EE 321 Chapter 10 Solutions

Solution of Homework problems 2 in Section 10.2

Chapter 10, Solution 1.

Known quantities:

Transistor diagrams, as shown in Figure P10.1:

(a) pnp, VEB = 0.6 V and VEC = 4.0 V

(b) npn, VCB = 0.7 V and VCE = 0.2 V

(c) npn, VBE = 0.7 V and VCE = 0.3 V

(d) pnp, VBC = 0.6 V and VEC = 5.4 V

Find:

For each transistor shown in Figure P10.1, determine

whether the BE and BC junctions are forward or

reverse biased, and determine the operating region.

Analysis:

(a) VBE = - 0.6 V for a pnp transistor implies that the BE junction is forward-biased.

VBC = VEC - VEB = 3.4 V. The CB junction is reverse-biased. Therefore, the transistor is in the

active region.

(b) VBC = - 0.7 V for a npn transistor implies that the CB junction is reverse-biased.

VBE = VBC - VEC = -0.5 V. The BE junction is reverse-biased. Therefore, the transistor is in the

cutoff region.

(c) VBE = 0.7 V for a npn transistor implies that the BE junction is forward-biased.

VBC = VEC - VEB = 0.4 V. The CB junction is forward-biased. Therefore, the transistor is in the

saturation region.

(d) VBC = 0.6 V for a pnp transistor implies that the CB junction is reverse-biased.

VBE = VBC – VEC = - 4.8 V. The BE junction is forward-biased. Therefore, the transistor is in

the active region.


Known quantities:

Transistor type and operating characteristics:

a) npn, VBE = 0.8 V and VCE = 0.4 V

b) npn, VCB = 1.4 V and VCE = 2.1 V

c) pnp, VCB = 0.9 V and VCE = 0.4 V

d) npn, VBE = - 1.2 V and VCB = 0.6 V

Find:

The region of operation for each transistor.

Analysis:

a) Since VBE = 0.8 V, the BE junction is forward-biased. VCB = VCE + VEB = - 0.4 V. Thus,

the CB junction is forward-biased. Therefore, the transistor is in the saturation region.

b) VBE = VBC + VCE = 0.7 V. The BE junction is forward-biased.

VCB = 1.4 V. The CB junction is reverse-biased. Therefore, the transistor is in the active

region.

c) VCB = 0.9 V for a pnp transistor implies that the CB junction is forward-biased.

VBE = VBC – VCE = - 1.3 V. The BE junction is forward-biased. Therefore, the transistor is in

the saturation region.

d) With VBE = - 1.2 V, the BE junction is reverse-biased.

VCB = - 0.6 V. The CB junction is reverse-biased. Therefore, the transistor is in the cutoff

region.


Known quantities:

The circuit of Figure P10.3: 100B

C

I

I .

Find:

The operating point and the state of the transistor.

Analysis:

V 6.0BEV and the BE junction is forward biased.

AVV

I

IIIIIVIV

BECC

B

BBCEEBEBCC

5.12911910

6.012

10191010820

101&91010820

3

3

mAIIBC

25.1

Writing KVL around the right-hand side of the circuit:

0 EECECCCC RIVRIV

V RIIRIVVEBCCCCCCE

1.8)910.0)(0125.025.1()2.2)(25.1(12

V VVVCEBEBC

5.71.86.0 : the BC junction is reverse biased

BECE VV

The transistor is in the active region.


Known quantities:

The magnitude of a pnp transistor's emitter and base current, and the magnitudes of the voltages

across the emitter-base and collector-base junctions:

IE = 6 mA, IB = 0.1 mA and VEB = 0.65 V, VBC = 7.3 V.

Find:

a) VCE.

b) IC.

c) The total power dissipated in the transistor, defined as BEBCEC

IVIVP .

Analysis:

a) VEC = VBC + VEB = 7.3 + 0.65 = 7.95 V.

b) IC = IE - IB = 6 - 0.1 = 5.9 mA.

c) The total power dissipated in the transistor can be found to be:

mW IVIVPBEBCEC

97.46101.065.0109.595.7 33


Known quantities:

The circuit of Figure P10.5, assuming the BJT has

V = 0.6 V.

