487

CHAPTER 13

Exercises

E13.1 The emitter current is given by the Shockley equation:

−

= 1exp

T

BEESE V

vIi

For operation with 1exp have we , >>

>>

T

BEESE V

vIi , and we can write

≅

T

BEESE V

vIi exp

Solving for BEv , we have

mV 4.7181010ln26ln 14

2

=

=

≅ −

−

ES

ETBE I

iVv

V2816.457184.0 −=−=−= CEBEBC vvv

9804.05150

1==

+=ββα

mA804.9== EC ii α

A 1.196 µβ

== CB

ii

E13.2 α

αβ−

=1

α β 0.9 9 0.99 99 0.999 999

E13.3 mA 5.0=−= CEB iii 95.0/ == EC iiα 19/ == BC iiβ

E13.4 The base current is given by Equation 13.8:

−

×=

−

−= − 1

026.0exp10961.11exp)1( 16 BE

T

BEESB

vVvIi α

which can be plotted to obtain the input characteristic shown in Figure 13.6a. For the output characteristic, we have BC ii β= provided that

488

V.0.2 elyapproximat ≥CEv For V,0.2 CCE iv ≤ falls rapidly to zero at .0=CEv The output characteristics are shown in Figure 13.6b.

E13.5 The load lines for V0.8-andV 8.0in =v are shown:

As shown on the output load line, we find

V.0.1andV, 5 V, 9 minmax ≅≅≅ CECEQCE VVV

489

E13.6 The load lines for the new values are shown:

As shown on the output load line, we have

V.0.3andV,7 V, 8.9 minmax ≅≅≅ CECEQCE VVV

490

E13.7 Refer to the characteristics shown in Figure 13.7 in the book. Select a point in the active region of the output characteristics. For example, we could choose the point defined by mA 5.2andV 6 =−= CCE iv at which we find A. 50 µ=Bi Then we have .50/ == BC iiβ (For many transistors the value found for β depends slightly on the point selected.)

E13.8 (a) Writing a KVL equation around the input loop we have the equation for the input load lines: 08000)(8.0 in =+−− BEB vitv The load lines are shown:

Then we write a KCL equation for the output circuit: CEC vi =+ 30009 The resulting load line is:

From these load lines we find

491

,A5A,24 A, 48 minmax µµµ ≅≅≅ BBQB III V 3.8,V 3.5V, 8.1 minmax −≅−≅−≅ CECEQCE VVV (b) Inspecting the load lines, we see that the maximum of vin corresponds to IBmin which in turn corresponds to VCEmin. Because the maximum of vin corresponds to minimum VCE, the amplifier is inverting. This may be a little confusing because VCE takes on negative values, so the minimum value has the largest magnitude.

E13.9 (a) Cutoff because we have V 5.4andV 5.0 −=−=< CEBEBCBE VVVV which is

less than 0.5 V. (b) Saturation because we have .BC II β< (c) Active because we have V.2.0and0 >> CEB VI

E13.10 (a) In this case ( )50=β the BJT operates in the active region. Thus the equivalent circuit is shown in Figure 13.18d. We have

A 5.717.0µ=

−=

B

CCB R

VI mA575.3== BC II β

=−= CCCCCE IRVV 11.43 V Because we have ,2.0>CEV we are justified in assuming that the

transistor operates in the active region. (b) In this case ( )250=β ,the BJT operates in the saturation region.

Thus the equivalent circuit is shown in Figure 13.18c. We have

V 2.0=CEV A 5.717.0µ=

−=

B

CCB R

VI mA 8.142.0=

−=

C

CCC R

VI

Because we have ,CB II >β we are justified in assuming that the

transistor operates in the saturation region.

E13.11 For the operating point to be in the middle of the load line, we want

V 102/ == CCCE VV and mA 2=−

=C

CECCC R

VVI . Then we have

(a) A 20/ µβ == CB II Ω=−

= k 9657.0B

CCB I

VR

(b) A667.6/ µβ == CB II Ω=−

= M 985.27.0B

CCB I

VR

492

E13.12 Notice that a pnp BJT appears in this circuit. (a) For ,50=β it turns out that the BJT operates in the active region.

A 3.197.020 µ=−=B

B RI mA 965.0== BC II β

V35.1020 −=−= CCCE IRV (b) For ,250=β it turns out that the BJT operates in the saturation region.

V 2.0−=CEV A 3.197.020 µ=−=B

B RI mA 98.12.020 =−=

CC R

I

Because we have ,CB II >β we are assured that the transistor operates in the active region.

