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 transistor is replaced with another one with transconductance $0.02$ $mho$ and current gain $20$, the voltage gain will be :

Question

What is the output of a NAND gate when both inputs are HIGH?

Answer 1

To determine the new voltage gain when the transistor in a common emitter (CE) amplifier is replaced, we need to understand the relationship of the voltage gain with the transconductance and current gain of the transistor.

The voltage gain $ G $ of a common emitter amplifier can be expressed as:

G = -g_m \cdot R_C

where:

g_m is the transconductance of the transistor.
R_C is the load resistance in the collector.

However, since explicitly R_C isn't involved in this comparison and the amplifier characteristics are identical except for the transistor, we look into the effective amplification relation:

In terms of the transconductance g_m and the current gain $ \beta $ of the transistor, it can be described as proportional to:

G \propto g_m \cdot \beta

Initially, the transistor has:

Transconductance, g_{m1} = 0.03 \, \text{mho}
Current gain, \beta_1 = 25

The effective product for this transistor is:

g_{m1} \cdot \beta_1 = 0.03 \times 25 = 0.75

When replaced, the new transistor has:

Transconductance, g_{m2} = 0.02 \, \text{mho}
Current gain, \beta_2 = 20

The effective product for the new transistor is:

g_{m2} \cdot \beta_2 = 0.02 \times 20 = 0.40

The ratio of the new voltage gain G' to the original voltage gain G will be:

\frac{G'}{G} = \frac{g_{m2} \cdot \beta_2}{g_{m1} \cdot \beta_1} = \frac{0.40}{0.75} = \frac{2}{3}

Thus, the new voltage gain is:

G' = \frac{2}{3} G

Therefore, the correct option is:

\frac{2}{3} G

Answer 2