The document contains solutions to 20 example problems (labeled EX5.1 through EX5.20) relating to bipolar junction transistors. Each problem involves calculating various transistor parameters such as bias currents, voltages, and resistor values given information such as supply voltages, transistor betas, and resistor/collector resistor values. The problems cover topics such as calculating bias points, determining operating modes, and designing transistor circuits.

1. Chapter 5
Exercise Problems
EX5.1
α
β=
1−α
0.980
For α = 0.980, β = = 49
1 − 0.980
0.995
For α = 0.995, β = = 199
1 − 0.995
So 49 ≤ β ≤ 199
EX5.2
BVCBO 200
BVCEO = = or BVCEO = 40.5 V
n β 3
120
EX5.3
V − VBE ( on ) 2 − 0.7
I B = BB = ⇒ 6.5 μ A
RB 200
I C = β I = (120 )( 6.5 μ A ) ⇒ I C = 0.78 mA
VCE = VCC − I C RC = 5 − ( 0.78 )( 4 ) or VCE = 1.88 V
P = I BVBE ( on ) + I CVCE = ( 0.0065 )( 0.7 ) + ( 0.78 )(1.88 ) ≅ 1.47 mW
EX5.4
V + − VEB ( on ) − VBB 5 − 0.7 − 2.8
IB = = or I B = 4.62 μ A
RB 325
I C = β I B = ( 80 )( 4.615 ) ⇒ I C = 0.369 mA
VEC = 2 = 5 − ( 0.369 ) RC which yields RC = 8.13 k Ω
EX5.5
VBB − VBE ( on ) 2 − 0.7
(a) IB = = ⇒ 5.91 μ A
RB 220
I C = β I B = (100 )( 5.91 μ A ) ⇒ 0.591 mA
I E = (1 + β ) I B = 0.597 mA
VCE = VCC − I C RC = 10 − ( 0.591)( 4 ) or VCE = 7.64 V
6.5 − 0.7
(b) IB = ⇒ 26.4 μ A
220
Transistor is biased in saturation mode, so
VCE = VCE ( sat ) = 0.2 V
VCC − VCE ( sat ) 10 − 0.2
IC = = or I C = 2.45 mA
RC 4
I E = I C + I B = 2.45 + 0.0264 ≅ 2.48 mA
EX5.6
For 0 ≤ VI < 0.7 V , Qn is cutoff, VO = 9 V
(100 )(VI − 0.7 )( 4 )
When Qn is biased in saturation, we have 0.2 = 9 − ⇒ VI = 5.1 V
200
So, for VI ≥ 5.1 V , VO = 0.2 V
EX5.7

2. VBB − VBE ( on ) 8 − 0.7
IB = =
RB + (1 + β ) RE 30 + ( 76 )(1.2 )
or I B = 60.2 μ A
I C = β I B = ( 75 )( 60.23 μ A ) ⇒ 4.52 mA
I E = (1 + β ) I B = 4.58 mA
VCE = VCC − I C RC − I E RE
= 12 − ( 4.517 )( 0.4 ) − ( 4.577 )(1.2 )
or VCE = 4.70 V
EX5.8
For VC = 4 V and ICQ = 1.5 mA,
V + − VC 10 − 4
RC = = ⇒ RC = 4 k Ω
I CQ 1.5
⎛ 1+ β ⎞ ⎛ 101 ⎞
IE = ⎜ ⎟ IC = ⎜ ⎟ (1.5 ) = 1.515 mA
⎝ β ⎠ ⎝ 100 ⎠
−V ( on ) − V −
also I E = BE
RE
−0.7 − ( −10 )
Then RE = ⇒ RE = 6.14 k Ω
1.515
EX5.9
3 − 0.7
I EQ = = 0.25
RE
RE = 9.2 kΩ
⎛ 75 ⎞
I CQ = ⎜ ⎟ ( 0.25 ) = 0.2467 mA
⎝ 76 ⎠
−0.7 + VCEQ + I CQ RC − 3 = 0
3 + 0.7 − 2
RC = ⇒ RC = 6.89 kΩ
0.2467
EX5.10
5 = I E RE + VEB ( on ) + I B RB − 2
⎛ 180 ⎞
(a) 5 + 2 − 0.7 = I E ⎜ 2 + ⎟ I E = 0.9859 mA
⎝ 41 ⎠
I C = 0.962 mA
⎛ 180 ⎞
(b) 6.3 = I E ⎜ 2 + ⎟ I E = 1.2725 mA
⎝ 61 ⎠
I C = 1.25 mA
⎛ 180 ⎞
(c) 6.3 = I E ⎜ 2 + ⎟ I E = 1.6657 mA
⎝ 101 ⎠
I C = 1.64 mA
⎛ 180 ⎞
(d) 6.3 = I E ⎜ 2 + ⎟ I E = 1.97365 mA
⎝ 151 ⎠
I C = 1.94 mA
EX5.11

