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UNIT 10
Basic audio frequency amplifiers
10.1 Introduction
Linear (or analogue) circuits are amplifier-type circuits. They handle signals that are electrical
representations (analogues) of physical quantities which vary continuously with time and take on all
values between a certain maximum and minimum. The information carried by such signals (e.g. the
loudness and pitch of a sound) is in the amplitude and shape of their waveforms. The circuit design
results in a linear relationship between the input and the output i.e. doubling the input doubles the output
and so on.
In general the job of an amplifier is to produce an output which is an enlarged copy of the input. The
symbol for an amplifier is shown in Fig. 10.01. Amplifiers can be classified according to their function
and the frequency range they cover. In a voltage (or small-signal) amplifier the output voltage is greater
than the input voltage. A power (or large-signal) amplifier is designed to deliver power to an output
transducer (although it may also have a voltage gain).
An audiofrequency (a.f) amplifier amplifies a.c. signals in the audio frequency range, i.e. 20 Hz to 20
kHz. Radio frequency (r.f.) amplifiers operate above 20 kHz, at radio and television signal frequencies.
A complete audio amplifier for, say, a CD player consists of several amplifier stages coupled together
so that the output of one stage becomes the input to the next stage. The early stages are voltage
amplifiers and the last one a power amplifier. Each stage is built round one or more transistors.
This Unit gives an introduction to a.f. voltage amplifiers. Unit 11 deals with more advanced aspects,
including a.f. power amplifiers, while in Unit 12 radio frequency amplifiers are considered.
10.2 Voltage amplifier using a junction transistor
Although a junction transistor in the common-emitter mode is basically a current amplifier (see section
8.3), it can act as a voltage amplifier if a suitable resistor, called the load, is connected in the collector
circuit. The small alternating voltage to be amplified, i.e. the input voltage vi is applied to the baseemitter circuit and causes small changes of base current which produce large changes in the collector
current flowing through the load. The load converts these current changes into voltage changes which
form the alternating output voltage v0v0 being much greater than vi. (Note the use of small italic letters to
represent instantaneous values of varying quantities.)
The simplest circuit for a transistor amplifier is shown in Fig. 10.02. To see in more detail just why
voltage amplification occurs, consider first the situation when there is no input (i.e. vi = 0)—-called the
quiescent state (quiescent means 'quiet').
(a) Quiescent state
For transistor action to take place, the base-emitter junction must be forward biased (and has to remain so
even when vi is applied and goes negative). As we saw in section 8.3, a simple way of doing this is to
connect a resistor RB called the base bias resistor, as shown. A steady (d.c.) base current IB flows from +
Vcc through RB into the base and back to 0 V via the emitter. The value of RB can be calculated (see
section 10.3) once the value of IB for the best amplifier performance has been decided.
If VCC is the power supply voltage and VBE is the base-emitter junction voltage (always about +0.6 V
for an n-p-n silicon transistor), then for the base-emitter circuit, since d.c. voltages add up, we can write:
VCC = IB x RB + VBE
(1)
IB causes a much larger collector current IC which produces a voltage drop IC x RL across the load RL.
The voltage at the end of RL joined to + VCC is fixed and so the voltage drop must be at the end connected
to the collector. If VCE is the collector-emitter voltage, then for the collector-emitter circuit we can write:
VCC = IC x RL + VCE
(2)
Component values are chosen so that the steady base bias current IB makes the quiescent collectoremitter voltage VCE about half the power supply voltage VCC. This allows v0 to have its maximum swing
capability, in theory from 0 V to VCC.
(b) Applied input signal
When vi is applied and goes positive, it increases the base-emitter voltage slightly (e.g. to +0.61 V).
When vi swings negative, the base-emitter voltage decreases slightly (e.g. to +0.59 V). As a result a small
alternating current is superimposed on the quiescent base current IB which in effect becomes a varying
d.c.
