2.11 (a) 2. k = 8. 8 .617 × 10
−5
eV/ eV /K 3/2
15
ni(T = 300 K) = 1. 1 .66 × 10 (300 K)
0.66 eV exp − 2 (8 (8..617 × 10 5 eV eV/ /K) (30 (3000 K) −
cm
cm
−3
= 2.465 × 1013 cm
−3
3/2
15
ni(T = 600 K) = 1. 1 .66 × 10 (600 K)
0.66 eV exp − 2 (8 (8..617 × 10 5 eV eV/ /K) (60 (6000 K) −
−3
= 4.124 × 1016 cm
−3
Compared to the values obtained in Example 2.1, we can see that the intrinsic carrier concentration 10 in Ge at T = 300 K is 21..465 = 2282 times higher than the intrinsic carrier concentration in 08 10 ×
×
13
10
Si at T = 300 K. Similarly, at T = 600 K, the intrinsic carrier concentration in Ge is 26..8 times higher than that in Si. 26
16
4.124×10 1.54×1015
(b) Since phosphorus is a Group V element, it is a donor, meaning N D = 5 × 1016 cm n-type material, we have:
−3
n = N D = 5 × 1016 cm
−3
2
[ni (T = 300 K)] p((T = 300 K) = p = 1.215 × 1010 cm n [ni (T = 600 K)]2 p((T = 600 K) = p = 3.401 × 1016 cm n
−3
−3
=
. For an
2.3 (a) Since the doping doping is uniform, uniform, we have have no diffusion curren current. t. Thus Thus,, the total current current is due only only to the drift component. I tot tot = I drift d rift = q (nµn + pµ p )AE n = 1017 cm−3 p = n2i /n = (1 (1..08 × 1010 )2 /1017 = 1.17 × 103 cm−3 µn = 1350 cm2 /V · s µ p = 480 cm2 /V · s 1V E = V /d = 0.1 µ m = 105 V/cm A = 0.05 µ m × 0.05 µ m = 2.5 × 10−11 cm2 Since nµn
≫ pµ p ,
we can write I tot tot
≈
qnµn AE
= 54 54..1 µ A (b) All of the parameters are the same except ni , whic which h mean meanss we must re-calculate re-calculate p. ni (T = 400 K) = 3. 3 .657 × 1012 cm−3 p = n2i /n = 1.337 × 108 cm−3 Since nµn ≫ pµ p sti still ll hol holds ds (no (note te tha thatt n is 9 orders of magnitude larger than p), the hole concentration once again drops out of the equation and we have I tot tot
≈
qnµn AE
= 54 54..1 µ A
2.4 (a) From Problem 1, we can calculate ni for Ge. ni (T = 300 K) = 2.465 × 1013 cm−3 I tot = q (nµn + pµ p )AE n = 1017 cm−3 p = n2i /n = 6.076 × 109 cm−3 µn = 3900 cm2 /V · s µ p = 1900 cm2 /V · s 1V E = V /d = 0.1 µ m 5 = 10 V/cm A = 0.05 µ m × 0.05 µ m = 2.5 × 10−11 cm2 Since nµn
≫ pµ p ,
we can write I tot
≈
qnµn AE
= 156 µ A (b) All of the parameters are the same except ni , which means we must re-calculate p. ni (T = 400 K) = 9.230 × 1014 cm−3 p = n2i /n = 8.520 × 1012 cm−3 Since nµn ≫ pµ p still holds (note that n is 5 orders of magnitude larger than p), the hole concentration once again drops out of the equation and we have I tot
≈
qnµn AE
= 156 µ A
2.5 Since there’s no electric field, the current is due entirely to diffusion. If we define the current as positive when flowing in the positive x direction, we can write I tot
dn = I diff = AJ diff = Aq Dn dx
dp − D p dx
A = 1 µ m × 1 µ m = 10−8 cm2 Dn = 34 cm2 /s D p = 12 cm2 /s dn 5 × 1016 cm−3 =− = −2.5 × 1020 cm−4 − 4 dx 2 × 10 cm dp 2 × 1016 cm−3 = = 1020 cm−4 dx 2 × 10−4 cm I tot = 10−8 cm2 1.602 × 10−19 C 34 cm2 /s =
−15.54 µ A
−2.5 ×
1020 cm−4
2
12 cm /s10 −
20
cm−4
2.8 Assume the diffusion lengths Ln and L p are associated with the electrons and holes, respectively, in this material and that Ln , L p ≪ 2 µ m. We can express the electron and hole concentrations as functions of x as follows: n(x) = N e−x/L
n
p(x) = P e(x−2)/L
p
# of electrons =
2
an(x)dx
0
2
=
aN e−x/L dx n
0
2
= −aN Ln e−x/L
n
0
= −aN Ln e−2/L
# of holes =
2
n
−
1
ap(x)dx
0
2
=
aP e(x−2)/L dx p
0
1
= aP L p e(x−2)/L = aP L p
p
−
2
0
e−2/L
Due to our assumption that Ln , L p ≪ 2 µ m, we can write e−2/L
n
≈
0
e
−2/Lp
≈
0
# of electrons
≈
aN Ln
# of holes
≈
aP L p
p
2.10 (a) nn = N D = 5 × 1017 cm−3 pn = n2i /nn = 233 cm−3 p p = N A = 4 × 1016 cm−3 n p = n2i /p p = 2916 cm−3 (b) We can express the formula for V 0 in its full form, showing its temperature dependence:
kT N A N D V 0 (T ) = ln q (5.2 × 1015 )2 T 3 e−E /kT g
V 0 (T = 250 K) = 906 mV V 0 (T = 300 K) = 849 mV V 0 (T = 350 K) = 789 mV Looking at the expression for V 0 (T ), we can expand it as follows: V 0 (T ) =
kT ln(N A ) + ln(N D ) − 2 ln 5.2 × 1015 q
−
3ln(T ) + E g /kT
Let’s take the derivative of this expression to get a better idea of how V 0 varies with temperature. dV 0 (T ) k = ln(N A ) + ln(N D ) − 2 ln 5.2 × 1015 dT q
−
3ln(T ) − 3
From this expression, we can see that if ln(N A ) + ln(N D ) < 2 ln 5.2 × 1015 + 3 ln(T ) + 3, or
2
equivalently, if ln(N A N D ) < ln 5.2 × 1015 T 3 − 3, then V 0 will decrease with temperature, which we observe in this case. In order for this not to b e true (i.e., in order for V 0 to increase with temperature), we must have either very high doping concentrations or very low temperatures.
