8.1
) V ( t u o
2
V
1
−5
−4
−3
−2
−1
0
0
1
2
3
4 V in
−1
−2
5 (mV)
8.11 V
−
V
−
V out V in
= V + = V in R4 (R2 + R3 ) R2 = V out = V in R1 + R4 (R2 + R3 ) R2 + R3
(R2 + R3 ) R2 = R1 + R4 (R2 + R3 ) R2 + R3 =
R4
1
−
(R2 + R3 ) [R1 + R4 (R2 + R3 )] R2 [R4 (R2 + R3 )]
If R1 → 0, we expect the result to be: =
V in V out V in
=
R2
+ R3 R2 + R3 R2
R1 =0
Taking limit of the original expression as
R1
R2
V out
=1 +
R3 R2
→ 0, we have:
(R2 + R3 ) [R1 + R4 (R2 + R3 )] (R2 + R3 ) [R4 (R2 + R3 )] = 0 R2 [R4 (R2 + R3 )] R2 [R4 (R2 + R3 )]
lim
R1 →
= 1+
R3 R2
This agrees with the expected result. Likewise, if R3 → 0, we expect the result to be: V in V out V in
R3 =0
Taking the limit of the original expression as
R4 V out R1 + R2 R4 R1 + R2 R4 = R2 R4 =
R2
=1 + R3
R1 R2
R4
→ 0, we have:
(R2 + R3 ) [R1 + R4 (R2 + R3 )] = 0 R2 [R4 (R2 + R3 )]
lim
R3 →
R2 (R1
+ R2 R4 ) R2 (R2 R4 ) R1 + R2 R4 = R2 R4 = 1+
This agrees with the expected result.
R1 R2
R4
8.14 We need to derive the closed-loop gain of the following circuit: R1 R2
Rout vout
+
+ vin
vX
−
vX vout
−A0 vX −
−
= ( vout − vin )
R2 R1
= (−A0 vX − vin ) =
+
+ R2
+ vin
+ R2 + vin Rout + R1 + R2 R1
−A0 (vout − vin )
R2 R1
+ R2
+ vin − vin
+ R2 + vin Rout + R1 + R2 R1
Grouping terms, we have: vout
+ R2 1 + A0 = vin R1 + R2 Rout + R1 + R2 R2
R1
R2 Rout + R1 + R2 + R2 A0 − A0 − 1 + Rout + R1 + R2 R1 + R2 R1 + R2 R1 + R2 Rout + R1 + R2 R1 = vin − A0 −1 Rout + R1 + R2 R1 + R2 R1 + R2 1 = vin [Rout + R1 + R2 − A0 R1 − R1 − R2 ] Rout + R1 + R2 A0 R1 + R1 + R2 = vin 1 − Rout + R1 + R2 R1
vout vin
= = =
1−
A0 R1 +R1 +R2 Rout +R1 +R2
1+
A0 R2 Rout +R1 +R2
Rout
+ R1 + R2 − A0 R1 − R1 − R2 Rout + R1 + R2 + A0 R2
− A0 R1 Rout + R1 + (1 + A0 ) R2
To find the output impedance, we must find
Rout
Z out
=
vt it
for the following circuit:
R1 R2
Rout
+
+
vX
−
+ −A0 vX
−
it
vt
−
it
=
vX
=
it
=
vt
+ A0 vX Rout R2
+
R1
+ R2
vt
+ R2 R vt + A0 R +R R1
2
1
Rout
vt
vt
+
vt R1
+ R2
1 = vt + + Rout Rout (R1 + R2 ) R1 + R2 R1 + (1 + A0 ) R2 + Rout = vt Rout (R1 + R2 ) Z out
=
vt it
=
1
2
A0 R2
Rout (R1
+ R2 ) R1 + (1 + A0 ) R2 + Rout
8.15 Refer to the analysis for Fig. 8.42. V out
V in
=
R1 R2
=4
= 10 kΩ = 4 R2 = 40 kΩ
Rin ≈ R2 R1
From Eq. (8.99), we have E =
1−
A0 −
1+
= 1000 Rout = 1 kΩ A0
E =
0.51 %
Rout R2
Rout R1
+ A0 +
R1 R2
8.17 V + V in R2 V out V in
If R1 → 0 or
R3
= V (since
= ∞) V out R3 R4 =− R3 R1 + R3 R4 −
R3 R1
= −
R2
A0
+ R3 R4 R3 R4
→ 0, we expect the amplifier to reduce to the standard inverting amplifier. V out V in V out V in
The gain reduces to the expected expressions.
