(A) The amplification factor $(\mu)$ is given by $\mu = g_m \times r_p$.Substituting the given values: $\mu = 2.5 \times 10^{-3} \text{ A/V} \times 20

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(A) The amplification factor $(\mu)$ is given by $\mu = g_m \times r_p$.Substituting the given values: $\mu = 2.5 \times 10^{-3} \text{ A/V} \times 20 \times 10^3 \, \Omega = 50$.The voltage gain $(A)$ of a triode amplifier is given by $A = \frac{\mu R_L}{r_p + R_L}$.Given $A = 10$, we have $10 = \frac{50 R_L}{20 + R_L}$.Multiplying both sides by $(20 + R_L)$, we get $10(20 + R_L) = 50 R_L$.$200 + 10 R_L = 50 R_L$.$200 = 40 R_L$.$R_L = \frac{200}{40} = 5 \text{ k}\Omega$.

Class 12 Physics Semiconductor Electronics Valve Electronics

Medium

$A$ triode has a transconductance $(g_m)$ of $2.5 \text{ mA/V}$ and an anode resistance $(r_p)$ of $20 \text{ k}\Omega$. If it is used as an amplifier with an amplification factor of $10$, what is the load resistance $(R_L)$ connected to the plate circuit in $\text{k}\Omega$?

A
$5$
B
$20$
C
$10$
D
$1$

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Semiconductor Electronics

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Valve Electronics

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$A$ triode has a transconductance $(g_m)$ of $2.5 \text{ mA/V}$ and an anode resistance $(r_p)$ of $20 \text{ k}\Omega$. If it is used as an amplifier with an amplification factor of $10$, what is the load resistance $(R_L)$ connected to the plate circuit in $\text{k}\Omega$?

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The plate voltage of a triode is increased from $200 \, V$ to $225 \, V$. To maintain the plate current,a change in grid voltage from $5 \, V$ to $5.75 \, V$ is required. The amplification factor is:

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