Analysis of Small-signal
Transistor Amplifiers
On completion of this chapter you should be able to predict the behaviour of given transistor
amplifier circuits by using equations and/or equivalent circuits that represent the transistor’s
a.c. parameters.
1
Reasons for Adopting this Technique
The gains of an amplifier 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 (RIN), output resistance (ROUT), and current gain (hFE), and
their relationship to the transistor’s output characteristics. In addition,
an a.c. amplifier circuit may be redrawn in terms of the appearance of
the circuit to a.c. signals. This is illustrated in Fig. 1.
67
68
Analysis of Small-signal Transistor Amplifiers
VCC
C
C
RC
RB
RS
b
RS
b
RC
RL
RB Vbe
RL
Vce
VS
Vce
VS
Vbe
RL
(b) a.c. equivalent
(a) circuit
Fig. 1
The a.c. equivalent circuit of Fig. 1(b) is useful in that the current flow
paths of the a.c. signal and the effective a.c. load can be appreciated,
but in order to analyse the complete amplifier 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 amplifiers 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
sufficiently 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.
i2
i1
ν1
Linear
network
ν2
Fig. 2
The relationships between the four quantities of a linear network can
be expressed by a number of equations, two of which are:
1 Ai1 B 2 ..................[1]
i2 C i1 D2 ..................[2]
Analysis of Small-signal Transistor Amplifiers
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 hi ohm;
B hr ;
C hf ;
D ho siemen
If the transistor is conected in common emitter configuration the two
equations would be written as follows
1 hie i1 hre 2
i2 h fe i1 hoe 2
If the transistor is connected in common base configuration then the
parameters would be hib, hrb, hfb and hob respectively.
The h-parameters are defined as follows:
hi: is the input impedance with the output short-circuited to a.c.
Thus, hi
1
ohm
i1
hr: is the reverse voltage feedback ratio with the input open-circuited
to a.c.
Thus, hr
1
2
ho: is the output admittance with the input open-circuited to a.c.
Thus, h0
i2
siemen
v2
hf: is the forward current gain with the output short-circuited to a.c.
Thus, h f i2
i1
Notes:
1 In modern transistors hr is very small (104) 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/ho) in parallel.
69
70
Analysis of Small-signal Transistor Amplifiers
Considering the amplifier circuit of Fig. 1, the complete h-parameter
equivalent circuit would be as shown in Fig. 3.
c
b
i1
hie
RS
i2
hfei1
RB
1/hoe
ν1
ν2
RL
νS
e
e
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. hi RIN ;
h f hF ;
1 / ho ROUT
3 h-parameter Equations
Ignoring hr the original two equations may be written as:
1 hi i1…………...........…[1]
i2 h f i1 ho 2 ……………[2]
and using these equations the following results can be obtained.
Amplifier current gain, Ai Amplifier voltage gain, A hf
(1)
1 ho RL
h f RL
hi (1 ho RL )
Ai RL
hi
(2)
Thus, knowing the values for a transistor’s h-parameters, the prediction
of amplifier 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 amplifier circuit of Fig. 4, (a) sketch the h-parameter equivalent circuit and, (b) determine the
amplifier current and voltage gains using (i) network analysis, and (ii) h-parameter equations.
The h-parameters are hie 1.5 k; hfe 90; hoe 50 ␮S
Analysis of Small-signal Transistor Amplifiers
VCC 12 V
RC
2.2 kΩ
RB
68 kΩ
RS
VS
RL
600 Ω
10 kΩ
100 mV
rms
Fig. 4
A
hie 1.5 k; hfe 90; hoe 50 106 S
(a) The h-parameter circuit will be as shown in Fig. 5.