When the base-current increases, large proportionate increases occur in the collector current. From
equation (2) it follows that there is a corresponding large decrease in the collector-emitter voltage (since
VCC is fixed). A decrease of base current causes a large increase of collector-emitter voltage. In practice,
positive and negative swings of a few millivolts in vi can result in a fall or rise of several volts in the
voltage across RL and therefore in the collector-emitter voltage as well.
The collector-emitter voltage is a varying direct voltage and may be regarded as an alternating voltage
superimposed on a steady direct voltage, i.e. on the quiescent value of VCE—see section 3.3. Only the
alternating part is wanted and capacitor С blocks the direct part but allows the alternating part, i.e. the
output v0 to pass on to the next stage.
Summing up, a transistor will act as a voltage amplifier if (i) it has a suitable collector load, and (ii) it is
biased so that the quiescent value of VCE ~ 0.5VCE known as the class A biasing condition.
(c) Further points
The transistor and load together bring about voltage amplification.
The output is 180° out of phase with the input, i.e. when the input has its maximum positive value, the
output has its maximum negative value, as the graphs in Fig. 10.02 show, i.e. the amplifier is an inverter.
The emitter is common to the input, output and power supply circuits and is usually taken as the
reference point for all voltages, i.e. 0 V. It is called 'common' or 'ground', or 'earth' if connected to earth.
10.3 Worked example
A silicon transistor in the simple voltage amplifier circuit of Fig. 10.02 operates satisfactorily on a
quiescent (no input) collector current (IC ) of 3 mA. If the power supply (VCC ) is 6 V, what must be the
value of (a) the load resistor (RL ) and (b) the base bias resistor (RB), for the quiescent collector-emitter
voltage (VCE ) to be half the power supply voltage? The transistor d.c. current gain (hFE ) is 100.
a) The collector-emitter circuit equation is:
VCC = IC x RL + VCE
Rearranging we get:
IC x RL = VCC - VCE
That is:
RL = (VCC - VCE )/ IC
Substituting VCC = 6 V, VCE = 3 V and IC = 3 mA gives:
RL = 1 k
b) The d.c. current gain is given by:
hFE = IC / IB
IB is the quiescent base current to produce the quiescent collector current IC .
Rearranging:
IB = IC / hFE
Substituting IC = 3 mA and hFE = 100, we get:
IB = 0.03 mA (30 А)
The base-emitter circuit equation is:
VCC = IB x RB + VBE
VBE is the base-emitter voltage. Rearranging gives:
RB = (VCC – VBE )/ IB
Substituting VCC = 6 V, VBE = 0.6 V (for a silicon transistor) and IB = 0.03 mA
gives:
RB = 180 k
10.4 Load lines, operating point and voltage gain
When designing a voltage amplifier, the aim is usually to obtain:
(i) a certain voltage gain;
(ii) minimum distortion of the output so that it is a good copy of the input;
(iii) operation within the current, voltage and power limits for the transistor.
The choice of the quiescent or d.c. operating point (i.e. the values of IC and VCE determines whether these
requirements will be met. This choice is made by constructing a load line, as we will now see.
If you look back to Fig. 8.08(b) (p. 95), you will see the output characteristics of a transistor showing
the relation between VCE and IC with no load in the collector circuit. With a load RL the equation
connecting them is (see section 10.2):
VCC = IC x RL + VCE
where VCC is the power supply voltage.
Rearranging we get:
VCE = VCC - IC x RL
(3)
Knowing VCC and RL this equation enables us to calculate VCE for different values of IC. If a graph of IC
(on the у-axis) is plotted against VCE (on the x-axis) we get a straight line, called a load line. The line can
be drawn if we know just two points. The easiest to find are the end points where the line cuts the V CEand IC-axes. We shall call these points A and В respectively.
For A we put IC = 0 in (3) and get VCE = VCC = 6 V (say).
For В we put VCE = 0 in (3) and get IC == VCC/RL. If RL == 1 k say, then IC = 6V/1 k = 6 mA.
