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The current gain of a BJT is
Answer & Solution
Correct Answer:
Option
C
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 ($$r_\pi$$).
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 ($$r_\pi$$).
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LoginWe know, ri= β/gm.
From this, β=rigm.