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2.1 (a)
k = 8 .617 × 10− 5 eV/ K
n i (T = 300 K) = 1 .66 × 1015 (300 K) 3 / 2 exp −
0.66 eV2 (8.617 × 10− 5 eV/ K) (300 K)
cm− 3
= 2 .465×
1013
cm− 3
n i (T = 600 K) = 1 .66 × 1015 (600 K) 3 / 2 exp −
0.66 eV2 (8.617 × 10− 5 eV/ K) (600 K)
cm− 3
= 4 .124 × 1016 cm− 3
Compared to the values obtained in Example 2.1, we can see that the intrinsic carrier concentrationin Ge at T = 300 K is 2 .465 × 10
13
1 .08 × 10 10 = 2282 times higher than the intrinsic carrier concentration inSi at T = 300 K. Similarly, at T = 600 K, the intrinsic carrier concentration in Ge is 4 .124 × 10
16
1 .54 × 10 15 =26.8 times higher than that in Si.
(b) Since phosphorus is a Group V element, it is a donor, meaning N D = 5 × 1016 cm− 3 . For ann-type material, we have:
n = N D = 5 × 1016 cm− 3
p(T = 300 K) = [n i (T = 300 K)]
2
n = 1 .215 × 1010 cm− 3
p(T = 600 K) = [n i (T = 600 K)]
2
n = 3 .401 × 1016 cm− 3
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2.3 (a) Since the doping is uniform, we have no diffusion current. Thus, the total current is due only tothe drift component.
I tot = I drift= q (nµ n + pµ p )AE
n = 10 17 cm− 3
p = n 2i /n = (1 .08 × 1010 )2 / 1017 = 1 .17 × 103 cm− 3
µn = 1350 cm2 / V · s
µ p = 480 cm2 / V · s
E = V /d = 1 V0.1 µ m
= 10 5 V/ cmA = 0 .05 µ m × 0.05 µ m
= 2 .5 × 10− 11 cm2
Since nµ n ≫ pµ p , we can write
I tot ≈ qnµ n AE
= 54 .1 µ A
(b) All of the parameters are the same except n i , which means we must re-calculate p.
n i (T = 400 K) = 3 .657 × 1012 cm− 3
p = n 2i /n = 1 .337 × 108 cm− 3
Since nµn ≫ pµ p still holds (note that n is 9 orders of magnitude larger than p), the holeconcentration once again drops out of the equation and we have
I tot ≈ qnµ n AE
= 54 .1 µ A
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2.4 (a) From Problem 1, we can calculate n i for Ge.
n i (T = 300 K) = 2 .465 × 1013 cm− 3
I tot = q (nµ n + pµ p)AE
n = 10 17 cm− 3
p = n2
i /n = 6 .076×
109
cm− 3
µn = 3900 cm2 / V · s
µ p = 1900 cm2 / V · s
E = V /d = 1 V0.1 µ m
= 10 5 V/ cmA = 0 .05 µ m × 0.05 µ m
= 2 .5 × 10− 11 cm2
Since nµ n ≫ pµ p , we can write
I tot ≈ qnµ n AE
= 156 µ A
(b) All of the parameters are the same except n i , which means we must re-calculate p.
n i (T = 400 K) = 9 .230 × 1014 cm− 3
p = n 2i /n = 8 .520 × 1012 cm− 3
Since nµn ≫ pµ p still holds (note that n is 5 orders of magnitude larger than p), the holeconcentration once again drops out of the equation and we have
I tot ≈ qnµ n AE
= 156 µ
A
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2.5 Since there’s no electric eld, the current is due entirely to diffusion. If we dene the current as positivewhen owing in the positive x direction, we can write
I tot = I diff = AJ diff = Aq D ndndx
− D pdpdx
A = 1 µ m × 1 µ m = 10 − 8 cm2
D n = 34 cm 2 / sD p = 12 cm
2 / sdndx
= −5 × 1016 cm− 3
2 × 10− 4 cm = − 2.5 × 1020 cm− 4
dpdx
= 2 × 1016 cm− 3
2 × 10− 4 cm = 10 20 cm− 4
I tot = 10− 8 cm2 1.602 × 10− 19 C 34 cm2 / s − 2.5 × 1020 cm− 4 − 12 cm2 / s 1020 cm− 4
= − 15.54 µ A
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2.8 Assume the diffusion lengths L n and L p are associated with the electrons and holes, respectively, in thismaterial and that Ln , L p ≪ 2 µ m. We can express the electron and hole concentrations as functionsof x as follows:
n (x ) = N e − x/L n
p(x ) = P e (x − 2) /L p
# of electrons = 20
an (x )dx
= 2
0aNe
− x/L n dx
= − aNL n e− x/L n
2
0
= − aNL n e− 2/L n − 1
# of holes = 2
0ap (x )dx
=
2
0aP e (x
− 2) /L p dx
= aP L p e (x− 2) /L p
2
0
= aP L p 1 − e− 2/L p
Due to our assumption that L n , L p ≪ 2 µ m, we can write
e− 2/L n ≈ 0
e− 2/L p ≈ 0
# of electrons ≈ aNL n
# of holes ≈ aP L p
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2.10 (a)
nn = N D = 5 × 1017 cm− 3
pn = n2
i /n n = 233 cm− 3
p p = N A = 4 × 1016 cm− 3
n p = n2
i /p p = 2916 cm− 3
(b) We can express the formula for V 0 in its full form, showing its temperature dependence:
V 0 (T ) = kT
q ln
N A N D(5.2 × 1015 )
2T 3 e− E g /kT
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
q ln(N A ) + ln( N D ) − 2 ln 5.2 × 10
15− 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 )dT
= kq
ln(N A ) + ln( N D ) − 2 ln 5.2 × 1015
− 3ln(T ) − 3
From this expression, we can see that if ln( N A ) + ln( N D ) < 2 ln 5.2 × 1015 + 3ln( T ) + 3, orequivalently, if ln( N A N D ) < ln 5.2 × 1015
2T 3 − 3, then V 0 will decrease with temperature,
which we observe in this case. In order for this not to be true (i.e., in order for V 0 to increase with
temperature), we must have either very high doping concentrations or very low temperatures.
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2.11 Since the p-type side of the junction is undoped, its electron and hole concentrations are equal to theintrinsic carrier concentration.
nn = N D = 3 × 1016 cm− 3
p p = n i = 1 .08 × 1010 cm− 3
V 0 = V T lnN D n i
n 2i
= (26 mV) lnN Dn i
= 386 mV
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2.12 (a)
C j 0 = qǫSi2 N A N DN A + N D 1V 0C j =
C j 0
1 − V R /V 0
N A = 2 × 1015 cm− 3
N D = 3 × 1016 cm− 3
V R = − 1.6 V
V 0 = V T lnN A N D
n 2i= 701 mV
C j 0 = 14 .9 nF / cm2
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 = 2 C j
qǫSi2 N ′A N DN ′A + N D 1V T ln(N ′A N D /n 2i ) − V R = 2 C jqǫSi
2N ′A N D
N ′A + N D1
V T ln(N ′A N D /n2i ) − V R
= 4 C 2j
qǫSi N ′A N D = 8 C 2j (N
′
A + N D )(V T ln(N ′
A N D /n2i ) − V R )
N ′A qǫSi N D − 8C 2j (V T ln( N
′
A N D /n2i ) − V R ) = 8 C
2j N D (V T ln(N
′
A N D /n2i ) − V R )
N ′A =8C 2j N D (V T ln(N
′
A N D /n2i ) − V R )
qǫSi N D − 8C 2j (V T ln(N ′
A N D /n2i ) − V R )
We can solve this by iteration (you could use a numerical solver if you have one available). Startingwith an initial guess of N ′A = 2 × 10
15 cm− 3 , we plug this into the right hand side and solve tond a new value of N ′A = 9 .9976 × 10
15 cm− 3 . Iterating twice more, the solution converges toN ′A = 1 .025 × 10
16 cm− 3 . Thus, we must increase the N A by a factor of N ′A /N A = 5 .125 ≈ 5 .
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2.16 (a) The following gure shows the series diodes.
I D
D 1
D2−
V D
+
Let V D 1 be the voltage drop across D 1 and V D 2 be the voltage drop across D 2 . Let I S 1 = I S 2 = I S ,since the diodes are identical.
V D = V D 1 + V D 2
= V T lnI DI S
+ V T lnI DI S
= 2 V T lnI DI S
I D = I S eV D / 2 V T
Thus, the diodes in series act like a single device with an exponential characteristic described byI D = I S e V D / 2 V T .
(b) Let V D be the amount of voltage required to get a current I D and V ′
D the amount of voltagerequired to get a current 10 I D .
