For a Common-Emitter amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. Suppose the current amplification factor of the transistor is 100. The base resistance is 1 kΩ. The input signal voltage will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.01 V
Given,
$R_C$ Resistance = $2kΩ=2×10^3Ω$ ($1kΩ=1000Ω$)
$V_o$, Audio signal voltage = $2V$
$β$ (current amplification factor) = 100
$R_b$ (base resistance) = $1kΩ=1×10^3Ω$
$I_B=\frac{V_o}{β×R_c}$
where, $I_B$ = Base current
$I_B=\frac{V_o}{β×R_c}=\frac{2}{100×2000}=10^{-5}A$
$V_i=I_B×R_b=10^{-5}×10^3$
$=0.01V$