Rizzoni5eSM_CH10

G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10

10.1 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 10: Bipolar Junction Transistors: Operation, Circuit Models, and Applications – Instructor Notes

Chapter 10 introduces bipolar junction transistors. The material on transistors is divided into two independent chapters, one on bipolar devices, and one on field-effect devices. The two chapters are functionally independent, except for the fact that Section 10.1, introducing the concept of transistors as amplifiers and

switches, can be covered prior to starting Chapter 11 if the instructor decides to only teach field-effect

devices, or to cover them before bipolar devices. Section 10.2 introduces the fundamental ideas behind the operation of bipolar transistors, and illustrates the

calculation of the state and operating point of basic transistor circuits. The discussion of the properties of the BJT in Section 10.2 is centered around a description of the base and collector characteristics, and purposely avoids a detailed description of the physics of the device, with the intent of providing an intuitive understanding of the transistor as an amplifier and electronic switch.

The second part of the chapter has been reorganized for clarity. Section 10.3 introduces large-signal models of the BJT, and also includes the box Focus on Methodology: Using device data sheets (pp. 559-561). Example 10.4 (LED Driver) and the box Focus on Measurements: Large Signal Amplifier for Diode Thermometer (pp. 566-568) provide two application examples. New to the 5th Edition are examples 10.5 and 10.6, that present simple but practically useful battery charger and DC motor drive BJT circuits. These examples are accompanied by related homework problems (10.25-10.27). Section 10.4 defines the concept of operating point and illustrates the selection of a bias point, introducing the idea of a small-signal amplifier in the most basic way. Finally, Section 10.5 introduces the analysis of BJT switches and presents TTL gates. The end-of-chapter problems are straightforward applications of the concepts illustrated in the chapter. The 5th Edition of this book includes 17 new problems; some of the 4th Edition problems were removed, increasing the end-of-chapter problem count from 40 to 51.

Learning Objectives

1. Understand the basic principles of amplification and switching. Section 10.1. 2. Understand the physical operation of bipolar transistors, and identify their state. Section

10.2 3. Understand the large-signal model of the bipolar transistor, and apply it to simple

amplifier circuits. Section 10.3. 4. Determine and select the operating point of a bipolar transistor circuit; understand the

principle of small signal amplifiers. Section 10.4. 5. Understand the operation of bipolar transistor as a switch and analyze basic analog and

digital gate circuits. Section 10.5.



Section 10.2: Operation of the Bipolar Junction Transistor

Problem 10.1

Solution:

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.

Problem 10.2

Solution:

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.

Problem 10.3

Solution:

Known quantities:

The circuit of Figure P10.3: 100==B

C

I

Iβ .

Find:

The operating point and the state of the transistor.

Analysis:

V 6.0=BEV and the BE junction is forward biased.

A 9.13820

6.012

1

µ=−

=−

=R

VVI BECCB

mA39.1=⋅= BC II β

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

0=+++− EECECCCC RIVRIV

VCE = VCC − IC RC − IC + IB( )RE = 12− (1.39)(2.2)− (1.39+ 0.0139)(0.910) = 7.664 V

VBC = VBE +VCE = 0.6 _ 7.664 = 8.264 V

⇒> BECE VV

The transistor is in the active region.



Problem 10.4

Solution:

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, I

B = 0.1 mA and V

EB = 0.65 V, V

CB = 7.3 V.

Find:

a) VCE.

b) IC.

c) The total power dissipated in the transistor, defined as BBECCE IVIVP += .

Analysis:

a) VCE = V

CB - V

EB = 7.3 - 0.65 = 6.65 V.

b) IC = I

E - I

B = 6 - 0.1 = 5.9 mA.

c) The total power dissipated in the transistor can be found to be: mW 39109.565.6 3 =××=≈ −CCE IVP

Problem 10.5

Solution:

Known quantities:

The circuit of Figure P10.5, assuming the BJT has Vγ = 0.6 V.

Find:

The emitter current and the collector-base voltage.