iS
i2
i1
iL
90i
RS
600 Ω
ν1
νS
1/hoe
20 kΩ
RB
hie
68 kΩ 1.5 kΩ
ν2
RC
2.2 kΩ
0.1 V
rms
Rin
RL
Fig. 5
(i)
1/hoe RC RL
2.2 10
106
20 k; RL’ ohm k 1.8 k
R
R
2.2 10
50
C
L
Input circuit: Rin hie RB
68 1.5
1.47 k
hie RB
68 1.5
Using potential divider technique:
v1 Rin
1.47
v s volt 0.1 V
Rs Rin
1.47 0.6
v1 71 mV
i1 v1
71 103
amp 47.3 A
hie
1 .5 1 03
Output circuit: 90i1 90 47.3 106 4.26 mA
Using current divider technique:
i2 20
1 / hoe
90i1 amp 4.26
20 1.8
1 / hoe RL
i2 3.91 mA
RL
10 kΩ
71
72
Analysis of Small-signal Transistor Amplifiers
v2 i2 RL’ volt 4 103 1.8 103
v2 7.2 V
(ii)
Ai i2
3.91 103
82.7 Ans
i1
47.3 106
Av v2
7.04
99 Ans
v1 71 103
Ai 90
hfe
1 hoe RL’
1 (50 106 1800 )
Ai 90
82.6 Ans
1.09
Av Ai RL’
82.6 1.8
1.5
hie
Av 99 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 flow in the load of the previous example
will not in fact be i2, but only a fraction of that, and is shown in Fig. 5
as iL. 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 amplifier
circuit output terminals.
Worked Example 2
Q
The transistor used in the circuit of Fig. 6 has the following h-parameters hie 2 k; hoe 60 ␮S;
hfe 100. Calculate (a) the amplifier 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.
VCC
RC
4.7 kΩ
R1
120 kΩ
RS
600 Ω
R2
20 kΩ
RE
1 kΩ
VS
0.2 Vp-p
Fig. 6
A
hie 2 k; hoe 60 106 S; hfe 100
RL
5 kΩ
Analysis of Small-signal Transistor Amplifiers
(a) The h-parameter equivalent circuit is shown in Fig. 7.
i1
iS
i2
iL
100i1
Rs
600 Ω
R1
120 kΩ
ν1
R2
20 kΩ
hie
2 kΩ
1/hoe
16.7 kΩ
RC
4.7 kΩ
ν2
νS
0.2 Vp-p
Rin
RL
Fig. 7
Input circuit:
1
1
1
1
1
1
1
siemen mS
Rin
R1
R2
hie
120 20 2
67
1
1 6 60
mS
Rin
120
120
so, Rin 1.79 k v1 Rin
1.79
vs 200 mV pk-pk
Rs Rin
0.6 1.79
v1 150 mV pk-pk
i1 Output circuit: RL’ v1
0.15
75 µA pk-pk
amp hie
2000
RC RL
4.7 5
k
RC RL
4.7 5
RL’ 2.42 k 100i1 7.5 mA pk-pk
i2 1 / hoe
16.7
100i1 7.5 mA pk-pk
1 / hoe RL’
16.7 2.42
i2 6.55 mA pk-pk
Ai i2
6.55 103
i1
75 106
Ai 87.3 Ans
Check: Ai (b) iL 100
hfe
87.3
1 hoe RL’
1 (60 106 2420 )
RC
4.7
i2 6.55 mA pk-pk
RC RL
9.7
iL 3.17 mA pk-pk
PL I L2RL watt, where I L is the r.m.s. value
so, I L iL
3.17
mA 1.12 mA
2 2
2 2
RL
5 kΩ
73
74
Analysis of Small-signal Transistor Amplifiers
PL (1.12 103 )2 5000
PL 6.3 mW Ans
(c) For maximum power transfer, RL must match the parallel combination of
1/hoe and Rc—call this Rp.
Rp Rp so,
Np
Ns
Np
Ns
16.7 4.7
k 3.67 k 16 .7 4 .7
N p2
Ns
RL ohm
Rp
RL
3.67
5
0.856 : 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 (rds) in series with it.
The small-signal equivalent circuit will therefore be as shown in Fig. 8.
The FET parameters rds and gm should already be familiar to you.
