Solution manual microelectronics; circuit analysis desing 3rd edition

For V PS max and I L min, then max 13.6 9 0 max 230 20 A The characteristic of the minimum Zener current being one-tenth of the maximum value is violated.. The proper circuit operation i

www.elsolucionario.net

1.42.1 10 300 exp

6

0.661.66 10 300 exp

1.6 10 10 10

16 /10

(10 12)exp

0.026

D D

1.15.23 10 400 exp

( ) ( ) ( ) ( )

6

0.661.66 10 400 exp

6

1.42.1 10 400 exp

(a)

2 10 2

15

8 10 1.5 10

0.033 0.225 1.54 Load line

P=I V ⇒ =I so I D =1 5 mA

19

1.15.23 10 250 exp

2 86 10 2502.067 10 exp 25.58

6

19

1.15.23 10 350 exp

2 86 10 3503.425 10 exp 18.27

6

17

1.42.10 10 250 exp

2 86 10 2508.301 10 exp 32.56

6

18

1.42.10 10 350 exp

2 86 10 3501.375 10 exp 23.26

(a) ( ) ( ) ( ) ( )

3 / 2 15

6

18

1.15.23 10 100 exp

2 86 10 1005.23 10 exp 63.95

6

19

1.15.23 10 300 exp

2 86 10 3002.718 10 exp 21.32

6

19

1.15.23 10 500 exp

2 86 10 5005.847 10 exp 12.79

2(86 10 )(300)2.10 10 300 1.65 10

5 10 cm1.5 10

2 86 10 3001.66 10 300 2.79 10

17

2 10 cm1.5 10

x J

0 8

/ j

0 8

/ j

0.6(10 ) exp 0.105 A

0.026

0.7(10 ) exp 49.3 mA

b ΔV D =V Tln 100( )⇒ ΔV D =119.7 mV 120 mV≈

1.31

(a) (i) ln (0 026 ln) 150 1015 6

10 = 0.669 V

D

D t

S D

I

I V

0.018652.147 10 1.237 10

9.374 10(100)

1 2 0 7

50 10

We have I1=I2+I D

8 3(0.65)

3.025 mA2

3.025 0.652.375 mA

V

0 02610

v v

Z Z

Assuming all diodes are conducting, we have V B = −0.7 and V A = 0

Summing currents, we have

(b) I R =0.233 A, I L =( )(0.9 0.233)=0.21 A

So 0.21 12 L 57.1

L

R R

2

r M

V t V

Percent time 100% 18.1%

r M

V t V

r M

V t V

(Minimum Zener current is zero.)

For V PS (max) and I L (min), then (max) 13.6 9 0 (max) 230

20

A

The characteristic of the minimum Zener current being one-tenth of the maximum value is violated

The proper circuit operation is questionable

TYU2.5

(min) PS(min) Z (max)

As v S goes negative, D turns on and v O = +5 V As v S goes positive, D turns off Output is a square

wave oscillating between +5 and +35 volts

5

1 0.029.22

From part (a) PIV =2v S(max)−Vγ =2 26.4( )−0.7

or PIV =52.1 V or, from part (b) PIV =2 101.4( )−0.7 or PIV =202.1 V

(c) For the half-wave rectifier

x x

1

2 1 1

x x

x

x

x x

L i L L

R R

(a) For − ≤ ≤10 v I 0, both diodes are conducting ⇒v O =0

For 0≤ ≤v I 3, Zener not in breakdown, so i1=0, v O =0

( )

1

3For 3

(b) For v I <0, both diodes forward biased

Then for v I >5.7 V

3.422.5

I I

I O

D

v v

v

i = − For v

I = 15, i D = 5.58 mA

i D(mA)