V D = 2 V T lnI DI S
V ′D = 2 V T ln10I D
I S
V ′D − V D = 2 V T ln10I D
I S
− lnI D
I S = 2 V T ln (10)
= 120 mV
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2.19
V X = I X R 1 + V D 1
= I X R 1 + V T lnI XI S
I X = V XR 1 −
V T R 1 ln
I XI S
For each value of V X , we can solve this equation for I X by iteration. Doing so, we nd
I X (V X = 0 .5 V) = 0 .435 µ AI X (V X = 0 .8 V) = 82 .3 µ A
I X (V X = 1 V) = 173 µ AI 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 nd
V D (V X = 0 .5 V) = 499 mV
V D (V X = 0 .8 V) = 635 mVV 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 experiencesan increase by a factor of over 3, V D changes by about 5 %). This is due to the exponential behaviorof the diode. As a result, a diode can allow very large currents to ow once it turns on, up until itbegins to overheat.
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2.22
V X / 2 = I X R 1 = V D 1 = V T ln( I X /I S )
I X = V T R 1
ln( I X /I S )
I X = 367 µ A (using iteration)V X = 2 I X R 1
= 1 .47 V
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3.1 (a)
I X =V XR 1 V X < 00 V X > 0
V X (V)
I X
Slope = 1/R 1
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3.2
I X =V XR 1 V X < 00 V X > 0
Plotting I X (t), we have
0
− V 0 /R 1 I X ( t ) f o r V
B =
1 V ( S o l i d )
− π/ω 0 π/ωt
− V 0
0
V 0
V X
( t ) ( D o t t e d
)
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3.3
I X =0 V X < V BV X − V B
R 1 V X > V B
Plotting I X vs. V X for V B = − 1 V and V B = 1 V, we get:
− 1 1V X (V)
I X
V B = − 1 VV B = 1 V
Slope = 1/R 1 Slope = 1/R 1
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3.4
I X =0 V X < V BV X − V B
R 1 V X > V B
Let’s assume V 0 > 1 V. Plotting I X (t) for V B = − 1 V, we get
0
(V 0 − V B )/R 1
I X ( t ) f o r V B = −
1 V ( S o l i d )
− π/ω 0 π/ωt
− V 0
0
V B
V 0
V X
( t ) ( D o t t e d
)
Plotting I X (t) for V B = 1 V, we get
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0
(V 0 − V B )/R 1
I X ( t ) f o r
V B =
1 V ( S o l i d )
− π/ω 0 π/ωt
− V 0
0
V B
V 0
V X
( t ) ( D o t t e d
)
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3.5
I X =V X − V B
R 1 V X < 0∞ V X > 0
Plotting I X vs. V X for V B = − 1 V and V B = 1 V, we get:
− 1V X (V)
− 1/R 1
1/R 1
I X
Slope = 1/R 1
Slope = 1/R 1
I X for V B = − 1 VI X for V B = 1 V
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3.6 First, note that I D 1 = 0 always, since D 1 is reverse biased by V B (due to the assumption that V B > 0).We can write I X as
I X = ( V X − V B )/R 1
Plotting this, we get:
V BV X (V)
I X
Slope = 1/R 1
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3.7
I X =V X − V B
R 1 V X < V BV X − V BR 1 R 2 V X > V B
I R 1 = V X − V B
R 1
Plotting I X and I R 1 for V B = − 1 V, we get:
− 1 V X (V)
I X for V B = − 1 VI R 1 for V B = − 1 V
Slope = 1/R 1
Slope = 1/R 1 + 1 /R 2
Plotting I X and I R 1 for V B = 1 V, we get:
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1V X (V)
I X for V B = 1 VI R 1 for V B = 1 V
Slope = 1/R 1
Slope = 1/R 1 + 1 /R 2
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3.8
I X =0 V X < V BR 1 + R 2 R 1V XR 1 +
V X − V BR 2 V X >
V BR 1 + R 2 R 1
I R 1 = V B
R 1 + R 2 V X < V B
R 1 + R 2 R 1V X
R 1 V X > V B
R 1 + R 2 R 1
Plotting I X and I R 1 for V B = − 1 V, we get:
V BR 1 + R 2 R1
V X (V)V BR 1 + R 2
− V B /R 2
I X for V B = − 1 VI R 1 for V B = − 1 V
Slope = 1/R 1
Slope = 1/R 1 + 1 /R 2
Plotting I X and I R 1 for V B = 1 V, we get:
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V BR 1 + R 2 R1
V X (V)
V BR 1 + R 2
I X for V B = 1 VI R 1 for V B = 1 V
Slope = 1/R 1
Slope = 1/R 1 + 1 /R 2
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3.9 (a)
V out =V B V in < V BV in V in > V B
− 5 − 4 − 3 − 2 − 1 0 1 2 3 4 5V in (V)
0
1
2
3
4
5 V
o u
t ( V )
Slope = 1
(b)
V out = V in − V B V in < V B
0 V in > V B
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− 5 − 4 − 3 − 2 − 1 0 1 2 3 4 5V in (V)
− 7
− 6
− 5
− 4
− 3
− 2
− 1
0
1
2 V
o u
t
( V )
Slope = 1
(c)
V out = V in − V B
− 5 − 4 − 3 − 2 − 1 0 1 2 3 4 5V in (V)
− 7
− 6
− 5
− 4
− 3
− 2
− 1
0
1
2
3 V
o u
t
( V )
Slope = 1
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(d)
V out =V in V in < V BV B V in > V B
− 5 − 4 − 3 − 2 − 1 0 1 2 3 4 5V in (V)
− 5
− 4
− 3
− 2
− 1
0
1
2 V
o u
t ( V )
Slope = 1
(e)
V out = 0 V in < V B
V in − V B V in > V B
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− 5 − 4 − 3 − 2 − 1 0 1 2 3 4 5V in (V)
0
1
2
3 V
o u
t
( V )
Slope = 1
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3.11 For each part, the dotted line indicates V in (t), while the solid line indicates V out (t). Assume V 0 > V B .
(a)
V out =V B V in < V BV in V in > V B
− π/ω π/ωt
− V 0
V B
V 0
V o u
t ( t ) ( V )
(b)
V out =V in − V B V in < V B0 V in > V B
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− π/ω π/ωt
− V 0 − V B
− V 0
V B
V 0
V o u
t ( t ) ( V )
(c)
V out = V in − V B
− π/ω π/ωt
− V 0 − V B
− V 0
V 0 − V B
V B
V 0
V o u t
( t ) ( V )
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(d)
V out =V in V in < V BV B V in > V B
−
π/ω π/ω t
− V 0
V B
V 0
V o u
t ( t ) ( V )
(e)
V out = 0 V in < V B
V in − V B V in > V B
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− π/ω π/ωt
− V 0
V 0 − V B
V B
V 0
V o u
t ( t ) ( V )
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3.12 For each part, the dotted line indicates V in (t), while the solid line indicates V out (t). Assume V 0 > V B .
