Bird J. Electrical Circuit Theory and Technology

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Transistors 155

the base of the transistor, the base current varies and can instantaneously

take any of the values between the extremes shown. Only two points

are necessary to draw the line and these can be found conveniently by

considering extreme conditions. From the equation:

D V

 I

(i) when I

D 0,V

D V

(ii) when V

D 0,I

D V

Thus the points A and B respectively are located on the axes of the

characteristics. This line is called the load line and it is dependent

for its position upon the value of V

and for its gradient upon R

.As

the gradient is given by 1/R

, the slope of the line is negative.

For every value assigned to R

in a particular circuit there will be a

corresponding (and different) load line. If V

is maintained constant, all

the possible lines will start at the same point (B) but will cut the I

axis

at different points A. Increasing R

will reduce the gradient of the line

and vice-versa. Quite clearly the collector voltage can never exceed V

(point B) and equally the collector current can never be greater than that

value which would make V

zero (point A).

Using the circuit example of Figure 12.15, we have

D V

D 12V, when I

D 0

1000

D 12mA, when V

D 0

The load line is drawn on the characteristics shown in Figure 12.21, which

we assume are characteristics for the transistor used in the circuit of

Figure 12.15 earlier. Notice that the load line passes through the oper-

ating point X, as it should, since every position on the line represents a

relationship between V

and I

for the particular values of V

and R

given. Suppose that the base current is caused to vary š0.1 mA about the

d.c. base bias of 0.1 mA. The result is I

changes from 0 mA to 0.2mA

and back again to 0mA during the course of each input cycle. Hence the

operating point moves up and down the load line in phase with the input

current and hence the input voltage. A sinusoidal input cycle is shown on

Figure 12.21.

12.8 Current and voltage

gains

The output signal voltage (v

) and current (i

) can be obtained by project-

ing vertically from the load line on to V

and I

axes respectively. When

the input current i

varies sinusoidally as shown in Figure 12.21, then v

varies sinusoidally if the points E and F at the extremities of the input

variations are equally spaced on either side of X.

The peak to peak output voltage is seen to be 8.5 V, giving an r.m.s.

value of 3V (i.e. 0.707 ð 8.5/2). The peak to peak output current is

8.75mA, giving an r.m.s. value of 3.1 mA. From these ﬁgures the voltage

and current ampliﬁcations can be obtained.

156 Electrical Circuit Theory and Technology

2 4 6 8 10 120

(V)

8.5 V pk−pk

8.75 mA

pk−pk

(mA)

Maximum positive excursion

Maximum negative excursion

Mean base bias

= 0.1 mA

= O

= 0.2 mA

Input current

variation is 0.1 mA

peak

Figure 12.21

The dynamic current gain A

(D ˛

as opposed to the static gain ˛

is given by:

change in collector current

change in base current

This always leads to a different ﬁgure from that obtained by the direct

division of I

which assumes that the collector load resistor is zero.

From Figure 12.21 the peak input current is 0.1 mA and the peak output

current is 4.375mA. Hence

4.375 ð10

3

0.1 ð10

3

D 43.75

The voltage gain A

is given by:

change in collector voltage

change in base voltage

This cannot be calculated from the data available, but if we assume that

the base current ﬂows in the input resistance, then the base voltage can

be determined. The input resistance can be determined from an input

characteristic such as was shown earlier.

Then R

change in V

change in I

and v

D i

and v

D i

Transistors 157

and A

D ˛

For a resistive load, power gain, A

, is given by

= A

× A

Problem 4. An n-p-n transistor has the following characteristics,

which may be assumed to be linear between the values of collector

voltage stated.

Base current Collector current (mA) for

(

µA) collector voltages of

1V 5V

30 1.4 1.6

50 3.0 3.5

70 4.6 5.2

The transistor is used as a common-emitter ampliﬁer with load

resistor R

D 1.2k and a collector supply of 7 V. The signal input

resistance is 1k. Estimate the voltage gain A

, the current gain A

and the power gain A

when an input current of 20µA peak varies

sinusoidally about a mean bias of 50

µA.