␯ I(V) 5.7

10(b) For v I ≤4.03 ,V i D =0

4.03,

o o

v v

L will tend to block the transient signals

D z will limit the voltage to +14 V and −0.7 V

Power ratings depends on number of pulses per second and duration of pulse

07.6 0.589 7.01 mA

03.6 0.589 3.01 mA

0.6 51.5 2.93

For v I >0. when D1 and D4 turn off

1.4 15

2.72 mA5

2.72 1.071.65 mA

(a) D1 on, D2 off

R R

SGA SGA

24.3 K0.1

R

R R

500

1 2.5 1.5 V1.5 1.507

40.838 V

0.838 6 5.16220

DQ DSS

P GS

28.2 k0.1

2 2

2 2

2 2

0.475 2.88 0.165

2

165 2

9.4335

(sat) 0.0640(3.5 1.5) (sat) 0.256 mA for 3.5 V

(sat) 0.0640(4.5 1.5) (sat) 0.576 mA for 4.5 V

3.9 8.85 10

400 108.63 10 F/cm

V V V

4 V, 0.2 4 1.2 1.57 mA

1

63.7 k0.01 1.57

100 V0.01

(a) ( ) 0.08 ( )2

Sat Region 15 2.5 1.39

20.739 mA

D D

D D

3.6 2.046

0.777 mA2

2.81 5

4.38 K0.5

3.35

( ) ( )( )

2 2

5 3 1.755

R

R R

2 2

60

4 120 A/V2

60

1 30 A/V2

2 2

2 2 1

0.75 2 1.25 V

2.5 1.25

12.5 K0.1

GS GS

D

V V

2

2

sat2

0.04

2

10019.5 7.81250

p

A

SGA SGA

2

2 1 1

2 1

sat2

V V

3.57

( ) ( )( )

( ) ( )( )

5.85 mA0.2

V

V V

2

2 2

1 2 1 V1

R R

x

x x

S S

( )

( )( )( ) ( )

5 1.71

0.822 mA4

2

2 1

0.015 2 0.030 /2

3.8 0.388Then middle of saturation region 0.388 2.094 V

(a)

( )( )

1

1 1

3 1

G

R V

( )( ) ( )( )

( ) ( )( )

Kn I

TYU4.12

(a)

( )( ) ( )( )

2 2

2

1

2

0.020 80 1.6 /2

0.020 1 0.020 /2

120

24

100 K0.02

o

V I

2%

500

DS D

o D

D

V I

r I

( )

2

cos 22

Ratio of signal at 2 to that at :

2 2

R

R R

||

||

||

200 || 3001.92 3 || 3 2.88 0.9836

200 || 300 22.83

D DS m

Si

v

I V g

AC load line Slope  3兩兩31 1.5 K1 0.92

2.76 V peak-to-peak output voltage swing

g R

A

g R

R R

m S

g R A

g R g

D

g g

1 10.6 K

v

S S

A

R R

500 K0.02 0.1

1 0.1

40 500 37.04

40 500 37.0413.7 K

6 0.293 6.293

10 6.293

1.85 k2

2 mA



1 1.85兩兩2  0.146

AC load line Slope 

b

( )( ) ( )( )

1.059 1.5 0.441 V0.441 3

10.2 K0.25

12.2

11

1 2.2 1.33

2.2

1 2.93 2.2

13.93 2.2

D D D

100 18 3.4 529 k529

0.614 3 2.386 V2.386 10

15.2 K0.5

L o m

( )

0 0

++

40.95

24.75 2 0.030 60 3.13 mA

20 k0.01 5

5 4.09 2

0.870

m m

i i

L

R R

( )( )

0.112 k8.94

2 0

0.5 0 10.5

4.0830

0

1.67 0.526 || 1.671.90

133.3 K0.075 0.1

2 1

200 K0.05 0.1

133.3 K0.075 0.1

1.085

0.9560.050 1.085 0.050

0.956 0.922 200 133.370.5

4.56

(a)

( )( ) ( )( ) ( )( )

(1) 2 2 2

2

x sg x

2

2 2

R

R R

5 1.81 3.19 Max output swing = 6.38 V peak-to-peak

2 2

4 4.4 V1

R

R R

2 2

R

R R

500332.5 250 k

10 6.12

38.8 k0.1

P VGS GS

V V V

5 k265

265

R

R R

1

1.754.5 10 1 1.143 0.3265

1 5兩兩10

AC load line Slope 

100 5.91 0.591

10 0.591 4 or 7.64

BB BE B

1.5 1.515 100

1.1 151 0.2which yields 0.1806

TH

BQ

R V

For 3 V CEO = V, then 2.3 V CO = V

400.25 0.2439

5 2.3

11.07 0.2439

2.5

1.21 2.5 1.29 1.29 5

1.97 1.88

(a) For V I =0.2 0,V <V BE( )on ⇒I B =I C = V O =5 and V P=0

(b) For V I =3.6 ,V transistor is driven into saturation, so

( ) 3.6 0.7

4.53 0.64

CE C

I BE B

−0.9 9

β

1

βαβ

=+

801.2 1.185 mA

8180

0.987781

800.80 0.790 mA

8180

0.987781

600.75 0.738 mA

6160

0.983661

10 0.738 5 106.31 V

601.5 1.475 mA

6160

0.983661

5.8

( )10 10 1.2

1.76 mA5

15 1

1.69 10Ratio of areas Ratio 24.4

6.94 10

Eo Eo

I I

( ) ( )