(a)
V out =V in − V B V in < V B0 V in > V B
− π/ω π/ω
t
− V 0 − V B
− V 0
V B
V 0
V o u
t ( t ) ( V )
(b)
V out =V in V in < V BV B V in > V B
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− π/ω π/ωt
− V 0
V B
V 0
V o u
t ( t ) ( V )
(c)
V out =0 V in < V BV in − V B V in > V B
− π/ω π/ω
t
− V 0
V 0 − V B
V B
V 0
V o u
t ( t ) ( V )
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(d)
V out = V in − V B
− π/ω π/ωt
− V 0 − V B
− V 0
V 0 − V B
V B
V 0
V o u
t ( t )
( V )
(e)
V out =V B V in < V BV in V in > V B
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− π/ω π/ωt
− V 0
V B
V 0
V o u
t ( t ) ( V )
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Slope = 1(V D,on + V B ) /R 1
I in
(V D,on + V B ) /R 1
I R 1
(c)
I R 1 =I in I in <
V D,on − V BR 1
V D,on − V BR 1 I in >
V D,on − V BR 1
Slope = 1
(V D,on − V B ) /R 1
(V D,on − V B ) /R 1I in
I R 1
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(d)
I R 1 =I in I in <
V D,onR 1
V D,onR 1 I in >
V D,onR 1
Slope = 1
V D,on /R 1I in
V D,on /R 1I R 1
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3.17 (a)
V out =I in R 1 I in <
V D,onR 1
V D,on I in > V D,on
R 1
− I 0 R1
0V D,on
V o u
t ( t ) ( S o l i d )
− π/ω 0 π/ωt
− I 0
0
V D,on /R 1
I 0
I i n
( t ) ( D o t t e d
)
(b)
V out =I in R 1 I in < V D,on + V BR 1V D,on + V B I in >
V D,on + V BR 1
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(d)
V out =I in R 1 + V B I in <
V D,onR 1
V D,on + V B I in > V D,on
R 1
− I 0 R1 + V B
0
V D,on + V B
V o u
t ( t ) ( S o l i d )
− π/ω 0 π/ωt
− I 0
0
V D,on /R 1
I 0
I i n
( t ) ( D o t t e d
)
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3.20 (a)
V out =I in R 1 I in >
V B − V D,onR 1
V B − V D,on I in < V B − V D,on
R 1
V B − V D,on
0
I 0 R1
V o u
t ( t ) ( S o l i d )
− π/ω 0 π/ωt
− I 0
0
(V B − V D,on ) /R 1
I 0
I i n
( t ) ( D o t t e d
)
(b)
V out =I in R 1 + V B I in > − V D,on + V BR 1− V D,on I in < −
V D,on + V BR 1
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− V D,on0
I 0 R1 + V B
V o u
t ( t ) ( S o l i d )
− π/ω 0 π/ωt
− I 0
0
− (V D,on + V B )/R 1
I 0
I i n
( t ) ( D o t t e d
)
(c)
V out =I in R 1 + V B I in > − V D,onR 1V B − V D,on I in < − V D,onR 1
V B − V D,on
0
I 0 R1 + V B
V o u
t ( t ) ( S o l i d )
− π/ω 0 π/ωt
− I 0
− V D,on /R 1
0
I 0
I i n
( t ) ( D o t t e d
)
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3.23 (a)
V out = R 2
R 1 + R 2V in V in <
R 1 + R 2R 2
V D,on
V D,on V in > R 1 + R 2
R 2V D,on
R 1 + R 2R 2 V D,on
V in (V)
V D,on V
o u
t ( V )
Slope = R2 / (R1 + R2 )
(b)
V out = R 2
R 1 + R 2 V in V in < R1 + R 2R 1 V D,on
V in − V D,on V in > R 1 + R 2R 1 V D,on
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R 1 + R 2R 1 V D,on
V in (V)
R 2R 1 V D,on
V o u
t
( V )
Slope = R2 / (R1 + R2 )
Slope = 1
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3.24 (a)
I R 1 = V in
R 1 + R 2V in <
R 1 + R 2R 2
V D,onV in − V D,on
R 1V in >
R 1 + R 2R 2
V D,on
I D 1 =0 V in < R 1 + R 2R 2 V D,onV in − V D,on
R 1 − V D,on
R 2 V in > R 1 + R 2
R 2 V D,on
R 1 + R 2R 2 V D,on
V in (V)
V D,on /R 2
Slope = 1/ (R1 + R2 )
Slope = 1/R 1
Slope = 1/R 1
I R 1I D 1
(b)
I R 1 = V in
R 1 + R 2V in <
R 1 + R 2R 1
V D,onV D,on
R 1V in >
R 1 + R 2R 1
V D,on
I D 1 =0 V in < R 1 + R 2R 1 V D,onV in − V D,on
R 2− V D,on
R 1V in >
R 1 + R 2R 1
V D,on
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R 1 + R 2R 1 V D,on
V in (V)
V D,on /R 1
Slope = 1/ (R1 + R2 )
Slope = 1/R 2
I R 1I D 1
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3.25 (a)
V out =V B + R 2R 1 + R 2 (V in − V B ) V in < V B +
R 1 + R 2R 1
V D,on
V in − V D,on V in > V B + R 1 + R 2R 1 V D,on
V B + R 1 + R 2R 1 V D,onV in (V)
V B + R 2R 1 V D,on
V o u
t ( V )
Slope = R2 / (R1 + R2 )
Slope = 1
(b)
V out = R 2
R 1 + R 2 V in V in < R1 + R 2R 1 (V D,on + V B )
V in − V D,on − V B V in > R 1 + R 2R 1 (V D,on + V B )
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V B + R 1 + R 2R 1 (V D,on + V B )V in (V)
R2
R 1 (V D,on + V B )
V o u
t
( V )
Slope = R2 / (R1 + R2 )
Slope = 1
(c)
V out = R 2
R 1 + R 2 (V in − V B ) V in > V B + R 1 + R 2
R 1V D,on
V in + V D,on − V B V in < V B + R 1 + R 2R 1 V D,on
V B + R 1 + R 2R 1 V D,onV in (V)
R 2R 1 V D,on
V o u
t
( V )
Slope = 1
Slope = R2 / (R1 + R2 )
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(d)
V out = R 2
R 1 + R 2 (V in − V B ) V in < V B + R 1 + R 2
R 1 (V D,on − V B )V in − V D,on V in > V B + R 1 + R 2R 1 (V D,on − V B )
V B + R 1 + R 2R 1 (V D,on − V B )V in (V)
R 2R 1 V D,on
V o u
t ( V )
Slope = R2 / (R1 + R2 )
Slope = 1
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3.26 (a)
I R 1 =V in − V BR 1 + R 2
V in < V B + R 1 + R 2R 1 V D,onV D,on
R 1V in > V B + R 1 + R 2R 1 V D,on
I D 1 =0 V in < V B + R 1 + R 2R 1 V D,onV in − V D,on − V B
R 2 − V D,on
R 1 V in > V B + R 1 + R 2
R 1 V D,on
V B + R 1 + R 2R 1 V D,on
V in (V)
V D,on /R 1
Slope = 1/ (R1 + R2 )
Slope = 1/R 2
I R 1I D 1
(b)
I R 1 = V in
R 1 + R 2V in <
R 1 + R 2R 1 (V D,on + V B )
V D,on + V BR 1
V in > R 1 + R 2
R 1 (V D,on + V B )
I D 1 =0 V in < R 1 + R 2R 1 (V D,on + V B )V in − V D,on − V B
R 2− V D,on + V B
R 1V in >
R 1 + R 2R 1 (V D,on + V B )
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R 1 + R 2R 1 (V D,on + V B )
V in (V)
(V D,on + V B ) /R 1
Slope = 1/ (R1 + R2 )
Slope = 1/R 2
I R 1I D 1
(c)
I R 1 =V in − V BR 1 + R 2
V in > V B − R 1 + R 2R 1 V D,on− V D,on
R 1V in < V B − R 1 + R 2R 1 V D,on
I D 1 =0 V in > V B − R 1 + R 2R 1 V D,on− V in + V D,on + V B
R 2− V D,on
R 1V in < V B − R 1 + R 2R 1 V D,on
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V B + R 1 + R 2R 1 (V D,on − V B )V in (V)
(V D,on − V B ) /R 1
I R 1I D 1
Slope = 1/ (R1 + R2 )
Slope = 1/R 2
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3.27 (a)
V out =0 V in < V D,on
R 2R 1 + R 2 (V in − V D,on ) V in > V D,on
V D,onV in (V)
V o u
t ( V )
Slope = R 2 / (R1 + R2 )
(b)
V out =− V D,on V in < − R 1 + R 2R 2 V D,on
R 2R 1 + R 2
V in − R 1 + R 2R 2 V D,on < V in < R 1 + R 2
R 1V D,on
V in − V D,on V in > R 1 + R 2R 1 V D,on
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− R 1 + R 2R 2 V D,on R1 + R 2R 1 V D,on
V in (V)− V D,on
R 2R 1 V D,on
V o u
t
( V )
Slope = R 2 / (R1 + R2 )
Slope = 1
(c)
V out = R 2
R 1 + R 2 (V in + V D,on ) − V D,on V in < − V D,onV in V in > − V D,on
− V D,onV in (V)− V D,on
V o u
t
( V )
Slope = R2 / (R1 + R2 )
Slope = 1
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(d)
V out =0 V in < V D,on
R 2R 1 + R 2 (V in − V D,on ) V in > V D,on
V D,onV in (V)
V D,on
V o u
t ( V )
Slope = R 2 / (R1 + R2 )
(e)
V out = R 2
R 1 + R 2 (V in + V D,on ) V in < − V D,on0 V in > − V D,on
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−V D,on V in (V)
V o u
t
( V )
Slope = R2 / (R1 + R2 )
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− R 1 + R 2R 2 V D,on R 1 + R 2R 1 V D,onV in (V)− V D,on /R 2
V D,on /R 1
Slope = 1/R 1
Slope = 1/ (R1 + R2 )Slope = 1
I R 1I D 1
(c)
I R 1 =V in + V D,on
R 1 + R 2V in < − V D,on
0 V in > − V D,on
I D 1 =0 V in < − V D,on0 V in > − V D,on
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−V D,on V in (V)
Slope = 1/ (R1 + R2 )
I R 1I D 1
(d)
I R 1 =0 V in < V D,onV in − V D,on
R 1 + R 2V in > V D,on
I D
1 =0 V in < V D,onV in
−V D,onR 1 + R 2 V in > V D,on
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V D,onV in (V)
Slope = 1/ (R1 + R2 )
I R 1I D 1
(e)
I R 1 =V in + V D,on
R 1 + R 2V in < − V D,on
0 V in > − V D,on
I D 1 =0 V in < − V D,on0 V in > − V D,on
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−V D,on V in (V)
Slope = 1/ (R1 + R2 )
