Q.The idea of amplification is clearly understood in the example illustrated in figure 3.1.2a. Let the microphone convert audio frequency sound into electrical signal of 25 mV and is connected to CB amplifier having input impedance of 25 $\Omega$ and output impedance of 5 k$\Omega$. Find the voltage amplification (Av).

In a common base (CB) amplifier the weak signal is applied to the low-resistance emitter-base junction while the output is taken across the high-resistance collector load, so a small input voltage produces a much larger output voltage. The emitter (input) current $I_E = v_i/Z_{in} = 25\,\text{mV}/25\,\Omega = 1\,\text{mA}$ flows almost unchanged through the $5\,\text{k}\Omega$ load to give $v_o = 5\,\text{V}$.

[!ANSWER]

Voltage amplification $A_v = v_o/v_i = 5\,\text{V}/25\,\text{mV} = 200$.

[!TLDR]

The CB amplifier passes the emitter current almost unchanged into a large collector load, turning a 25 mV microphone signal into a 5 V output — a voltage gain of 200.

A common base amplifier is a good voltage amplifier because its input (emitter-base) side has a very low impedance while its output (collector-base) side has a very high impedance. The signal current established in the low-impedance input is delivered almost entirely to the high-impedance output, so the voltage is stepped up even though the current gain $\alpha$ is slightly less than 1 (there is voltage amplification but no current amplification, since $I_E \approx I_C$).

Given: input signal $v_i = 25\,\text{mV}$, input impedance $Z_{in} = 25\,\Omega$, collector load $R_L = 5\,\text{k}\Omega$.

Step 1 — input (emitter) current:

Step 2 — output voltage. In a CB stage $I_C \approx I_E$, so the same 1 mA flows through the load:

Step 3 — voltage amplification:

The weak microphone signal is amplified 200 times, a classic Karnataka 2nd PUC Electronics illustration of how a CB amplifier gives voltage amplification without current amplification.

[!ANSWER]

Voltage amplification $A_v = 200$.