74

The transistor.

We have already discussed the physical principles behind the operation of the bipolar

junction transistor and so we will now move on to show how it can be used in both

digital and analogue circuits. We know that the bipolar junction transistor is a three

layer structure consisting of a sandwich of p and n type semiconductors. It is a three

terminal device with connections to the two outer layers as well as to the central

control region. There are two possible arrangements, npn and pnp. We shall

concentrate on the npn type although our general conclusions apply equally to the pnp

device except that the polarities of the voltages and currents must be reversed. The

basic npn device together with its circuit symbol is show below.

The arrow on the emitter indicates that the transistor is an npn device. The symbol for

a pnp transistor is similar except that the direction of the arrow is reversed. As we

know the structure looks rather like two pn junctions connected back to back with a

common intermediate connection called the base terminal.

Although the structure resembles two pn junctions the common base region is made

sufficiently thin so as to lead to behaviour which is completely different from that

n

p

n

Collector

Emitter

Base

E

I C

I E

I BB

C

C

B

E

75

which would be expected from two isolated diodes. Let us imagine that the current-

voltage-temperature-etc, dependence in a diode is given by

€

I = f V( )where for

example the functional relationship might be of the form

€

f V( ) = I0 exp eV kT( ) −1( ). Thus in the collector arm the current is given by

€

IC = fC VBC( )and in the emitter arm by

€

IE = fE VBE( ). Note that in general the two functions are not the same,

€

fE .( ) ≠ fC .( )

since the diodes have different geometries etc. The remarkable property of the

transistor is that a large fraction of the current generated in each diode also flows (in

the 'wrong' direction) through the other. This leads to the Ebers-Moll model of

transistor action

where, in general,

€

αE ≠αC . As we have seen we generally arrange to use the

transistor when

€

VC >VB >VE . This very important since it means that

€

VBC =VB −VC

will be negative and the collector diode will be reverse biased and not conduct. This

essentially sets

€

fC VBC( ) = 0 .

€

VBE on the other hand is positive and so a current

€

IE = fE VBE( )will flow. This permits us to simplify the model of the transistor model

dramatically to obtain

C

B

E

!EfE(VBE)

!CfC(VBC)

VBC

VBE

BIB

IE

IC

C

E

! "E

76

where we have written

€

αE =α which, as we have seen, usually has a value very close

to unity, 0.98 or 0.99 being typical values. The equations describing the transistor are

essentially

€

IE = fE VBE( ) ≈ I0 exp eVkT

−1

IC = α IE

There must of course be current balance and so

€

IB + IC = IE . Alternatively we can

substitute

€

IE = IC α and so obtain

€

IC =α1−α

IB = βIB

where β is the current gain of the transistor which of course depends on the material

properties of the semiconductors and the device geometry. It's value can vary

dramatically from transistor to transistor and lies in the range 50 -200. We often take

100 as a typical value. However the important point is, as we have seen, that the

collector current is directly proportional to the base current and so we can think of the

device as being 'current controlled'.

The basic properties of the BJT may be summarised as follows.

(i) When the base-emitter junction is forward biased it drops a nominal voltage of,

say, 0.6 V for a silicon pn junction.

(ii) Since the base-collector junction is reverse biased

€

IC is almost independent of

€

VBC .

(iii)

€

IC is an exponential function of

€

VBEand directly proportional to

€

IB .

These properties are illustrated on the characteristics below. The relationship between

€

IB and VBE may be regarded as the input characteristics and that between

€

IC and VCE

as the output characteristics. We note that there is a very weak VCE dependence of the

input characteristics but that disappears for VCE greater than about one volt. We also

note that it is conventional to relate all voltages to that of the emitter.

77

It is clear that the

€

IB /VBE characteristics are essentially independent of VCE for VCE

greater than about one volt, at which time it is reasonable to set VBE ~0.6 V for all IB.

Similarly it is clear that for VCE greater than about one volt the IC characteristics are

essentially independent of VCE and that IC is directly proportional to IB ,

€

IC − βIB .

This permits us to simplify the characteristics to

0.5 mA5 V

20 V

I B

1 VVBE0

VCE

0 2 4 6 8 10 12

10

20

30

050

100

150

200

250

I B = 300I C (mA)

0VCE

!A

IB

VBE0

IC

VCE

0.6 V

for all VCE

0.5V, say

IB

IC = ! IB

78

A simple transistor circuit.

Consider the following circuit and suppose that we want to know how Vout changes as

VS varies between -10 and +10 Volts. We'll also assume that β = 100 and that VCE >

1.0V.