Find: Change 15 V to 15 V

The emitter current and the collector-base voltage.

Analysis:

Applying KVL to the right-hand side of the circuit, A V

I BE

E480

30000

6.015

30000

15

Then, on the left-hand side, assuming >> 1:

V RIV

VRI

CCCB

CBCC

8.21015104801010

010

36


Known quantities:


V 6.0BEV and =150.

Find:

The operating point and the region in which the

transistor operates.

Analysis:

Define

RC 3.3 k, RE 1.2 k, R1 62 k, R2 15 k, VCC 18 V

By applying Thevenin’s theorem from base and mass, we have

VIRIRVV

mAII

μARR

VVI

VVRR

RV

kΩRRR

EECCCCCE

BC

EB

BEBB

B

CCBB

B

857.7101515112001025.2330018

25.2

15)1(

5.3

078.12||

63

21

2

21

From the value of VCE it is clear that the BJT is in the active region.


Known quantities:


V6.0V .

Find:

The emitter current and the collector-base voltage.

Analysis:

Applying KVL to the right-hand side of the circuit,

0EBEEBB

VRIV

μAR

VVI

E

EBBB

E4.497

1039

6.0203

. Since 1 , μA4.497 EC II

VCC = 20V VBB = 20V

Applying KVL to the left-hand side: 0CCCCCB

VRIV

VVRIVCCCCCB

05.102010204.497 3


Known quantities:

The collector characteristics for a certain transistor,

as shown in Figure P10.9.

Find:

a) The ratio IC/IB for VCE = 10 V and A 600 and A,200 A, 100 BI

b) VCE, assuming the maximum allowable

collector power dissipation is 0.5 W for

A 500 BI .

Analysis:

a) For IB = 100 µA and VCE = 10 V, from the characteristics, we have IC = 17 mA. The ratio IC /

IB is 170.

For IB = 200 µA and VCE = 10 V, from the characteristics, we have IC = 33 mA. The ratio IC /

IB is 165.

For IB = 600 µA and VCE = 10 V, from the characteristics, we have IC = 86 mA. The ratio IC /

IB is 143.

b) For IB = 500 A, and if we consider an average from a., we have IC = 159·500 10-3

= 79.5

mA. The power dissipated by the transistor is CCEBBECCE IVIVIVP , therefore:

VCE P

IC

0.5

79.5 103 6.29 V.


Known quantities:

Figure P10.10, assuming both transistors are

silicon-based with 100 .

Find:

a) IC1, VC1, VCE1.

b) IC2, VC2, VCE2.

Analysis:

a) From KVL: 030 111 BEBB VRI

μA07.3910750

7.03031

BI

mA907.311 BC II

V779.52.6907.33030 111 CCC IRV

V779.511 CCE VV .

b) Again, from KVL: 0779.5 222 EEBE RIV mA081.1107.4

7.0779.532

EI

and mA07.1101

100081.1

122

EC II .

Also, 0)(30 2222 CEECC VRRI V574.3)7.420()07.1(302 CEV .

Finally, V603.8)20()07.1(3030

22

22

C

C

CC V

R

VI .


Known quantities:

Collector characteristics of the 2N3904 npn

transistor, see data sheet pg. 560.

Find:

The operating point of the transistor in Figure

P10.11, and the value of at this point.

Analysis:

Construct a load line. Writing KVL, we have: 0500050 CEC VI .

Then, if 0CI , V50CEV ; and if 0CEV , mA10CI . The load line is shown superimposed on

the collector characteristic below:

The operating point is at the intersection of the

load line and the A 20BI line of the

characteristic. Therefore,

mA 5CQI and V 20CEQV .

Under these conditions, an A 5 increase in BI

yields an increase in CI of approximately

mA 156 . Therefore,

200105

1016

3

B

C

I

I

The same result can be obtained by checking the

hFE gain from the data-sheets corresponding to 5

mA.