Analysis:

Applying KVL to the right-hand side of the circuit, A 52030000

156.0

30000

15µ−=

+−=

+−= BE

EV

I

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

( ) V 8.171010105201010

010

36 =×××−−=−=

⇒=++−−

CCCB

CBCC

RIV

VRI



Problem 10.6

Solution:

Known quantities:

The circuit of Figure P10.6, assuming the BJT has

V 6.0=BEV and β =150.

Find:

The operating point and the region in which the transistor

operates.

Analysis:

Define RC = 3.3 kΩ, RE = 1.2 kΩ, R1 = 62 kΩ, R2 = 15 kΩ, VCC = 18 V

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

V857.7101515112001025.2330018

mA25.2

µA15)1(

V5.3

kΩ078.12||

63

21

2

21

=⋅⋅⋅−⋅⋅−=−−=

==

≅++

−=

≅+

=

==

−−EECCCCCE

BC

EB

BEBBB

CCBB

B

IRIRVV

II

RR

VVI

VRR

RV

RRR

β

β

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

Problem 10.7

Solution:

Known quantities:

The circuit of Figure P10.7, assuming the BJT has

V6.0=γV .

Find:

The emitter current and the collector-base voltage.

Analysis:

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

0=++− EBEECC VRIV

µA4.4971039

6.0203

=⋅

−=

−=

E

EBCCE

R

VVI . Since 1>>β , µA4.497=≈ EC II

Applying KVL to the left-hand side: 0=−+ DDCCCB VRIV

VCB = VDD − IC RC = 20− 497.4 ⋅20 ⋅10−3 = 10.05V



Problem 10.8

Solution:

Known quantities:

The circuit of Figure P10.7, assuming the emitter resistor is changed to Ωk 22 and the BJT has V6.0=γV .

Find:

The operating point of the transistor.

Analysis:

Aµ8.8811022

6.0203

=⋅

−=

−=

E

EBCCE

R

VVI , µA8.881=≈ EC II

V364.2310208.88120 =−⋅⋅−=−= CCDDCB RIVV

Problem 10.9

Solution:

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 µµµ=BI

b) VCE, assuming the maximum allowable collector power

dissipation is 0.5 W for A 500 µ=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 µA, and if we consider an average β 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 ⋅10−3= 6.29 V.



Problem 10.10

Solution:

Known quantities:

Figure P10.10, assuming both transistors are silicon-based with 100=β .

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 β

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 =

⋅=

+=

ββ

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 .

Problem 10.11

Solution:

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 β at this point.

Analysis:

Construct a load line. Writing KVL, we have: 0500050 =+⋅+− CEC VI .

Then, if 0=CI , V50=CEV ; and if 0=CEV , mA10=CI . 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 20 µ=BI line of the characteristic. Therefore,

mA 5≈CQI and V 20≈CEQV .

Under these conditions, an A 5 µ increase in BI yields an

increase in CI of approximately mA 156 =− .

Therefore,

200105

1016

3

=⋅

⋅=

∆∆

≈−

−

B

C

I

Iβ

The same result can be obtained by checking the hFE gain

from the data-sheets corresponding to 5 mA.

Problem 10.12

Solution:

Known quantities:

The circuit shown in Figure P10.12. With reference to Figure 10.20, assume V6.0=γV , V2.0=satV .

Find:

The operating point of the transistor, by computing the ratio of collector current to base current.

Analysis:

V2.0== satCE VV , therefore mA8.92.010

=−

=C

CR

I V6.0== γVVBE , therefore µA1026.07.5

=−

=B

BR

I

β<<=⋅

⋅=

−

−08.96

10102

108.96

3

B

C

I

I

Load line



Problem 10.13

Solution:

Known quantities:

The circuit in Figure 10.28 in the text. VCC=20 V, RC=5kΩ, RE=1kΩ.

Find:

The region of operation of the transistor.

Analysis:

(a) IE = IC + IB = 1 mA + 20µA = 1.02 mA

VE = 1000 IE = 1.02 V

VRC = 5000 IC = 5 V

VCB = 20 - VRC -VBE -VE

= 20 - 5 - 0.7 - 1.02 = 13.28 V

The CB junction is reverse-biased. Therefore, the transistor is operating in the active region.