It is clear from our previous discussion that transistor action does not begin, i.e. the

transistor does not switch on, until VBE reaches 0.6V. VBE thereafter remains fixed at

0.6V. Thus

€

−10 <VS < 0.6 IB = IC = 0 and Vout = 20

VS > 0.6V IB =VS − 0.6

100mA and IC = β

VS − 0.6100

mA

and

€

Vout = 20 − 5IC = 20 − 5β VS − 0.6100

However neither IC or Vout can increase indefinitely. The limit is set by the maximum

value IC can take which is determined primarily by the 5 kΩ load resistor, the supply

voltage and the requirement that VCE > 1V. The limit is therefore given by

5 k!

VS

100 k!

Vout

20V

VBE

IC

IB

79

IC MAX =

€

20 −15

mA = 3.8mA. Any further increase would make the transistor appear

to be a source of energy whereas, of course, all the transistor can actually do is to act

as a variable obstruction to current flow.

The limit of IC = 3.8 mA occurs when VS = 4.4V (β = 100). Thus we may write

€

0.6 <VS < 4.4 Vout = 23− 5VS

and

€

dVout

dVS = − 5

and so, in this region it is reasonable to regard the circuit as an amplifier with gain of

minus 5.

Outside this range IC is either zero – the transistor is cut-off – or limited by the 5 kΩ

resistor --the transistor is saturated --and so could be used as a voltage controlled

SWITCH.

Let's now analyse the circuit again graphically with the aid, this time, of our idealised

characteristics. We know that when the transistor is switched on

€

20 = 5IC +VCE or IC = 4 − VCE5

20

1.0

0.6VS

active (linear)

cut-off

saturated

Vout

4.4

80

We can superimpose this load line on the transistor characteristics in just the same

way that we did for the diode case.

It is clear that as IB increases from zero the values of (IC, VCE) are given by the

intersection of the load line with the horizontal characteristic corresponding to the

particular value of IB of interest. Consequently (IC, VCE) varies from (0, 20V) at point

A to (3.8 mA, lV) at point B.

The transistor switch --the inverter.

It is probably fair to say that more transistors are used as switches in computers and

calculators than are used as linear amplification deices, such as in hi-fi amplifiers and

radios. The fundamental switching circuit is actually the one we have just discussed

and is often called an inverter for reasons which will become apparent.

I C

1V 20V VCE

A

4mA

IB = 38 !AB

RB

RC

0 Input Output

E

VHIE

81

When the input to the circuit is low, 0V, the transistor is switched off, no current

flows and V0 = 20V. On the other hand when the input is high, RB and RC are chosen

such that the maximum possible collector current flows, i.e. the transistor is saturated.

This sets V0 = 0V (if we neglect the small vCE drop). If we set VHI=E then the output

pulse is simply an inverted form of the input pulse -- hence the name.

In terms of the load line construction the change in state of the input corresponds to

an abrupt change in operating point from A (off) to B (on).

I C

VCE

B

A

IB when transistor ‘on’

82

The BJT as an amplifier.

Our intention is to build an amplifier which will be capable of amplifying small a.c.

signals. These signals can be thought of as containing many sinusoidal components

such as

€

vin t( ) = asin ωt( ). Unfortunately, as we have just seen, our transistor cannot

cope with negative signals --it cuts off and transistor action ceases. The way around

this is to bias the transistor by introducing a constant input so as to set a dc output

level somewhere in the middle of the active range of transistor action so that a small

ac waveform can be superimposed on it without the transistor becoming cut-off. The

output will now take the form

€

vout =V0 + Asin ωt( )

where V0 is the constant output voltage caused by the biasing. It is also referred to as

the operating point or quiescent point of the transistor. The small ac signal

€

Asin ωt( )is

superimposed on this. As we have said VO (and a) must be chosen so that VO+a does

not drive the transistor into saturation and VO-a does not cause it to cut-off. Often a is

quite small and in any case, the biasing is usually designed so as to set VO roughly in

the middle of the output range.

We will now begin by discussing methods to reliably set the operating point, i.e. how

to bias the transistor (XIN and XOUT) before moving on to introduce simple techniques

to anaylse the small signal behaviour (xin and xout).

We shall follow the same general method that we developed for the diode to derive

simple relationships between (xin and xout).

XIN + xin sin !t XOUT + xout sin !t

83

Biasing --defining the operating point.

