Small-signal Transistor Amplifiers

Analysis of Small-signal Transistor Amplifi ers

On completion of this chapter you should be able to predict the behaviour of given transistor amplifi er circuits by using equations and/or equivalent circuits that represent the transistor s a.c. parameters.

67

1 Reasons for Adopting this Technique

The gains of an amplifi er circuit may be obtained by drawing the load lines on the plotted output characteristics. However, for a number of reasons, this is not a truly practical method.

(a) Manufacturers do not provide graphs or data to enable the characteristics to be plotted.

(b) Even if such data were available, the process would be very time consuming.

(c) Obtaining results from plotted graphs is not always very accuratemuch depends upon the skill and interpretation of the individual concerned.

For these reasons an alternative method, which involves the use of equations and/or simple network analysis, is preferred. This method involves the use of the transistor parameters, the data for which is provided by manufacturers. This information is most commonly obtained from component catalogues produced by suppliers such as Radio Spares and Maplin Electronics.

2 BJT Parameters

You should already be familiar with the d.c. parameters such as input resistance ( R IN ), output resistance ( R OUT ), and current gain ( h FE ), and their relationship to the transistor s output characteristics. In addition, an a.c. amplifi er circuit may be redrawn in terms of the appearance of the circuit to a.c. signals. This is illustrated in Fig. 1.

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68 Analysis of Small-signal Transistor Amplifi ers

The a.c. equivalent circuit of Fig. 1(b) is useful in that the current fl ow paths of the a.c. signal and the effective a.c. load can be appreciated, but in order to analyse the complete amplifi er circuit the load lines would still need to be drawn on the characteristics. What is required is a simple network representation of the transistor itself, which can then be inserted into Fig. l(b) in place of the transistor symbol. There are a variety of transistor parameters that may be used in this way. Amongst these are Z-parameters, Y-parameters, hybrid parameters, and h-parameters. For the analysis of small-signal audio frequency amplifi ers the use of h-parameters is the most convenient, and will be the method adopted here.

Provided that the transistor is correctly biased and the input signal is suffi ciently small so as to cause excursions of currents and voltages that remain within the linear portions of the characteristics, then the transistor itself may be considered as a simple four-terminal network as shown in Fig. 2 .

Linear network

i1

1 2

i2

Fig. 2

Fig. 1

RS

RB RC

RL

b

C

VS Vbe

VCC

Vce

(a) circuit

RS

RB

RC RL

b

C

VS

Vbe

Vce

(b) a.c. equivalentRL

The relationships between the four quantities of a linear network can be expressed by a number of equations, two of which are:

1 1 2

2 1 2

A B ..................[1]C D ...............

ii i ....[2]

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Analysis of Small-signal Transistor Amplifi ers 69

Examination of the units involved in these two equations reveals that A must be an impedance (ohm), B and C are dimensionless (ratios), and D must be an admittance (siemen). Since there is a mixture or hybrid of units involved, they are known as the hybrid or h-parameters, having the following symbols:

A ohm; B ; C ; D siemen h h h hi r f o

If the transistor is conected in common emitter confi guration the two equations would be written as follows

1 1 2

2 1 2

h i hi h i h

ie re

fe oe

If the transistor is connected in common base confi guration then the parameters would be h ib , h rb , h fb and h ob respectively.

The h-parameters are defi ned as follows:

h i : is the input impedance with the output short-circuited to a.c.

Thus, h i

11i

ohm

h r : is the reverse voltage feedback ratio with the input open-circuited to a.c.

Thus, h r

1

2

h o : is the output admittance with the input open-circuited to a.c.

Thus, h 0

iv

2

2

siemen

h f : is the forward current gain with the output short-circuited to a.c.

Thus, h f

ii2

1

Notes:

1 In modern transistors h r is very small ( 10 4 ) so this parameter will be ignored.

2 Just as conductance G 1/ R siemen, so admittance, Y 1/ Z siemen. 3 The h-parameters will vary with temperature, ageing and frequency.

For the analysis at this level we shall consider that they remain constant.

4 Since the transistor is a current-operated device it is convenient to represent its collector circuit as a current generator with its internal impedance (1/ h o ) in parallel.

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Considering the amplifi er circuit of Fig. 1 , the complete h-parameter equivalent circuit would be as shown in Fig. 3 .

RSRB

hie hfei1i1

b

ee

1/hoe1

S

2

i2

RL

c

Fig. 3

For practical purposes it may be assumed that the h-parameters will have the same numerical values as their d.c. counterparts

i.e. h R h h h Ri IN f F o OUT ; ; /1

3 h-parameter Equations

Ignoring h r the original two equations may be written as:

1 1

2 1 2

12

h ii h i h

i

f o

...........

