EE 321 Chapter 10 Solutions

(1)

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.

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:

(2)

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.

The circuit of Figure P10.3:  100 B C I I  . Find:

The operating point and the state of the transistor.

Analysis: V 6 . 0  BE

V and the BE junction is forward biased.

A

V

I

V

I

V

BE CC B B B C E E BE B CC



5 .

12 911910

6 .

0

12

101

910

10

820

101 &

910

10

820

3 3



























mA

I

_C







_B



1 .

25

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

0     V_CC I_CR_C V_CE I_ER_E



I



R

V

R

I

V

CE



CC



C C



C



B E



12 

(

1 .

25 )(

2 .

2 )



(

1 .

25 

0 .

0125

)(

0 .

910 )



8 .

1 V

V

_BC



_BE



_CE



0 .

6 

8 .

1 



7 .

5

: the BC junction is reverse biased

  _BE

CE V V

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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:

I_E = 6 mA, I_B = 0.1 mA and V_EB = 0.65 V, V_BC = 7.3 V. Find:

a) VCE. b) IC.

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

P



V

_EC

I

_C



V

_EB

I

_B.

Analysis:

a) V_EC = V_BC + V_EB = 7.3 + 0.65 = 7.95 V.

b) I_C = I_E - I_B = 6 - 0.1 = 5.9 mA.

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

mW

I

V

I

V

P

EC C EB B

7 .

95

5 .

9

10

0 .

65

0 .

1

10

46 .

97

3 3















  Chapter 10, Solution 5. 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,

I

V

BE

A

E

480 

30000

6 .

0

15 30000

15 

































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





V

R

I

V

R

I

C C CB CB C C

8 .

2

10

15

10

480

10

0

10

3 6























(4)

V 6 . 0  BE V 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

V

I

R

I

R

V

mA

I

μA

R

V

I

V

R

V

kΩ

R

E E C C CC CE B C E B BE BB B CC BB B

857 .

7

10

15

151 1200

10

25 .

2 3300

18

25 .

2

15 )

1 (

5 .

3

078 .

12 ||

6 3 2 1 2 2 1































 



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

V 6 . 0   V . Find:

The emitter current and the collector-base voltage.

Analysis:

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

0 





V

BB

I

E

R

E

V

EB

μA

R

V

I

E EB BB E

497 .

4

10

39

6 .

0

20

3













. Since 1, I_C I_E497.4μA VCC = 20V VBB = 20V

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Applying KVL to the left-hand side:

V

_CB



I

_C

R

_C



V

_CC



0 V

V

R

I

V

_CB



_C _C



_CC



497 .

4 

20 

10

3



20 



10 .

05

Chapter 10, Solution 9. 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    B I

b) VCE, assuming the maximum allowable collector power dissipation is 0.5 W for

A 500   B I . 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 PV_CEI_C V_BEI_BV_CEI_C, therefore:

 V_CE  P I_C  0.5 79.5103 6.29 V. Chapter 10, Solution 10. 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: 30I_B₁R_B₁V_BE₁0  μA 07 . 39 10 750 7 . 0 30 3 1     B I

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    ₁ 3.907mA 1 B C I I  V 779 . 5 2 . 6 907 . 3 30 30 ₁ ₁ 1  C C     C R I V V 779 . 5 1 1 C  CE V V . b) Again, from KVL: 5.779V_BE₂I_E₂R_E₂0  1.081mA 10 7 . 4 7 . 0 779 . 5 3 2     E I and 1.07mA 101 100 081 . 1 1 2 2                   E C I I . Also, 30I_C₂(R_C₂R_E₂)V_CE₂ 0  V_CE₂30(1.07)(204.7)3.574V. Finally, 30 ₂ 30 (1.07) (20) 8.603V 2 2 2        _C C C C V R V I . Chapter 10, Solution 11. 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: 505000I_CV_CE 0.

Then, if I_C 0, V_CE 50V; and if V_CE 0, I_C10mA. The load line is shown superimposed on the collector characteristic below:

The operating point is at the intersection of the load line and the I_B20A line of the

characteristic. Therefore, mA 5  CQ I andV_CEQ20V.

Under these conditions, an 5A increase in I_B yields an increase in I_C of approximately

mA 1 5 6  . Therefore, 200 10 5 10 1 6 3        _ B C I I 

The same result can be obtained by checking the hFE gain from the data-sheets corresponding to 5

mA.