We recall that in order that the transistor operates in the active region it is necessary

that

(i) The base-emitter junction is forward biased so that a current flows into the base.

(ii) The collector-emitter voltage is sufficiently positive that the collector-base

junction is reverse biased.

If these conditions hold then

(iii) The base-emitter voltage is almost constant at -0.6V.

(iv)

€

IC is independent of

€

VCEand

€

IC = βIB .

These simplifications are equivalent to approximating the real characteristics by the

straight line versions we have already discussed. We can now think of biasing as

establishing steady state currents and voltages such that conditions (i) and (ii) are met.

We shall divide the biasing process into two parts. The first will deal with the input

requirements and the second with the output requirements. We shall concentrate on

the common emitter configuration we have already met (it is called common emitter

because the emitter can be thought of as being the common connection between the

input and the output). Let's now try and set condition (i).

Plan A. Why not use a constant voltage source to set VBE to the precise

value required to achieve conduction? This is not sensible because

(i) The exact value of VBE will vary from transistor to transistor.

(ii) IB and IC change very rapidly with VBE and so are likely to be ill defined.

(iii) If VBE really is set to be constant it would have the removing (effectively short-

circuiting) any signal voltage we attempted to impress on VBE ! Let's think again.

84

Plan B. Why not set IBand let VBE have whatever value it likes. The simple circuit

below gives

€

IB =V −VBE

RB

which, since VBE ~ 0.6V, can be well defined if V is large enough. This input biasing

has also set IC via IC = β IB. It now remains to set the collector voltage such that

condition (ii) is met. Again we cannot use a perfect voltage source since this would

effectively short-circuit any signal voltage developed at the collector. We note that we

may write VCE as

€

VCE = E − ICRC

We note that it is usually convenient to use the same supply voltage for the base and

collector currents and so V = E.

These steady state voltages and currents exist in the absence of any input signal. If a

small signal voltage is impressed on this quiescent value of VBE then the base current

will change and so will IC which, in its turn, will cause a change in the output voltage

developed across RC. This is easy to visualise from the load line construction shown

RCRB

IC

+V +E

85

below where we have superimposed the load line (equation above) on the device

characteristics

The quiescent operating point Q is determined by the biasing and lies at the

intersection of the load line with the transistor characteristic for the particular value of

current IB. As the input signal, IB, varies the operating point moves along the load

line. The corresponding changes in IC and VBE can be obtained by projection from the

current and voltage axes. The relationship between the input voltage, VBE and the

input current, IB, can also be seen graphically as shown below.

If the output signal is relatively small then the exact position of Q is not important.

However as we have said before we usually try to position it roughly in the middle of

the output range so as to allow for large output swings and also biasing tolerances, i.e.

we might not get the quiescent point exactly right in practice!

Having said all this I am rather reluctant to admit that this biasing plan is also a bad

idea. It relies on defining IC via IB and we know that this is a rather unreliable

parameter which varies from transistor to transistor. The operating point is not well

defined by this approach and we should try to do better.

Plan C. Since IC is what we are actually trying to set why don't we stop messing

around and try to devise a circuit which would set IC and VCE directly rather than in

some indirect manner. The ideal circuit should be insensitive to any device parameter

which is likely to be ill-defined as well as to temperature. As you might reasonably

imagine many people have given a good deal of thought to this problem and the

I B

IC

VCE

Q

86

circuit we show below is pretty much the standard method of biasing single stage

amplifiers.

The improved circuit provides nearly constant base voltage by potential division from

a reference source, usually E. This means that the emitter voltage is well defined since

VBE is nearly constant. This then gives the emitter current as

€

IE =VE

RE

=VE −VBE

RE

Also

€

IE = IB + IC and

€

IC = βIB and so

€

IC = β1+ β

IE ≈ IE since β is large.

In order to analyse the circuit further let's replace the potential divider by it's

Thevenin equivalent

RC

R2 RE

R1

IC

IE

E

87

where

€

VT =R2

R1 + R2

E and RT = R1R2

R1 + R2

We can now write

€

VT = IBRT +VBE + ICRE

and hence the collector current is given by

€

IC =VT −VBE

RE + RT β

For large β and a proper choice of RT (R1 and R2) and RE , we can arrange that

€

RE >> RT β and hence make IC independent of β.