(b) IE = 3.2 mA + 0.3 mA = 3.5 mA

VE = 3.5 V

VRC = 16 V

VCB = 20 -16 - 0.8 - 3.5 = -30 mV

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

(c) IE = 3 mA + 1.5 mA = 4.5 mA

VE = 4.5 V

VRC = 15 V

VCB = 20 - 15 - 0.85 - 4.5 = -0.35 V

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



Problem 10.14

Solution:

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

β=

11.9

50= 0.238 mA = 238µA

And since

mA 10

BEsatBBB

VVI

−=

Therefore,

V V 98.26.0k10mA 238.0 =+Ω⋅=BBV

Problem 10.15

Solution:

Known quantities:

β=60; VBE=0.6V; VCB=7.2V; |IE|=4mA.

Find:

a) IB; b) VCE;

Analysis:

(a) Since

BCBE IIII )1( +=+= β

we can compute

IB =IE

β +1=

4

61= 65.6 mA

(b) VCE = VCB - VBE

= 7.2 - 0.6 = 6.6 V



Problem 10.16

Solution:

Known quantities:

Collector characteristics of 2N3904 npn transistor; Transistor circuits;

Find:

The operating point;

Analysis:

From KVL,

or

If 0=CEV , mAk

IC 99.410

9.49== , and if 0=CI , 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 µ20=

line of the characteristic. Therefore, mAICQ 3≈ and VVCEQ 8≈ .

Under these conditions, a Aµ10 increase in BI yields an increase in CI of

approximately mAmAmA 235 =− . Therefore,

20010

2==

∆∆

≈A

mA

I

I

B

C

µβ

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 µ

9.491.05010 =−=+ CCE kIV



Section 10.3: BJT Large-Signal Model

Problem 10.17

Solution:

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

γ

γ β

Find:

Range of RC.

Analysis:

Ω=−−

≤−−

= 34001.0

2.04.15

LED

CEsatLEDCCC

I

VVVR

γ

From the maximum power

Ω=−−

>

===

47

mA714.1

1.0

max

maxmax

LED

CEsatLEDCCC

LEDLED

I

VVVR

V

PI

γ

γ

Therefore, RC ∈[47, 340] Ω

Problem 10.18

Solution:

Known quantities:

For the circuit shown in Figure 10.18 in the text: Ω======= 500188.5,V,6V,75.0,V12,kΩ33,V1.1 SCEQBECCBD RVVVRV β

Find:

The resistance RC.

Analysis:

The current through the resistance RB is given by

µA6.1033000

75.01.1=

−=

−=

B

BEQDB

R

VVI

The current through RS is: mA8.9500

1.16=

−=

−=

S

DCEQS

R

VVI

It follows that the current through the resistance RC is mA8.11=+= SBCQ III β

Finally, Ω=−

=−

= 5.5080118.0

612

CQ

CEQCCC

I

VVR



Problem 10.19

Solution:

Known quantities:


mW100,mA10V,4.1

95,V,2.0V,7.0,V5,Ω340,mA5,V5,V0

max

max

=≥=

========

PIV

VVVRIVV

LEDLED

CEsatCCCBonoff

γ

γ β

Find: Range of RB.

Analysis:

If the BJT is in saturation

mA10=−−

=C

CEsatLEDCCC

R

VVVI

γ

In order to guarantee that the BJT is in saturation

Ω=−

≥

Ω=−

=−

≤

860

k85.40

95

01.0

7.05

/

maxB

onB

C

onB

I

VVR

I

VVR

γ

γ

β

Problem 10.20

Solution:

Known quantities:


mW100,mA10V,4.1

V,2.0V,7.0,V5,Ω340,kΩ10,mA5,V5,V0

max

max

=≥=

========

PIV

VVVRRIVV

LEDLED

CEsatCCCBBonoff

γ

γ

Find: Minimum value of β that will ensure the correct operation of the LED.

