Question

In a common emitter (CE) amplifier having a voltage gain G, the transistor used has trans conductance 0.03 mho and current gain 25. If the above transistor is replaced with another one with trans conductance 0.02 mho and current gain 20, the voltage gain will be?

Answer 1

Hint: For the vacuum tubes, trans conductance can be defined as the change in the plate current divided by the corresponding change in the grid voltage, with a constant plate to cathode voltage applied. To solve this question we have to write the formula for initial voltage gain from the given data and write the new voltage gain in terms of the given data and then compare the two values.

Complete step-by-step answer:
Voltage gain G is the ratio of output voltage and the input voltage that is voltage of collector to the voltage of base.
$$A_v=\beta {\displaystyle \frac{R_{in}}{R_{out}}}$$
Where $$\beta$$ is the ratio of collector current to the base current. A trans conductor is the ratio of collector current to base voltage.

$$G=25{\displaystyle \frac{R_{in}}{R_{out}}}$$ (1)
Now, we know
$$G_m={\displaystyle \frac{\beta }{R_1}}$$
$$={\displaystyle \frac{\beta }{G_m}}$$
$${\displaystyle \frac{\beta }{G_m}}={\displaystyle \frac{25}{0.03}}$$
$$G=25\times {\displaystyle \frac{R_{out}}{25}}\times 0.03$$ -------(i)
$$G^{\prime }=20\times {\displaystyle \frac{R_{out}}{20}}\times 0.02$$ -------(ii)
From (i) and (ii) we get,
$$G^{\prime }={\displaystyle \frac{2}{3}}G$$
Hence, the voltage gain will be $$G^{\prime }={\displaystyle \frac{2}{3}}G$$.

Note: The current gain for the common-base configuration can be defined as the change in collector current divided by the change in emitter current when the base-to-collector voltage is kept constant. It’s value is generally closer to unity. The voltage gain can be defined as the product of the current gain and the ratio of the output resistance of the collector to the input resistance of the base circuits.