=−
R1 →0
R3 →0
=−
R3 R2 R1 R2
8.18
V out
V out
R3
−
R2
V +
= V (since
V X
=
R3 R3
+ R4
A0
V out
= ∞) =
R2 R1
+ R2
(V out − V in ) + V in
= 1− ) +
+ R4 R1 + R2 R3 (R1 + R2 ) − R2 (R3 + R4 (R1 + R2 ) (R3 + R4 ) R3
−
V out V in
R2
V in
= V in =
R1
R2
R1
R1
+ R2
R1 (R3
+ R4 ) R3 (R1 + R2 ) − R2 (R3 + R4 )
8.22 We must find the transfer transfer func function tion of the following following circuit: C 1 R1
vout
+ vin
Rin
−
+ vX −
+
−A0 vX −
vout = −A0 vX vX vX
1
1 1+ + sRin C 1 sR1 C 1
vX vout vout
sR1 Rin C 1 1 + A0 sR1 Rin C 1 + R1 + Rin vout vin
1 vX vX − vin = vout − + sC 1 Rin R1 vin = vout + sR1 C 1 sR1 Rin C 1 vout + Rin vin = sR1 Rin C 1 + R1 + Rin sR1 Rin C 1 vout + Rin vin = −A0 sR1 Rin C 1 + R1 + Rin Rin = −A0 vin sR1 Rin C 1 + R1 + Rin −A0 Rin sR1 Rin C 1 + R1 + Rin = · sR1 Rin C 1 + R1 + Rin sR1 Rin C 1 + R1 + Rin + sR1 Rin C 1 A0 −A0 Rin = sR1 Rin C 1 + R1 + Rin + sR1 Rin C 1 A0 −A0 Rin = sR1 Rin C 1 (1 + A0 ) + R1 + Rin −A0 Rin = R R C (1+A ) 1+s R +R 1
in
1
=
1
0
in
−A0 Rin / (R1 + Rin ) 1 + s (R1 Rin ) C 1 (1 + A0 )
s p = −
1 (R1 Rin ) C 1 (1 + A0 )
Comparing this to the result in Eq. (8.37) Comparing (8.37),, we can see that we can simply replace replace R1 with R1 Rin , effectively increasing the pole frequency (since R1 Rin < R1 for finite Rin ). We can also write the result as 1 s p = − R1 C 1 (1 + A0 )
R1 1+ Rin
In this form, it’s clear that the pole frequency increases by 1 + R1 /Rin .
8.23 We must find the transfer transfer func function tion of the following following circuit: C 1 R1
Rout +
+ vin
−A0 vX
vX
−
vout
+
−
−
vout = −A0 vX +
vin − vout 1 Rout R1 + sC 1
vX
R1 = vin + 1 (vout − vin ) R1 + sC 1
vout = −A0 vin +
vout 1 + vout
R1 +
A0 R1 + Rout 1 R1 + sC 1
1 sC 1
+ A0 R1 + Rout
R1 +
1
= vin −A0 +
R1 vin − vout 1 (vout − vin ) + 1 Rout R1 + sC R1 + sC 1
A0 R1 + Rout 1 R1 + sC 1
1
1
= vin
sC 1
−A0 R1 − A0 sC + A0 R1 + Rout 1
R1 +
1 sC 1
vout {1 + sC 1 [(1 + A0 ) R1 + Rout ]} = −vin {A0 − sC 1 Rout } vout A0 − sC 1 Rout = − vin 1 + sC 1 [(1 + A0 ) R1 + Rout ] s p = −
1 C 1 [(1 + A0 ) R1 + Rout ]
Comparing this to the result in Eq. (8.37), we can see that the pole gets reduced in magnitude due to Rout .