€

IC =VT −VBE

RE

We now re-write the loop equation for VT above as

RC

RE

IC

IE~IC

E

RT

VT

IB

88

€

VT = IBRT + ICRE +VBE

= ICRTβ

+ RE

+VBE

and see that our decision to set

€

RE >> RT β is equivalent to ignoring the voltage drop

across RT due to the very small base current. Further we can simplify the equation to

€

VT = ICRE +VBE = VE +VBE = VB

which is to be expected as our choice of

€

RE >> RT β is, as we have said, equivalent to

ignoring the small voltage drip across RT.

Finally

€

VCE = E − ICRC −VE

An aside. Although it is not obvious from the above discussion this circuit also

compensates for variations in VBE from device to device or due to temperature

change. We can get a rough idea about operating point 'stability' by differentiating to

obtain

€

dICdβ

=RT VT −VBE( )βRE + RT( )2

and

€

dICdVBE

=−1

RE + RT /β( )

Good stability is achieved if RE and β are high and RT is low.

Another Aside. This may well be the first circuit you have met where negative

feedback --an extremely powerful technique of which you will learn a great deal more

later -- has been used. The resistor, RE, has been used to 'feed back' information about

the output circuit, IC, IE etc. to the input circuit.

89

The base-emitter voltage may be regarded as the input voltage minus a sample of the

output,

€

IERE ≈ ICRE . If , for example, the transistor were to be replaced by one with a

lower β then IC, IE, and VE will fall. However since VT is held constant the fall in VE

will be compensated by an increase in VBE and hence IB. Thus IC recovers to a level

very close to the original.

Choosing the bias resistors --circuit design.

If values are given for the resistors then it is easy to determine the operating point. It

is less easy, but more fun, to choose the values, i,e. to design the circuit since

compromises have to be made and so there are often more than one set

of resistor values which will result in acceptable performance. As an example let's

assume that we want to choose R1, R2, RE and RC so as to set the operating point

nominally at IC = 2mA, VCE = 7V and E = 12V for a transistor with a minimum

current gain, β, of 100.

Since we want to make the base voltage as well defined as possible let's begin by

allowing, say, a l V drop across RE. This is probably reasonable since it won't

dramatically reduce the output swing and is also sufficiently large to swamp any

variations in VBE from transistor to transistor so that VT -VBE is reasonably constant.

RC

RE

12 V

RT

VT

2 mA

7 V

90

This choice of emitter voltage allows us to calculate RE and RC remembering, IC = IE,

as

€

RE = 500 Ω

and

€

RC = 12 − 7 −1( ) /2mA = 2 kΩ

We also know that VB = 1+0.6 = 1.6V if we assume VBE = 0.6. It now remains to find

Rl and R2.

We recall from our previous analysis that

€

IC =VT −VBE

RE + RT β

which will be reasonably independent of β if we choose, say, RE, > 10RT/β. In our

case we will set RT = 10 RE = 5 kΩ. It now remains to find Rl and R2.

We now have one equation relating the two resistors R1 and R2 via, RT = 5 kΩ but we

still need another. We may obtain this equation, in the spirit of our approximation by

setting VT = VB as we have already discussed. Our two equations are now

€

VT =R2

R1 + R2

12 = VB = 1.6 V and RT = R1R2

R1 + R2

= 5 kΩ

from which we find

€

R1 = 37.5 kΩ and

€

R2 = 5.7 kΩ. The closest standard values of

resistor available would suggest that we might choose Rl = 36 kΩ (37.5), R2 = 5.6 kΩ

(5.7), RE= 510 Ω, (500) and RC = 2 Ω (2). This gives rise to IC = 1.82 mA rather than

the nominal value of 2mA. Why is this so different and does it matter? The main

reasons are that we took RE = 10 RT and so we were prepared to accept a 10% ish

error in the analysis. We also ignored the base current. It's not too surprising then that

if we use the actual design values we chose in a rigorous circuit calculation that we

won't get quite the nominal values we aimed for. Whether the discrepency matters or

not depends on the application. In most cases a design values such as those we have

obtained are perfectly acceptable.

91

We emphasise that this is not a unique solution and that the design of circuits and the

choice of component values is not an exact science. The resistors we use will all have

tolerances as will the transistor parameters. It doesn't matter if the actual operating

point isn't exactly at the nominal point since for small signals the swings are unlikely

to cause the transistor either to cut-off or to saturate. We have tried to present a

plausible design philosophy. Naturally many other people have considered the

problem and another rule of thumb which has been found to provide well behaved

circuits is to set

€

VE = 0.1E

R2 =βRE

10

This approach gives, if we again ignore the base current, RE = 600 W , RC = 1.9 kΩ

and Rl = 34 kΩ which gives substantially similar component values to the ones found

from our design.