The current gain of a BJT is
A. $${{\text{g}}_{\text{m}}}{\text{ }}{{\text{r}}_{\text{o}}}$$
B. $$\frac{{{{\text{g}}_{\text{m}}}}}{{{{\text{r}}_{\text{o}}}}}$$
C. $${{\text{g}}_{\text{m}}}{\text{ }}{{\text{r}}_\pi }$$
D. $$\frac{{{{\text{g}}_{\text{m}}}}}{{{{\text{r}}_\pi }}}$$
Answer: Option C
Solution (By Examveda Team)
Option A: $$g_m r_o$$ is incorrect because this expression represents the voltage gain in a transistor, not the current gain. The term $$r_o$$ is the output resistance, and multiplying $$g_m$$ (transconductance) with $$r_o$$ gives the voltage gain, not the current gain.Option B: $$\frac{g_m}{r_o}$$ is incorrect because this expression represents the inverse relationship between transconductance and output resistance, which is related to the voltage gain of the amplifier, not the current gain.
Option C: $$g_m r_\pi$$ is correct because this is the standard expression for the current gain of a BJT in terms of the transconductance ($$g_m$$) and the base-emitter resistance ($$r_\pi$$). The current gain is the product of these two parameters.
Option D: $$\frac{g_m}{r_\pi}$$ is incorrect because this expression does not correctly represent the current gain of a BJT. Instead, it gives a ratio that relates to the voltage gain and other characteristics of the transistor, but not the current gain.
Conclusion: The correct answer is Option C: $$g_m r_\pi$$ because the current gain of a BJT is given by the product of transconductance ($$g_m$$) and base-emitter resistance ($$r_\pi$$).
This Question Belongs to Electronics And Communications Engineering >> Analog Electronics
We know, current gain AV=hfe. In π model, hfe is referred to β.
We know, ri= β/gm.
From this, β=rigm.