Load line


Known quantities:

The circuit of Figure P10.14, VCEsat=0.1V, VBEsat=0.6V,

and β=50.

Find:

The base voltage required to saturate the transistor.

Analysis: The collector current is

mA 9.111

1.012

CI

The base current is

IB IC

11.9

50 0.238 mA 238A

And since

mA 10

BEsatBBB

VVI

Therefore,

V V 98.26.0k10mA 238.0 BBV


Known quantities:

Collector characteristics of 2N3904 npn transistor; Transistor

circuits;

Find:

The operating point;

Analysis:

From KVL,

or

If 0CEV , mAk

IC 99.410

9.49 , and if 0CI , VVCE 9.49 . The load

line is shown superimposed on the collector characteristic below:

The operating point is at the intersection of the load line and the

AIB 20 line of the characteristic. Therefore, mAICQ 3 and

VVCEQ 8 .

Under these conditions, a A10 increase in BI yields an increase

in CI of approximately mAmAmA 235 . Therefore,

20010

2

A

mA

I

I

B

C

Addition of the emitter resistor effectively increased the current

gain by decreasing the magnitude of the slope of the load line.

0)20(5550 AIkVkI CCEC

9.491.05010 CCE kIV


Known quantities:

For the circuit shown in Figure 10.14 in the text:

mW100,mA10V,4.1

95,V,2.0V,7.0,V5,kΩ1,mA5,V5,V0

max

PIV

VVVRIVV

LEDLED

CEsatCCBBonoff

Find:

Range of RC.

Analysis:

34001.0

2.04.15

LED

CEsatLEDCCC

I

VVVR

From the maximum power

47

mA714.1

1.0

max

maxmax

LED

CEsatLEDCCC

LEDLED

I

VVVR

V

PI

Therefore, RC [47, 340]


Known quantities:

For the circuit shown in Figure 10.14 in the text:

A1

V,1V,7.0,V13,Ω12,kΩ1,mA1,V5,V0 max

C

CEsatCCBBonoff

I

VVVRRIVV

Find:

Minimum value of that will ensure the correct operation

of the fuel injector.

Analysis:

A112

113

R

VVI CEsatCCC

1000101

13

maxmin

B

C

I

I


Known quantities:

The circuit of Figure P10.25: IC = 40 mA; Transistor

large signal parameters.

Find: Design a constant-current battery charging circuit, that is,

find the values of VCC, R1, R2 that will cause the transistor

Q1 to act as a 40-mA constant current source.

Assumptions:

Assume that the transistor is forward biased. Use the

large-signal model with = 100.

Analysis:

The battery charging current is 40 mA, IC = 40 mA.

Thus, the emitter current must be mA4.401

EE II

.

Since the base-emitter junction voltage is assumed to be 0.6 V, then resistor R2 has a voltage:

V 56.06.52 VVV z , so the required value of R2 to be:

8.1230404.0

52

EI

VR

Since the only purpose of R1 is to bias the Zener diode, we can select a value that will supply

enough current fro the Zener to operate, for example R1 > 100 , so that there will be as little

current flow through this resistance as possible.

Finally, we need to select an appropriate supply voltage. VCC must be greater than or equal to the

sum of the battery voltage, the CE junction voltage and the voltage across R2. That is,

59 CECC VV . A collector supply of 24 V will be more than adequate for this task.


Known quantities:

The circuit of Figure of P10.26.

Find: Analyze the operation of the circuit and explain how EI is

decreasing until the battery is full.

Find the values of VCC, R1 that will result in a practical

design.

Assumptions:

Assume that the transistor is forward biased.

Analysis:

When the Zener Diode works in its reverse breakdown area, it provides a constant voltage: V 11zV . That means:

V 11 ZB VV .

When the transistor is forward biased, according to KVL,

batteryBEBEZ VVRIV , where BER is the base resistance.

As the battery gets charged, the actual battery charging voltage batteryV will increase from 9.6 V to

10.4 V.

As batteryV increases gradually, ZV and V stay unchanged, then we can see that BEI will decrease

gradually.

So BEE II 1 will also decrease at the same time.