0

3 3

751.86 1.836 mA

760.7 4 3.3 V

0.5 0.493 mA

76

8.11 K0.493

1 0.7 1.7 V

3 1.7 3

0.2708 mA4.8

0.2708

0.0020.99261

24 0.7

699 k0.0333

24 12

6 k2

CC BE BQ

B CQ BQ

100 0.0333 3.33 mA

CC BE BQ

10 53.33 V

BB

L CC

0.091 mA30

0.7 12

0.127 mA100

0.127 0.091 0.218 mA0.7 0.218 15 0.7 3.97 V

5 0.7

1032 K0.00417

R R

=

For 1 M 10% 1.1 M, 5.1 k 10% 5.61 K

5 0.7

3.91 A 0.469 mA1.1

2 2

( )( ) ( )( ) ( )

So 1.6, 4.8

O C

V O(V)

4.8

I(V)5.39

(a) For 4.3,V I ≥ Q is off and V O = 0

When transistor enters saturation, 5 101 ( )1 0.2 ( )4 0.958 mA

( ) ( )

( ) ( ) ( )

1014

I V

8 255.82 V

From this quadratic, we find R2 =48 64.5 kΩ ⇒ R1= kΩ

(b) Standard resistor values:

I V

0.008 100

10.1 12.0 63.6

71.8

10.6 12.4 71.8

( )

12 2 0.26

R

R

R R

R R

1.84 20.0 1.685 k

1.84

12 6 4.99 V1.84 20.0

4.99 0.7 6 0.31

0.0173 mA1.685 81 0.2 17.89

Specifications are met

(b) For β =100, we find I BQ =0.0161 , 1.61 , mA I CQ = mA V ECQ =15.13 , 1.63 V I EQ = mA

R R

2

2

1

on1

R R

8.18

20 2.908.18 48.2

2

48.5 6.197

72.572.5 k , 30.0

72.551.2 k

R R

R R

C

I

I I

which gives 97.3 , and 48.4

2

1.67 mA3

10 0.7 1.67 7.63

0.0593 mA6.67 61 2 128.7

( )( ) ( )( )

( ) ( )

R R

2

2 2

For a bias stable circuit

0.1 12.1 2.5 30.25 K

30.25 12 0.00667 30.25 0.7 21

R

R R

R R

R R

2

2 2

For a bias stable circuit

R

R

R R

( ) ( ) ( )

0.7 0.7 1.4

1.4 5

3.6 mA1

0.0444 mA3.56 mA

Current through V− source =I E1+I E2 and I E1=I E2 = +(1 β)I B1=( )(51 8.26 )μA

So total current =2 51 8.26 A=843 A( )( ) μ μ

5.21 k0.792

BB BE BQ

150 0.026

11.8 0.33

200

606 0.33

Now

0.801

30.8 /0.026

90 0.026

2.92 0.801

120

150 0.801

100

476 0.21

C L o

400 0.25

+which yields

120 0.026

1.36 2.29

100

43.7 2.29

V V

11

0.967 1

1

E o o

TH BQ

V I

− − −

=+ +

( )(100 0.026)

2.08 1.25

125

100 1.25

1

4.2105 0.7 5

2.91 5.53 126 0.2

1

1

0.7 55

1

C C

( ) ( )( )

0.364

14.0 /0.026

125 0.026

0.674 4.82

4.82

185 /0.026

120 0.026

12.5 0.25

150

600 0.25

As a first approximation, C

v E

R A R

R A R

β

−

= −+ +

0.84

32.3 /0.026

200

238 0.84

100 0.026

3.28 0.793

125

158 0.793

15.2 4.93

1000.93 0.921

0.921

35.42 /0.026

π π

32.23 /0.026

10 1

C L o

T CQ

0.377 0.026

100 0.026

2.63 0.990

0.990

38.1 /0.026

( )

1

2 1

2652.63

180 0.026

2.34 k2

150

75 k2

( )