The characteristics are drawn in Figure 12.22. The load line equation is

D V

 I

which enables the extreme points of the line to be

calculated.

When I

D 0,V

D V

D 7.0V

and when V

D 0, jI

1200

D 5.83mA

The load line is shown superimposed on the characteristic curves with

the operating point marked X at the intersection of the line and the 50

µA

characteristic.

From the diagram, the output voltage swing is 3.6V peak to peak. The

input voltage swing is i

where i

is the base current swing and R

the input resistance.

Therefore

D 70  30 ð 10

6

ð 1 ð 10

D 40mV peak to peak

Hence, voltage gain, A

output volts

input volts

3.6

40 ð 10

3

D 90

Note that peak to peak values are taken at both input and output. There

is no need to convert to r.m.s. as only ratios are involved.

158 Electrical Circuit Theory and Technology

3.0mA

(V)

pk−pk

3.6V

pk−pk

70µA

50µA

30µA

(mA)

234567

Figure 12.22

From the diagram, the output current swing is 3.0mA peak to peak.

The input base current swing is 40

µA peak to peak.

Hence, current gain, A

output current

input current

3 ð 10

3

40 ð 10

6

D 75

For a resistive load R

the power gain, A

, is given by:

D voltage gain ð current gain D A

ð A

D 90 ð 75 D 6750

12.9 Thermal runaway

When a transistor is used as an ampliﬁer it is necessary to ensure that it

does not overheat. Overheating can arise from causes outside of the tran-

sistor itself, such as the proximity of radiators or hot resistors, or within

the transistor as the result of dissipation by the passage of current through

it. Power dissipated within the transistor, which is given approximately by

the product I

, is wasted power; it contributes nothing to the signal

output power and merely raises the temperature of the transistor. Such

overheating can lead to very undesirable results.

The increase in the temperature of a transistor will give rise to the

production of hole electron pairs, hence an increase in leakage current

represented by the additional minority carriers. In turn, this leakage current

leads to an increase in collector current and this increases the product

. The whole effect thus becomes self-perpetuating and results in

thermal runaway. This rapidly leads to the destruction of the transistor.

Problem 5. Explain how thermal runaway might be prevented in

a transistor.

Transistors 159

+ V

Figure 12.23

Two basic methods are available and either or both may be used in a

particular application.

Method 1 is in the circuit design itself. The use of a single biasing

resistor R

as shown earlier in Figure 12.18 is not particularly good prac-

tice. If the temperature of the transistor increases, the leakage current

also increases. The collector current, collector voltage and base current

are thereby changed, the base current decreasing as I

increases. An alter-

native is shown in Figure 12.23. Here the resistor R

is returned, not to

the V

line, but to the collector itself.

If the collector current increases for any reason, the collector voltage

will fall. Therefore, the d.c. base current I

will fall, since I

. Hence the collector current I

D ˛

will also fall and compen-

sate for the original increase.

A commonly used bias arrangement is shown in Figure 12.24. If the

total resistance value of resistors R

and R

is such that the current ﬂowing

through the divider is large compared with the d.c. bias current I

, then

the base voltage V

will remain substantially constant regardless of vari-

ations in collector current. The emitter resistor R

in turn determines the

value of emitter current which ﬂows for a given base voltage at the junc-

tion of R

and R

. Any increase in I

produces an increase in I

and

a corresponding increase in the voltage drop across R

. This reduces

the forward bias voltage V

and leads to a compensating reduction

in I

Figure 12.24

Method 2 concerns some means of keeping the transistor temperature

down by external cooling. For this purpose, a heat sink is employed, as

shown in Figure 12.25. If the transistor is clipped or bolted to a large

conducting area of aluminium or copper plate (which may have cooling

ﬁns), cooling is achieved by convection and radiation.

