Exercise Solns Chapter5

7/23/2019 Exercise Solns Chapter5

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R. C. Jaeger and T. N. Blalock

08/09/10

1

Microelectronic Circuit DesignFourth Edition

Solutions to ExercisesCHAPTER 5

Page 223

a( ) "F =#

F

1$#F

=

0.970

1$ 0.970= 32.3 | "

F=

0.993

1$ 0.993=142 | "

F=

0.250

1$ .250= 0.333

b( ) #F ="F

"F+1

=

40

41= 0.976 | #

F=

200

201= 0.995 | #

F=

3

4= 0.750

Page 225

iC=10"15A exp

0.700

0.025

#

$%

&

'(" exp

"9.30

0.025

#

$%

&

'(

)

*+

,

-."

10"15A

0.5exp

"9.30

0.025

#

$%

&

'("1

)

*+

,

-.=1.45 mA

iE=10

"15A exp

0.700

0.025

#

$%

&

'(" exp

"9.30

0.025

#

$%

&

'(

)

*+

,

-.+

10"15A

100exp

0.700

0.025

#

$%

&

'("1

)

*+

,

-.=1.46 mA

iB=

10"15A

100exp

0.700

0.025

#

$%

&

'("1

)

*+

,

-.+

10"15A

0.5exp

"9.30

0.025

#

$%

&

'("1

)

*+

,

-.=14.5 A


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Page 227

iC=10

"16A exp

0.750

0.025

#

$%

&

'(" exp

0.700

0.025

#

$%

&

'(

)

*+

,

-."

10"16A

0.4exp

0.700

0.025

#

$%

&

'("1

)

*+

,

-.= 563 A

iE=10"16A exp0.750

0.025#

$% &

'(" exp0.700

0.025#

$% &

'()

*+ ,

-.+ 10

"16

A

75exp0.750

0.025#

$% &

'("1)

*+ ,

-.= 938 A

iB=

10"16A

75exp

0.750

0.025

#

$%

&

'("1

)

*+

,

-.+

10"16A

0.4exp

0.700

0.025

#

$%

&

'("1

)

*+

,

-.= 376 A

" " "

iT=10"15A exp

0.750

0.025

#

$%

&

'(" exp

"2

0.025

#

$%

&

'(

)

*+

,

-.=10.7 mA

iT=10

"16A exp

0.750

0.025

#

$%

&

'(" exp

"4.25

0.025

#

$%

&

'(

)

*+

,

-.=1.07 mA


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Page 230

VBE

=VTln

IC

IS

+1"

#$

%

&'= 0.025Vln

10(4A

10(16A

+1

"

#$

%

&'= 0.691 V

VBE

=VTln

IC

IS

+1"

#$

%

&'= 0.025Vln

10(3A

10(16

A

+1

"

#$

%

&'= 0.748 V

Page 231

npn :VBE > 0, VBC < 0" Forward #Active Region | pnp :VEB > 0, VCB > 0" Saturation Region

Page 233

"F =#F

1$#F=

0.95

0.05=19 "R =

#R

1$#R=

0.25

0.75=

1

3

VBE = 0, VBC


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Page 236

a( )The currents do not depend upon VCCas long as the collector - base junction is reversebiased by more than 0.1 V. (Later when Early voltage V

Ais discussed, one should revisit

this problem.)

b( )Forward - active region : IB =100 A | IE = "F +1( )IB = 5.10 mA | IC = "FIB = 5.00 mA

VBE

=VT

ln I

C

IS

+1#

$%

&

'(= 0.025Vln

5.00mA

10)16A

+1#

$%

&

'(= 0.789 V | Checking : VBC = )5+ 0.789 = )4.21

) ) )

Forward - active region with VCB

* 0 requires VCC

*VBE

or VCC

* 0.764 V

Page 238

a( )Resistor R is changed.

IE ="0.7V " "9V

( )5.6k# =1.48 mA | IB =IE

$F +1=

IE

51= 29.1 A | IC = $FIB = 50IB =1.45 mA

VCE =VC "VE = 9 " 4300IC( ) " "0.7( ) = 3.47 V | Q -Po int : 1.45 mA, 3.47 V( )

b( )IE =$F +1

$FIC =

51

50100A =102A | R =

"0.7V " "9V( )102A

=

8.3V

102A= 81.4 k#

The nearest 5% value is 82 k#.

Page 239

IE ="0.7V" "9V( )

5.6k#

=1.48 mA | IB =IE

$F +1

=

IE

50

= 29.1 A | IC = $FIB = 50IB =1.45 mA

" " "

IE =$F +1

$FIC =

51

50IC =1.02IC VBE =VT ln

IC

IS+1

%

&'

(

)*= 0.025ln 2x10

15IC +1( )

VBE+ 8200 1.02 5x10"16( )exp

VBE

0.025

%

&'

(

)*"1

+

,-

.