0.0260.026

120 0.026

5 0.624

0.00482 mA120

on0.00482 25 0.70 0.820 V

120 0.026

5.40 k0.578

0.578

22.2 mA/V0.026

100

173 k0.578

5sin mV

2.89sin A1.73 k

20.60sin mA

ω μβ

2

2 2

β+

For min 1.9 kΩ and max 1.05 k

100 1.9

1.762.18 101 1.05

p

22

2

6.14

R R

100 11.3

55.110.4 101 0.1

R R

2.54 95.230.3

30.77 10.9 5105

E E

1.408 mA

80 0.0261.408 54.15 mA/V

π π

β

βπ

S S

o C L o

Assume an npn transistor with b=100 and V A = ∞ Let V CC =10 V

Bias at I CQ =1 mA and let R E =1 kΩ

For a bias stable circuit

10 1 2 6 1

or 0 674

S s

With this R C , the dc bias is OK

Finish Design, Set R C =2 K R E =1 K

Design specification met

6.26

Let R S = Need an input resistance of 0.

3

3 6

β

ββ

R A

bFirst approximation:

2

2 2

R

R

R R

80 6.6

11.12.97 81 0.55

R R

( ) ( ) ( )( )

EC

v v

Max output swing 3 24 V peak-to-peak= .

0

Max output swing 2.85 V peak-to-peak

2 85Swing in current

4

0 712 mA peak-to-peak

i

( ) ( )

0.11



AC load line Slope  11.21

For maximum symmetrical swing

( )( ) ( )( ) ( )( )( )

1

1 1

2

2 2

R

R R

6.39

a

( ) ( )( )



AC load line Slope  0.50.3 1

( ) ( )( )

( ) ( )( )



AC load line Slope  511

L o o

β

ωω

4 4.02 V peak-to-peak at base0.995 0.995

( ) ( )( )

& & &

& & &

151 0.833 20.3

0.8113.82 151 0.833 20.3 4

& & & &

& & & &

( ) ( )

1

9 0.71001

8.3

100 1

518.3

100

1201

E

E

E E

E

I

R I

I

β

ββ

201 1 1 50.3200

50.3 100.943 201 1 1

s

b

L L v

L L

180 0.026

9.36 0.5

on 11

( )

b

E L TH

i

TH TH TH

TH TH

R R

ββ

Now min min 3.73sin

max max 3.86sin

With R S =10 ,kΩ we will not be able to meet this voltage gain requirement Need to insert a buffer or an

op-amp voltage follower (see Chapter 9) between R S and C C1

( )

( ) ( )

15.2 0.7

0.0117 7.40 126 3

10 0.1

35.42 4.545

0.02180.1

( )

10.984

14 35 5 0 00833 14 35 0 7 1 008 2which yields 25 3

( ) ( )( )

42.7 2 1.77B40.17

0.00477 mA10.68 121 2

0.572 mA

0.572

22 mA/V0.026

120 0.026

5.45 k0.572

0.0405 mA11.25 121 1.6

12 5.13

22 10 7.92 97.21

1

121 0.216

0.9760.642 121 0.216

1

15 45 10 5.29 k

0.642 5.29

1.6 0.049 1.6 47.6 121

100 0.026

0.0372 69.9

V A V

( )( ) ( )( )

r r V

A

V

A

π π π π

100 0 026

2 6 1

12 6.31

0.949

AC load line Slope  61051

2 2

20 6.53

0.894

AC load line Slope  521

πω

πω

πω

πω

πω

3

2 3

17.96 104

101.4 m0.026

3.265 5

2.17 k0.8

DQ P SG TP

SG SG S

0.97864.57 0.1

2.30 mA

100 0.026

1.13 k2.30

2.30

88.46 mA / V0.026

2.79 k0.932

32.3 mA/V0.026

200

238 k0.84

& & & & &

& & &

58.4 MHz

113.110101

100 0.026

5.2 K0.5

0.5

19.23 mA/V0.026

6.40 5.2 12.86919.23 1.579

2.869 122.5

i S C

m L

S C m

S C L

dB

S C

m L S

11Then

L

i S

L L m

L L L

L L

sC r

=

max

Z 3

3.18 MH

75 1.5 51.5 0.5281

( )( )( )

π

π π

TYU7.4 Computer Analysis

TYU7.5 Computer Analysis

1

5

50Phase tan Phase 84.3

1

1

111

P P i

11/ 1

P

S P

o i

2 1

2 2 2

2

1

11

1

11

o

i

R R

( ) ( )