Since the only purpose of R1 is to bias the Zener diode, we can select a value that will supply

enough current fro the Zener to operate, for example R1 > 100 , so that there will be as little

current flow through this resistance as possible.

Finally, we need to select an appropriate supply voltage. VCC must be greater than or equal to the

sum of the battery voltage, the CE junction voltage. That is, CECC VV 11 . A collector supply

of 12 V should be adequate for this task.


Known quantities:

For the circuit shown in Figure P10.32: V 12CCV 130 Ωk821 R Ωk222 R Ωk5.0ER

Ω16LR .

Find:

CEQV at the DC operating point.

Analysis:

Simplify the circuit by obtaining the Thèvenin

equivalent of the biasing network (R1,, R2, VCC) in the base circuit:

Ωk35.172282

2282Suppress

V538.22282

2212:VD

= = R + R

R R = R = R:V

= = R + R

R V = V = V = V

21

21eqBCC

21

2CCOCTHBB

Redraw the circuit using the Thèvenin equivalent. The "DC blocking" or "AC coupling" capacitors act as open

circuits for DC; therefore, the signal source and load can be neglected since this is a DC problem. Specify directions

of current and polarities of voltages.

Assume the transistor is operating in its active region. Then, the base-emitter junction is forward

biased.

μA18.22

500113017350

7.0538.2

1 = =

R + + R

V - V = I

EB

BEQBBBQ

V55.105.0906.212

:KVL

mA906.21018.2211301 6

= = R I - V = V

0 = V + V - R I -

= + = I + = I

EEQCCCEQ

CCCEQEEQ

BQEQ

The collector-emitter voltage is greater than its saturation value (0.3 V for Silicon). Therefore

the initial assumption (operation in the active region) was correct and the solution is valid.


Known quantities:

For the circuit shown in Figure P10.33: V12CCV 100 V 4EEV Ωk100BR

Ωk3CR Ωk3ER

Ωk6LR Ωk6.0SR mV )1028.6cos( 1 3tvS .

Find:

CEQV and the region of operation.

Analysis: The "DC blocking" or "AC coupling" capacitors act as open circuits for

DC; therefore, the signal source and load can be neglected since this is a

DC problem. Specify directions of current and polarities of voltages.

Assume the transistor is operating in its active region; then, the base-

emitter junction is forward biased and:

0 = R I] + [ + V + R I + V -

0 = R I + V + R I + V -

I] + [ = I [Si] 700 V

EBQBEQBBQBB

EEQBEQBBQBB

BQEQBEQ

1

:KVL

1mV

V 06.11

300010100.8273000109.818124

0 :KVL

A 0.82710189.8)1100()1(

A9.81810189.8)100(

A189.8)3000)(1100(100000

7.04

1

0 :KVL

)1(

][ mV 700

66

6

6

EEQCCQCCEECEQ

CCCCQCEQEEQEE

BQEQ

BQCQ

EB

BEQEEBQ

EEQBEQBBQEE

BQEQBQCQ

BEQ

RIRIVVV

VRIVRIV

II

II

RR

VVI

RIVRIV

IIII

SiV

The collector-emitter voltage is greater (more positive) than its saturation

value (+ 0.3 V for Silicon). Therefore the initial assumption (operation in

the active region) was correct and the solution is valid.

Notes:

1. DC power may be supplied to an npn BJT circuit by connecting the

positive terminal of a DC source to the collector circuit, or, by

connecting the negative terminal of a DC source to the emitter circuit,

or, as was done here, both.

2. In a pnp BJT circuit the polarities of the sources must be reversed. Negative to collector and

positive to emitter.


Known quantities:

For the circuit shown in Figure P10.35: V3Sv 100 Ωk60BR

Find:

a) The value of ER so that EI is 1 mA.

b) CR so that CV is 5 V.

c) The small-signal equivalent circuit of the amplifier

for k5LR

d) The voltage gain.

