In a triode amplifier,the value of maximum ga | vedclass

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(B) The voltage gain $(A_v)$ of a triode amplifier is given by the formula: $A_v = \frac{\mu R_L}{r_p + R_L}$,where $\mu$ is the amplification factor,

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Amplification factor

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(B) The voltage gain $(A_v)$ of a triode amplifier is given by the formula: $A_v = \frac{\mu R_L}{r_p + R_L}$,where $\mu$ is the amplification factor,$r_p$ is the plate resistance,and $R_L$ is the load resistance.To find the maximum gain,we consider the limit as the load resistance $R_L$ approaches infinity $(R_L 	o \infty)$.$(A_v)_{max} = \lim_{R_L 	o \infty} \frac{\mu R_L}{r_p + R_L} = \lim_{R_L 	o \infty} \frac{\mu}{\frac{r_p}{R_L} + 1} = \frac{\mu}{0 + 1} = \mu$.Therefore,the maximum gain is equal to the amplification factor $\mu$.

In a triode amplifier,the value of maximum gain is equal to

Similar Questions

$A$ triode has a plate resistance of $10 \, k\Omega$ and an amplification factor of $24$. If the input signal voltage is $0.4 \, V$ $(r.m.s.)$ and the load resistance is $10 \, k\Omega$,then the output voltage $(r.m.s.)$ is ...... $V$.

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The correct relation for a triode is:

The relation between $I_p$ and $V_p$ for a triode is $I_p = (0.125V_p - 7.5) \text{ mA}$. Keeping the grid potential constant at $1 \text{ V}$,the value of $r_p$ will be . . . . . . $\text{k}\Omega$.

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