Analysis:

mA43.010000

3.4==

−=

B

onB

R

VVI

γ

25.231043.0

01.03

minmin =

⋅==

−B

LED

I

Iβ



Problem 10.21

Solution:

Known quantities:


mW100,mA10V,4.1

V,2.0V,7.0,V5,Ω340,kΩ10,mA5,V3.3,V0

max

max

=≥=

========

PIV

VVVRRIVV

LEDLED

CEsatCCCBBonoff

γ

γ

Find:

Minimum value of β that will ensure the correct operation of the LED.

Analysis:

mA26.010000

7.03.3=

−=

−=

B

onB

R

VVI

γ

5.381026.0

01.03

minmin =

⋅==

−B

LED

I

Iβ

Problem 10.22

Solution:

Known quantities:


A1

V,1V,7.0,V13,Ω12,kΩ1,mA1,V5,V0 max

≥

========

C

CEsatCCBBonoff

I

VVVRRIVV γ

Find: Minimum value of β that will ensure the correct operation of the fuel injector.

Analysis:

A112

113=

−=

−=

R

VVI CEsatCCC

1000101

13

maxmin =

⋅==

−B

C

I

Iβ



Problem 10.23

Solution:

Known quantities:


A1

V,1V,7.0,V13,Ω12,2000,mA1,V5,V0 max

≥

========

C

CEsatCCBonoff

I

VVVRIVV γβ

Find: The range of RB that will ensure the correct operation of the fuel injector.

Analysis:


A1=−

=R

VVI CEsatCCC

Because this is the minimum value allowed for the current to drive the fuel injector, it is necessary to guarantee that the BJT is in saturation. In order to guarantee that the BJT is in saturation

Ω=−

≥

Ω=−

=−

≤

k3.4

k6.8

2000

1

7.05

/

maxB

onB

C

onB

I

VVR

I

VVR

γ

γ

β

Problem 10.24

Solution:

Known quantities:


A1

V,1V,7.0,V13,Ω12,2000,mA1,V3.3,V0 max

≥

========

C

CEsatCCBonoff

I

VVVRIVV γβ

Find: The range of RB that will ensure the correct operation of the fuel injector.

Analysis:


A1=−

=R

VVI CEsatCCC

Because this is the minimum value allowed for the current to drive the fuel injector, it is necessary to guarantee that the BJT is in saturation. In order to guarantee that the BJT is in saturation

Ω=−

≥

Ω=−

=−

≤

k6.2

k2.5

2000

1

7.03.3

/

maxB

onB

C

onB

I

VVR

I

VVR

γ

γ

β



Problem 10.25

Solution:

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 β = 100.

Analysis:

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

Thus, the emitter current must be mA4.401

=+

= EE IIβ

β.

Since the base-emitter junction voltage is assumed to be 0.6 V, then resistor R2 has a voltage: V 56.06.52 =−=−= γVVV z , so the required value of R2 to be:

Ω=== 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 Ω, 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.



Problem 10.26

Solution:

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 11=zV . That means:

V 11== ZB VV . When the transistor is forward biased, according to KVL,

batteryBEBEZ VVRIV ++⋅= γ , 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 γV stay unchanged, then we can see that BEI will decrease gradually.

So ( ) BEE II 1+= β 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 Ω, 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.



Problem 10.27

Solution:

Known quantities:

The circuit of Figure P10.27: Vin= 5 V Find: Values of Rb.

Assumptions:

Assume that the transistors are in the active region. Use the large-signal model with β = 40 for each transistor.

Analysis:

The emitter current from Q1, iE1 = (β+1) iB1 becomes the base current for Q2, and therefore, iC2 = β iE1 = β (β+1) iB1. The Q1 base current is given by the expression

b

inB

R

VVVi

γγ −−=1

Therefore the motor current will reach maximum when Vin= 5 V:

( ) A 34.01max =

−−+=

b

inC

R

VVVi

γγββ

So, ( ) ( ) Ω=−

⋅=−

+= 329,18)2.15(

34.0

41402

34.0

1γ

ββVVR inb

Since 18.33 kΩ is a standard resistor value, we should select Rb = 18.33 kΩ, which will result in a slightly lower maximum current.



