 V VCE
VCE =
Almost the entire area of Fig. 17.57 falls within the power limits. (b) dc =
IC I 4 mA 4 mA = 0.4, Fig. 17.56 C = 0.4 10 mA I F 10 mA IF
The fact that the IF characteristics of Fig. 17.57 are fairly horizontal reveals that the level of IC is somewhat unaffected by the level of VCE except for very low or high values. Therefore, a plot of IC vs. IF as shown in Fig. 17.56 can be provided without any reference to the value of VCE. As noted above, the results are essentially the same.
206
30.
(a) IC 3 mA (b) At IC = 6 mA; RL = 1 k, t = 8.6 s RL = 100 ; t = 2 s 1 k:100 = 10:1 8.6 s:2 s = 4.3:1 R:t 2.3:1 3RB2
3 = 0.75, VG = VBB = 0.75(20 V) = 15 V 4
31.
=
32.
VP = 8.7 V, IP = 100 A
ZP =
VP 8.7 V = 87 k ( open) I P 100 A
VV = 1 V, IV = 5.5 mA
ZV =
VV 1V = 181.8 (relatively low) IV 5.5 mA
3RB2 RB2
87 k: 181.8 = 478.55:1 500:1 33.
Eq. 17.23: T = RC loge
VBB VBB RC log e VBB VP VBB (VBB VD )
Assuming VBB VD, T = RC loge
= RC loge
34.
RB1 RB2 RB2
RC log e 1
VBB VBB (1 ) RB1 RB2
= RC loge(1/1 ) = RC loge
Eq. 17.24
(a) Minimum VBB: V VP 20 k Rmax = BB IP VBB (VBB VD ) = 20 k IP VBB VBB VD = IP 20 k VBB(1 ) = IP 20 k + VD I 20 k VD VBB = P 1 (100 A)(20 k) 0.7 V = 1 0.67 = 8.18 V 10 V OK
207
1
1 RB1 RB1 RB2
VBB VV 12 V 1 V = 2 k 5.5 mA IV R < 2 k
(b) R <
(c) T RC loge 1
RB1 RB2
2 103 = R(1 106)loge 1 10 k 5 k loge3 = 1.0986 3
2 10 (1 10 6 )(1.0986) R = 1.82 k
R=
208