From Fig. 8: Vo RL
gm rdsVi volt
RL rds
Vo
g r R
A m ds L
Vi
RL rds
but in practice, rds
(3)
RL , so
Vo
g r R
m ds L
Vi
rds
and, A gm RL
(4)
d D
G
rds
Vi gmrdsVi
S
V0
S
Fig. 8
RL
Analysis of Small-signal Transistor Amplifiers
Worked Example 3
Q
The FET used in the amplifier circuit of Fig. 9 has parameter values of rds 80 k and gm 4 mS.
Calculate (a) the amplifier voltage gain, and (b) the effective input resistance of the amplifier circuit.
VDD
R1
56 kΩ
RD
2 kΩ
RG
1 MΩ
Vi
RS
1 kΩ
R2
4.6 kΩ
V0
RL
3 kΩ
Fig. 9
A
rds 80 103 ; gm 4 103 S; RL 3 k
RD RL
32
ohm k
RD RL
5
RL 1.2 k (a) For this circuit, the effective a.c. load, RL’ and since rds
RL , then equation (4) may be used
so, Av gmRL 4 103 1.2 103
Av 4.8 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, Av gmrds RL’
4 103 80 103 1.2 103 384 000
’
03
81 200
rds R L
80 103 1.2 10
Av 4.73, which confirms the validity of equaation (4)
Note that a FET amplifier provides very much less voltage gain than a
comparable BJT amplifier.
(b) Looking in at the input terminals, for a.c. signals, the gate resistor
RG is in series with the parallel combination of R1 and R2, as shown
in Fig. 10.
Rin RG R1 R2
56 4.7
ohm 106 R1 R2
60.7
Rin 1.0043 M Ans (say 1 M )
75
76
Analysis of Small-signal Transistor Amplifiers
G
RG
Rin
R1
R2
S
Fig. 10
Thus, the inherently high input resistance of the FET is preserved in the amplifier
circuit by the inclusion of RG.
5 Practical Implications
It should be borne in mind that when designing an amplifier circuit, the
results of the equations as shown in this chapter give only theoretical
answers. If an amplifier circuit thus analysed is then constructed and
tested, the actual gain figures 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 hf. In calculations the typical value is normally used. Thus the
mathematical analysis should be considered as only the first step in the
design process, and component values will then need to be adjusted in
the light of practical tests.
Summary of Equations
BJT amplifier: Current gain, Ai hf
1 ho RL
Ai RL
hi
Power gain, Ap Ai Av
Voltage gain, Av FET amplifier: Approx. voltage gain, Av gmRL
or, more accurately, Av gm rds RL
rds RL
Analysis of Small-signal Transistor Amplifiers
Assignment Questions
1
The h-parameters for the transistor used in the
circuit of Fig. 11 are hfe 250, hie 5 k, and
hoe 40 S.
VCC
3.9 kΩ
56 kΩ
(a) sketch the h-parameter equivalent circuit
and hence, or otherwise,
(b) calculate the amplifier current, voltage
and power gains.
0.2 V V
1
pk-pk
VCC
4.7 kΩ
7.5 kΩ
500 Ω
4.7 kΩ
120 kΩ
Fig. 13
V0
0.25 V V1
pk-pk
5
The parameters for the FET in Fig. 14 are
rds 85 k and gm 4.1 mS.
(a) calculate the amplifier voltage gain, and
(b) the power dissipated in the external 15 k
load.
Fig. 11
VDD 30 V
2
The circuit of Fig. 11 is now reconnected so that
the transistor is connected in common base
configuration. If the common base parameters
hib and hob are 100 and 20 S respectively,
1.2 MΩ
(a) sketch the equivalent circuit, and
V1
(b) calculate the amplifier current, voltage
and power gains.
3
10 kΩ
56 kΩ
Figure 12 shows a simply biased common
source FET amplifier, where the transistor
parameters are gm 3 mS, and rds 75 k.
Calculate the amplifier voltage gain.
