The transconductance of a triode amplifier is | vedclass

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(A) The voltage amplification $A_v$ of a triode amplifier is given by the formula: $A_v = \frac{\mu R_L}{r_p + R_L}$, where $\mu = g_m 	imes r_p$.Giv

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$5$

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(A) The voltage amplification $A_v$ of a triode amplifier is given by the formula: $A_v = \frac{\mu R_L}{r_p + R_L}$, where $\mu = g_m 	imes r_p$.Given: Transconductance $g_m = 2.5 	imes 10^{-3} \, \Omega^{-1}$, Plate resistance $r_p = 20 	imes 10^3 \, \Omega$, and Voltage amplification $A_v = 10$.First, calculate the amplification factor $\mu$: $\mu = g_m 	imes r_p = (2.5 	imes 10^{-3}) 	imes (20 	imes 10^3) = 50$.Now, substitute the values into the amplification formula: $10 = \frac{50 	imes R_L}{20 	imes 10^3 + R_L}$.$10(20 	imes 10^3 + R_L) = 50 R_L$.$200 	imes 10^3 + 10 R_L = 50 R_L$.$40 R_L = 200 	imes 10^3$.$R_L = \frac{200 	imes 10^3}{40} = 5 	imes 10^3 \, \Omega = 5 \, 	ext{k}\Omega$.

The transconductance of a triode amplifier is $2.5 \, \text{mS}$ (milli-mho) and its plate resistance is $20 \, \text{k}\Omega$. If the voltage amplification is $10$, find the load resistance in $\text{k}\Omega$.

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