In a common emitter transistor amplifier circ | vedclass

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(C) Given:Input resistance,$R_i = 1.8 	ext{ k}\Omega = 1800 \Omega$Load resistance,$R_L = 9 	ext{ k}\Omega = 9000 \Omega$Current gain,$\beta = 70$In

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$2.1$

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(C) Given:Input resistance,$R_i = 1.8 	ext{ k}\Omega = 1800 \Omega$Load resistance,$R_L = 9 	ext{ k}\Omega = 9000 \Omega$Current gain,$\beta = 70$Input voltage,$V_i = 6 	ext{ mV} = 6 	imes 10^{-3} 	ext{ V}$Step $1$: Calculate the input current $(I_b)$:$I_b = \frac{V_i}{R_i} = \frac{6 	imes 10^{-3}}{1800} = \frac{1}{3} 	imes 10^{-5} 	ext{ A}$Step $2$: Calculate the output current $(I_c)$:$I_c = \beta 	imes I_b = 70 	imes \frac{1}{3} 	imes 10^{-5} = \frac{7}{3} 	imes 10^{-4} 	ext{ A}$Step $3$: Calculate the output voltage $(V_o)$:$V_o = I_c 	imes R_L = (\frac{7}{3} 	imes 10^{-4}) 	imes 9000 = 7 	imes 3 	imes 10^{-1} = 2.1 	ext{ V}$Alternatively,Voltage Gain $A_v = \beta 	imes \frac{R_L}{R_i} = 70 	imes \frac{9000}{1800} = 70 	imes 5 = 350$.$V_o = A_v 	imes V_i = 350 	imes 6 	ext{ mV} = 2100 	ext{ mV} = 2.1 	ext{ V}$.

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