I R 1I D 1
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3.29 (a)
V out =V in V in < V D,onV D,on + R 2R 1 + R 2 (V in − V D,on ) V D,on < V in < V D,on +
R 1 + R 2R 1 (V D,on + V B )
V in − V D,on − V B V in > V D,on + R 1 + R 2R 1 (V D,on + V B )
V D,on V D,on + R1 + R 2R 1 (V D,on + V B )V in (V)
V D,on
V D,on + R2R 1 (V D,on + V B )
V o u
t
( V )
Slope = 1
Slope = R 2 / (R1 + R2 )
Slope = 1
(b)
V out =V in + V D,on − V B V in < V D,on + R 1 + R 2R 1 (V B − 2V D,on )
R 2R 1 + R 2 (V in − V D,on ) V in > V D,on +
R 1 + R 2R 1 (V B − 2V D,on )
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(d)
V out =0 V in < V D,on
R 2R 1 + R 2 (V in − V D,on ) V D,on < V in < V D,on +
R 1 + R 2R 2 (V B + V D,on )
V D,on + V B V in > V D,on + R 1 + R 2R 2 (V B + V D,on )
V D,on V D,on + R 1 + R 2R 2 (V B + V D,on )V in (V)
V D,on + V B V
o u
t
( V )
Slope = R2 / (R1 + R2 )
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3.30 (a)
I R 1 =0 V in < V D,onV in − V D,on
R 1 + R 2 V D,on < V in < V D,on + R 1 + R 2
R 1 (V D,on + V B )V D,on + V B
R 1 V in > V D,on + R 1 + R 2
R 1 (V D,on + V B )
I D 1 = 0 V in < V D,on + R 1 + R 2
R 1 (V D,on + V B )V in − 2 V D,on − V BR 2 −
V D,on + V BR 1 V in > V D,on +
R 1 + R 2R 1 (V D,on + V B )
V D,on V D,on + R 1 + R 2R 1 (V D,on + V B )V in (V)
V D,on + V B
Slope = 1/ (R1 + R2 )
Slope = 1/R 2
I R 1I D 1
(b) If V B < 2V D,on :
I R 1 = I D 1 =0 V in < V D,onV in − V D,on
R 1 + R 2 V in > V D,on
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V D,onV in (V)
Slope = 1/ (R1 + R2 )
I R 1I D 1
If V B > 2V D,on :
I R 1 = I D 1 =V B − 2 V D,on
R 1 V in < V D,on + R 1 + R 2
R 1 (V B − 2V D,on )V in − V D,on
R 1 + R 2 V in > V D,on + R 1 + R 2
R 1 (V B − 2V D,on )
V D,on + R 1 + R 2R 1 (V B − 2V D,on )V in (V)
V B − 2 V D,onR 1
Slope = 1/ (R1 + R2 )
I R 1I D 1
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V D,on V D,on + R 1 + R 2R 2 (V B + V D,on )V in (V)
V B + V D,onR 2 Slope = 1/ (R1 + R2 )
Slope = 1/R 2
I R 1I D 1
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3.31 (a)
I D 1 = V in − V D,on
R 1= 1 .6 mA
r d 1 = V T I D 1
= 16 .25 Ω
∆ V out = R 1r d + R 1∆ V in = 98 .40 mV
(b)
I D 1 = I D 2 = V in − 2V D,on
R 1= 0 .8 mA
r d 1 = r d 2 = V T I D 1
= 32 .5 Ω
∆ V out = R 1 + r d 2
R 1 + r d 1 + r d 2∆ V in = 96 .95 mV
(c)
I D 1 = I D 2 = V in − 2V D,on
R 1= 0 .8 mA
r d 1 = r d 2 = V T I D 1
= 32 .5 Ω
∆ V out = rd 2
r d 1 + R 1 + r d 2∆ V in = 3 .05 mV
(d)
I D 2 = V in − V D,on
R 1−
V D,onR 2
= 1 .2 mA
r d 2 = V T
I D 2= 21 .67 Ω
∆ V out = R 2 r d 2
R 1 + R 2 r d 2∆ V in = 2.10 mV
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3.32 (a)
∆ V out = ∆ I in R 1 = 100 mV
(b)
I D 1 = I D 2 = I in = 3 mAr d 1 = r d 2 =
V T I D 1
= 8 .67 Ω
∆ V out = ∆ I in (R 1 + r d 2 ) = 100 .867 mV
(c)
I D 1 = I D 2 = I in = 3 mA
r d 1 = r d 2 = V T I D 1
= 8 .67 Ω
∆ V out = ∆ I in r d 2 = 0 .867 mV
(d)
I D 2 = I in − V D,on
R 2= 2 .6 mA
r d 2 = V T I D 2
= 10 Ω
∆ V out = ∆ I in (R 2 r d 2 ) = 0 .995 mV
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3.34
π/ω 2π/ωt
− V p
0.5 V
V D,on + 0 .5 V
V p − V D,on
V p
V in (t)V out (t)
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3.35
π/ω 2π/ωt
− V p
0.5 V
− V D,on + 0 .5 V
− V p + V D,on
V p
V in (t)V out (t)
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3.36
V R ≈ V p − V D,on
R L C 1 f inV p = 3 .5 V
R L = 100 Ω
C 1 = 1000 µ Ff in = 60 Hz
V R = 0 .45 V
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3.37
V R = I LC 1 f in
≤ 300 mV
f in = 60 HzI L = 0 .5 A
C 1 ≥ I L(300 mV) f in
= 27 .78 mF
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3.38 Shorting the input and output grounds of a full-wave rectier shorts out the diode D 4 from Fig. 3.38(b).Redrawing the modied circuit, we have:
+
V in−
D 2
D 3
R L
+
V out−
D 1
On the positive half-cycle, D 3 turns on and forms a half-wave rectier along with R L (and C L , if included). On the negative half-cycle, D 2 shorts the input (which could cause a dangerously largecurrent to ow) and the output remains at zero. Thus, the circuit behaves like a half-wave recier.The plots of V out (t) are shown below.
π/ω 2π/ωt
− V 0
V D,on
V 0 − V D,on
V 0
V in (t) = V 0 sin(ωt)
V out (t) (without a load capacitor)V out (t) (with a load capacitor)
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3.39 Note that the waveforms for V D 1 and V D 2 are identical, as are the waveforms for V D 3 and V D 4 .
π/ω 2π/ωt
− V 0
− V 0 + V D,on
− V 0 + 2V D,on
− 2V D,on
− V D,on
V D,on
2V D,on
V 0 − 2V D,on
V 0
V in (t) = V 0 sin(ωt)V out (t)
V D 1 (t), V D 2 (t)V D 3 (t), V D 4 (t)
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3.40 During the positive half-cycle, D 2 and D 3 will remain reverse-biased, causing V out to be zero asno current will ow through RL . During the negative half-cycle, D 1 and D 3 will short the input(potentially causing damage to the devices), and once again, no current will ow through R L (eventhough D 2 will turn on, there will be no voltage drop across R L ). Thus, V out always remains at zero,and the circuit fails to act as a rectier.
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3.42 Shorting the negative terminals of V in and V out of a full-wave rectier shorts out the diode D 4 fromFig. 3.38(b). Redrawing the modied circuit, we have:
+
V in−
D 2
D 3
R L
+
V out−
D 1
On the positive half-cycle, D 3 turns on and forms a half-wave rectier along with R L (and C L , if included). On the negative half-cycle, D 2 shorts the input (which could cause a dangerously largecurrent to ow) and the output remains at zero. Thus, the circuit behaves like a half-wave recier.The plots of V out (t) are shown below.
π/ω 2π/ωt
− V 0
V 0
V in (t) = V 0 sin(ωt)
V out (t) (without a load capacitor)V out (t) (with a load capacitor)
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3.44 (a) We know that when a capacitor is discharged by a constant current at a certain frequency, theripple voltage is given by I Cf in , where I is the constant current. In this case, we can calculate thecurrent as approximately V p
− 5 V D,onR 1 (since V p − 5V D,on is the voltage drop across R 1 , assuming
R 1 carries a constant current). This gives us the following:
V R ≈ 1
2
V p − 5V D,on
R L C 1 f inV p = 5 V
R L = 1 kΩC 1 = 100 µ Ff in = 60 Hz
V R = 166 .67 mV
(b) The bias current through the diodes is the same as the bias current through R 1 , which isV p − 5 V D,on
R 1 = 1 mA. Thus, we have:
r d = V T
I D= 26 Ω
V R,load = 3r d
R 1 + 3 r dV R = 12 .06 mV
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4.4 According to Equation (4.8), we have
I C = AE qD n n
2i
N B W BeV BE /V T − 1
∝
1W B
We can see that if W B increases by a factor of two, then I C decreases by a factor of two .
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4.11
V BE = 1 .5 V − I E (1 kΩ)≈ 1.5 V − I C (1 kΩ) (assuming β ≫ 1)
= V T lnI C I S
I C = 775 µ AV X ≈ I C (1 kΩ)
= 775 mV
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4.12 Since we have only integer multiples of a unit transistor, we need to nd the largest number thatdivides both I 1 and I 2 evenly (i.e., we need to nd the largest x such that I 1 /x and I 2 /x are integers).This will ensure that we use the fewest transistors possible. In this case, it’s easy to see that we shouldpick x = 0 .5 mA, meaning each transistor should have 0 .5 mA owing through it. Therefore, I 1 shouldbe made up of 1 mA / 0.5 mA = 2 parallel transistors, and I 2 should be made up of 1 .5 mA/ 0.5 mA = 3parallel transistors. This is shown in the following circuit diagram.