[ ][ ]

and using these equations the following results can be obtained.

Amplifier current gain, Ahh Rif

o L

1

(1)

Amplifier voltage gain, Ah R

h h RA R

hf L

i o L

i L

i

( )1

(2)

Thus, knowing the values for a transistor s h-parameters, the prediction of amplifi er gains can simply be obtained by either using the above equations or by simple network analysis using the h-parameter equivalent circuit.

Worked Example 1

Q For the amplifi er circuit of Fig. 4 , (a) sketch the h-parameter equivalent circuit and, (b) determine the amplifi er current and voltage gains using (i) network analysis, and (ii) h-parameter equations.

The h-parameters are h ie 1.5 k ; h fe 90; h oe 50 S

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A

h ie 1.5 k ; h fe 90; h oe 50 10 6 S

(a) The h-parameter circuit will be as shown in Fig. 5 .

RS

RL

VCC 12 V

10 k600

RC2.2 k

RB68 k

VS

100 mV rms

Fig. 4

i2 iL

90i

RL

iS i1

S1 2

hie 1.5 k

1/hoe 20 k

0.1 V rms

RS 600

RB 68 k

RC 2.2 k

RL 10 k

Rin

Fig. 5

(i) 1/h oe

1050

6

20 k ;

RL

R RR R

C L

C L ohm

2 2 02 2 0

.

.

11

k 1.8 k

Input circuit: R in

h Rh R

ie B

ie B

68 568 5

11

.

. 1.47 k

Using potential divider technique:

vR

R Rv

v

in

s ins1

1

11

1

1

volt V

mV

.. .

.47

47 0 60

7

i 1 vhie

1

7 05 0

3

3

1 11 1

. amp 47.3 A

Output circuit: 90 i 1 90 47.3 10 6 4.26 mA Using current divider technique:

ih

h Ri

i

oe

oe L2

2

9020

20 84 26

3 9

11 1

1

1/

/ ..

.

amp

mA

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v 2 i 2 RL volt 4 10 3 1.8 10 3 v 2 7.2 V

A i

ii2

3

6

3 9 047 3 01

1 11

..

82.7 Ans

A v

vv

23

7 047 01 1 1

. 99 Ans

(ii)

Ahh R

A

AA R

ife

oe L

i

vi L

1 1 1 1

1

..

9050 0 800

9009

82 6

6( )

Ans

. ..h

Aie

v

82 6 85

99

11

Ans

Thus, allowing for the cumulation of rounding errors in part (i), the results from the equations agree with those from the network analysis.

The actual current that will fl ow in the load of the previous example will not in fact be i 2 , but only a fraction of that, and is shown in Fig. 5 as i L . Thus the power delivered to the external load will be less than the maximum possible. This problem may be minimised by the use of a matching transformer connected between the load and the amplifi er circuit output terminals.

Worked Example 2

Q The transistor used in the circuit of Fig. 6 has the following h-parameters h ie 2 k ; h oe 60 S; h fe 100. Calculate (a) the amplifi er current gain, (b) the actual power delivered to the external load, and (c) the turns ratio required for a matching transformer in order to maximise the power delivered to the load.

RS600

VS0.2 Vp-p

VCC

R220 k

R1120 k

RE1 k

RL5 k

RC4.7 k

Fig. 6

A

h ie 2 k ; h oe 60 10 6 S; h fe 100

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(a) The h-parameter equivalent circuit is shown in Fig. 7 .

S

1 2

i1iS

hie2 k

1/hoe16.7 k

i2 iL

RL

100i1Rs600 R1

120 kR220 k

RC4.7 k

RL5 k

Rin

0.2 Vp-p

Fig. 7

Input circuit:

1 1 1 11R R R hin ie

2

siemen

11

1 120 20 2

mS

1 11 1

1

R

R

in

in

6 6020

6720

79

mS

so, k.

vR

R Rv

v

i

in

s ins1

1

1

11

1

.. .

790 6 79

200

50

mV pk-pk

mV pk-pk

vhie

1 1amp A pk-pk0 52000

75.

Output circuit:

R

R RR RL

C L

C L

..

4 7 54 7 5

k

R

iL .

.

2 42

00 7 5

k

mA pk-pk

1 1

ih

h Ri

i

oe

oe L2

2

006 7

6 7 2 427 5

6 55

11

1 111

//

.. .

.

.

mA pk-pk

mmA pk-pk

Aii

A

i

i

2

3

6

6 55 075 0

87 3

1

11

.

. Ans

Check:

A

hh Rife

oe L

11

1 1.