2.11 Since the p-type side of the junction is undoped, its electron and hole concentrations are equal to the intrinsic carrier concentration. nn = N D = 3 × 1016 cm
−3
p p = ni = 1.08 × 1010 cm
−3
V 0 = V T ln
N D ni n2i
= (26 mV) ln = 386 mV
N D ni
2.12 (a) C j 0 = C j =
qǫSi N A N D 1 2 N A + N D V 0 C j0 1 − V R /V 0
N A = 2 × 1015 cm
−3
N D = 3 × 1016 cm V R = −1.6 V
−3
V 0 = V T ln
N AN D n2i
C j 0 = 14.9 nF/cm2
= 701 mV
C j = 8.22 nF/cm2 = 0.082 fF/cm2 ′
′
′
(b) Let’s write an equation for C j in terms of C j assuming that C j has an acceptor doping of N A. ′
C j = 2C j
qǫ Si N AN D 1 = 2C j 2 N A + N D V T ln(N A N D /n2i ) − V R ′
′
′
qǫ Si N AN D 1 = 4C j2 2 N A + N D V T ln(N A N D /n2i ) − V R ′
′
′
qǫSi N A N D = 8C j2 (N A + N D )(V T ln(N A N D /n2i ) − V R ) ′
′
′
N A qǫ Si N D − 8C j2 (V T ln(N A N D /n2i ) − V R ) = 8C j2 N D (V T ln(N A N D /n2i ) − V R ) ′
′
′
8C j2 N D (V T ln(N A N D /n2i ) − V R ) N A = qǫSi N D − 8C j2 (V T ln(N AN D /n2i ) − V R ) ′
′
′
We can solve this by iteration (you could use a numerical solver if you have one available). Starting with an initial guess of N A = 2 × 1015 cm 3 , we plug this into the right hand side and solve to find a new value of N A = 9.9976 × 1015 cm 3 . Iterating twice more, the solution converges to N A = 1.025 × 1016 cm 3 . Thus, we must increase the N A by a factor of N A /N A = 5.125 ≈ 5 . ′
′
′
−
−
−
′
2.16 (a) The following figure shows the series diodes. I D
+
D1
V D D2
−
Let V D1 be the voltage drop across D1 and V D2 be the voltage drop across D2 . Let I S 1 = I S 2 = I S , since the diodes are identical. V D
= V D1 + V D2 = V T
ln + I D
I S
= 2V T ln I D
=
V T ln
I D
I S
I D
I S V /2V T I S e D
Thus, the diodes in series act like a single device with an exponential characteristic described by V /2V . I D = I S e D
T
(b) Let V D be the amount of voltage required to get a current required to get a current 10 I D . V D ′
V D ′
V D
= 2V T = 2V T
I D
and
ln 10 ln 10 I D
I S
I D
I S
− V D = 2V T ln
= 2V T ln (10) = 120 mV
I D
I S
− ln
I D
I S
′
V D
the amount of voltage
2.19 V X = I X R1 + V D1
I = I R + V ln I V V I X
X
T
1
S
I X =
X
R1
T
−
R1
ln
X
I S
For each value of V X , we can solve this equation for I X by iteration. Doing so, we find I X (V X = 0.5 V) = 0.435 µ A I X (V X = 0.8 V) = 82.3 µ A I X (V X = 1 V) = 173 µ A I X (V X = 1.2 V) = 267 µ A Once we have I X , we can compute V D via the equation V D = V T ln(I X /I S ). Doing so, we find V D (V X = 0.5 V) = 499 mV V D (V X = 0.8 V) = 635 mV V D (V X = 1 V) = 655 mV V D (V X = 1.2 V) = 666 mV As expected, V D varies very little despite rather large changes in I D (in particular, as I D experiences an increase by a factor of over 3, V D changes by about 5 %). This is due to the exponential behavior of the diode. As a result, a diode can allow very large currents to flow once it turns on, up until it begins to overheat.
2.22 V X /2 = I X R1 = V D1 = V T ln(I X /I S ) V T I X = ln(I X /I S ) R1 I X = 367 µ A (using iteration) V X = 2I X R1 = 1.47 V
