Prove that in a common-emitter amplifier, the output and input differ in phase by 180^{o}. In a transistor, the change of base current by 30 \mu A produces change of 0·02 V in the base-emitter voltage and a change of 4 mA in the collector current. Calculate the current amplification factor and the load resistance used, if the voltage gain of the amplifier is 400.

 

 
 
 

Answers (1)

Let us consider that input signals,

V_{i}=\Delta I_{B}r_{i}

and the output signal,

V_{o}=-\Delta I_{c}R_{L}

The voltage amplification,

A_{v}=\frac{V_{o}}{V_{i}}

Since, A_{v}=\frac{-\Delta I_{B}}{\Delta I_{C}}\times \frac{r_{i}}{R_{L}}

\therefore A_{v}=-\beta \times resistance \; gain

here (-) negative sign indicated that output is 180^{o} out of phase with respect to the input signal.

Now, calculating the Amplification factor

\beta =\frac{\Delta I_{C}}{\Delta I_{B}}=\frac{4\times 10^{-3}}{30\times 10^{-6}}=\frac{400}{3}

r_i =\frac{\Delta V_{BE}}{\Delta I_{B}}=\frac{0.02}{30\times 10^{-6}}=\frac{2\times 10^{-2}}{3\times 10^{-5}}

r_i=\frac{2}{3}\times 10^{3}\Omega

Hence, A_{v}=\beta \frac{R_{L}}{r_{i}}

Now, calculating the lead resistance (R_{L})

R_{L}= \frac{A_{v}\times r_i}{\beta }=\frac{400\times 2\times 10^{3}\times 3}{400\times 3}

R_{L}= 2\times 10^{3}\Omega

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