8.26 We must find the transfer function of the following circuit: R1 C 1 vout
+
+
vin
Rin
+
−A0 vX
vX
−
−
−
vout vX
vX
1 + sRin C 1 +
Rin R1
vX
vout
vout
vout
vout
1+
A0
Rin R1
1 + sRin C 1 +
Rin R1
1 + sRin C 1 + (1 + A0 ) RR
in
1
1 + sRin C 1 +
Rin R1
1 + sRin C 1 + (1 + A0 )
Rin R1
vout vin
lim
A0 →∞
vout vin
= −A0 vX
= (vin − vX ) sC 1 − = vin sRin C 1 + vout =
vin sRin C 1
= −vin = −vin
− vout R1
Rin
Rin
R1
+ vout RR
in
1
1 + sRin C 1 +
= −A0
vX
vin sRin C 1
Rin R1
+ vout RR
in
1
1 + sRin C 1 +
Rin R1
sRin C 1 A0
1 + sRin C 1 +
Rin R1
sRin C 1 A0
1 + sRin C 1 +
Rin R1
= −vin sRin C 1 A0 = −
sR1 Rin C 1 A0 R1
+ sR1 Rin C 1 + (1 + A0 ) Rin
= −sR1 C 1
Comparing this to Eq. (8.42), we can see that if we let (8.42).
A0
→ ∞, the result actually reduces to Eq.
8.27 We must find the transfer function of the following circuit: R1 C 1
Rout vout
+
+ vin
+
−A0 vX
vX
−
−
−
vout
= −A0 vX +
− vout Rout 1 R1 + sC
vin
1
1
vX
vout
vout
vout
R1
+
1
1+ 1
sC 1
vout {1 + A0
A0 sC
1
R1
+ Rout
+
1
sC 1
1
+
= −A0
sC 1
R1
1
+
vin
+
sC 1 sC 1
R1
(vout − vin ) +
1
+
sC 1
1
= vin −A0 + = vin
(vout − vin )
1
1
1 + A0 sC + Rout
R1
= vin +
A0 sC
1
R1
+ Rout
+
1
sC 1
− vout Rout 1 R1 + sC
vin
1
1 1 −A0 R1 − A0 sC + A0 sC + Rout 1
R1
sC 1
+
1
1
sC 1
+ sC 1 (R1 + Rout )} = −vin {sC 1 (A0 R1 − Rout )} vout vin
lim
A0 →∞
vout vin
= −
− Rout ) 1 + A0 + sC 1 (R1 + Rout ) sC 1 (A0 R1
= −sR1 C 1
Comparing this to Eq. (8.42), we can see that if we let (8.42).
A0
→ ∞, the result actually reduces to Eq.