THICK ALUMINIUM

OR COPPER PLATE

POWER TRANSISTOR

BOLTED TO THE PLATE

Figure 12.25

Heat sinks are usually blackened to assist radiation and are normally

used where large power dissipation’s are involved. With small transistors,

heat sinks are unnecessary. Silicon transistors particularly have such small

leakage currents that thermal problems rarely arise.

12.10 Further problems

on transistors

1 Explain with the aid of sketches, the operation of an n-p-n transistor

and also explain why the collector current is very nearly equal to the

emitter current.

2 Explain what is meant by the term ‘transistor action’.

3 Describe the basic principle of operation of a bipolar junction tran-

sistor including why majority carriers crossing into the base from the

emitter pass to the collector and why the collector current is almost

unaffected by the collector potential.

4 For a transistor connected in common-emitter conﬁguration, sketch

the output characteristics relating collector current and the collector-

emitter voltage, for various values of base current. Explain the shape

of the characteristics.

160 Electrical Circuit Theory and Technology

5 Sketch the input characteristic relating emitter current and the emit-

ter-base voltage for a transistor connected in common-base conﬁgu-

ration, and explain its shape.

6 With the aid of a circuit diagram, explain how the output character-

istics of an n-p-n transistor having common-base conﬁguration may

be obtained and any special precautions which should be taken.

7 Draw sketches to show the direction of the ﬂow of leakage current

in both n-p-n and p-n-p transistors. Explain the effect of leakage

current on a transistor connected in common-base conﬁguration.

8 Using the circuit symbols for transistors show how (a) common-base,

and (b) common-emitter conﬁguration can be achieved. Mark on the

symbols the inputs, the outputs, polarities under normal operating

conditions to give correct biasing and current directions.

9 Draw a diagram showing how a transistor can be used in common

emitter conﬁguration. Mark on the sketch the input and output polar-

ities under normal operating conditions, and current directions.

10 Sketch the circuit symbols for (a) a p-n-p and (b) an n-p-n transistor.

Mark on the emitter electrodes the direction of conventional current

ﬂow and explain why the current ﬂows in the direction indicated.

11 State whether the following statements are true or false:

(a) The purpose of a transistor ampliﬁer is to increase the frequency

of an input signal

(b) The gain of an ampliﬁer is the ratio of the output signal ampli-

tude to the input signal amplitude

current to the base voltage.

(d) The equation of the load line is V

D V

 I

(e) If the load resistor value is increased the load line gradient is

reduced

(f) In a common-emitter ampliﬁer, the output voltage is shifted

through 180

with reference to the input voltage

(g) In a common-emitter ampliﬁer, the input and output currents

are in phase

(h) If the temperature of a transistor increases, V

, I

and ˛

all

increase

(i) A heat sink operates by artiﬁcially increasing the surface area

of a transistor

(j) The dynamic current gain of a transistor is always greater than

the static current.

[(a) false (b) true (c) false (d) true (e) true (f) true

(g) true (h) false (V

decreases) (i) true (j) true]

12 An ampliﬁer has A

D 40 and A

D 30. What is the power gain ?

[1200]

Transistors 161

13 What will be the gradient of a load line for a load resistor of value

4k? What unit is the gradient measured in? [1/4000 siemen]

14 A transistor ampliﬁer, supplied from a 9V battery, requires a d.c. bias

current of 100

µA. What value of bias resistor would be connected

from base to the V

line (a) if V

is ignored (b) if V

is 0.6V?

[(a) 90 k (b) 84 k]

15 The output characteristics of a transistor in common-emitter conﬁg-

uration can be regarded as straight lines connecting the following

points

D 20µA50µA80µA

(V) 1.0 8.0 1.0 8.0 1.0 8.0

(mA) 1.2 1.4 3.4 4.2 6.1 8.1

Plot the characteristics and superimpose the load line for a 1k load,

given that the supply voltage is 9 V and the d.c. base bias is 50

µA.