( ) ( )

1.565 0.7

0.0759 mA1.30 101 0.1

7.585 mA

100 0.026

0.343 k7.59

2 2

0.5 mA 0.5 0.5 0.25 V

0.51.5 3.08 V0.2

3.08 0.25 3.33 V

13.331

R

R R

2.414 V

2.414 5

2.59 k1

3 2.414 0.586 V

5 0.59

4.41 k1

DQ

n S

10 1.064 2 1.075 4

3.57 V

1.064

40.92 mA/V0.026

100 0.026

2.44 K1.064

v m

1

2250.2 0.2 0.2

( ) ( ) ( )( )

2.5 0.7

0.319 0.6 101 0.05

31.9

100 0.026

0.0815 31.9

1 and so that

2 25 s R C eq C

τπ

1010.0063712.64 504

12 F663.7

( )

( )

2

0.50.8 1.8 0.5

S m

g A

( ) ( )

For 0

10.7 10

1.86 mA5

r

C C

v w

E C

v w

R A

8

3 0

1 and

21

2 5 103.18 10

121 pF0.282 4 10

1000 2.3 12.29

2.29 143.7

where

1

11

1

11

m L

b b

r r

r r

sr C sC

π π

π π

So ( ) 0( ) ( ) ( ) 1

11

1

Let 1

m L

B S

i

B S b i

r R

g R

π π

π π

192.3 mA/V0.026

1.26 0.7

0.0222 mA5.04 101 0.2

2.22 mA

100 0.026

1.17 k2.22

( ) ( )

3 6

0.00222 mA50.4 101 2

0.2217 mA

100 0.026

11.73 K0.222

2

3 2

7.47

( ) ( )( )

I I

L W

W

L f

μ

π

μπ

μπ

m S

I R V

g r

=+Then 0

1

m i i

120 0.026

9.94 k0.3138

R r

0.00633 mA4.44 121 0.5

0.760 mA

120 0.026

4.11 k0.760

0.405 10 15.6 149 10

6.67 10 s

1

2.39 MHz2

2.64 10 4.7 10 1.24 10 s

1

12.8 Hz2

29.23 0.8102 5 2.5

39.5

i S

1.53 10 s

1

10.4 MHz2

19.0 mA/V

0.026

100 0.026

5.25 k0.495

1

161 MHz2

i m

1

10.7 MHz2

Since 3 freq dominated by

( )

20 0.7

1.93 mA10

1

56.3 MHz2

i m

1

3.75 MHz2

Since , 3 frequency is dominated by

17.9 MHz2

4 0.51.810.492

T P

D D

EX8.6 Computer Analysis

EX8.7 No Exercise Problem

( )( ) ( )( )

1 1

0.7 1

1.53 18 0.225 1.53 0.7 26.4 kΩ26.4

I V

0.0260.6225

5 10 exp

0.02612.5 mA

65.2 1.59 mA

411.59 0.244 1.35 mA

max

0.6 A max 12 0.6 7.2 Watts2

If β4 and β are large, then 5 I E ≅β β β3 4 5I B3

So that composite current gain is β β β β≅ 3 4 5 www.elsolucionario.net

0.625 A40

80 40

64 0.625

D D DS T

D

D S

V V P

1.671.67 20.8 mA

80

24 0.7

1.12 k20.8

1.67 7.2

4620.026

0.1

250 mA

25 12.5

125 mA0.1

21.56 W P125

1.25 mA

100

25 0.7

19.4 k1.25

Point (b): Maximum power delivered to load

Point (a): Will obtain maximum signal current output

Point (c): Will obtain maximum signal voltage output

3.81 A, 7.16 W

D S

D S D

P

V V

( )( ) ( )( )( )

( )

0

0

2 2

( )

case amb case amb

P P

( )( ) ( )

V V

L L

V P

R

R R

V P

R V

V V

80 mA

80 10

ln 0.026 ln

5 100.6708 V

0.5987 0.6708 8 8.072 V2

0.5265

0.311 mA0.026

1 V cutoff 0

BB

BB SGp o I

Approximation for i D is okay

Diodes and transistors matched ⇒i N = =i P 545 mA

o L

o I

I

v i

Dn BB

n

I V

We can write g m =2 K I n DQ =190 mA/V

This is the required transconductance for the output transistor This implies a very large transistor