3. VBB − VEB ( on ) 4 − 0.7
IE = ⇒ RE =
RE 1.0
or RE = 3.3 k Ω
I C = α I E = ( 0.992 )(1) = 0.992 mA
I B = I E − I C = 1.0 − 0.992 or I B = 8 μ A
VCB = I C RC − VCC = ( 0.992 )(1) − 5
or VCB = −4.01 V
EX5.12
V + − (Vγ + VCE ( sat ) ) 5 − (1.5 + 0.2 )
R= =
IC 15
or R = 220 Ω
15
IB = = 1 mA
15
v − VBE ( on ) 5 − 0.7
RB = I = = 4.3 k Ω
IB 1
P = I BVBE (on) + I CVCE
= (1)( 0.7 ) + (15 )( 0.2 ) = 3.7 mW
EX5.13
(a) For V1 = V2 = 0, All currents are zero and VO = 5 V.
(b) For V1 = 5 V, V2 = 0; IB2 = IC2 = 0,
5 − 0.7
I B1 = = 4.53 mA
0.95
5 − 0.2
I C1 = ⇒ I C1 = I R = 8 mA
0.6
VO = 0.2 V
(c) For V1 = V2 = 5 V, IB1 = IB2 = 4.53 mA; IR = 8 mA, IC1 = IC2 = IR/2 = 4 mA, VO = 0.2 V
EX5.14
vO = 5 − iC RC = 5 − β iB RC
ΔvO = − βΔiB RC
VBB + ΔvI − VBE ( on )
iB =
RB
ΔvI
ΔiB =
RB
ΔvO − β RC
Then =
ΔvI RB
Let β = 100, RC = 5 k Ω, RB = 100 k Ω
Δvo − (100 )( 5 )
So = = −5
ΔvI 100
Want Q-point to be vo ( Q − pt ) = 2.5 = 5 − (100 ) I BQ ( 5 )
VBB − 0.7
Then I BQ = 0.005 mA, I BQ = 0.005 =
100
so VBB = 1.2 V
Also, I CQ = β I BQ = (100 )( 0.005 ) = 0.5 mA
EX5.15

4. VCEQ = 2.5 = 5 − I CQ RC
5 − 2.5
or RC = = 10 k Ω
0.25
I CQ 0.25
I BQ = = = 0.002083 mA
β 120
5 − 0.7
Then RB = ⇒ RB = 2.06 M Ω
0.002083
EX5.16
(a) RTH = R1 R2 = 9 2.25 = 1.8 k Ω
⎛ R2 ⎞ ⎛ 2.25 ⎞
VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ ( 5)
⎝ R1 + R2 ⎠ ⎝ 9 + 2.25 ⎠
or VTH = 1.0 V
VTH − VBE ( on ) 1 − 0.7
(b) I BQ = =
RTH + (1 + β ) RE 1.8 + (151)( 0.2 )
or I BQ = 9.375 μ A
I CQ = β I BQ = (150 )( 9.375 μ A )
or I CQ = 1.41 mA
I EQ = (1 + β ) I BQ ⇒ I EQ = 1.42 mA
VCEQ = 5 − I CQ RC − I EQ RE
= 5 − (1.41)(1) − (1.42 )( 0.2 )
or VCEQ = 3.31 V
(c) For β = 75
1 − 0.7
I BQ = = 17.6 μ A
1.8 + ( 76 )( 0.2 )
I CQ = β I BQ = ( 75 )(17.6 μ A )
or I CQ = 1.32 mA
I EQ = (1 + β ) I BQ = ( 76 )(17.6 μ A )
or I EQ = 1.34 mA
VCEQ = 5 − (1.32 )(1) − (1.34 )( 0.2 )
or VCEQ = 3.41 V
EX5.17
VCEQ ≅ VCC − I CQ ( RC + RE )
or 2.5 ≅ 5 − I CQ (1 + 0.2 )
which yields
I CQ = 2.08 mA,
I CQ 2.08
I BQ = = = 0.0139 mA
β 150
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(151)( 0.2 )
or RTH = 3.02 k Ω
⎛ R2 ⎞ 1
Now VTH = ⎜ ⎟ ⋅ VCC = ⋅ RTH ⋅ VCC
⎝ R1 + R2 ⎠ R1
1
so VTH = ( 3.02 )( 5 )
R1
We can write VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE

5. 1
or ( 3.02 )( 5 ) = ( 0.0139 )( 3.02 ) + 0.7 + (151)( 0.0139 )( 0.2 )
R1
We obtain R1 = 13 k Ω and then R2 = 3.93 k Ω
EX5.18
β = 150, RTH = R1 || R2
5−0
I CQ = = 5 mA
1
I CQ 5
I BQ = = = 0.0333 mA
β 150
VTH − VBE ( on ) − ( −5 )
I BQ =
RTH + (1 + β ) RE
Set RTH = ( 0.1)(1 + β ) RE
⎛ R2 ⎞
We have VTH = ⎜ ⎟ (10 ) − 5
⎝ R1 + R2 ⎠
⎛ R2 ⎞
⎜ ⎟ (10 ) − 0.7
R + R2 ⎠
Then I BQ = 0.0333 = ⎝ 1
(1.1)(151)( 0.2 )
⎛ R2 ⎞
which yields ⎜ ⎟ = 0.1806
⎝ R1 + R2 ⎠
⎛ RR ⎞
Now RTH = ⎜ 1 2 ⎟ = ( 0.1)(151)( 0.2 ) = 3.02 k Ω
⎝ R1 + R2 ⎠
so R1 ( 0.1806 ) = 3.02 k Ω
We obtain R1 = 16.7 k Ω and R2 = 3.69 k Ω
EX5.19
V + − V − − VECQ 5 − ( −5 ) − 5
I CQ ≅ =
RC + RE 4.5 + 0.5
so I CQ = 1 mA, and
I CQ 1
I BQ = = = 0.00833 mA
β 120
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 0.5 )
or RTH = 6.05 k Ω
We can write V + = I EQ RE + VEB ( on ) + I BQ RTH + VTH
1
We have VTH = ⋅ RTH (10 ) − 5 and if we let I EQ ≅ I CQ = 1 mA,
R1
1
then we have 5 = (1)( 0.5 ) + 0.7 + ( 0.00833)( 6.05 ) + ( 6.05 )(10 ) − 5
R1
which yields R1 = 6.91 k Ω and R2 = 48.6 k Ω
EX5.20
⎛ 2⎞ ⎛ 2 ⎞
I1 = I Q ⎜ 1 + ⎟ = ( 0.25 ) ⎜1 + ⎟ = 0.2625 mA
⎝ β⎠ ⎝ 40 ⎠
0 − VBE ( on ) − V −
0 − 0.7 − ( −5 )
R1 = =
I1 0.2625
or R1 = 16.38 k Ω

