Fundamentals_of_Microelectronics (Manual) by Razavi

8/18/2019 Fundamentals_of_Microelectronics (Manual) by Razavi

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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)


− 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)


− 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)


−

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)



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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)


− 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)



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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)


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)


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


gm = I C /V T = 38 .5 mS

rπ = β/g m = 2 .6 kΩro = ∞


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


gm = I C /V T = 38 .5 mS

r π = β/g m = 2 .6 kΩr o = ∞


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


gm = I C /V T = 38 .5 mS

r π = β/g m = 2 .6 kΩr o = ∞


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



gm = I C /V T = 48 .3 mS

r π = β/g m = 2 .07 kΩr o = ∞



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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



gm = I C /V T = 38 .1 mS

r π = β/g m = 2 .63 kΩr o = ∞


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



gm = I C /V T = 38 .1 mS

r π = β/g m = 2 .63 kΩr o = ∞



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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


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


gm = I C /V T = 7 .69 mS

r π = β/g m = 13 kΩ

r o = ∞


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


gm = I C /V T = 13 .1 mS

r π = β/g m = 7 .64 kΩr o = ∞



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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


gm = I C /V T = 19 .0 mS

r π = β/g m = 5 .25 kΩr o = ∞


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 = ∞


−

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 = ∞


−

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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(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


(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


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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(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


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−


= 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−


= 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

Documents

Fundamentals_of_Microelectronics (Manual) by Razavi