V B−
+
I 1 I 2
Now we have to pick V B so that I C = 0 .5 mA for each transistor.
V B = V T lnI C I S
= (26 mV) ln 5 × 10− 4 A3 × 10− 16 A
= 732 mV
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4.15
V B − V BE R 1
= I B
= I C
β
I C = β R 1 [V B − V T ln(I C /I S )]
I C = 786 µ A
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4.17 First, note that V BE 1 = V BE 2 = V BE .
V B = ( I B 1 + I B 2 )R 1 + V BE
= R1
β (I X + I Y ) + V T ln( I X /I S 1 )
I S 2
=
5
3I S 1
⇒ I Y = 53
I X
V B = 8R 1
3β I X + V T ln( I X /I S 1 )
I X = 509 µ A
I Y = 848 µ A
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4.21 (a)
V BE = 0 .8 V
I C = I S eV BE /V T
= 18 .5 mA
V CE = V CC − I C RC = 1 .58 V
Q1 is operating in forward active. Its small-signal parameters are
gm = I C /V T = 710 mS
r π = β/g m = 141 Ωr o = ∞
The small-signal model is shown below.
B
r π
+vπ
−
E
gm vπ
C
(b)
I B = 10 µ A
I C = βI B = 1 mA
V BE = V T ln( I C /I S ) = 724 mVV CE = V CC − I C RC
= 1 .5 V
Q1 is operating in forward active. Its small-signal parameters are
gm = I C /V T = 38 .5 mS
rπ = β/g m = 2 .6 kΩro = ∞
The small-signal model is shown below.
B
r π+vπ
−
E
gm vπ
C
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(c)
I E = V CC − V BE
RC =
1 + β β
I C
I C = β 1 + β
V CC − V T ln(I C /I S )RC
I C = 1 .74 mA
V BE = V T ln( I C /I S ) = 739 mV
V CE = V BE = 739 mV
Q1 is operating in forward active. Its small-signal parameters are
gm = I C /V T = 38 .5 mS
r π = β/g m = 2 .6 kΩr o = ∞
The small-signal model is shown below.
B
r π+vπ
−
E
gm vπ
C
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4.22 (a)
I B = 10 µ A
I C = βI B = 1 mA
V BE = V T ln( I C /I S ) = 739 mV
V CE = V CC − I E (1 kΩ)= V CC −
1 + β β
(1 kΩ)
= 0 .99 V
Q1 is operating in forward active. Its small-signal parameters are
gm = I C /V T = 38 .5 mS
r π = β/g m = 2 .6 kΩr o = ∞
The small-signal model is shown below.
B
r π+vπ
−
E
gm vπ
C
(b)
I E = V CC − V BE
1 kΩ =
1 + β β
I C
I C = β 1 + β
V CC − V T ln(I C /I S )1 kΩ
I C = 1 .26 mA
V BE = V T ln( I C /I S ) = 730 mV
V CE = V BE = 730 mV
Q1 is operating in forward active. Its small-signal parameters are
gm = I C /V T = 48 .3 mS
r π = β/g m = 2 .07 kΩr o = ∞
The small-signal model is shown below.
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B
r π+vπ
−
E
gm vπ
C
(c)
I E = 1 mA
I C = β 1 + β
I E = 0 .99 mA
V BE = V T ln( I C /I S ) = 724 mV
V CE = V BE = 724 mV
Q1 is operating in forward active. Its small-signal parameters are
gm = I C /V T = 38 .1 mS
r π = β/g m = 2 .63 kΩr o = ∞
The small-signal model is shown below.
B
r π+vπ
−
E
gm vπ
C
(d)
I E = 1 mA
I C = β 1 + β
I E = 0 .99 mA
V BE = V T ln( I C /I S ) = 724 mV
V CE = V BE = 724 mV
Q1 is operating in forward active. Its small-signal parameters are
gm = I C /V T = 38 .1 mS
r π = β/g m = 2 .63 kΩr o = ∞
The small-signal model is shown below.
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B
r π
+
v π
−
E
g m v π
C
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4.31
I C = I S eV BE /V T 1 + V CE
V AI C,Total = nI C
= nI S eV BE /V T 1 + V CE
V A
gm,Total = ∂I C ∂V BE
= n I S V T
eV BE /V T
≈ nI C V T
= ng m
= n × 0.4435 S
I B,Total = 1β
I C,Total
r π,Total = ∂I B,Total∂V BE
− 1
≈I C,Total
βV T
− 1
=nI C βV T
− 1
= rπ
n
= 225.5 Ω
n (assuming β = 100)
r o,Total =∂I C,Total
∂V CE
− 1
≈I C,Total
V A
− 1
= V AnI C
= ron
= 693.8 Ω
n
The small-signal model is shown below.
B
r π,Total
+vπ
−
E
gm,Total vπ
C
r o,Total
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4.32 (a)
V BE = V CE (for Q 1 to operate at the edge of saturation)V T ln( I C /I S ) = V CC − I C RC
I C = 885 .7 µ A
V B = V BE = 728 .5 mV(b) Let I ′C , V
′
B , V ′
BE , and V ′
CE correspond to the values where the collector-base junction is forwardbiased by 200 mV.
V ′BE = V ′
CE + 200 mVV T ln( I ′C /I S ) = V CC − I
′
C RC + 200 mVI ′C = 984 .4 µ AV ′B = 731 .3 mV
Thus, V B can increase by V ′B − V B = 2 .8 mV if we allow soft saturation.
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4.34
V BE = V CC − I B RBV T ln( I C /I S ) = V CC − I C RB /β
I C = 1 .67 mAV BC = V CC − I B RB − (V CC − I C RC )
< 200 mVI C RC − I B RB < 200 mV
RC < 200 mV + I B RB
I C
= 200 mV + I C RB /β
I C RC < 1.12 kΩ
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4.41
V EB = V EC (for Q 1 to operate at the edge of saturation)V CC − I B RB = V CC − I C RC
I C RB /β = I C RC RB /β = RC
β = RB /R C
= 100
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4.44 (a)
I B = 2 µ AI C = βI B
= 200 µ A
V EB = V T ln( I C /I S )= 768 mV
V EC = V CC − I E (2 kΩ)
= V CC −1 + β
β I C (2 kΩ)
= 2 .1 V
Q1 is operating in forward active. Its small-signal parameters are
gm = I C /V T = 7 .69 mS
r π = β/g m = 13 kΩ
r o = ∞
The small-signal model is shown below.
B
r π+vπ
−
E
gm vπ
C
(b)
I E = V CC − V EB
5 kΩ1 + β
β I C =
V CC − V T ln( I C /I S )5 kΩ
I C = 340 µ A
V EB = 782 mV
V EC = V EB = 782 mV
Q1 is operating in forward active. Its small-signal parameters are
gm = I C /V T = 13 .1 mS
r π = β/g m = 7 .64 kΩr o = ∞
The small-signal model is shown below.
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B
r π+vπ
−
E
gm vπ
C
(c)
I E = 1 + β
β I C = 0 .5 mA
I C = 495 µ A
V EB = 971 mV
V EC = V EB = 971 mV
Q1 is operating in forward active. Its small-signal parameters are
gm = I C /V T = 19 .0 mS
r π = β/g m = 5 .25 kΩr o = ∞
The small-signal model is shown below.
B
r π+vπ
−
E
gm vπ
C
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4.49 The direction of current ow in the large-signal model (Fig. 4.40) indicates the direction of positivecurrent ow when the transistor is properly biased.The direction of current ow in the small-signal model (Fig. 4.43) indicates the direction of positivechange in current ow when the base-emitter voltage v be increases. For example, when v be increases,the current owing into the collector increases, which is why i c is shown owing into the collector inFig. 4.43. Similar reasoning can be applied to the direction of ow of i b and i e in Fig. 4.43.
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4.53 (a)
V CB 2 < 200 mVI C 2 RC < 200 mV
I C 2 < 400 µ AV EB 2 = V E 2
= V T ln( I C 2 /I S 2 )< 741 mV
β 21 + β 2
I E 2 RC < 200 mV
β 21 + β 2
1 + β 1β 1
I C 1 RC < 200 mV
I C 1 < 396 µ AV BE 1 = V T ln( I C 1 /I S 1 )
< 712 mVV in = V BE 1 + V EB 2
< 1.453 V(b)
I C 1 = 396 µ AI C 2 = 400 µ A
gm 1 = 15 .2 mS
r π 1 = 6 .56 kΩr o 1 = ∞
gm 2 = 15 .4 mS
r π 2 = 3 .25 kΩ
r o 2 = ∞
The small-signal model is shown below.