/0= 91VBE = 0.7079 V using a calculator solver

or spreadsheet. IC = 5x10"16 exp

0.7079

0.025

%

&'

(

)*= 992 A | VCE = 9 " 4300IC" "0.708( ) = 5.44 V

" " "ISD =

ISBJT2F

=

2x10"14A

0.95= 21.0fA


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Page 242

Resistor R is changed :

"IC=

"0.7V " "9V( )5.6k#

=1.48 mA | IB=

"IC

$R+1

=

"IC

2= 0.741 mA | "I

E= $

RIB= 1( )IB = 0.741 mA

Page 244

VCESAT

= 0.025V( ) ln1

0.5

"

#$

%

&'

1+1mA

2 40A( )

1(1mA

50 40A( )

)

*

++++

,

-

.

.

.

.

= 99.7 mV

( ( (

VBESAT

= 0.025V( ) ln0.1mA+ 1( 0.5( )1mA

10(15A 150

+1( 0.5"

#$ %

&'

)

*

++

++

,

-

.

.

..

= 0.694 mV

VBCSAT

= 0.025V( ) ln0.1mA(

1mA

50

10(15A

1

0.5

"

#$

%

&'

1

50+1( 0.5

"

#$

%

&'

)

*

++++

,

-

.

.

.

.

= 0.627 mV | VCESAT

=VBESAT

(VBCSAT

= 67.7mV

Page 247

a( )Dn =kT

q

n = 0.025V 500cm2/V"s( ) =12.5cm2 /s

b( )IS =qADnni

2

NABW=

1.6x10"19C50m

2( ) 10"4cm /m( ) 12.5cm2 /s( ) 1020 / cm6( )10

18/ cm3( ) 1m( )

=10"18A =1 aA

Page 250

VT =1.38x10"

23J/K( ) 373K( )

1.60x10"19C

= 32.2mV | CD =IC

VT#F =

10A

0.0322V4x10"9s( ) =1.24 F

"""

f$ =fT

$F=

300MHz

125= 2.40MHz


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Page 251

IC=10

"15Aexp

0.7

0.025

#

$%

&

'(1+

10

50

#

$%

&

'(=1.74 mA | )F = 75 1+

10

50

#

$%

&

'(= 90.0 | IB =

1.74mA

90.0=19.3 A

IC=10"15Aexp

0.7

0.025

#

$%

&

'(=1.45 mA | )F = 75 | IB =

1.45mA

75

=19.3 A

Page 253

gm =40

V10

"4A( ) = 4.00 mS | gm =

40

V10

"3A( ) = 40.0 mS

CD = 4.00mS 25ps( ) = 0.100pF | CD = 40.0mS 25ps( ) =1.00pF

Page 256

VT =kT

q=

1.38x10"23

300( )1.60x10

"19= 25.9 mV | IS=

IC

exp

VBEVT

#

$%

&

'(

=

350A

exp

0.68

0.0259

#

$%

&

'(

=1.39fA

BF = 80 | VAF = 70 V

Page 260

VEQ=18k"

18k"+ 36k"12V= 4.00 V | REQ= 18k" 36k" =12 k"

IB=4.00 # 0.7

12 + 75+1( )16V

k"= 2.687 A | IC= 75IB= 202 A | IE= 76IB= 204 A

VCE= 12# 22000IC#16000IE= 4.29 V | Q -po int : 202 A, 4.29 V( )###

VEQ=180k"

180k"+ 360k"12V= 4.00 V | REQ=180k" 360k" =120 k"

IB=4.00 # 0.7

120 + 75+1( )16V

k"= 2.470 A | IC= 75IB= 185 A | IE= 76IB= 188 A

VCE= 12# 22000IC#16000IE= 4.29 V | Q -po int : 185 A, 4.93 V( )


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Page 261

I2 =IC

5=

50IB

5=10IB

"""

VEQ =18k#

18k#+ 36k#12V= 4.00 V | REQ =18k# 36k# =12 k#

IB =4.00 "0.7

12 + 500+1( )16V

k#= 0.4111 A | IC = 500IB = 205.6 A | IE = 76IB = 206.0 A

VCE =12" 22000IC"16000IE = 4.18 V | Q- point : 206 A, 4.18 V( )

Page 262

The voltages all remain the same, and the currents are reduced by a factor of 10. Hence all

the resistors are just scaled up by a factor of 10.

120 k"#1.2 M" 82 k"# 820 k" 6.8 k"# 68 k"

Page 264

VCE

= 0.7 Vat the edge of saturation. 12V- RC+

76

7516k"

#

$%

&

'(205A( )) 0.7V*RC+ 38.9 k"

, , ,

VBESAT

= 4 ,12k" 24A( ),16k" 184A( ) = 0.768 V

VCESAT

=12 , 56k" 160A( ),16k" 184A( ) = 0.096 V

Page 265

IB=

9" 0.7

36+ 50+1( )1V

k#= 95.4 A | I

C= 50I

B= 4.77 mA | I

E= 51I

B= 4.87 mA

VCE= 9"1000 I

C+I

B( ) = 4.13 V | Q- point : 4.77 mA, 4.13 V( )

Page 266

VBE (V) IC (A) V'BE (V)

0.70000 2.0155E-04 0.67156

0.67156 2.0328E-04 0.67178

0.67178 2.0327E-04 0.671780.67178 2.0327E-04 0.67178

Exercise Solns Chapter5

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