6. For VCEO = 3 V , then VCO = 2.3 V
⎛ β ⎞ ⎛ 40 ⎞
I CO = ⎜ ⎟ ⋅ I Q = ⎜ ⎟ ( 0.25 ) = 0.2439 mA
⎝ 1+ β ⎠ ⎝ 41 ⎠
+
V − VCO 5 − 2.3
RC = = = 11.07 k Ω
I CO 0.2439
EX5.21
RTH 50 ||100 = 33.3 k Ω
⎛ 50 ⎞
VTH = VTH = ⎜ ⎟ (10 ) − 5 = −1.67 V
⎝ 50 + 100 ⎠
−1.67 − 0.7 − ( −5 )
I B1 = ⇒ 11.2 μ A
33.3 + (101)( 2 )
I C1 = 1.12 mA, I E1 = 1.13 mA
VE1 = I E1 RE1 − 5 = (1.13)( 2 ) − 5 = −2.74 V
VCE1 = 3.25 V ⇒ VC1 = 0.51 V
Now VE 2 = 0.51 + 0.7 = 1.21 V
5 − 1.21
IE2 = = 1.90 mA ⇒ I B 2 = 18.8 μ A
2
IC 2 = 1.88 mA
I R1 = I C1 − I B 2 = 1.12 − 0.0188 = 1.10 mA
5 − 0.51
RC1 = = 4.08 k Ω
1.10
VEC 2 = 2.5 ⇒ VC 2 = VE 2 − VEC 2
= 1.21 − 2.5 = −1.29 V
−1.29 − ( −5 )
RC 2 = = 1.97 k Ω
1.88
EX5.22
12
We find = 240 k Ω = R1 + R2 + R3
0.05
Then VB1 = ( 0.5)( 2 ) + 0.7 = 1.7 V
1.7
R3 = = 34 k Ω
0.05
Also VB 2 = ( 0.5 )( 2 ) + 4 + 0.7 = 5.7 V
ΔVR 2 = 5.7 − 1.7 = 4 V
4
so R2 = = 80 k Ω
0.05
and R1 = 240 − 80 − 34 = 126 k Ω
VC 2 = 1 + 4 + 4 = 9 V
V + − VC 2 12 − 9
Then RC = = = 6 kΩ
I CQ 0.5
Test Your Understanding Exercises
TYU5.1
β
α=
1+ β
75
For β = 75, α = = 0.9868
76

7. 125
For β = 125, α = = 0.9921
126
TYU5.2
I E = (1 + β ) I B
IE 0.780
so 1 + β = = = 81.25 then β = 80.3
I B 0.00960
Now
β 80.25
α= = = 0.9877
1+ β 81.25
I C = α I E = ( 0.9877 )( 0.78 ) = 0.770 mA
TYU5.3
α 0.990
β= = = 99
1 − α 1 − 0.990
IE 2.15
Now I B = = ⇒ 21.5 μ A and I C = α I E = ( 0.990 )( 2.15 ) = 2.13 mA
1 + β 100
TYU5.4
V 150
ro = A =
IC IC
For I C = 0.1 mA ⇒ ro = 1.5 M Ω
For I C = 1.0 mA ⇒ ro = 150 k Ω
For I C = 10 mA ⇒ ro = 15 k Ω
TYU5.5
⎛ V ⎞
I C = I O ⎜1 + CE ⎟
⎝ VA ⎠
At VCE = 1 V , I C = 1 mA
⎛ 1 ⎞
(a) For VA = 75 V , I C = 1 = I O ⎜1 + ⎟ ⇒ I O = 0.9868 mA
⎝ 75 ⎠
Then, at VCE = 10 V
⎛ 10 ⎞
I C = ( 0.9868 ) ⎜ 1 + ⎟ = 1.12 mA
⎝ 75 ⎠
⎛ 1 ⎞
(b) For VA = 150 V , I C = 1 = I O ⎜ 1 + ⎟ ⇒ I O = 0.9934 mA
⎝ 150 ⎠
⎛ 10 ⎞
At VCE = 10 V , I C = ( 0.9934 ) ⎜ 1 + ⎟ = 1.06 mA
⎝ 150 ⎠
TYU5.6
BVCBO
BVCEO = so BVCBO = 3 100 ( 30 ) = 139 V
n β
TYU5.7
(a) For VI = 0.2 V < VBE ( on ) ⇒ I B = I C = 0, VO = 5 V and P = 0
(b) For VI = 3.6 V , transistor is driven into saturation, so
VI − VBE ( on ) 3.6 − 0.7 V + − VCE ( sat ) 5 − 0.2
IB = = = 4.53 mA and I C = = = 10.9 mA
RB 0.64 RC 0.440