In Fig. 10.03, AB is the load line for VCC = 6 V and RL = 1 k. It is shown superimposed on the output
characteristics of the transistor used in the circuit of Fig. 10.02. We can regard a load line as the output
characteristic of the transistor and load for particular values of VCC and RL. Different values of either give
a different load line; for example, a smaller value of RL gives a steeper line.
The choice of load line and d.c. operating point affects the shape and size of the output waveform.
Choosing a line which cuts the characteristics where they are not linear (straight) or where they are not
equally spaced can cause distortion. Selecting an operating point too near either the IC- or the VCE-axis
can have the same effect, shown in Figs. 10.04(a) and (b) respectively. The best position for the d.c.
operating point is near the middle of the chosen load line, e.g. at Q in Fig. 10.03. The 'swing' capability
of the output is then a maximum (from near V^ to near 0 V) and distortion a minimum as shown in Fig.
10.04(c). But also note that too large an input can cause distortion even if the operating point has been
correctly chosen, as in Fig. 10.04(d). In severe cases a sine wave input would be 'clipped' so much as to
give a square wave output.
Having chosen Q, the quiescent values of VCE and IC can be read off. In Fig. 10.03 they are VCE = 3 V
(i.e. half VCE) and IC = 3 mA. The value of IB which gives these values is obtained from the transistor
output characteristics passing through Q (since IC and VCE have to satisfy at the same time both the
characteristic and the load line). Here it is the 30 A characteristic. RB can then be calculated as in
section 10.3.
The voltage gain A is given by:
A = (output voltage) \ (input voltage) = (change in VCE ) \ (change in VBE ) =VCE \VBE
It can be obtained from the load line by noting that when the input causes IB to vary from 10 to 50 A
(from R to P), VCE varies from 4.5 to 1.5 V (from Y to X). From the input characteristic of the transistor
in Fig. 8.08(a) we can find the change in VBE to cause this change of 40 A in IB. From the graph, it is
approximately 40 mV (0.04 V). We have: A = 75
10.5 Stability and bias
(a) Thermal runaway
If the temperature of a transistor rises, there is greater vibration of the semiconductor atoms, resulting in
the production of more free electrons and holes. The collector current increases causing further heating of
the transistor until eventually the transistor is damaged or destroyed. Initial temperature rise may be due
to the heating effect of the collector current or to a rise in the surrounding temperature.
To stop this 'thermal runaway' effect and stabilize the d.c. operating point, special bias circuits have
been designed which automatically compensate for variations of collector current. The simplest of these
will now be considered; a more complex one is discussed in section 11.1.
(b) Collector-to-base bias
The basic circuit of Fig. 10.02 for a voltage amplifier can be adequately stabilized for many applications
by halving the value of RB and connecting it between the collector and base as in Fig. 10.05, rather than
between + VCC and base.
For the circuit in Fig. 10.05 we can write (since 1^ is much greater than /g):
VCC = IC x RL + VCE
where:
(4)
VCE = IB x RB + VBE
(5)
From (4) you can see that if IC increases for any reason, VCE decreases since VCC is fixed. From (5) it
therefore follows that since VBE is constant (0.6 V or so), IB must also decrease and in so doing tends to
bring back IC to its original value. Taking the quiescent conditions to be (as in the worked example in
section 10.3):
VCE = 0.5VCC = 3 V, IC = 3 mA, IB = 0.03 mA of RB in Fig. 10.05 is found by rearranging equation (5) to
give:
RB = (VCE – VBE )/IB = 82 k (preferred value)
This is about half the value of 180 k for RB in the unstabilized circuit of Fig. 10.02. The collector-tobase bias circuit is a useful general purpose voltage amplifier circuit.
10.6 Simple two-stage voltage amplifier
When greater gain is required, two (or more) amplifier stages are coupled. The circuit in Fig. 10.06 is for
a two-stage capacitor-coupled a.f. voltage amplifier using collector-to-base bias.