8.28 vout
= −A0 v
−
1
v−
vout
vout
1 + A0
1
sC 1
1
vout
sC 1
R1 + 1
sC 1
vout
(1 + A0 )
1
sC 2
1
sC 1
R1
sC 1
1
R1 +
sC 2
1
R2
1
R2 + A0
R1 + 1
sC 1
R1
R2
sC 2
R1 +
1
sC 2
R2
vout vin
= −A0
= −vin A0
1
sC 1
R1 +
1
sC 1
1
R1 + 1
sC 2
R1
1
R1 +
1
1
sC 2
sC 2
sC 2
R2
R2
R2 − 1
R1 +
sC 1
1
1
R2
R1
R1 +
sC 1
1
sC 2
sC 1
1
sC 1
1
sC 1
= −vin A0 1 −
= −vin A0
+ (vout − vin )
vin
R1
sC 1
= vin + (vout − vin )
sC 2
1
sC 1
R1
R2
R2
1
sC 2
= −A0 (1 + A0 )
1
sC 1
R2
R1 +
1
sC 2
R2
Unity gain occurs when the numerator and denominator are the same (note that we can drop the negative sign since we only care about the magnitude of the gain): A0
(A0 − 1)
1
sC 2
1
sC 2 1
sC 2 1
sC 1
R2 R2 R2 R1
= (1 + A0 ) = (1 + A0 ) =
1 sC 1
1 sC 1
R1 + R1
1 sC 2
R2
+1 A0 − 1 A0
It is possible to obtain unity gain by choosing the resistors and capacitors according to the above formula.
8.31 vout v1
− vX R2 vout RF
+
+
v2
v1 R2
− vX R1
+
v2
R1
vout vout
1 + A0
(R1 R2 RF ) RF
= −A0 vX vX − vout = =
RF vX
R1
R2 RF
= −A0 (R1 R2 RF )
= −A0 (R1 R2 RF )
vout
= −A0 (R1 R2 RF )
vout RF v1 R2
= −
v1 R2
+
v2 R1
+ v1 R2
R2
v2 R1
+
1 + A0 (R
= −A0 RF (R1 R2 RF )
+
v1
1
+
R1
v2 R1
R2 RF ) RF v1 R2
RF
v2
+
v2 R1
+ A0 (R1 R2 RF )
[RF A0 (R1 R2 RF )]
8.32 For A0 = ∞, we know that no effect on vout .
v+
= v− , meaning that no current flows through
vout
For
A0 <
= −RF
vX
R2
+
v2 R1
,
A0
RP
will have
=∞
1 RP
+
1 R1
+
1 R2
+
1+
A0 RF
= −A0 vX
vX
=
=
vX
=
1 RF
vout vout
v1
Thus,
∞, we have to include the effects of RP . vout
RP .
R2
+
v1 R2
= −A0
(R1 R2 RF RP ) = −A0 = −A0 =−
= −
R1
+
vout
v2
+
v2 R1
v1 v1 v1
R2 v1 R2
+
+
v2 R1
+
− vX
RF
RP
RF
+
R2
+
vout
vout
+
R2
v1
R1
+
R2
− vX
v2
+
R2
v1
− vX
v1
RF
v2 R1 v2 R1 v2 R1
v2
R1
vout
+
(R1 R2 RF RP )
vout RF
(R1 R2 RF RP )
(R1 R2 RF RP ) (R1 R2 RF RP ) A 1+ R (R1 R2 RF RP ) 0
F
RF A0 (R1
R2 RF RP ) RF + A0 (R1 R2 RF RP ) [RF A0 (R1 R2 RF RP )] ,
A0 <
∞
8.33 We must find
vout
for the following circuit: R2
v1
RF Rout
+
R1
v2
−A0 vX
vX
−
−
vout
= −A0 vX + = −vX
vX vX
1 RF
+
1 R1
+
=
vX
=
R2
vout
= vout +
1
vout RF
=−
RF
+
vout
+
A0
v1
R2
vout RF
− vX
Rout
− vX +
v1
+
R2 v2
Rout
+ Rout
− vX R1
v2
v1
+
R2
v2 R1
RF
R1
+
R2
+
R1
Rout
+
R1
− vX
v2
+
R2
R2
v1
+
v1
vout
+
v1 R2
v2 R1
+
v2
R1
(R1 R2 RF )
(R1 R2 RF )
A0
+
Rout R1