Section 10.4: Selecting an Operating Point for a BJT

Problem 10.28

Solution:

Known quantities:

The circuit of Figure of 10.22 in the text, V.10 ,50 V,5 , K1 min ===Ω= CCBBC VVR β

Find: The range of BR to make the transistor in the saturation state.

Analysis:

Assuming V 2.0=CEsatV , the current CI is:

mA 8.9=−

=C

CEsatCCC

R

VVI

Therefore, mA 196.0==b

II CB

Assuming V 6.0== BEsatVVγ , we have

Ω=−

= k 45.22B

BEBBB

I

VVR

That is Ω<< k 45.220 BR

Problem 10.29

Solution:

Known quantities:

The circuit of Figure of 10.22 in the text, V.5 ,50 ,10K , K1 min ==Ω=Ω= CCBC VRR β

Find: The range of BBV to make the transistor in the saturation state.

Analysis:

Assume V 2.0=CEsatV , the current CI can be found as

mA 8.4=−

=C

CEsatCCC

R

VVI

Therefore, A 96mA 096.0 µ===b

II CB

Assuming V 6.0== BEsatVVγ , we have

V 56.1=+= BEsatBBBB VRIV

That is V 56.1>BBV



Problem 10.30

Solution:

Known quantities:

The circuit of Figure 10.20 in the text, V.10 ,100 A,20 , k2 ===Ω= CCBBC VIR βµ

Find: .,,, CBCEEC VVII

Analysis:

V 4.56.06 Then

V 6.0 Assume

V 62210

mA 02.21020101

)1(

mA 21020100

6

6

=−=−=

=

=×−=−=

=××=

+=+=

=××==

−

−

BECECB

BE

CCCCCE

BCBE

BC

VVV

V

RIVV

IIII

II

β

β

Problem 10.31

Solution:

Known quantities:

For the circuit shown in Figure P10.31: V20=CCV 130=β MΩ8.11 =R Ωk3002 =R

Ωk3=CR Ωk1=ER

Ωk1=LR Ωk6.0=SR mV )1028.6cos( 1 3tvS ×= .

Find:

The Thèvenin equivalent of the part of the circuit containing 1R , 2R , and

CCV with respect to the terminals of 2R . Redraw the schematic using the

Thèvenin equivalent.

Analysis:

Extracting the part of the circuit specified, the Thèvenin equivalent voltage is the open circuit voltage. The equivalent resistance is obtained by suppressing the ideal independent voltage source:

Note that CCV must remain in the circuit because it supplies current to

other parts of the circuit:



Problem 10.32

Solution:

Known quantities:

For the circuit shown in Figure P10.32: V 12=CCV 130=β Ωk821 =R Ωk222 =R Ωk5.0=ER Ω16=LR .

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:

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.

( ) ( ) µA18.22500113017350

7.0538.2

1 = =

R + + R

V - V = I

EB

BEQBBBQ ⋅++

−⋅β

( ) ( )

V55.105.0906.212

:KVL

mA906.21018.2211301 6

= = R I - V = V

0 = V + V - R I -

= + = I + = I

EEQCCCEQ

CCCEQEEQ

BQEQ

⋅−

⋅⋅ −β

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.

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

+⋅

+⋅

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

β

β≈



Problem 10.33

Solution:

Known quantities:

For the circuit shown in Figure P10.33: V12=CCV 100=β V 4=EEV Ωk100=BR

Ωk3=CR Ωk3=ER


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:

[ ]

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

µβ

µβ

µβ

ββ

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.



Problem 10.34

Solution:

Known quantities:

For the circuit shown in Figure P10.34: V12=CCV 130=β Ωk325=BR Ωk9.1=CR

Ωk3.2=ER


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. The base and collector currents both flow through the collector

resistor in this circuit.

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.

