Question 19.4: Determine the small-signal behaviour of the following feedba......
Determine the small-signal behaviour of the following feedback circuit.
This circuit is identical to that in Example 19.3, but we are now considering the smallsignal (or AC) behaviour.
Small-signal voltage gain
From the circuit diagram it is clear that the input signal is applied to the base of the transistor through the coupling capacitor C. Normally C would be chosen to have negligible impedance at the frequencies of interest and so it can be ignored. We will discuss later how to determine the effects of C if these are not negligible, but for the moment we will assume that the base voltage v_{b} is equal to the input voltage v_{i} .
From the second of our assumptions, v_{be} is considered to be very small, so the small-signal voltage on the emitter is effectively equal to that on the base. That is,
v_{e} \approx v_{b} \approx v_{i} (19.19)
Now, from Ohm’s law we know that
i_{e} = \frac{v_{e}}{R_{E}}and as
i_{c} \approx i_{e}it follows that
v_{o} = -i_{c}R_{C} \approx -i_{e}R_{C} = – \frac{v_{e}}{R_{E}}R_{C}where the minus sign reflects the fact that the output voltage goes down when the current increases. If you expected to see V_{CC} in this expression, remember that the supply rail has no small-signal voltages on it.
Substituting from Equation 19.19 gives
v_{o} = -v_{e} \frac{R_{C}}{R_{E}} ≈ -v_{i} \frac{R_{C}}{R_{E}}and so the voltage gain of the circuit is given by
voltage gain = \frac{v_{o}}{v_{i}} \approx – \frac{R_{C}}{R_{E}} (19.20)
For the component values used this gives
voltage gain \approx – \frac{2.2 kΩ}{1.0 kΩ} \approx -2.2