−
vin+
B1
r π 1+vπ 1
−
C1
gm 1 vπ 1
E1 /E 2
r π 2
−
vπ 2+B2
gm 2 vπ 2
C2 vout
RC
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4.55 (a)
V BC 2 < 200 mVV BE 2 − (V CC − I C 2 RC ) < 200 mV
V T ln( I C 2 /I S 2 ) + I C 2 RC − V CC < 200 mVI C 2 < 3.80 mA
V BE 2 < 799.7 mV
I E 1 = 1 + β 1
β 1I C 1 = I B 2 = I C 2 /β 2
I C 1 < 75.3 µ AV BE 1 < 669.2 mV
V in = V BE 1 + V BE 2
< 1.469 V
(b)
I C 1 = 75 .3 µ AI C 2 = 3 .80 mA
gm 1 = 2 .90 mS
r π 1 = 34 .5 kΩr o 1 = ∞
gm 2 = 146 .2 mS
r π 2 = 342 Ωr o 2 = ∞
The small-signal model is shown below.
−
vin+ B
1
r π 1+vπ 1
−
C1
gm 1 vπ 1
E 1 /B 2
r π 2+vπ 2
−
C2
gm 2 vπ 2
E 2
vout
RC
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R in
Q 1
V CC
r π 2
R in = r π 1 + (1 + β 1 )r π 2
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5.4 (a) Looking into the collector of Q1 we see an equivalent resistance of ro1 , so we can draw the followingequivalent circuit for nding Rout :
Rout
R1r o1
Rout = ro1 R1
(b) Let’s draw the small-signal model and apply a test source at the output.
RB
r π 1+vπ 1
−gm 1vπ 1 ro1
−vt
+i t
i t = gm 1vπ 1 + vtro1
vπ 1 = 0
i t = vtr o1
Rout = vti t = ro1
(c) Looking down from the emitter of Q1 we see an equivalent resistance of 1gm 2 rπ 2 ro2 , so wecan draw the following equivalent circuit for nding Rout :
Q1
Rout
1gm 2 r π 2 r o2
Rout = ro1 + (1 + gm 1ro1) r π 1 1gm 2
r π 2 ro2
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(d) Looking into the base of Q2 we see an equivalent resistance of r π 2 , so we can draw the followingequivalent circuit for nding Rout :
Q1
Rout
r π 2
Rout = ro1 + (1 + gm 1r o1 ) ( r π 1 r π 2 )
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5.5 (a) Looking into the base of Q1 we see an equivalent resistance of r π 1 , so we can draw the followingequivalent circuit for nding R in :
R in
R1
R2 rπ 1
R in = R1 + R2 r π 1
(b) Let’s draw the small-signal model and apply a test source at the input.
r π 1
+
vπ 1−
−vt
+i t
gm 1vπ 1 R1
i t = −vπ 1
r π 1− gm 1vπ 1
vπ 1 = − vt
i t = vtr π 1
+ gm 1vt
i t = vt gm 1 + 1r π 1
R in = vti t
= 1gm 1
r π 1
(c) From our analysis in part (b), we know that looking into the emitter we see a resistance of 1
gm 2 r π 2 . Thus, we can draw the following equivalent circuit for nding R in :
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R in
Q1
V CC
1gm 2 rπ 2
R in = rπ 1 + (1 + β 1) 1gm 2
r π 2
(d) Looking up from the emitter of Q1 we see an equivalent resistance of 1gm 2 r π 2 , so we can drawthe following equivalent circuit for nding R in :
R in
Q1
1gm 2 rπ 2
V CC
R in = rπ 1 + (1 + β 1) 1gm 2
r π 2
(e) We know that looking into the base of Q2 we see Rin = rπ 2 if the emitter is grounded. Thus,transistor Q1 does not affect the input impedance of this circuit.
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5.6 (a) Looking into the collector of Q1 we see an equivalent resistance of ro1 , so we can draw the followingequivalent circuit for nding Rout :
Rout
RC r o1
Rout = RC r o1
(b) Looking into the emitter of Q2 we see an equivalent resistance of 1gm 2 r π 2 r o2 , so we can drawthe following equivalent circuit for nding Rout :
Q1
Rout
RE = 1gm 2 r π 2 r o2
Rout = ro1 + (1 + gm 1ro1) r π 1 1gm 2
r π 2 ro2
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5.7 (a)
V CC − I B (100 kΩ) = V BE = V T ln(I C /I S )
V CC − 1β
I C (100 kΩ) = V T ln( I C /I S )
I C = 1 .754 mA
V BE = V T ln( I C /I S ) = 746 mV
V CE = V CC − I C (500 Ω) = 1 .62 V
Q1 is operating in forward active.(b)
I E 1 = I E 2 ⇒ V BE 1 = V BE 2V CC − I B 1 (100 kΩ) = 2 V BE 1
V CC − 1β
I C 1(100 kΩ) = 2 V T ln( I C 1 /I S )
I C 1 = I C 2 = 1 .035 mA
V BE 1 = V BE 2 = 733 mV
V CE 2 = V BE 2 = 733 mVV CE 1 = V CC − I C (1 kΩ) − V CE 2
= 733 mV
Both Q1 and Q2 are at the edge of saturation.(c)
V CC − I B (100 kΩ) = V BE + 0 .5 V
V CC − 1
β I C (100 kΩ) = V T ln( I C /I S ) + 0 .5 V
I C = 1 .262 mA
V BE = 738 mVV CE = V CC − I C (1 kΩ) − 0.5 V
= 738 mV
Q1 is operating at the edge of saturation.
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5.8 See Problem 7 for the derivation of I C for each part of this problem.
(a)
I C 1 = 1 .754 mA
gm 1 = I C 1 /V T = 67 .5 mS
rπ 1 = β/g m 1 = 1 .482 kΩ
100 kΩ
r π 1+vπ 1
−gm 1vπ 1 500 Ω
(b)
I C 1 = I C 2 = 1 .034 mAgm 1 = gm 2 = I C 1 /V T = 39 .8 mS
r π 1 = rπ 2 = β/g m 1 = 2 .515 kΩ
100 kΩ
r π 1+vπ 1
−gm 1vπ 1 1 kΩ
r π 2+vπ 2
−gm 2vπ 2
(c)
I C 1 = 1 .26 mA
gm 1 = I C 1 /V T = 48 .5 mS
rπ 1 = β/g m 1 = 2 .063 kΩ
100 kΩ
r π 1
+vπ 1
−gm 1vπ 1 1 kΩ
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5.9 (a)
V CC − V BE 34 kΩ
− V BE 16 kΩ
= I B = I C
β
I C = β V CC − V T ln( I C /I S )
34 kΩ − β
V T ln( I C /I S )16 kΩ
I C = 677 µ AV BE = 726 mV
V CE = V CC − I C (3 kΩ) = 468 mV
Q1 is in soft saturation.(b)
I E 1 = I E 2⇒ I C 1 = I C 2⇒ V BE 1 = V BE 2 = V BE
V CC − 2V BE 9 kΩ −
2V BE 16 kΩ = I B 1 =
I C 1β
I C 1 = β V CC − 2V T ln( I C 1 /I S )
9 kΩ − β
2V T ln(I C 1 /I S )16 kΩ
I C 1 = I C 2 = 1 .72 mA
V BE 1 = V BE 2 = V CE 2 = 751 mV
V CE 1 = V CC − I C 1(500 Ω) − V CE 2 = 890 mV
Q1 is in forward active and Q2 is on the edge of saturation.(c)
V CC − V BE − 0.5 V12 kΩ
− V BE + 0 .5 V13 kΩ
= I B = I C β
I C = β V CC − V T ln(I C /I S ) − 0.5 V
12 kΩ − β
V T ln( I C /I S ) + 0 .5 V13 kΩ
I C = 1 .01 mA
V BE = 737 mV
V CE = V CC − I C (1 kΩ) − 0.5 V = 987 mV
Q1 is in forward active.
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5.10 See Problem 9 for the derivation of I C for each part of this problem.
(a)
I C = 677 µ A
gm = I C /V T = 26 .0 mS
rπ = β/g m = 3 .84 kΩ
(34 kΩ) (16 kΩ) rπ
+vπ
−gm vπ 3 kΩ
(b)
I C 1 = I C 2 = 1 .72 mAgm 1 = gm 2 = I C 1 /V T = 66 .2 mS
r π 1 = rπ 2 = β/g m 1 = 1 .51 kΩ
(9 kΩ) (16 kΩ) rπ 1
+vπ 1
−gm 1vπ 1 500 Ω
gm 2vπ 2r π 2−vπ 2
+
(c)
I C = 1 .01 mA
gm = I C /V T = 38 .8 mS
rπ = β/g m = 2 .57 kΩ
(12 kΩ) (13 kΩ) rπ+vπ
−gm vπ 1 kΩ
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5.13 We know the input resistance is Rin = R 1 R 2 r π . Since we want the minimum values of R1 andR2 such that R in > 10 kΩ, we should pick the maximum value allowable for r π , which means pickingthe minimum value allowable for gm (since r π ∝ 1/g m ), which is gm = 1 / 260 S.
gm = 1260
S
I C = gm V T = 100 µ AV BE = V T ln( I C /I S ) = 760 mV
I B = I C
β = 1 µ A
V CC − V BE R1
− V BE
R2= I B
R1 = V CC − V BE
I B + V BER 2
rπ = β gm
= 26 kΩ
R in = R1 R2 rπ
= V CC − V BE I B + V BER 2
R2 r π
> 10 kΩ
R2 > 23.57 kΩ
R1 > 52.32 kΩ
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5.14
gm = I C V T
≥ 126
S
rπ = β gm
= 2 .6 kΩ
R in = R1 R2 r π≤ r π
According to the above analysis, R in cannot be greater than 2 .6 kΩ. This means that the requirementthat Rin ≥ 10 kΩ cannot be met. Qualitatively, the requirement for gm to be large forces rπ to besmall, and since R in is bounded by r π , it puts an upper bound on R in that, in this case, is below therequired 10 kΩ.