0060 0 2420

87 36( )

(b)

iR

R RiL

C

C L

2

4 79 7

6 55..

. mA pk-pk

iP RL

L L L L

3 72

.1 mA pk-pk watt, where is the valuI I r.m.s. ee

so,

IL L

i

2 23 72 2

2.

.1 1 1mA mA

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P

P

L

L

( )

mW

1 1 1.

.

2 0 5000

6 3

3 2

Ans (c) For maximum power transfer, R L must match the parallel combination of

1/ h oe and R c call this R p .

R

RN

NR

N

N

R

R

p

pp

sL

p

s

p

L

11

6 7 4 76 7 4 7

3 67

3

2

. .

. ..k k

ohm

so,

..

.

675

0 856N

Np

s : 1 Ans

4 FET Parameters and Equivalent Circuits

Since a FET has an extremely high input impedance then its input circuit may be represented simply as an open circuit. Also, being a voltage operated device it is convenient to represent the output circuit as a voltage source with the internal resistance ( r ds ) in series with it. The small-signal equivalent circuit will therefore be as shown in Fig. 8 . The FET parameters r ds and g m should already be familiar to you.

From Fig. 8 :

V RR r

g r Vo LL ds

m ds i

volt

VV

Ag r RR r

o

i

m ds L

L ds

(3)

but in practice, r ds RL , so

VV

g r Rr

A g R

o

i

m ds L

ds

m L

and,

(4)

S

G D

S

RL

rds

V0Vi

d

gmrdsVi

Fig. 8

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Worked Example 3

Q The FET used in the amplifi er circuit of Fig. 9 has parameter values of r ds 80 k and g m 4 mS. Calculate (a) the amplifi er voltage gain, and (b) the eff ective input resistance of the amplifi er circuit.

R1RD

RL

ViV0

VDD

56 k2 k

RG1 M

3 k

R24.6 k

RS1 k

Fig. 9

A

r ds 80 10 3 ; g m 4 10 3 S; R L 3 k

(a) For this circuit, the eff ective a.c. load,

RR R

R RR

LD L

D L

L

.

ohm k

k

3 25

2

1 and since r ds RL , then equation (4) may be used

so,

A g R

A

v m L

v

4 0 2 0

4 8

3 31 1 1.

. Ans

In order to check the validity of using the approximation of equation (4), we can also calculate the gain using equation (3) and compare the two answers.

Thus, Ag r Rr Rv

m ds L

ds L

..

4 0 80 0 2 080 0 2

3 3 3

3

1 1 1 11 1 100

384 0008 200

4 73

3

1

Av . , which confirms the validity of equaation (4)

Note that a FET amplifi er provides very much less voltage gain than a comparable BJT amplifi er.

(b) Looking in at the input terminals, for a.c. signals, the gate resistor R G is in series with the parallel combination of R 1 and R 2 , as shown in Fig. 10 .

R RR R

R R

R

in G

in

1

11

1

2

2

6056 4 7

60 7

0043

ohm

M (say

..

. Ans M )1

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Thus, the inherently high input resistance of the FET is preserved in the amplifi er circuit by the inclusion of R G .

5 Practical Implications

It should be borne in mind that when designing an amplifi er circuit, the results of the equations as shown in this chapter give only theoretical answers. If an amplifi er circuit thus analysed is then constructed and tested, the actual gain fi gures achieved may well be different to those predicted. There are a number of reasons for this: the resistors will have actual values depending upon how close to tolerance they are, and the transistor parameters cannot be guaranteed to be exactly those quoted by the manufacturer. Indeed, manufacturers recognise this by quoting minimum, maximum and typical values for such parameters as h f . In calculations the typical value is normally used. Thus the mathematical analysis should be considered as only the fi rst step in the design process, and component values will then need to be adjusted in the light of practical tests.

Rin

G

S

RG

R1 R2

Fig. 10

Summary of Equations

BJT amplifi er: Current gain,

Ahh Rif

o L

1

Voltage gain,

Power gain,

AA R

hA A A

vi L

i

p i v

FET amplifi er: Approx. voltage gain, A v g m R L

or, more accurately,

Ag r Rr Rv

m ds L

ds L

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Assignment Questions

1 The h-parameters for the transistor used in the circuit of Fig. 11 are h fe 250, h ie 5 k , and h oe 40 S.

(a) sketch the h-parameter equivalent circuit and hence, or otherwise,

(b) calculate the amplifi er current, voltage and power gains.

4 The transistor of the amplifi er circuit shown in Fig. 13 has the following parameters: h ie 2.5 k , h fe 120, and h oe 100 S. Sketch the equivalent circuit and determine the amplifi er current and voltage gains, and the power dissipated in the external 7.5 k load.