E13.13 V 521

2 =+

=RR

RVV CCB EB

BEBB RR

VVI)1( ++

−=

β

BC II β= )( BCECCCCCE IIRIRVV +−−= β BI (µA) CI (mA) CEV (V) 100 32.01 3.201 8.566 300 12.86 3.858 7.271 For the larger values of 21 andRR used in this Exercise, the ratio of the collector currents for the two values of β is 1.205, whereas for the smaller values of 21 and RR used in Example 13.7, the ratio of the collector currents for the two values of β is 1.0213. In general in the four-resistor bias network smaller values for 21 andRR lead to more nearly constant collector currents with changes in β.

E13.14 Ω=+

= k 333.311

121 RR

RB V 521

2 =+

=RR

RVV CCB

( ) A 13.141

µβ

=++

−=

EB

BEBBQ RR

VVI mA 239.4== BQCQ II β

( ) Ω=== 1840mA 238.4mV26300

CQ

T

IVr β

π

493

Ω 7.66611

1 =+

=′CL

L RRR 7.108−=

′−=

π

βrRA L

v

0.163oc −==π

βr

RA Lv Ω=

++= 1186

1111

21 πrRRZin

43.64−==L

invi R

ZAA 7004== ivAAG

Ω== k 1Co RZ

)sin(46.76in

inin t

RZZvAvAv

ssvvo ω−=

+==

E13.15 First, we determine the bias point:

Ω=+

= k 00.5011

121 RR

RB V 1021

2 =+

=RR

RVV CCB

( ) A 26.141

µβ

=++

−=

EB

BEBBQ RR

VVI mA 279.4== BQCQ II β

Now we can compute πr and the ac performance.

( ) Ω=== 1823mA 279.4mV26300

CQ

T

IVr β

π Ω 7.66611

1 =+

=′EL

L RRR

( )( ) 9910.0

11

=′++

+′=

L

Lv Rr

RAββ

π

( )( ) 9970.0

11

oc =+++

=E

Ev Rr

RAββ

π

( ) Ωk 10.40]1[11

1 =′+++

=LB

in RrRZ

βπ

74.39==L

invi R

ZAA

38.39== ivAAG Ωk 333.811

1 =+

=′sB

s RRR

( ) 18.33111 Ω=

++′+

=

Es

o

RrR

Z

π

β

Problems

P13.1

20

P13.37 For the Darlington pair, we have 21eq BEBEBE VVVFor the Sziklai pair, we have 1eq BEBE VV

P13.38 (a) Active Region

(b) BC II β than less is ). (c)

Saturation region (because

P13.39 (a) Active region. (b) Cutoff region. (c) Cutoff region (because V 4.0 for 0 BEB VI at room temperature). (d) Saturation region (because BC II β than less is ).

P13.40 See Figure 13.16 in the text.

P13.41* 1. Assume operation in saturation, cutoff, or active region.

2. Use the corresponding equivalent circuit to solve for currents andvoltages.

3. Check to see if the results are consistent with the assumption made instep 1. If so, the circuit is solved. If not, repeat with a different assumption.

P13.42* The results are given in the table:

Circuit βRegion of operation

CI(mA)

CEV(volts)

(a) 100 active 1.93 10.9 (a) 300 saturation 4.21 0.2 (b) 100 active 1.47 5.00 (b) 300 saturation 2.18 0.2 (c) 100 cutoff 0 15 (c) 300 cutoff 0 15 (d) 100 active 6.5 8.5 (d) 300 saturation 14.8 0.2

© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publicationis protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publicationis protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

1.4 V

0.7V.

Cutoff region (IB > 0 and VCE < 0.2V)

27

P13.61 The common-emitter amplifier is inverting. Both the voltage gain and current gain magnitudes are potentially greater than unity.

P13.62* The solution is similar to that of Problem P13.64. The results are: High impedance amplifier

(Problem 13.62) Low impedance amplifier

(Problem 13.64) ICQ 0.0393 mA 3.93 mA

66.2 kΩ 662 Ω vA -75.5 -75.5 ocvA -151 -151 inZ 54.8 kΩ 548 Ω

iA -41.4 -41.4 G 3124 3124

oZ 100 kΩ 1 kΩ

P13.63 (a) The small-signal equivalent circuit is:

in which we have defined

LCL RR

R/1/1

1

(b) From the equivalent circuit we can write: bEbin iRirv )1( βπ (1) bLo iRv β Then we have

E

L

bEb

bL

in

ov Rr

RiRir

iRvvA

)1()1(

ββ

ββ

ππ

(c) From the equivalent circuit we can write:

© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publicationis protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publicationis protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.