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5.15
Rout = RC = R0
Av = − gm RC = − gm R0 = −I C V T
R0 = − A0
I C =
A0R0 V T
r π = β V T I C
= β R0A0
V BE = V T ln( I C /I S ) = V T lnA0 V T R0I S
V CC − V BE R1
− V BE
R2= I B =
I C β
R1 = V CC − V BE
I Cβ +
V BER 2
R in = R1 R2 rπ
=V CC − V T ln A 0 V T
R 0 I SI Cβ +
V T R 2 ln
A 0 V T R 0 I S
R2 β R0A0
In order to maximize R in , we can let R2 → ∞ . This gives us
R in,max = β V CC − V T ln A 0 V T R 0 I S
I C β
R0A0
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5.16 (a)
I C = 0 .25 mAV BE = 696 mV
V CC − V BE − I E RE R1
− V BE + I E RE
R2= I B =
I C β
R1 =V CC − V BE − 1+ ββ I C RE
I Cβ +
V BE + 1+ ββ I C R ER 2
= 22 .74 kΩ
(b) First, consider a 5 % increase in RE .
RE = 210 ΩV CC − V BE − I E RE
R1−
V BE + I E RE R2
= I B = I C
β V CC − V T ln( I C /I S ) − 1+ ββ I C RE
R1−
V T ln( I C /I S ) + 1+ ββ I C RE
R2= I B =
I C
β I C = 243 µ A
I C − I C,nomI C,nom
× 100 = − 2.6 %
Now, consider a 5 % decrease in R E .
RE = 190 ΩI C = 257 µ A
I C − I C,nomI C,nom
× 100 = +2 .8 %
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5.17
V CE ≥ V BE (in order to guarantee operation in the active mode)V CC − I C RC ≥ V T ln( I C /I S )
I C ≤ 833 µ AV CC − V BE − I E RE
30 kΩ − V BE + I E RE
R2 = I B = I C
β
R2 = V BE + I E RE
V CC − V BE − I E R E30 kΩ −
I Cβ
=V T ln( I C /I S ) + 1+ ββ I C RE
V CC − V T ln( I C /I S )− 1+ ββ I C R E30 kΩ −
I Cβ
R2 ≤ 20.66 kΩ
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5.18 (a) First, note that V BE 1 = V BE 2 = V BE , but since I S 1 = 2 I S 2 , I C 1 = 2 I C 2 . Also note thatβ 1 = β 2 = β = 100.
I B 1 = I C 1
β =
V CC − V BE − (I E 1 + I E 2 )RE R1
− V BE + ( I E 1 + I E 2 )RE
R2
I C 1 = β
V CC − V T ln( I C 1/I S 1 ) − 321+ β
β I C 1RE
R1 −
V T ln( I C 1 /I S 1) + 321+ β
β I C 1RE
R2I C 1 = 707 µ A
I C 2 = I C 1
2 = 354 µ A
(b) The small-signal model is shown below.
R1 R2 rπ 1+vπ 1
−r π 2
+vπ 2
−gm 1vπ 1
RC
gm 2vπ 2
RE
We can simplify the small-signal model as follows:
R1 R2 rπ 1 r π 2
+vπ
−gm 1vπ
RC
gm 2vπ 2
RE
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gm 1 = I C 1/V T = 27 .2 mS
r π 1 = β 1 /g m 1 = 3 .677 kΩ
gm 2 = I C 2/V T = 13 .6 mS
r π 2 = β 2 /g m 2 = 7 .355 kΩ
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5.19 (a)
I E 1 = I E 2 ⇒ V BE 1 = V BE 2V CC − 2V BE 1
9 kΩ −
2V BE 116 kΩ
= I B 1 = I C 1
β 1
I C 1 = β 1V CC − 2V T ln( I C 1 /I S 1)
9 kΩ − β 12V T ln(I C 1 /I S 1)
16 kΩI C 1 = I C 2 = 1 .588 mA
V BE 1 = V BE 2 = V T ln( I C 1 /I S 1) = 754 mV
V CE 2 = V BE 2 = 754 mV
V CE 1 = V CC − I C 1 (100 Ω) − V CE 2 = 1 .587 V
(b) The small-signal model is shown below.
(9 kΩ) (16 kΩ) rπ 1+vπ 1
−
gm 1vπ 1
r π 2+vπ 2
−gm 2vπ 2
100 Ω
gm 1 = gm 2 = I C 1V T
= 61 .1 mS
r π 1 = rπ 2 = β 1gm 1
= 1 .637 kΩ
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5.22
V CC − I E (500 Ω) − I B (20 kΩ) − I E (400 Ω) = V BE
V CC −1 + β
β I C (500 Ω + 400 Ω) −
1β
I C (20 kΩ) = V T ln( I C /I S )
I C = 1 .584 mA
V BE = V T ln( I C /I S ) = 754 mVV CE = V CC − I E (500 Ω) − I E (400 Ω)
= V CC −1 + β
β I C (500 Ω + 400 Ω) = 1 .060 V
Q1 is operating in forward active.
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5.23
V BC ≤ 200 mVV CC − I E (1 kΩ) − I B RB − (V CC − I E (1 kΩ) − I C (500 Ω)) ≤ 200 mV
I C (500 Ω) − I B RB ≤ 200 mVI B RB ≥ I C (500 Ω) − 200 mV
V CC − I E (1 kΩ) − I B RB = V BE = V T ln( I C /I S )
V CC −1 + β
β I C (1 kΩ) − I C (500 Ω) + 200 mV ≤ V T ln(I C /I S )
I C ≥ 1.29 mA
RB ≥I C (500 Ω) − 200 mV
I Cβ
≥ 34.46 kΩ
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5.25 (a)
I C 1 = 1 mAV CC − (I E 1 + I E 2 )(500 Ω) = V T ln( I C 2 /I S 2 )
V CC −1 + β
β I C 1 +
1 + β β
I C 2 (500 Ω) = V T ln( I C 2 /I S 2 )
I C 2 = 2 .42 mAV B − (I E 1 + I E 2 )(500 Ω) = V T ln( I C 1 /I S 1 )
V B −1 + β
β I C 1 +
1 + β β
I C 2 (500 Ω) = V T ln( I C 1 /I S 1 )
V B = 2 .68 V
(b) The small-signal model is shown below.
r π 1+vπ 1
−r π 2
+vπ 2
−
200 Ω
gm 1 vπ 1 gm 2 vπ 2
500 Ω
gm 1 = I C 1 /V T = 38 .5 mS
r π 1 = β 1 /g m 1 = 2 .6 kΩ
gm 2 = I C 2 /V T = 93 .1 mS
r π 2 = β 2 /g m 2 = 1 .074 kΩ
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5.26 (a)
V CC − I B (60 kΩ) = V EB
V CC − 1β pnp
I C (60 kΩ) = V T ln( I C /I S )
I C = 1 .474 mA
V EB = V T ln( I C /I S ) = 731 mV
V EC = V CC − I C (200 Ω) = 2 .205 V
Q1 is operating in forward active.(b)
V CC − V BE 1 − I B 2 (80 kΩ) = V EB 2V CC − V T ln(I C 1 /I S ) − I B 2 (80 kΩ) = V T ln(I C 2 /I S )
I C 1 = β npn1 + β npn
I E 1
= β npn1 + β npn I E 2
= β npn1 + β npn
· 1 + β pnpβ pnp
I C 2
V CC − V T ln β npn
1 + β npn· 1 + β pnp
β pnp· I C 2
I S − 1
β pnpI C 2 (80 kΩ) = V T ln(I C 2 /I S )
I C 2 = 674 µ A
V BE 2 = V T ln(I C 2 /I S ) = 711 mV
I C 1 = 680 µ A
V BE 1 = V T ln(I C 1 /I S ) = 711 mV
V CE 1 = V BE 1 = 711 mVV CE 2 = V CC − V CE 1 − I C 2 (300 Ω)
= 1 .585 V
Q1 is operating on the edge of saturation. Q2 is operating in forward active.