VCC

V0V1

4.7 k120 k

0.25 V pk-pk

Fig. 11

2 The circuit of Fig. 11 is now reconnected so that the transistor is connected in common base confi guration. If the common base parameters h ib and h ob are 100 and 20 S respectively, (a) sketch the equivalent circuit, and

(b) calculate the amplifi er current, voltage and power gains.

3 Figure 12 shows a simply biased common source FET amplifi er, where the transistor parameters are g m 3 mS, and r ds 75 k . Calculate the amplifi er voltage gain.

1 M

15 k

20 k

RS

VDD

Fig. 12

5 The parameters for the FET in Fig. 14 are r ds 85 k and g m 4.1 mS. (a) calculate the amplifi er voltage gain, and

(b) the power dissipated in the external 15 k load.

VCC

56 k3.9 k

500 7.5 k4.7 k0.2 V

pk-pkV1

Fig. 13

V1

VDD 30 V

56 k 10 k

1.2 M

15 k

4.7 k 4.7 k3 V

pk-pk

Fig. 14

6 For the two equivalent circuits shown in Figs. 15(a) and (b) , sketch the amplifi er circuits that they represent, showing component values, and also identify the values for the transistor parameters in each case.

V1

120 V1

6.8 k

5 k90 k

1.2 M

(a)

Fig. 15

i1

600

V1

82 k 10 k 3 k25 k 4.7 k 10 k

50i1

(b)


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78

Supplementary Worked Example 1

Q Calculate the minimum value of h fe required for the transistor in Fig. 16 in order that a power of 3.5 mW is dissipated in the 10 k load resistor. The values for h ie and h oe are 4 k and 50 S respectively.

VCC

4.7 k100 k

10 k600 0.25 V pk-pk

Fig.16

A

The h-parameter equivalent circuit is shown in Fig. 17 . Since R B h ie then the shunting eff ect of R B will be negligible, and it has therefore been omitted from the calculation.

h ie 4000 ; h oe 50 10 6 S; V i 0.25 V pk-pk; P o 3.5 10 3 W

RC RL1/hoeV220 k4 k100 k 4.7 k 10 k

i1

VSRB hie

0.25 V pk-pk

V1

hfei1

Fig. 17

PVR

V P R

V

Vh

h

0L

0 L

ie

22

2

2

watt so volt 3.5 0 0

5.9 6 V

3 41 1

1

1iie s

sRV

V

V

volt pk-pk 4

4.60.25

0.2 7V pk-pk

so, 0.2

1

1

1177

76.8mV r.m.s.2 2

Voltage gain required,

AVV

A

v

v

2 5.9 60.768

771

1

Analysis of Small-signal Transistor Amplifi ers

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Ah R

h h RR

h Rh Rv

fe L

ie oe LL

oe L

oe L

( where

/ 474.7

31

11 1) /

..2 k

so, { ( )} 77{4( 50 0 3.2 06 3

hA h h R

Rfev ie oe L

L

1 1 1 1 )}33.2

2 hie 11 Ans

Supplementary Worked Example 2

Q The FET in the circuit of Fig. 18 has r ds 50 k and g m 5 mS. Determine the value of the output voltage, V 2 , and the power developed in the 25 k load.

A r ds 50 10 3 ; g m 5 10 3 S

V2V1 4 V

pk-pk

VDD 40 V

2.2 k

100 k39 k

25 k2 M

Fig. 18

RR R

R RLD L

D L

ohm

25 3.925 3.9

5.2 k1

Now, since r ds is NOT RL , then the approximate equation for voltage gain should not be used, hence

Ag r Rr R

A

vm ds L

ds L

v

1 1 1 1 1

1

.5 0 50 0 5.2 065.2

7.48

Thu

3 3 3

ss, 7.48 4 V pk-pk 70 V pk-pk

so, 24.75 V

w

V

V

PVRL

2

2

022

1

Ans

aatt24.75

25 0

0.2 mW

2

3

1

P0 Ans

Note that had the approximate equation A v g m RL been used in this case an error of about 22% would have resulted in the value for A v . This would be an unacceptably large error.

The approximate form of the equation should be used only when r ds is at least 10 times larger than RL .

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Answers to Assignment Questions

1 (b) A i 210; A v 197; A p 41 370 2 (b) A i 0.91; A v 42.8; A p 39 3 25.7 4 A i 95.5; A v 98.1; P o 6.5 mW 5 A v 24.6; P o 45.4 mW


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Small-signal Transistor Amplifiers