VDD
15 kΩ
15 kΩ
3V
pk-pk
4.7 kΩ
4.7 kΩ
Fig. 14
6
For the two equivalent circuits shown in
Figs. 15(a) and (b), sketch the amplifier circuits
that they represent, showing component
values, and also identify the values for the
transistor parameters in each case.
90 kΩ
120 V1
20 kΩ
5 kΩ
V1
1.2 MΩ
1 MΩ
6.8 kΩ
RS
(a)
i1
Fig. 12
600 Ω 82 kΩ
10 kΩ
4
The transistor of the amplifier circuit shown
in Fig. 13 has the following parameters:
hie 2.5 k, hfe 120, and hoe 100 S.
Sketch the equivalent circuit and determine
the amplifier current and voltage gains, and the
power dissipated in the external 7.5 k load.
50i1
3 kΩ
V1
(b)
Fig. 15
25 kΩ 4.7 kΩ 10 kΩ
77
78
Analysis of Small-signal Transistor Amplifiers
Supplementary Worked Example 1
Q
Calculate the minimum value of hfe 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 hie and hoe are 4 k and 50 µS
respectively.
VCC
100 kΩ
0.25 V
pk-pk
4.7 kΩ
600 Ω
10 kΩ
Fig.16
A
The h-parameter equivalent circuit is shown in Fig. 17. Since RB
hie then the
shunting effect of RB will be negligible, and it has therefore been omitted from
the calculation.
hie 4000 ; hoe 50 106 S; Vi 0.25 V pk-pk; Po 3.5 103 W
i1
hfei1
VS
RB
V1 100 kΩ
1/hoe
20 kΩ
hie
4 kΩ
V2
RC
4.7 kΩ
0.25 V
pk-pk
Fig. 17
P0 V22
watt so V2 P0 RL volt 3.5 103 10 4
RL
V2 5.916 V
V1 4
hie
Vs volt pk-pk 0.25
4.6
hie Rs
V1 0.21 7V pk-pk
so, V1 0.217
7
76.8mV r.m.s.
2 2
V2
5.916
Voltage gain required, Av V 0.768
1
Av 77
RL
10 kΩ
Analysis of Small-signal Transistor Amplifiers
Av so, hfe 1/hoe RL
47
hfe RL
where RL 3.2 k 1 / hoe RL
14.7
hie (1 hoe RL )
Av {hie (1 hoe RL )} 77{4(1 50 106 3.2 103 )}
3.2
RL
hie 112 Ans
Supplementary Worked Example 2
Q
The FET in the circuit of Fig. 18 has rds ⴝ 50 k and gm ⴝ 5 mS. Determine the value of the output
voltage, V2, and the power developed in the 25 k load.
A
rds 50 103 ; gm 5 103 S
VDD 40 V
39 kΩ
100 kΩ
V2
V1
4V
pk-pk
25 kΩ
2 MΩ
2.2 kΩ
Fig. 18
RL 25 3.9
RD RL
ohm 15.2 k RD RL
25 3.9
RL , then the approximate equation for voltage gain
Now, since rds is NOT
should not be used, hence
Av gmrds RL
1.5 103 50 103 15.2 103
rds RL
65.2
Av 17.48
Thus, V2 17.48 4 V pk-pk 70 V pk-pk
so, V2 24.75 V Ans
P0 V22
24.752
watt RL
25 103
P0 0.2 mW Ans
Note that had the approximate equation Av gm RL been used in this case an
error of about 22% would have resulted in the value for Av. This would be an
unacceptably large error.
The approximate form of the equation should be used only when rds is at least
10 times larger than RL .
79
80
Analysis of Small-signal Transistor Amplifiers
Answers to Assignment Questions
1
2
3
4
5
(b) Ai 210; Av 197; Ap 41 370
(b) Ai 0.91; Av 42.8; Ap 39
25.7
Ai 95.5; Av 98.1; Po 6.5 mW
Av 24.6; Po 45.4 mW