(a) Capacitor coupling
The output from Tr1 is applied to the input of Tr2 via C2 which connects the collector of Tr1 (it must have
a quiescent d.c. voltage of +4.5 V for correct operation) and the base of Tr2 (it is +0.6 V above the emitter
at 0 V, being a forward biased junction). A direct connection, without C2 would have the disastrous effect
of fixing the collector of Tr1 at only +0.6 V above ground and would also send a base current of several
mA into Tr2 (through the 1 k Tr1 collector load) that would saturate Tr2 permanently. At most audio
frequencies the reactances of C2 (Xc= 1/(2fC2)) and also of the input and output coupling capacitors C1
and C3 are small and so the alternating part of the output voltage from Tr1 is transferred with little loss to
the base of Tr2.
C1 C2 and C3 are electrolytics and must be connected with the correct polarities. For example, as the
collector voltage of is more positive than the base voltage of Tr2, the + terminal of C2 therefore goes to
the collector of Tr1.
In an amplifier with an even number of stages, as in Fig. 10.06, the output and input are in phase, i.e. it
is a non-inverting amplifier. An inverting amplifier has an odd number of stages.
(b) Frequency response
The voltage gain A of a capacitor-coupled amplifier is fairly constant over most of the a.f. range but it
falls off at the lower and upper limits. At low frequencies, the reactances of the coupling capacitors
increase and less of the low frequency part of the input is passed on. At high frequencies various stray
capacitances in the active devices and between the connections can cause the fall.
A typical voltage gain-frequency curve is shown in Fig. 10.07. The bandwidth is the range of
frequencies within which the voltage gain does not fall below 0.7 of its maximum value A max. (To fit in
the large frequency range, frequencies are not plotted on the usual linear scale but on a logarithmic
scale in which equal divisions represent equal changes in the 'log of the frequency f) The two points P
and Q at which this happens are called the '3 dB points'. The decibel (dB) scale compares signal levels
and is explained in section 11.9.
10.7 Voltage amplifier using a FET
In a FET voltage amplifier, changes in the gate (input) voltage causes changes in the drain current
which are converted into larger voltage changes by a load resistor in the drain (output) circuit. The load
line and d.c. operating point are selected as for a junction transistor voltage amplifier—see section 10.4.
The chosen operating point is realized in practice by applying the correct quiescent bias voltage to the
gate.
The circuit for an n-channel JUGFET voltage amplifier in common-source connection is shown in
Fig. 10.08. The values of the load resistor RL and the supply voltage VDD (note the symbol: VCC is used
for the supply voltage to a junction transistor) are both higher than for a bipolar transistor to obtain a
reasonable gain. Negative (quiescent) bias is required for the gate since the FET works in the depletion
mode—see section 8.7. It is provided as follows.
(i) Source and gate resistors (RS and RG) In the quiescent state, the source current
RS (=ID +IG ~ID since IG ~0) is teady and causes a voltage drop across resistor RS. The source
end of RS is therefore positive with respect to the other end connected by the high resistor RG
to the gate. RG ensures that the gate has the same voltage as the lower end of RS on the
diagram. This is so because there is negligible current through RG and hence practically no
voltage across it. Both ends of RG have the same voltage, namely that of the lower end of RS
i.e. 'ground' or 0 V. The source is therefore at a higher voltage than the gate, i.e. the gate is
negative with respect to the source. (If you find this difficult to understand, refer back to
section 2.4.)
The circuit automatically compensates for any change of IS and helps to stabilize the d.c.
operating point because any increase in IS increases the voltage VS ( = IS x RS across RS The
voltage of the source end of RS (and so of the source) rises and since the lower end of RS and
the gate are tied to 0 V, the gate-source voltage VGS must go more negative, tending to reduce
IS to its previous value.