0)(1)(-

0 :KVL

V 896.4

3.2691.19.1691.112

0 :KVL

mA 691.11096.12)1130()1(

A91.1210))9.13.2()1130(325(

7.012

))(1(

)1(

0 :KCL

)1(

][ mV 700

6

3

=+−−++

⇒=+−−−−

=

⋅−⋅−=−−=

⇒=+−−−

=⋅⋅+=⋅+==

=⋅+⋅++

−=

+++

−=

+=+=

⇒=−+

+=⋅=

≈

−

CCBBQBEQCEBQ

CCCRCBBQBEQEEQ

EEQCRCCCCEQ

CCCRCCEQEEQ

BQEQRC

CEB

BEQCCBQ

BQBQCQRC

RCCQBQ

BQEQBQCQ

BEQ

VRIVRRI

VRIRIVRI

RIRIVV

VRIVRI

III

RRR

VVI

IIII

III

IIII

SiV

β

β

µβ

β

ββ



Problem 10.35

Solution:

Known quantities:

For the circuit shown in Figure P10.35: V3=Sv 100=β Ωk60=BR

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 Ω= k5LR

d) The voltage gain.

Analysis:

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

EBBB RIRI )1(6.03 β+++=

EB

RI

101k60

4.2

+Ω=

mARk

IE

E 110160

4.2101 =

+Ω=

Therefore,

Ω=−⋅

= k 81.1101

604.2101ER

(b) EECCCE RIRIV −−=15

From (a), we have mA 99.01=

+=

ββ

EC II

Therefore, Ω=−−

= 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

Ω=×

=∂

=−

kI

Vh

BQIB

BEie 6.60

100099.0

6.03

∂

OUTvvS

+

-

∆ IB

C

E-

h ie

oeh1

I∆ C

R C

R L

+

BR B

∆ I Bh fe



Problem 10.36

Solution:

Known quantities:

For the circuit shown in Figure P10.36: Ωk200=CR

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

Ω== 87.3749|| 21 RRRB

Assuming V 6.0=BEV , we have

V 5.5=−= BEBEV VVV

mA 22==E

EE

R

VI

mA 088.01=

+=b

II EB

and

( )V 12.55.51021.912200-15

5.5

3-

.

=−⋅⋅=

−−=−= CCCCECCE IRVVVV

(b) The AC equivalent circuit is shown on the right:

Ω=×

≈=−

kI

Vh

BQIB

BEie 82.6

10088.0

6.03∂

∂

BCBEout IIIRv )1250(250)( +=+=

BieBoutieBin IhIvhIv ⋅⋅+=+= 251250

Therefore, the voltage gain is

902.0==in

outV

v

vA and

BCBout IIIi ⋅+++= )1(β

BBieBBB

inBin RIhII

R

vIi )251250( ⋅⋅++=+=

and the current gain is

84.12)251250(

)1(=

⋅⋅+++

=BBieBB

B

in

out

RIhII

I

i

i β



(c) To find the input resistance we compute:

BieBin IhIv ⋅⋅+= 251250

BBieBBin RIhIIi )251250( ⋅⋅++=

Therefore. the input resistance is

Ω== 3558in

ini

i

vr

(d) To find the output resistance we compute

BCBEout IIIRv )1250(250)( +=+=

BCBout IIIi ⋅+++= )1(β

Therefore, the output resistance is

Ω== 250out

outo

i

vr

Problem 10.37

Solution:

Known quantities:

The circuit shown in Figure P10.37(a), P10.37(b):

Find:

The duration of the fuel injector pulse.

Analysis:

(a) With VCE = 0.3 V, VBE = 0.9 V and VBATT = 13

V, TC = 100°, from Figure P9.6(d), we have KC = 0,

VCIT = 16/13= 1.23 ms. The signal duration is:

τ = 1×10-3×0 + 1.23×10-3 = 1.23 ms

When Vsignal is applied, the base current is

IB = VBATT /80= 0.1625 A

Thus, the transistor will be in the saturation region.

