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 accurate—much 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

ν1 ν2

i2

Fig. 2

Fig. 1

RS

RBRC

RL

b

C

VS Vbe

VCC

Vce

(a) circuit

RS

RB

RC RL

b

C

VS

Vbe

Vce

(b) a.c. equivalent

RL

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 ...............

i

i 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 h

i h i hie 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

1

1i 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

i

v2

2 siemen

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

Thus, h f

i

i2

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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70 Analysis of Small-signal Transistor Amplifi ers

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

RS

RB

hie hfei1i1

b

ee

1/hoeν1

νS

ν2

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

1

2

h i

i h i hi

f o

…………...........…

……………

[ ]

[ ]

and using these equations the following results can be obtained.

Amplifier current gain, A

h

h Rif

o L

1

(1)

Amplifier voltage gain, A

h R

h h R

A 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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Analysis of Small-signal Transistor Amplifi ers 71

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 kΩ600 Ω

RC2.2 kΩ

RB68 kΩ

VS

100 mV rms

Fig. 4

i2 iL

90i

RL

iS i1

νSν1 ν2

hie 1.5 kΩ

1/hoe 20 kΩ

0.1 V rms

RS 600 Ω

RB 68 kΩ

RC 2.2 kΩ

RL 10 kΩ

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

.

.

1

1 k 1.8 k

Input circuit: R in

h Rh R

ie B

ie B

68 568 5

1

1

.

. 1.47 k

Using potential divider technique:

vR

R Rv

v

in

s ins1

1

1

11

1

volt V

mV

.. .

.47

47 0 60

7

i 1 vhie

1

7 05 0

3

3

1 1

1 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

1

1 1

1

1/

/ ..

.

amp

mA

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72 Analysis of Small-signal Transistor Amplifi ers

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 1

1

..

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

1

1

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 Ω

VS0.2 Vp-p

VCC

R220 kΩ

R1120 kΩ

RE1 kΩ

RL5 kΩ

RC4.7 kΩ

Fig. 6

A

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

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Analysis of Small-signal Transistor Amplifi ers 73

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

νS

ν1 ν2

i1iS

hie2 kΩ

1/hoe16.7 kΩ

i2 iL

RL

100i1Rs600 Ω

R1120 kΩ

R220 kΩ

RC4.7 kΩ

RL5 kΩ

Rin

0.2 Vp-p

Fig. 7

Input circuit:

1 1 1 1

1R R R hin ie

2 siemen

1

1

1 1

20 20 2

mS

1 1

1 1

1

R

R

in

in

6 6020

6720

79

mS

so, k.

vR

R Rv

v

i

in

s ins1

1

1

1

1

1

.. .

790 6 79

200

50

mV pk-pk

mV pk-pk

vhie

1 1amp A pk-pk

0 52000

75.

µ

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

1

11

1

11/

/.

. ..

.

’ mA pk-pk

mmA pk-pk

Aii

A

i

i

2

3

6

6 55 075 0

87 3

1

1

1

.

. Ans

Check:

A

hh Rife

oe L

1

1

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

Li 2 2

3 72 2

2.

.1

1 1mA mA

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74 Analysis of Small-signal Transistor Amplifi ers

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

1

1

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

R

R rg r Vo

L

L dsm ds i

volt

V

VA

g r R

R ro

i

m ds L

L ds

(3)

but in practice, r ds RL , so

V

V

g r R

r

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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Analysis of Small-signal Transistor Amplifi ers 75

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

Vi

V0

VDD

56 kΩ2 kΩ

RG1 MΩ

3 kΩ

R2

4.6 kΩ

RS1 kΩ

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 1

1 1 100384 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

1

1

1

2

2

6056 4 7

60 7

0043

ohm

M (say

..

. Ans M )1

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76 Analysis of Small-signal Transistor Amplifi ers

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, A

h

h Rif

o L

1

Voltage gain,

Power gain,

AA R

h

A A A

vi L

i

p i v

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

or, more accurately,

A

g r R

r Rvm 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 kΩ120 kΩ

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 MΩ

15 kΩ

20 kΩ

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 kΩ3.9 kΩ

500 Ω 7.5 kΩ4.7 kΩ0.2 V pk-pk

V1

Fig. 13

V1

VDD 30 V

56 kΩ 10 kΩ

1.2 MΩ

15 kΩ

4.7 kΩ 4.7 kΩ3 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 kΩ

5 kΩ

90 kΩ

1.2 MΩ

(a)

Fig. 15

i1

600 Ω

V1

82 kΩ 10 kΩ3 kΩ

25 kΩ 4.7 kΩ 10 kΩ

50i1

(b)

Analysis of Small-signal Transistor Amplifi ers 77

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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 µ S respectively.

VCC

4.7 kΩ100 kΩ

10 kΩ600 Ω 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/hoe

V220 kΩ4 kΩ100 kΩ 4.7 kΩ 10 kΩ

i1

VS

RB 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

1

17776.8mV 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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Analysis of Small-signal Transistor Amplifi ers 79

Ah R

h h RR

h Rh Rv

fe L

ie oe LL

oe L

oe L

( where

/ 474.7

31

1

1 1) /..2 k

so, ( ) 774( 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 kΩ

100 kΩ39 kΩ

25 kΩ2 MΩ

Fig. 18

R

R RR RL

D 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

80 Analysis of Small-signal Transistor Amplifi ers

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