How a transistor works as amplifier — common emitter configuration

Question

How does a transistor amplify signals in the common emitter (CE) configuration? What determines the voltage and current gain?

Solution — Step by Step

In a CE amplifier, the emitter is common to both input and output circuits.

Input: Small AC signal applied between base and emitter (base-emitter junction is forward biased)
Output: Amplified signal taken between collector and emitter (collector-base junction is reverse biased)

The base current $I_B$ is very small (microamperes), while the collector current $I_C$ is much larger (milliamperes). The relationship is:

I_C = \beta \cdot I_B

where $\beta$ (current gain) is typically 50-200.

A small change in base current ( $\Delta I_B$ ) causes a large change in collector current ( $\Delta I_C = \beta \cdot \Delta I_B$ ).

This large collector current flows through the load resistance $R_L$ in the output circuit, producing a large voltage change:

\Delta V_{\text{out}} = \Delta I_C \times R_L = \beta \cdot \Delta I_B \times R_L

The input voltage change is: $\Delta V_{\text{in}} = \Delta I_B \times r_{\text{in}}$ (where $r_{\text{in}}$ is the input resistance)

Voltage gain:

A_v = \frac{\Delta V_{\text{out}}}{\Delta V_{\text{in}}} = \beta \times \frac{R_L}{r_{\text{in}}}

In CE configuration, when the input signal increases the base current, the collector current increases, and the voltage drop across $R_L$ increases. This means the collector voltage ( $V_{CE} = V_{CC} - I_C R_L$ ) decreases.

So when input goes up, output goes down — there is a 180-degree phase shift between input and output. This is a key characteristic of the CE amplifier.

graph LR
    A[Small AC input signal] --> B[Base-Emitter junction]
    B -->|Small delta IB| C[Transistor]
    C -->|Large delta IC = beta x delta IB| D[Load resistance RL]
    D --> E[Large amplified output]
    E --> F["180-degree phase shift"]

Why This Works

The transistor acts as a current-controlled current source. The thin, lightly doped base region is the key — most charge carriers injected from the emitter pass right through the base (only about 2-5% recombine there) and reach the collector. This is why $I_C \gg I_B$ .

Parameter	Symbol	Formula	Typical Value
Current gain	$\beta$	$I_C / I_B$	50-200
Voltage gain	$A_v$	$\beta \times R_L / r_{in}$	10-500
Power gain	$A_p$	$\beta \times A_v$	500-100000
Phase shift	-	-	180 degrees

The CE configuration is the most commonly used because it provides both current and voltage gain, giving the highest power gain among all three configurations (CE, CB, CC).

Alternative Method

For CBSE board exams, you may need to draw the circuit diagram. The essential components are:

NPN transistor with E, B, C labeled
$V_{BB}$ (base battery) with $R_B$ in the base circuit
$V_{CC}$ (collector battery) with $R_L$ in the collector circuit
Input signal source coupled to base
Output taken across $R_L$

For JEE numericals: if given $\beta = 100$ , $R_L = 5$ k $\Omega$ , and $r_{in} = 500$ $\Omega$ , the voltage gain is $A_v = 100 \times 5000/500 = 1000$ . Always check units — $R_L$ and $r_{in}$ must be in the same unit.

Common Mistake

Students forget the 180-degree phase inversion in CE amplifiers. When asked “what is the phase relationship between input and output in a CE amplifier?”, the answer is always “out of phase by 180 degrees” (or “inverted”). This is different from the common base (CB) configuration where there is no phase inversion. CBSE boards specifically ask this as a 1-mark question.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next