Question 19.6: Design a single-stage amplifier with a small-signal voltage ......
Design a single-stage amplifier with a small-signal voltage gain of -4 and a maximum output swing of 10 V peak to peak (when used with a high-impedance load) that operates from a 15 V supply line. The amplifier should be AC coupled, but have a gain that is approximately constant down to 100 Hz.
We will use an amplifier with an emitter resistor and potential divider biasing as in earlier examples.
Designing circuits is not a precise art. There is no unique ideal solution and often we use rules of thumb to simplify component choice.
Quiescent output voltage and collector current
One of our first tasks is to decide on appropriate values for the quiescent output voltage and the quiescent collector current. The first of these is constrained by the relatively large required output swing. In order to produce 10 V peak-to-peak output, it must be able to go above and below its quiescent value by at least 5 V. In order to leave a reasonable voltage across the emitter resistor (to increase stability), let us choose to make the quiescent output voltage about 5.5 V below V_{CC} , therefore
V_{C(quiescent)} \approx V_{CC} – 5.5 = 9.5 VThe choice of the quiescent collector current is fairly arbitrary, as the load impedance is high. Let us choose a value of 1 mA.
This immediately allows us to calculate an appropriate value for R_{C} as
V_{C(quiescent)} = V_{CC} – I_{C(quiescent)}R_{C}
and thus
R_{C} = \frac{V_{CC} – V_{C(quiescent)}}{I_{C(quiescent)}} = \frac{15.0 V – 9.5 V}{1 mA} = 5.5 kΩAs 5.5 kΩ is not a standard resistor value and the value is not critical, we choose the closest standard value, which is 5.6 kΩ. Therefore
R_{C} = 5.6 kΩ
Small-signal voltage gain
From our discussions in Example 19.5 we know that
voltage gain = -\frac{R_{C}}{R_{E} + r_{e}}
Therefore, in order that the gain is determined by the passive components in the circuit we require R_{E} ⪢ r_{e} and, in order to obtain a gain of -4, we require R_{C} = 4R_{E} .
With a quiescent collector current of 1 mA, the emitter current is also 1 mA and the emitter resistance is given by
r_{e} \approx \frac{1}{40 I_{E}} \approx \frac{1}{40 × 10^{-3}} \approx 25 ΩIf R_{C} is equal to 5.6 kΩ then R_{E} should be 5.6/4 = 1.4 kΩ to give a gain of -4. This value satisfies the condition that R_{E} ⪢ r_{e} .
Again, 1.4 kΩ is not a standard resistor value. If it is vital that the gain is very close to -4, this value could easily be achieved by combining two high-tolerance resistors of appropriate values. Alternatively, we choose the closest standard value and accept that the actual gain will be slightly above or below the specified value. In this case, let us choose R_{E} = 1.3 kΩ . This gives a value for the gain of
voltage gain = -\frac{R_{C}}{R_{E} + r_{e}} \approx -\frac{5.6 kΩ}{1.3 Ω + 25 Ω} \approx -4.2
Base-bias resistors
If the quiescent emitter current is to be about 1 mA and R_{E} is 1.3 kΩ, the quiescent emitter voltage must be given by
V_{E(quiescent)} = I_{E(quiescent)} × R_{E} = 10^{-3} × 1.3 × 10^{3} = 1.3 VTo achieve this emitter voltage the base must be biased to 1.3 – 0.7 = 2.0 V. The ratio of R_{1} to R_{2} must therefore be determined by the relationship
\frac{R_{2}}{R_{1} + R_{2}} V_{CC} = 2.0 VChoice of the absolute values of the base resistors is a compromise between high values, which give a high input resistance to the circuit, and low values, which make the current through the bias resistors large compared with the base current. As mentioned earlier, a common rule of thumb solution is to choose R_{2} as approximately ten times R_{E} . Therefore, R_{2} becomes 13 kΩ. Rearranging the above expression, we have
R_{1} = \frac{R_{2}(V_{CC} – 2.0)}{2.0} = -\frac{13 kΩ(15.0 – 2.0)}{2.0} = 84.5 kΩThe nearest standard value for R_{1} is 82 kΩ. The use of this value raises the base voltage slightly above 2.0 V, which in turn increases the emitter current and reduces the quiescent collector voltage. Calculation of these values is left as an exercise for the reader.
Input resistance and the choice of C
From our earlier discussions, we know that the input resistance is given approximately by the parallel combination of R_{1} and R_{2} . Therefore in this case
input resistance \approx R_{1}//R_{2} = 13 kΩ//82 kΩ \approx 11.2 kΩ
From Section 8.6 we know that, if the effects of the source resistance are neglected, the presence of the coupling capacitor will produce a low-frequency cut-off at a frequency given by
f_{c} = \frac{1}{2 \pi C R}where R is the input resistance of the amplifier.
In this example, we require the gain to be approximately constant down to 100 Hz. We therefore choose a lower cut-off frequency of 100/10 = 10 Hz, which gives a value for C of
C = \frac{1}{2 \pi f_{c} R} = \frac{1}{2 × \pi × 10 × 11.2 × 10^{3}} = 1.4 μFTherefore, a non-polarised capacitor of more than 1.4 μF would be used, for example, a 2.2 μF polyester type.
Thus our final design is as follows.