8. I C 10.9
Note that = = 2.41 < β which shows that the transistor is indeed driven into saturation. Now,
I B 4.53
P = I BVBE ( on ) + I CVCE ( sat )
= ( 4.53)( 0.7 ) + (10.9 )( 0.2 ) = 5.35 mW
TYU5.8
For VBC = 0 ⇒ VO = 0.7 V
5 − 0.7 I 9.77
Then I C = = 9.77 mA and I B = C = = 0.195 mA
0.44 β 50
Now VI = I B RB + VBE ( on ) = ( 0.195 )( 0.64 ) + 0.7 or VI = 0.825 V
Also P = I BVBE ( on ) + I CVCE
= ( 0.195 )( 0.7 ) + ( 9.77 )( 0.7 ) = 6.98 mW
TYU5.9
V + − VC 10 − 6.34
IC = = = 0.915 mA
RC 4
−VBE ( on ) − V − −0.7 − ( −10 )
And I E = = or IE = 0.930
RE 10
I C 0.915 α 0.9839
Now α = = = 0.9839 and β = = = 61
I E 0.930 1 − α 1 − 0.9839
Also I B = I E − I C = 0.930 − 0.915 ⇒ 15 μ A and VCE = VC − VE = 6.34 − ( −0.7 ) = 7.04 V
TYU5.10
10 − 0.7
IE = = 1.16 mA
8
1.1625
IB = = 22.8 μ A
51
I C = ( 50 )( 22.8 μ A ) = 1.14 mA
VE = V + − I E RE = 10 − (1.1625 )( 8 ) = 0.7 V
VC = I C RC − 10 = (1.1397 )( 4 ) − 10 = −5.44
VEC = 0.7 − ( −5.44 )
VEC = 6.14 V
TYU5.11
VBB = I B RB + VBE ( on ) + I E RE
or VBB = I B RB + VBE ( on ) + (1 + β ) I B RE
VBB − VBE ( on ) 2 − 0.7
Then I B = =
RB + (1 + β ) RE 10 + ( 76 )(1)
or I B = 15.1 μ A
Also I C = ( 75 )(15.1 μ A ) = 1.13 mA and I E = ( 76 )(15.1 μ A ) = 1.15 mA
Now VCE = VCC + VBB − I C RC − I E RE
= 8 + 2 − (1.13)( 2.5 ) − (1.15 )(1) = 6.03 V
TYU5.12
VCE = 2.5 V ⇒ VE = 2.5 V = I E RE
We have VBB = I B RB + VBE ( on ) + VE

9. VBB − VBE ( on ) − VE 5 − 0.7 − 2.5
so I B = =
RB 10
or I B = 0.18 mA
Then I E = (101)( 0.18 ) = 18.2 mA
2.5
And RE = = 0.138 k Ω ⇒ 138 Ω
18.18
TYU5.13
VBB = I E RE + VEB ( on ) + I B RB
2.2
I E = 2.2 mA ⇒ I B = = 0.0431 mA
51
⎛ β ⎞ ⎛ 50 ⎞
and I C = ⎜ ⎟ ⋅ I E = ⎜ ⎟ ( 2.2 ) = 2.16 mA
⎝ 1+ β ⎠ ⎝ 51 ⎠
Then VBB = ( 2.2 )(1) + 0.7 + ( 0.0431)( 50 )
or VBB = 5.06 V
Now VEC = 5 − I E RE = 5 − ( 2.2 )(1) = 2.8 V
TYU5.14
(a) For vI = 0, iB = iC = 0, vO = 12 V , P = 0
vI − VBE ( on ) 12 − 0.7
(b) For vI = 12 V , iB = = = 47.1 mA
RB 0.24
VCC − VCE ( sat ) 12 − 0.1
iC = = = 2.38 A
RC 5
vO = 0.1 V
and
P = iBVBE ( on ) + iCVCE ( sat )
= ( 0.0471)( 0.7 ) + ( 2.38 )( 0.1) = 0.271 W
TYU5.15
5 − 2.5
(a) For VCEQ = 2.5 V , I CQ = = 1.25 mA
2
I CQ 1.25
I BQ = = ⇒ I BQ = 12.5 μ A
β 100
5 − 0.7
Then RB = = 344 k Ω
0.0125
(b) IBQ is independent of β .
5 −1
For VCEQ = 1 V , I C = = 2 mA
2
IC 2
β= = ⇒ β = 160
I B 0.0125
5−4
For VCEQ = 4 V , I C = = 0.5 mA
2
IC 0.5
β= = ⇒ β = 40
I B 0.0125
So 40 ≤ β ≤ 160
TYU5.16
5 − 0.7
I BQ = = 0.005375 mA
800