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5.27 See Problem 26 for the derivation of I C for each part of this problem.
(a) The small-signal model is shown below.
60 kΩ
r π+
vπ−
gm vπ 200 Ω
I C = 1 .474 mA
gm = I C V T
= 56 .7 mS
r π = β g
m
= 1 .764 kΩ
(b) The small-signal model is shown below.
r π 1+vπ 1
−gm 1 vπ 1
r π 2
−
vπ 2
+
80 kΩ
gm 2 vπ 2
300 Ω
I C 1 = 680 µ A
gm 1 = I C 1V T = 26 .2 mS
r π 1 = β npngm 1
= 3 .824 kΩ
I C 2 = 674 µ A
gm 2 = I C 2V T
= 25 .9 mS
r π 2 = β pnpgm 2
= 1 .929 kΩ
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5.30
V CC − I C (1 kΩ) = V EC = V EB (in order for Q1 to operate at the edge of saturation)= V T ln(I C /I S )
I C = 1 .761 mAV EB = 739 mV
V CC − V EBRB
−V EB5 kΩ
= I B = I C
β RB = 9 .623 kΩ
First, let’s consider when RB is 5 % larger than its nominal value.
RB = 10 .104 kΩV CC − V T ln( I C /I S )
RB−
V T ln( I C /I S )5 kΩ
= I C
β I C = 1 .411 mA
V EB = 733 mV
V EC = V CC − I C (1 kΩ) = 1 .089 VV CB = − 355 mV (the collector-base junction is reverse biased)
Now, let’s consider when RB is 5 % smaller than its nominal value.
RB = 9 .142 kΩV CC − V T ln( I C /I S )
RB−
V T ln( I C /I S )5 kΩ
= I C
β I C = 2 .160 mA
V EB = 744 mVV EC = V CC − I C (1 kΩ) = 340 mV
V CB = 405 mV (the collector-base junction is forward biased)
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5.31
V BC + I C (5 kΩ)10 kΩ
−V CC − V BC − I C (5 kΩ)
10 kΩ = I B =
I C β
V BC = 300 mVI C = 194 µ A
V EB = V T ln(I C /I S ) = 682 mVV CC − I E RE − I C (5 kΩ) = V EC = V EB + 300 mV
V CC −1 + β
β I C RE − I C (5 kΩ) = V EB + 300 mV
RE = 2 .776 kΩ
Let’s look at what happens when RE is halved.
RE = 1 .388 kΩV CC − I E RE − V EB
10 kΩ −
V CC − (V CC − I E RE − V EB )10 kΩ
= I B = I C
β
β V CC − 1+ ββ I C RE − V T ln( I C /I S )
10 kΩ − β V
CC − V CC − 1+ β
β I C RE − V T ln( I C /I S )10 kΩ
= I C
I C = 364 µ AV EB = 698 µ VV EC = 164 µ V
Thus, when R E is halved, Q1 operates in deep saturation.
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5.32
V CC − I B (20 kΩ) − I E (1.6 kΩ) = V BE = V T ln( I C /I S )
V CC −I C β
(20 kΩ) −1 + β
β I C (1.6 kΩ) = V BE = V T ln( I C /I S )
I S = I C
ehV CC −I C
β (20 kΩ) − 1+ ββ I C (1 .6 kΩ) i/V T I C = 1 mA
I S = 3 × 10− 14 A
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5.38 (a)
Av = − gm 1 1gm 2
r π 2
R in = rπ 1
R out = 1gm 2 r π 2
(b)
Av = − gm 1 R 1 + 1gm 2
r π 2
R in = rπ 1
R out = R 1 + 1gm 2
r π 2
(c)
Av = − gm 1 R C + 1gm 2 r π 2
R in = rπ 1
R out = RC + 1gm 2
r π 2
(d) Let’s determine the equivalent resistance seen looking up from the output by drawing a small-signal model and applying a test source.
−vt
+
i t rπ 2
+
vπ 2−
R C
gm 2 vπ 2
i t = vπ 2r π 2
+ gm 2 vπ 2
vπ 2 = vt
i t = vt 1r π 2
+ gm 2
vti t
= 1gm 2
r π 2
Av = − gm 1 1gm 2
r π 2
R in = rπ 1
R out = 1gm 2
r π 2
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(e) From (d), we know the gain from the input to the collector of Q 1 is − gm 1 1gm 2 r π 2 . If we ndthe gain from the collector of Q1 to vout , we can multiply these expressions to nd the overallgain. Let’s draw the small-signal model to nd the gain from the collector of Q 1 to vout . I’ll referto the collector of Q 1 as node X in the following derivation.
−vX
+r π 2
+vπ 2
−
R C
vout
gm 2 vπ 2
vX − voutR C
= gm 2 vπ 2
vπ 2 = vXvX − vout
R C = gm 2 vX
vX 1R C
− gm 2 = voutR C
voutvX
= 1 − gm 2 R C
Thus, we have
Av = − gm 1 1gm 2
r π 2 (1 − gm 2 R C )
R in = rπ 1
To nd the output resistance, let’s draw the small-signal model and apply a test source at theoutput. Note that looking into the collector of Q 1 we see innite resistance, so we can exclude itfrom the small-signal model.
r π 2
+vπ 2
−
R C
−vt
+i tgm 2 vπ 2
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i t = gm 2 vπ 2 + vπ 2r π 2
vπ 2 = rπ 2
r π 2 + RC vt
i t = gm 2 + 1r π 2
rπ 2r π 2 + R C vt
R out = vti t
= 1gm 2
r π 2 r π 2 + R C
r π 2
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5.39 (a)
Av = − gm 1 r o1 1gm 2
r π 2 r o 2
R in = rπ 1
R out = ro1 1gm 2 r π 2 r o 2
(b)
Av = − gm 1 r o 1 R 1 + 1gm 2
r π 2 r o2
R in = rπ 1
R out = ro1 R 1 + 1gm 2
r π 2 r o2
(c)
Av = − gm 1 r o 1 R C + 1gm 2
r π 2 r o2
R in = rπ 1
R out = ro 1 R C + 1gm 2
r π 2 r o2
(d) Let’s determine the equivalent resistance seen looking up from the output by drawing a small-signal model and applying a test source.
−vt
+i t rπ 2
+vπ 2
−
R C X
r o2gm 2 vπ 2
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i t = vπ 2r π 2
+ vt − vX
R C vX − vt
R C + gm 2 vπ 2 +
vXr o2
= 0
vπ 2 = vtvX
1R C
+ 1r o2
= vt 1R C
− gm 2
vX = vt 1R C
− gm 2 (r o2 R C )
i t = vtr π 2
+ vtR C
− 1R C
vt 1R C
− gm 2 (r o 2 R C )
= vt 1r π 2
+ 1R C
− 1R C
1R C
− gm 2 (r o2 R C )
= vt 1r π 2
+ 1R C
+ gm 2 − 1R C
ro2r o2 + R C
vti t
= r π 2 R C r o2 + R C
r o 21
gm 2 − 1R C
Av = − gm 1 r o1 r π 2 R C r o2 + R C
r o21
gm 2 − 1R C
R in = rπ 1
R out = ro1 r π 2 R C r o 2 + RC
r o21
gm 2 − 1R C
(e) From (d), we know the gain from the input to the collector of Q1 is − gm 1 r o1 r π 2 R C r o 2 + R Cr o 21
gm 2 − 1R CIf we nd the gain from the collector of Q 1 to vout , we can multiply these expressions to nd theoverall gain. Let’s draw the small-signal model to nd the gain from the collector of Q 1 to vout .I’ll refer to the collector of Q 1 as node X in the following derivation.
−vX
+r π 2
+vπ 2
−
R C vout
r o 2gm 2 vπ 2
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vout − vXR C
+ gm 2 vπ 2 + voutr o2
= 0
vπ 2 = vXvout − vX
R C + gm 2 vX +
vout
r o2= 0
vout 1R C
+ 1r o2
= vX 1R C
− gm 2
voutvX
= 1R C
− gm 2 (R C r o2 )
Thus, we have
Av = − gm 1 r o1 r π 2 R C r o2 + R C
r o21
gm 2 − 1R C
1R C
− gm 2 (R C r o2 )
R in = rπ 1
To nd the output resistance, let’s draw the small-signal model and apply a test source at theoutput. Note that looking into the collector of Q 1 we see r o1 , so we replace Q 1 in the small-signalmodel with this equivalent resistance. Also note that r o2 appears from the output to ground, sowe can remove it from this analysis and add it in parallel at the end to nd R out .
r o1 rπ 2
+vπ 2
−
R C
−vt
+i tgm 2 vπ 2
i t = gm 2 vπ 2 + vπ 2
r π 2 r o 1
vπ 2 = rπ 2 r o1
r π 2 r o 1 + RC vt
i t = gm 2 + 1
r π 2 r o1 rπ 2 r o1
r π 2 r o1 + R C vt
R out = r o2 vti t
= ro2 1gm 2 r π 2 r o1