# 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.

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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