10. For β = 75, I CQ = β I BQ = ( 75 )( 0.005375 )
Or I CQ = 0.403 mA
For β = 150, I CQ = (150 )( 0.005375 )
Or I CQ = 0.806 mA
Largest I CQ ⇒ Smallest VCEQ
5 −1
For β = 150, RC = = 4.96 k Ω
0.806
5−4
For β = 75, RC = = 2.48 k Ω
0.403
5 − 2.5
For a nominal I CQ = 0.604 mA and VCEQ = 2.5 V , RC = = 4.14 k Ω
0.604
Now for I CQ = 0.403 mA, VCEQ = 5 − ( 0.403)( 4.14 ) = 3.33 V
For I CQ = 0.806 mA, VCEQ = 5 − ( 0.806 )( 4.14 ) = 1.66 V
So, for RC = 4.14 k Ω, 1.66 ≤ VCEQ ≤ 3.33 V
TYU5.17
(a)
⎛ β ⎞ ⎛ 100 ⎞
I CQ = ⎜ ⎟ ⋅ I EQ = ⎜ ⎟ (1) = 0.99 mA
⎝1+ β ⎠ ⎝ 101 ⎠
I EQ 1
I BQ = = ⇒ 9.90 μ A
1 + β 101
VB = − I BQ RB = − ( 0.0099 )( 50 )
or VB = −0.495 V
⎛ I CQ ⎞ ⎛ 0.99 × 10−3 ⎞
VBE = VT ln ⎜ ⎟ = ( 0.026 ) ln ⎜ −14 ⎟
⎝ IS ⎠ ⎝ 3 × 10 ⎠
or VBE = 0.630 V
Then VE = VB − VBE = −0.495 − 0.630 = −1.13 V
VC = 10 − ( 0.99 )( 5 ) = 5.05 V
Then VCEQ = VC − VE = 5.05 − ( −1.13) = 6.18 V
(b)
I EQ 1
I EQ = 1 mA, I BQ = = = 0.0196 mA
1 + β 51
VB = − ( 0.0196 )( 50 ) = −0.98 V
⎛ β ⎞ ⎛ 50 ⎞
I CQ = ⎜ ⎟ ⋅ I EQ = ⎜ ⎟ (1) = 0.98 mA
⎝1+ β ⎠ ⎝ 51 ⎠
⎛ 0.98 × 10−3 ⎞
VBE = ( 0.026 ) ln ⎜ −14 ⎟
= 0.629 V
⎝ 3 × 10 ⎠
VE = −0.98 − 0.629 = −1.61 V
VC = 10 − ( 0.98 )( 5 ) = 5.1 V
VCEQ = VC − VE = 5.1 − ( −1.61) = 6.71 V
TYU5.18
IQ IQ
IB = =
1 + β 121
and

11. ⎛ IQ ⎞
VB = ⎜ ⎟ ( 20 ) = I Q ( 0.165 )
⎝ 121 ⎠
VE = I Q ( 0.165 ) + 0.7
⎛ β ⎞ ⎛ 120 ⎞
I CQ = ⎜ ⎟ ⋅ I EQ = ⎜ ⎟ ⋅ I Q = ( 0.992 ) I Q
⎝ 1+ β ⎠ ⎝ 121 ⎠
VC = I CQ RC − 5 = ( 0.992 ) I Q ( 4 ) − 5
= 3.97 I Q − 5
VECQ = VE − VC
= ⎡ I Q ( 0.165 ) + 0.7 ⎤ − ⎡ I Q ( 3.97 ) − 5⎤
⎣ ⎦ ⎣ ⎦
= −3.805 I Q + 5.7
Then 3 = 5.7 − 3.805I Q
which yields I Q = 0.710 mA