Therefore,

Vinj = VBATT - VCE = 13 - 0.3 = 12.7 V

The time constant of the injector circuit is:

τ' = L/R= 0.1 ms << τ = 1.23 ms

As Vsignal = 0, the transistor is in the cut-off region. The differential equation governing the injector current is:

1×10-3 dIinj/dt+ 10Iinj = Vinj,

and Iinj(0) = 0. Thus,



Iinj = Vinj/10- Vinj/10∙e-10000t

The time when Iinj ≥ 0.1 is then found to be

tinj = -(ln(1.17/1.27))/104= 8.2 µs

That is, 8.2 ms after the Vsignal is applied, the fuel will be

injected into the intake manifold.

(b) From Figure P10.37(d), we have

KC = - 1/60∙TC + 5/3; at TC = 20, KC = 4/3. Also, VCIT =

16/8.6= 1.86 ms

The signal duration therefore is

τ = 1×10-3×4/3+ 1.86×10-3 = 3.19 ms

When Vsignal is applied, the base current is

IB = VBATT /80= 0.1075 A

Thus, the transistor will be in the saturation region. Therefore,

Vinj = VBATT - VCE = 8.6 - 0.3 = 8.3 V

The time constant of the injection circuit is:

τ' = L/R= 0.1 ms << τ = 3.19 ms

When Vsignal is 0, the transistor is in the cut-off region. Using the same differential equation and initial condition

as in part (a),

Iinj = Vinj/10- Vinj/10∙e-10000t

The time when Iinj ≥ 0.1 can be found as

t = 12.84 µs

That is, 12.84 µs after the Vsignal is applied, the fuel will be injected into the intake manifold.



Problem 10.38

Solution:

Known quantities:

For the circuit shown in Figure P10.38: V 8.0=γV

Find:

The maximum of frequency with which the light can be switched.

Analysis:

From the power dissipation of the relay,

W5.02

WR

VP == ,

we can compute

RW = 50 Ω

When vS is 5 V, the transistor is in the saturation region and the relay current at steady state is:

mA 10650

2.055=

−=RI

When vS is 0 V, the transistor will be in cut off, and we have the following equation

050005.0 =+ RR I

dt

dI

That is

010103 =⋅+ R

R Idt

dI

Solving the above equation with

IR(0) = 0.106 A,

IR(t) = 0.106e-10000t

vR(t) = - 25 IR = - 2.65e-10000t

The relay will be cut-off as soon as vS is 0.

Therefore, the switching frequency will be the frequency of the input signal vS.



Problem 10.39

Solution:

Known quantities:

For the circuit shown in Figure P10.39: 70: ;130: 21 == ββ QQ .

Find:

The overall current gain.

Analysis:

The AC circuit is shown on the right:

The current gain is

9300)1(hh

h

1

112fe1

1

12fe1

1

22fe1

2

2

1

1

1

21

1

=++=+=

+=

+=

+==

B

Bfefe

B

Efe

B

Bfe

B

C

B

C

B

CC

Bi

I

Ihh

I

Ih

I

Ih

I

I

I

I

I

II

I

IA

Problem 10.40

Solution:

Known quantities:

For the circuit shown in Figure P10.40: V 6.0=γV .

Find:

1R and 2R .

Analysis:

a) It’s given by the problem.

b)

For 5=CEQV , VVVEC RR 20=+ .

Assume EC II ≈ .

VV

VVmAk

I

B

EC

7.87.08

885.2

20

=+=

=⇒==

For 20=β , mAIB 4.020

8== .

hie1

IB1

I B2h ie2

IC1

IC2h fe1IB1

h fe2I B2

B'C'

E'

I



For 50=β , and mAIC 8.8≤ ,

VV

mAI

B

B

5.97.08.8

176.050

8.8

=+=

=≤

5.910176.0

7.8104.0

3

3

=×−

=×−−

−

BB

BB

RV

RV

Solving, Ω= kRB 35.5 .

VVB 84.9=

Since )25(21

1

21

1

RR

RV

RR

RV CCB +

=+

=

and 21

21

RR

RRRB +

= , we can solve for

Ω= kR 82.81 and Ω= kR 59.132 .

c)

Assume VVCE 23max

≈ and VVCE 3min

≈ .