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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 . 11 1 1 . 0 12 _  C I

The base current is

 I_BIC   11.9 50 0.238 mA238A And since mA 10 BEsat BB B V V I   Therefore, V V_BB0.238mA10k0.62.98 V

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Collector characteristics of 2N3904 npn transistor; Transistor circuits;

Find:

The operating point;

Analysis: From KVL, or If V_CE 0, mA k I_C 4.99 10 9 . 49 _

 , and if I_C 0, V_CE 49.9V. 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

I_B20 line of the characteristic. Therefore, I_CQ 3mA and V

V_CEQ 8 .

Under these conditions, a 10A increase in I_B yields an increase in I_C of approximately 5mA3mA2mA. Therefore, 200 10 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 ( 5 5 50      kI_C V_CE k I_C A 9 . 49 1 . 0 50 10     _C CE kI V

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For the circuit shown in Figure 10.14 in the text:

mW 100 , mA 10 V, 4 . 1 95, V, 2 . 0 V, 7 . 0 , V 5 , kΩ 1 , mA 5 , V 5 , V 0 max           P I V V V V R I V V LED LED CEsat CC B B on off    Find: Range of RC. Analysis:         340 01 . 0 2 . 0 4 . 1 5 LED CEsat LED CC C I V V V R 

From the maximum power

        47 mA 71 4 . 1 1 . 0 max max max LED CEsat LED CC C LED LED I V V V R V P I   Therefore, RC [47, 340]  Chapter 10, Solution 22. Known quantities:

For the circuit shown in Figure 10.14 in the text:

A 1 V, 1 V, 7 . 0 , V 13 , Ω 12 , kΩ 1 , mA 1 , V 5 , V 0 _max          C CEsat CC B B on off I V V V R R I V V _ Find:

Minimum value of  that will ensure the correct operation

of the fuel injector.

Analysis: A 1 12 1 13 _    R V V IC CC CEsat 1000 10 1 1 3 max min     _ B C I I 

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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 I_E  1I_E40.4mA

 

.

Since the base-emitter junction voltage is assumed to be 0.6 V, then resistor R2 has a voltage: V 5 6 . 0 6 . 5 2 V V   

V _z , so the required value of R2 to be:

    123.8 0404 . 0 5 2 E I V R

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,

5 9   _CE

CC V

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

The circuit of Figure of P10.26.

Find:

Analyze the operation of the circuit and explain how I_Eis decreasing until the battery is full.

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

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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  z V _{. That means:} V 11   _Z B V V _.

When the transistor is forward biased, according to KVL, battery

BE BE

Z I R V V

V    _  _{, where}R_BE_{is the base resistance.}

As the battery gets charged, the actual battery charging voltage Vbatterywill increase from 9.6 V to 10.4 V.

As Vbatteryincreases gradually, VZand Vstay unchanged, then we can see that IBEwill decrease gradually.

So IE 



1



IBEwill 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, V_CC 11V_CE. A collector supply of 12 V should be adequate for this task.

For the circuit shown in Figure P10.32:

V 12  CC V 130 R₁82kΩ R₂ 22kΩ R_E0.5kΩ Ω 16  L R . Find: CEQ

V 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:

Ω k 35 . 17 22 82 22 82 Suppress V 538 . 2 22 82 22 12 : VD = = R + R R R = R = R : V = = R + R R V = V = V = V 2 1 2 1 eq B CC 2 1 2 CC OC TH BB    

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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.









22.18μA 500 1 130 17350 7 . 0 538 . 2 1 R = = + + R V -V = I E B BEQ BB BQ _ _ _   









V 55 . 10 5 . 0 906 . 2 12 : KVL mA 906 . 2 10 18 . 22 1 130 1 6 = = R I -V = V 0 = V + V -R I = + = I + = I E EQ CC CEQ CC CEQ E EQ BQ EQ      

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.

