Prove that in a common-emitter amplifier, the output and input differ in phase by $180^\circ$. In a transistor, the change of base current by 30 $\mu$A produces change of $0\cdot02$ 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.

Prove that in a common-emitter amplifier, the output and input differ in phase by 180^°. In a transistor, the

Question

180^\circ

CBSECBSE Class XII Board 2019Subjective· 3mImportance★★★★★

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Concept understanding — Common Emitter Voltage Gain

Common Emitter Voltage Gain – From Intuition to Formula

Imagine you have a transistor in the common emitter configuration. The input signal goes into the base, and the output comes from the collector. The transistor is a current amplifier — a small change in base current produces a much larger change in collector current, scaled by $\beta$ (the current gain). But we care about voltage gain, not just current gain.

How does a current change turn into a voltage change? Through resistors. The collector current flows through the collector resistor $R_C$ (or the load resistor $R_L$ connected there). By Ohm's law, a change in collector current $\Delta I_C$ produces a change in voltage across that resistor: $\Delta V_{out} = -\Delta I_C \cdot R_C$ . The minus sign appears because an increase in collector current drops more voltage across $R_C$ , pulling the collector voltage down — the CE amplifier inverts.

On the input side, the base-emitter junction looks like a diode. A small change in base-emitter voltage $\Delta V_{BE}$ produces a change in base current $\Delta I_B$ , and the relationship is governed by the input resistance $r_{in}$ (often called $r_\pi$ or $h_{ie}$ ). So $\Delta V_{in} = \Delta I_B \cdot r_{in}$ .

Now chain it together. The collector current change is $\Delta I_C = \beta \cdot \Delta I_B$ . The output voltage change is $-\Delta I_C \cdot R_C = -\beta \cdot \Delta I_B \cdot R_C$ . The input voltage change is $\Delta I_B \cdot r_{in}$ . The voltage gain $A_V$ is the ratio of output to input voltage:

A_V = \frac{\Delta V_{out}}{\Delta V_{in}} = \frac{-\beta \cdot \Delta I_B \cdot R_C}{\Delta I_B \cdot r_{in}} = -\beta \cdot \frac{R_C}{r_{in}}

A_V = -\beta \frac{R_C}{r_{in}}

The negative sign tells you the output is inverted relative to the input. The magnitude of the gain is $\beta$ times the ratio of the output resistor to the input resistance.

Tip

Think of it as: the transistor multiplies the current by $\beta$ , and then the resistors convert that current change into a voltage change. The ratio $R_C / r_{in}$ is the "voltage conversion factor" — how many volts of output you get per volt of input, after the current amplification.

Why the ratio matters …

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In a common-emitter amplifier the output voltage is $180^\circ$ out of phase with the input because rising base current raises the collector current, increasing the drop across $R_L$ and lowering the collector voltage. From the data, the current amplification factor is $\beta = 133.3$ and the load resistance is $R_L = 2000\ \Omega$ .

Why the phase reverses

In the common-emitter configuration the transistor is a current-controlled current source: a small change $\Delta I_B$ in base current produces a large change $\Delta I_C = \beta\,\Delta I_B$ in collector current, which flows through the load $R_L$ tied to the supply $V_{CC}$ . By Kirchhoff's voltage law the collector (output) voltage is

V_{CE} = V_{CC} - I_C R_L.

When the input signal drives $I_B$ up (positive half-cycle), $I_C$ rises, the drop $I_C R_L$ grows, and $V_{CE}$ falls. A positive-going input thus gives a negative-going output — a phase difference of $180^\circ$ . On the negative half-cycle the reverse happens, so the output is a faithful, inverted copy of the input.

Note

The collector current stays in phase with the input; it is the collector voltage that inverts, because of the fixed supply minus the growing $I_C R_L$ drop.

Given data

$\Delta I_B = 30\ \mu\text{A} = 30\times10^{-6}\ \text{A}$
$\Delta V_{BE} = 0.02\ \text{V}$
$\Delta I_C = 4\ \text{mA} = 4\times10^{-3}\ \text{A}$
Voltage gain $A_V = 400$

Current amplification factor

\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{4\times10^{-3}}{30\times10^{-6}} = \frac{4000}{30} = 133.3.

Load resistance …

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Why the phase reverses

Given data

Current amplification factor

Load resistance …

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