(ii) Decoupling capacitor (CS) The large decoupling capacitor CS provides a bypass (i.e. a
low impedance) round RS for the a.c. part of the source current (which becomes a varying
d.c.) when an alternating input is applied. Otherwise the varying voltage developed across RS
would cause unwanted changes in the value (quiescent) of VGS required to give the chosen
operating point.
(iii) Blocking capacitor (C1) This blocks any d.c. voltage from the input which would affect
the operating point. With RG it forms a voltage divider across the input. The alternating
voltage developed across RG is applied to the gate for amplification and for this voltage to be
as large as possible, the value of RG should be large compared with the reactance of C1. Since
the resistance of RG is usually in the range of 2.2 to 10 М, the capacitance C1 can be small
(e.g. 0.1 F).
FET voltage amplifiers give lower gains than junction transistor types (typically ten compared
with up to a thousand). It can be shown that the voltage gain A is given approximately by A =
gm x RL where gm is the transconductance of the FET—see section 8.7. However, their much
greater input impedances make them better for certain applications, e.g. as impedancematching devices (see section 11.6) and as r.f. amplifiers (see section 12.4).
10.8 Revision questions
1. Explain the following terms:
a) voltage amplifier,
b) power amplifier,
c) a.f. amplifier,
d) r.f. amplifier.
2. Is a junction transistor in common-emitter connection basically a current or a voltage
amplifier?
3. Answer the following questions about the voltage amplifier circuit in Fig. 10.09.
a) Which are the input terminals?
b) Which are the output terminals?
c) Where is the power supply connected (give polarities)?
d) What is the purpose of RL?
e) What is the purpose of RB?
0 What function does C1 serve?
g) What does C2 do?
h) What is the phase relationship between the input and output voltages?
4. Explain the following terms as applied to an amplifier:
a) quiescent state,
b) d.c. operating point,
c) voltage gain.
5. When designing a voltage amplifier what are the three main aims?
6. a) What is the purpose of drawing a load line?
b) Name the factors which should be considered when choosing (i) a load line, (ii)
the d.c. operating point.
c) In Fig. 10.10, comment on the choice of (i) AB, (ii) AC, as load lines.
d) In Fig. 10.10, comment on the choice of
(i) P, (ii) Q, (iii) R, (iv) S, as d.c. operating points.
7. What is meant by thermal runaway? How is it prevented?
8. Answer the following questions about the FET voltage amplifier in Fig. 10.11,
a) What is the purpose of RS?
b) What does RG ensure?
c) What doesCS do?
10.9 Problems
1. In the circuit of Fig. 10.12, if VCC = 9 V, RL = 1 k, IC = 3 mA and hFE = 200 for the
transistor, calculate
a) VCE
b) IB and
c) RB if VBE=0.6V.
2. The load line for the amplifier in Fig. 10.12 is shown in Fig. 10.13.
a) Use it to find the values of VCC and RL.
b) If Q is the d.c. operating point, what are the quiescent values of VCE IC IB
c) Calculate RB if the transistor is a silicon type.
3. The output characteristics of a junction transistor in common-emitter connection are
shown in Fig. 10.14(a). The transistor is used in an amplifier with a 9 V supply and a load
resistor of 1.8 k.
a)Copy the graph and draw the load line.
b)Choose a suitable d.c. operating point and read off the quiescent values of IC, IB and
VCE.
c) What is the quiescent power consumption of the amplifier?
d) If an alternating input voltage varies the base current by ±20 А about its quiescent
value, what is
(i) the variation in the collector-emitter voltage, and (ii) the peak output
voltage?
e) An input characteristic of the transistor is given in Fig. 10.14(b). Use it to
find the base-emitter voltage variation which causes a change of ±20 A in
the base current.
0 Using your answers from d) and e) find the voltage gain of the amplifier. g) If the
amplifier uses the collector-to-base bias circuit of Fig. 10.12, calculate
the value of RB to give the quiescent value of IB. (Assume
VBE = 0.6 V.)