V 12  CC V 100 V_EE 4V R_B100kΩ Ω k 3  C R RE3kΩ Ω k 6  L R RS 0.6kΩ vS 1cos(6.28103t)mV. Find: CEQ

V 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 E BQ BEQ B BQ BB E EQ BEQ B BQ BB BQ EQ BEQ 1 : KVL 1 mV   

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



V 06 . 11 3000 10 10 0 . 827 3000 10 9 . 818 12 4 0 : KVL A 0 . 827 10 189 . 8 ) 1 100 ( ) 1 ( A 9 . 818 10 189 . 8 ) 100 ( A 189 . 8 ) 3000 )( 1 100 ( 100000 7 . 0 4 1 0 : KVL ) 1 ( ] [ mV 700 6 6 6 6                                                            E EQ C CQ CC EE CEQ CC C CQ CEQ E EQ EE BQ EQ BQ CQ E B BEQ EE BQ E EQ BEQ B BQ EE BQ EQ BQ CQ BEQ R I R I V V V V R I V R I V I I I I R R V V I R I V R I V I I I I Si V        

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.

V 3  S v 100 R_B60kΩ Find:

a) The value of R_Eso that I_E is 1 mA. b) R_Cso that V_Cis 5 V.

c) The small-signal equivalent circuit of the amplifier for R_L5k

d) The voltage gain.

Analysis:

(a) With RB60k and VB3 V, applying KVL, we have E B B BR I R I 0.6 (1 ) 3    E B R I 101 k 60 4 . 2   

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mA R k I E E 1 101 60 4 . 2 101     Therefore,      1.81k 101 60 4 . 2 101 E R (b) V_CE 15I_CR_C I_ER_E

From (a), we have 0.99 mA 1     E C I I Therefore,    8.27 k 99 . 0 81 . 1 5 15 C R

(c) The small signal equivalent circuit is shown below

(d) iw B S B h R V I               oe L C out h R I v 1  _fe _B oe out C _h h I V I     1

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

15 . 4 100        ie B L s out V _R _h R v v A Chapter 10, Solution 36. Known quantities:

Ω k 200  C R Find:

e) The operating point of the transistor. f) Voltage gain vout vin; current gain iout iin g) Input resistance r_i h) Output resistance r_o Analysis: (a) 6.1 V 2 1 2 _   R R R V V_B _CC R_BR₁||R₂3749.87  Assuming V_BE 0.6 V, we have V 5 . 5    _B _BE EV V V V       _ k I V h BQ I B BE ie 60.6 10 0099 . 0 6 . 0 3  OUT v v_S + -I_B C E -hie oe h 1 I  _C R C R L + B R B _{I B} h fe

(15)

mA 22   E E E R V I mA 088 . 0 1   b I I_B E and





V 12 . 5 5 . 5 10 21.912 200 -15 5 . 5 3 -.           _C _E _CC _C _C CE V V V R I V

(b) The AC equivalent circuit is shown on the right:      _ k I V h BQ I B BE ie 6.82 10 088 . 0 6 . 0 3   B C B E out R I I I v  (  )250(2501) B ie B out ie B in I h v I h I v    250251

Therefore, the voltage gain is

902 . 0   in out V v v A and B C B out I I I i   (1) B B ie B B B in B in I I h I R R v I i    ( 250251 )

and the current gain is

84 . 12 ) 251 250 ( ) 1 (        B B ie B B B in out R I h I I I i i 

(c) To find the input resistance we compute: B ie B in I h I v  250251 B B ie B B in I I h I R i  ( 250251 )

Therefore. the input resistance is

   3558 in in i _i v r

(d) To find the output resistance we compute B C B E out R I I I v  (  )250(2501) B C B out I I I i   (1)

Therefore, the output resistance is    250 out out o i v r

(16)

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 v_o₁.

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 Chapter 10, Solution 42. 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 v_o₂.

In the circuit given in Figure P10.45 the minimum value of vin for a high input is 2.0 V.

(17)

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: mA 4 . 2 2000 2 . 0 5 _  c i , therefore, i_B = i_C/ = 0.24 mA.

(v_in_{)min = 2.0 V and (v}_in_{)max = 5.0 V, therefore, applying KVL: -vin +R}B i_B + 0.6 = 0

or B in B i v

R  0.6. Substituting for (v_in_{)min and (v}_in_{)max , we find the following range for R}_B: Ω 333 . 18 Ω 833 . 5 k R_B k Chapter 10, Solution 46. Known quantities:

For the circuit given in Figure P10.46:

Ω k 27 , kΩ 10 ₁ ₂ 2 1CRC RBR B R . 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.