The signal input resistance is 800  . When a peak input current

of 30

µA varies sinusoidally about a mean bias of 50 µA, determine

(a) the quiescent collector current (b) the current gain (c) the voltage

gain (d) the power gain

[(a) 4 mA (b) 104 (c) 83 (d) 8632]

16 Explain brieﬂy what is meant by ‘thermal runaway’.

Assignment 3

This assignment covers the material contained in chapters 8

to 12.

The marks for each question are shown in brackets at the end of

each question.

1 A conductor, 25cm long, is situated at right angles to a magnetic

ﬁeld. Determine the strength of the magnetic ﬁeld if a current of

12A in the conductor produces a force on it of 4.5N. (3)

2 An electron in a television tube has a charge of 1.5 ð 10

19

Cand

travels at 3 ð 10

m/s perpendicular to a ﬁeld of ﬂux density 20 µT.

Calculate the force exerted on the electron in the ﬁeld. (3)

3 A lorry is travelling at 100km/h. Assuming the vertical component

of the earth’s magnetic ﬁeld is 40

µT and the back axle of the lorry

is 1.98 m, ﬁnd the e.m.f. generated in the axle due to motion. (5)

4 An e.m.f. of 2.5kV is induced in a coil when a current of 2 A

collapses to zero in 5 ms. Calculate the inductance of the coil. (4)

5 Two coils, P and Q, have a mutual inductance of 100mH. If a current

of 3 A in coil P is reversed in 20 ms, determine (a) the average e.m.f.

induced in coil Q, and (b) the ﬂux change linked with coil Q if it

wound with 200 turns. (5)

6 A moving coil instrument gives a f.s.d. when the current is 50mA and

has a resistance of 40. Determine the value of resistance required

to enable the instrument to be used (a) as a 0–5A ammeter, and

(b) as a 0–200V voltmeter. State the mode of connection in each

case. (6)

7 An ampliﬁer has a gain of 20 dB. It’s input power is 5mW. Calculate

it’s output power. (3)

8 A sinusoidal voltage trace displayed on a c.r.o. is shown in

Figure A3.1; the ‘time/cm’ switch is on 50ms and the ‘volts/cm’

switch is on 2V/cm. Determine for the waveform (a) the frequency

(b) the peak-to-peak voltage (c) the amplitude (d) the r.m.s. value.

(7)

Figure A3.1

9 With reference to a p-n junction, brieﬂy explain the terms:

(a) majority carriers (b) contact potential (c) depletion layer

(d) forward bias (e) reverse bias. (5)

10 The output characteristics of a common-emitter transistor ampliﬁer

are given below. Assume that the characteristics are linear between

Assignment 3 163

the values of collector voltage stated.

D 10µA40µA70µA

(V) 1.0 7.0 1.0 7.0 1.0 7.0

(mA) 0.6 0.7 2.5 2.9 4.6 5.35

Plot the characteristics and superimpose the load line for a 1.5k

load resistor and collector supply voltage of 8V. The signal input

resistance is 1.2k. Determine (a) the voltage gain, (b) the current

gain, and (c) the power gain when an input current of 30

µA peak

varies sinusoidally about a mean bias of 40

µA(9)

Main formulae for Part 1

General

Charge Q D It Force F D ma Work W D Fs Power P D

Energy W D Pt

Ohm’s law V D IR or I D

or R D

Conductance G D

Power P D VI D I

R D

Resistance R D

l

Resistance at 

C, R



D R

1 C ˛



Terminal p.d. of source, V D E  Ir

Series circuit R D R

C R

CÐÐÐ

Parallel network

CÐÐÐ

Capacitors and

capacitance

E D

C D

Q D It D D

D ε

C D

An  1

Capacitors in parallel C D C

C C

CÐÐÐ

Capacitors in series

CÐÐÐ W D

Magnetic circuits

B D



D NI H D

D 



S D

mmf





Electromagnetism F D Bil sinFD QvB

Electromagnetic

induction

E D Blv sinEDN

d

DL

W D

L D

N

DM

Measurements

Shunt R

Multiplier R

V  Ir

Power in decibels D 10 log

D 20 log

Wheatstone bridge R

Potentiometer E

D E