+
Rout R2
+ Rout
v1 R2
+
v2 R1
Grouping terms, we have: vout
1+
(R1 R2 RF )
RF
A0
+
Rout R1 R2
=− =−
vout
v1 R2 v1 R2
= −RF
+ +
v2 R1 v2 R1
v1 R2
+
(R1 R2 RF )
(R1 R2 RF )
v2 R1
Rout RF
A0
+
A0
Rout
+
+ Rout
R2
R1
Rout R1
+ (R1 R2 RF )
+ (R1 R2 RF )
R2 A0
A0
+
+
+ Rout
Rout R1 R2
Rout R1 R2
v1
R2
+
v2 R1
8.34 We must find
vout
for the following circuit: R2
v1
RF
vout
+
R1
v2
Rin
+ −A0 vX
vX
−
−
RP
vout
vX
= −A0 vX =
v1
− vX 1 +
RP Rin
R1
v2
+
RP Rin
− vX 1 + R2
vout
− vX 1 +
+
RP Rin
RF
Grouping terms, we have: vX
1 Rin vX
+ 1+ (R1
RP
1
=
R2 RF R2 RF ) + RP + Rin = Rin (R1 R2 RF ) Rin
R1
=
vX vout
v1 R2 v1 R2
+ +
v1 R2
= −A0
v2 R1 v2 R1
+
+ +
v2 R1
v1 R2
Rin
vout RF vout RF
+
+
vout RF
v2 R1
+
Rin (R1
(R1
vout RF
R2 RF ) R2 RF ) + RP + Rin Rin (R1 R2 RF ) (R1 R2 RF ) + RP + Rin
Grouping terms, we have:
vout
RF [(R1
R2 RF ) =− vout 1 + RF (R1 R2 RF ) + RP + Rin R2 RF ) + RP + Rin] + A0 Rin (R1 R2 RF ) =− RF [(R1 R2 RF ) + RP + Rin ] A0
Rin (R1
v1 R2 v1 R2
+ +
v2 R1 v2 R1
A0 Rin (R1
R2 RF ) (R1 R2 RF ) + RP + Rin A0 Rin (R1 R2 RF ) (R1 R2 RF ) + RP + Rin
Simplifying, we have: vout
= −
v1 R2
+
v2 R1
A0 RF Rin (R1
RF [(R1
R2
R2 RF ) RF ) + RP + Rin] + A0 Rin (R1 R2 RF )
8.35 I D1
Plotting
I D1 (t),
=
V in R1
V in >
0
V in
0 <0
we have
V 0
) d e t t o D ( )
V 0 /R1
) t (
1
D
0
0
I
t ω
( s o c
0
V = ) t ( n i
V
−π/ω
0
t
π/ω
−V 0
8.36 I D1
Plotting
I D1 (t),
=
V in R1
V in >
0
V in
0 <0
we have
V 0
) d e t t o D ( )
V 0 /R1
) t (
1
D
I
0
0
t ω
( s o c
0
V = ) t ( n i
V
−π/ω
0 t
π/ω
−V 0
8.37 V Y
Plotting
=
V Y (t)
V in
− V D,on
V DD
and
V out (t),
0 V out >0
V in < V in
=
0 I D1 = >0
V in
V in <
0
V in
V in R1
V in <
0
V in
0 >0
we have
V in(t) = V 0 cos(ωt) V Y (t) V out(t) V DD
V 0
0 −π/ω
0
π/ω t
−V 0
Plotting
I D1 (t),
we have:
I D1(t) 0
V 0 / R 1
−
π / ω
t
0
π / ω
−
V
0
0
V in(t) = V 0 cos(ωt) (Dotted)
V
0
8.38 Since the negative feedback loop is never broken (even when the diode is off, feedback), V + = V − will always hold, meaning V X = V in . We must determine when be off, and V X will follow
D1
turns on/off to determine V Y . We know that for V in < 0, the diode will As V in begins to go positive, the diode will remain off until
V Y
will be fixed at V X V Y
Plotting
V Y (t)
and
provides negative
V in .
