7/23/2019 Exercise Solns Chapter5
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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
7/23/2019 Exercise Solns Chapter5
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R. C. Jaeger and T. N. Blalock
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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
#
$%
&
'(
)
*+
,
-.=1.07 mA
7/23/2019 Exercise Solns Chapter5
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R. C. Jaeger and T. N. Blalock
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Page 230
VBE
=VTln
IC
IS
+1"
#$
%
&'= 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
7/23/2019 Exercise Solns Chapter5
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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