Analysis:

(a) With k 60BR and V 3BV , applying KVL, we have

EBBB RIRI )1(6.03

EB

RI

101k60

4.2

mARk

IE

E 110160

4.2101

Therefore,

k 81.1101

604.2101ER

(b) EECCCE RIRIV 15

From (a), we have mA 99.01

EC II

Therefore,

k 27.899.0

81.1515CR

(c) The small signal equivalent circuit is shown below

(d) iwB

SB

hR

VI

oeLCout

hRIv

1 Bfe

oe

outC Ih

h

VI

1

Since hoe is not given, we can reasonably assume that 1/hoe is very large. Therefore,

15.4100

ieB

L

s

outV

hR

R

v

vA


Known quantities:

For the circuit shown in Figure P10.36: Ωk200CR

Find:

e) The operating point of the transistor.

f) Voltage gain inout vv ; current gain inout ii

g) Input resistance ir

h) Output resistance or

Analysis:

(a) V 1.621

2

RR

RVV CCB

87.3749| | 21 RRRB

Assuming V 6.0BEV , we have

V 5.5 BEBEV VVV

kI

Vh

BQIB

BEie 6.60

100099.0

6.03

OUTvvS

+

-

IB

C

E

-

h ie

oeh

1

I C

R C

R L

+

BR B

I Bh fe

mA 22E

EE

R

VI

mA 088.01

b

II E

B

and

V 12.55.51021.912200-15

5.5

3-

.

CCCCECCE IRVVVV

(b) The AC equivalent circuit is shown on the

right:

kI

Vh

BQIB

BEie 82.6

10088.0

6.03

BCBEout IIIRv )1250(250)(

BieBoutieBin IhIvhIv 251250

Therefore, the voltage gain is

902.0in

outV

v

vA and

BCBout IIIi )1(

BBieBBB

inBin RIhII

R

vIi )251250(

and the current gain is

84.12)251250(

)1(

BBieBB

B

in

out

RIhII

I

i

i

(c) To find the input resistance we compute:

BieBin IhIv 251250

BBieBBin RIhIIi )251250(

Therefore. the input resistance is

3558in

ini

i

vr

(d) To find the output resistance we compute

BCBEout IIIRv )1250(250)(

BCBout IIIi )1(

Therefore, the output resistance is

250out

outo

i

vr


Known quantities:

The circuit given in Figure P10.41.

Find:

Show that the given circuit functions as an OR gate

if the output is taken at v01.

Analysis:

Construct a state table. This table clearly describes an AND gate when the output is taken at 1ov .

v1 v2 Q1 Q2 Q3 vo1 vo2

0 0 off off on 0 5V

0 5V off on off 5V 0

5V 0 on off off 5V 0

5V 5V on on off 5V 0


Known quantities:

The circuit given in Figure P10.41.

Find:

Show that the given circuit functions as a NOR gate if

the output is taken at v02.

Analysis:

See the state table constructed for Problem 10.41. This

table clearly describes a NOR gate when the output is

taken at 2ov .


Known quantities:

In the circuit given in Figure P10.45 the

minimum value of vin for a high input is 2.0 V.

Assume that the transistor Q1 has a of at least 10.

Find:

The range for resistor RB that can guarantee that the transistor is on.

Analysis:

mA4.22000

2.05

ci , therefore, iB = iC/ = 0.24 mA.

(vin)min = 2.0 V and (vin)max = 5.0 V, therefore, applying KVL: -vin +RB iB + 0.6 = 0

or B

inB

i

vR

6.0 . Substituting for (v

in)min and (v

in)max , we find the following range for R

B:

Ω333.18Ω833.5 kRk B


Known quantities:

For the circuit given in Figure P10.46:

Ωk27,kΩ10 2121 BBCC RRRR .

Find:

a) vB, vout, and the state of the transistor Q1 when

vin is low.

b) vB, vout, and the state of the transistor Q1 when

vin is high.

Analysis:

a) vin is low Q1 is cutoff vB = 5 V Q2 is in saturation vout = low = 0.2 V.

b) vin is high Q1 is in saturation vB = 0.2 V Q2 is cutoff vout = high = 5 V.

EE 321 Chapter 10 Solutions

Documents