V in
Once the diode turns on,
RP
V out (t),
RP R1
> V D,on
+ V D,on . Thus, we can write:
V in
= V in =
V in V in
1+
RP R1
+ V D,on
R
V in < V D,on R 1
P
R V in > V D,on R 1 P
we have
V in(t) = V 0 cos(ωt) V X (t) V Y (t) V 0 + V D,on V 0
0 −π/ω
0
π/ω t
−V 0
−V 0 (1 + RP /R1)
8.40 Note that although in theory the output is unbounded (i.e., by Eq. (8.66), we can take the logarithm of an arbitrarily small positive number), in reality the output will be limited by the positive supply rail, as shown in the following plot. V out V X
V DD
0 −1 −1
0
R1 I S
1 V in (V)
8.42 When V in > 0, the feedback loop will be broken, and the output will go to the positive rail. When V in < 0, we have: I C
V out
=−
V in R1
= I S eV
BE
= −V T ln −
/V T
V in R1 I S
= I S e−V
out
/V T
This gives us the following plot of V out vs. V in : ) V ( t V DD u o
V
−1
−R1 I S
0
0
Note that this circuit fails to behave as a non-inverting logarithmic amplifier.
1 V in (V)
8.44 (a) V out
= −V T
ln 1V V in
R1 I S
−0.2 V = −V T ln R1 I S
R1 I S
= 456 µ V
(b) Av
=
dV out dV in
=−
V T V in
V in =1 V
V in =1 V
= −0.026
8.45 When V in < V T H , the output goes to the positive rail. When V in > V T H , we have: I D =
V in − V T H R1
V GS = −V out = V T H +
V out = −V T H −
dV out 1 =− dV in 2
= −
2I D W µ C L n ox
2 (V in − V T H ) R1 W µ C L n ox
R1 W µ C L n ox
2 (V in −
2
V T H ) R1 W µ C L n ox
1 , V in > V T H (V in − V T H )
µ C 2R1 W L n ox
8.46 When V in > 0, the output goes to the negative rail. When V in < 0, we have: I D = −
V in R1
V SG = V out = |V T H | +
V out = V T H +
−
2 |I D |
W µ C L p ox
2V in , W R1 L µ p C ox
V in < 0
8.49 We model an input offset with a series voltage source at one of the inputs. R1 R2 −
V in + −
V out
+
V os + −
V in − V os (R1 + R2 ) R2 R1 + R2 R1 + R2 + V os 1− R2 R2
V out = V in − = V in
R1 R1 = − V in + 1 + R2 R2 Note that even when V in = 0, V out = (1 + R1 /R2 ) V os .
V os
8.54 Let
V in
= 0. V + V out
= −I B 1 (R1 R2 ) = − (I B 2 + ∆I ) (R1 R2 ) = = V − +
I B 2
+
V − R2
R1
= − (I B 2 + ∆I ) (R1 R2 ) +
I B 2
= − (I B 2 + ∆I ) (R1 R2 ) 1 +
−
R1 R2
(I B2 + ∆I ) (R1 R2 ) R2
+ I B 2 R1
= −∆I R1
If the magnitude of the error must be less than ∆ V , we have: ∆I R1
<
R1 <
Note that this does not depend on
R2 .
V −
∆V ∆V ∆I
R1
8.61 Let
E
refer to the gain error. R1 R2
=8
R1
= 8 kΩ
R2
= 1 kΩ
vout vin
=− =−
E =
A0 −
R1 R2 R1 R2
1−
1+
Rout R2
Rout R1
+ A0 +
R1 R2
(Eq. 8.99)
(1 − E ) A0 −
1+
Rout R2
Rout R1
+ A0 +
R1 R2
= 0.1 % A0
= 9103
Note that we can pick any R1 , R2 such that their ratio is 8 (i.e., this solution is not unique). However, A0 will change depending on the values chosen.