10.15 10 0.7 101 0.15 0.1

1.55 101.55 0.736.80 V0.73 1.55 0.496 k

( )( )

5.77 72 415

415 0.017 7.06 V7.06 2.35 V

( )

2 1

2

1

2 1

2

11

π

π

π π

n

π

ββ

⎟

′+

′+ +

R R

3

6149.18 41.11 8.07 A

3 3

3

3 1

30 10

ln 0.026 ln

5 100.6453 V

2.508 10

ln 0.026 ln

5 100.5807 V

0 3 1

0

6 0.6453 0.58076.0646 V

Voltage gain

6

0.9896.0646

50 0.026Now 0.4545

2.86

120 0.026

10.91 0.286

So

10.910.4545 32.5

π

π

ββ

Because of rπ1 and Z, neglect effect of r0 Then neglecting r01, r02 and r03, we find

3 3

0.4995 10

10 0.58062

10

100100

101.2694 0.7184 0.55099 V

0.55099

0.02610

10 10

2 100.2

2 100.5 10

20

max 1 A1

11

50.00272

60.002267 A

C E C

i i i

R A

CL

d

R A

280 80 k

3.5

280

28 k10

R

R = = and

1

20 210

cm d

dB

A

C M R R

A A

10 2

50 A160

b a I I

a

b a a

b a I I

R

R R

v v

R

R R

v v

Then we want 3

7 1.45

For common-mode input v I2=v I1⇒v0 = ⇒0 Common Gain 0, = C M R R= ∞

b A d(min)⇒R2′ min, maxR1

Let R3 = R4 so the difference amplifier gain is unity

Minimum Gain ⇒ Maximum R1

v

δδδδ

d

v A

1

12.8

R

R =Let R2 =150 k , Ω R1=11.7 kΩ

3 2

( )

15

151

150.2

10 A20

1.2

80 A15

1st op-amp: 90 Aμ into output terminal

2nd op-amp: 80 Aμ out of output terminal

v

od v

( ) ( ) ( )

od od

A

A

A

A A

125

5 103.9960

7.80 A6.41

v v

0 0

I

F I

od I

od

v

A v

v 0 must remain within the bias voltages of the op-amp; the larger the R2, the smaller the range of input

voltage v I in which the output is valid

9.54

(a)

Then 1.19

1For example, 119 K, 100

v x

2.50 2.40909

0.0045455 mA20

47.5 1 4.52381

9.63

0 1

( )

( ) ( )

3sin A

0.0030 sin 115 0.50 0.030 sin0.50 0.375sin

0.50 0.030 sin 0.003sin 1150.50 0.375sin

1

21

2 1575

10 102

2 1575

333

16.6520

750.604 mA

μμβ

150 0.026

156 kΩ0.025

0.176 A17,040

01

So

1110.90

11

10 3 7 2

V V

1 2 1

2

DSS DS

GS DSS

P GS

2 =0, 1 = −

1 1

X

m X D

m X

D

m

D

D m

R g

0.15 0.65

0.721 0.915 1.09 kΩ10.3 10.3

r

I

R R

T REF

EC AP CEo

80 100 kΩ0.8

100

139.5 kΩ0.717

0.0287 mA139.5

1 1

1 0.01 26

12 1 0.01 3.93

DS REF

+

++

2 1

5 5 1 1.787 5 54.7 kΩ

D

m DSS GS

( )

2

0 0

exp

EC CE

T

I T I T

V V

n I TN v

160 kΩ0.5

80

80 K1

9.615 100.026 ln

100.5378 V

604.03 A

BE REF

Advantage: Requires smaller resistance

(c) For part (a):

80 40 kΩ2

400.2325

1

5 0.7 4.3

20.5 K0.21

0.239 0.230 mA

21

1.3 0.00597 mA 0.236 mA217

0 2

2 0

and

12

2

21

0.70 1

80 81 81 100.700216 0.000864

2For 0.8 mA

The analysis is exactly the same as in the text We have

0

121

1 mA, 0.0133 mA

750.0133 0.0267 0.04 mA0.04

100 400

20 M2

o O

0.026 0.4624

1.50.46240.01733ln

( ) ( )

0.0068

0.2615 mA/V0.026

80 0.026

306 k0.0068

ln0.026ln 0.50

E

I

R R

I

I R

O

O O

O

V I

R

I

I

μΔ

I

I

I I

100.7 0.026 ln 2.03 10 A

0.3174 mA

BE REF