Question

In a common emitter (CE) amplifier having a voltage gain G, the transistor used has transconductance 0.03 mho and current gain 25. If the above transistors replaced with another one with transconductance 0.02 mho and current gain 20, the voltage gain will be -
A. $\dfrac{2}{3}G$
B. $1.5G$
C. $\dfrac{1}{3}G$
D. $$$$$$\dfrac{5}{4}G$$ ${G_{}}$

Answer 1

In this question we need to use the expression for ratio of voltage gain in terms of gran conductance and we know that Transconductance is an expression of the performance of a bipolar transistor or field-effect transistor (FET). In general, the larger the transconductance figure for a device, the greater the gain(amplification) it is capable of delivering, when all other factors are held constant.it is denoted by g.

Formula used:
${G_2} = {g_2}R_c$

Complete answer:
We have given voltage gain is G, transconductance of 0.03mho, current gain 25A
After replacement of transistor transconductance is 0.02mho current gain is of 20A
We know that the common emitter amplifier is a three basic single stage bipolar junction transistor and is used as a voltage amplifier.
The voltage gain is 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
Voltage gain= $G = \dfrac{{V_c}}{{V_b}} = \dfrac{{I_cR_c}}{{V_b}}$
Where $V_c$ is voltage of collector, $V_b$ is voltage of base, $I_c$ is current through the collector, $R_c$is resistance through collector.

We know that transconductance(g)= $\dfrac{I_c}{V_b}$
${G_1} = {g_1}R_c$
${G_2} = {g_2}R_c$
So the ratio of voltage gain can be written as,
$\dfrac{{{G_1}}}{{{G_2}}} = \dfrac{{{g_1}R_c}}{{{g_2}R_c}}$
Or $\dfrac{{{G_1}}}{{{G_2}}} = \dfrac{{{g_1}}}{{{g_2}}}$
Substituting the values transconductance
$\dfrac{{{G_1}}}{{{G_2}}} = \dfrac{{0.03}}{{0.02}}$
Or after elimination of decimal and cross multiplication
${G_2} = {G_1}\dfrac{2}{3}$
Here ${G_1}$is given as ${G_{}}$

Hence, the correct answer is option (A).

Note: All types of transistor amplifiers operate using AC signal inputs which alternate between a positive value and a negative value so some way of “presetting” the amplifier circuit to operate between these two peak values is required this is achieved due to biasing.