Then CEQV should be set to the middle of this range, or VVCEQ 132

323=

+= .

From KVL,

2515001000 =++ CQCEQEQ IVI

or,

25)(1500)1(1000 =+++ BQCEQBQ IVI ββ

Solving,

121325251000 =−=BQI

AIBQ µ81.47=∴

VIV BQEQ 829.4)1(1000 =+= β

VVVV BEEQBQ 529.5=+=

Again from KVL,

( ) 529.5)81.47(22 11 =−⇒=− AIRVIIR RBQBQR µ

And 471.19529.5252522 22 =−=⇒=+ RBQR IRVIR

Choose a typical value for 2R , say Ω= kR 102 . Then,

mAR

IR 947.1471.19

22

==

And Ω=−

= 291181.47947.1

529.51

AmAR

µ



Section 10.4: BJT Switches and Gates

Problem 10.41

Solution:

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

Problem 10.42

Solution:

Known quantities:


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 .



Problem 10.43

Solution:

Known quantities:


Find:

Show that the given circuit functions as an AND 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 on 0 5V

5V 0 on off on 0 5V

5V 5V on on off 5V 0

Problem 10.44

Solution:

Known quantities:


Find:

Show that the given circuit functions as a NAND gate if the output is taken at v02.

Analysis:

See the state table constructed for Problem 10.32. This table clearly describes a NAND gate when the output is taken at 2ov .



Problem 10.45

Solution:

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 β 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, i

B = i

C/β = 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 (vin)max , we find the following range for RB

:

Ω333.18Ω833.5 kRk B ≤≤

Problem 10.46

Solution:

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.



Problem 10.47

Solution:

Known quantities:

For the inverter given in Figure P10.47: Ωk5,Ωk221 === BCC RRR .

Find:

The minimum values of β1 and β2 to ensure that Q1 and Q2 saturate when vin is high.

Analysis:

mA4.22000

2.05=

−=ci , therefore, mA

5.2

β=ci . Applying

KVL: 06.06.06.05 1 =++++− BBiR

Therefore, iB1

= 0.64 mA. 2111 500600

BBE iii +=⋅= β or

21

5.22.164.0 ββ +=⋅

Choose β2 = 10 ⇒ β

1 = 2.27.

Problem 10.48

Solution:

Known quantities:

For the inverter given in Figure P10.47: 4,Ωk2,Ωk5.2 2121 ==== ββCC RR .

Find:

Show that Q1 saturates when vin is high. Find a condition for

2CR to ensure that Q2 also saturates.

Analysis:

mA2.3mA8.04000

2.311 =⇒== CB ii

Applying KCL:

mA8;mA22.3500

6002222 =⋅==⇒=+ BCBB iiii β

Applying KVL: Ω600008.02.05 22 =⇒⋅=− CC RR



Problem 10.49

Solution:

Known quantities:

The basic circuit of a TTL gate, shown in Figure P10.49.

Find:

The logic function performed by this circuit.

Analysis:

The circuit performs the function of a 2-input NAND gate. The analysis is similar to Example 10.10.

Problem 10.50

Solution:

Known quantities:

The circuit diagram of a three-input TTL NAND gate, given in Figure P10.50.

Find:

vB1, vB2, vB3, vC2, and vout, assuming that all the input voltages are high.

Analysis:

Q2 and Q3 conduct, while Q4 is cutoff. vB1 = 1.8 V, vB2 = 1.2 V, vB3 = 0.6 V, and vC2 = vout = 0.2 V.



Problem 10.51

Solution:

Known quantities:

Figure P10.51.

Find:

Show that when two or more emitter-follower outputs are connected to a common load, as shown in Figure P10.51, the OR operation results; that is, v0 = v1 OR v2.

Analysis:

v2 v

1 Q

1 Q

2 v

0

L L L L L

L H H L H

H L L H H

H H H H H L : Low; H : High.

Rizzoni5eSM_CH10

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total power

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pnp transistor

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thvenin equivalent