Question

Explain the working of a common base transistor amplifier with the help of a circuit diagram. The load resistance of output circuit is $600k\Omega$ and resistance of input circuit is $150p\Omega$ in a common base amplifier. If current amplification is $0.90$, then calculate the voltage amplification.

Answer 1

Common-Base amplifier using npn transistor:
In this type of amplifier, the input signal is applied between emitter and base terminals and the output signal is obtained across the collector and base terminals through the load resistance $\left(R_L\right)$. The base is common to both the input and output terminals.

The amplifier circuit using npn transistor is as shown:

The input a.c signal is applied across the emitter and base circuit which is forward biased by a battery of emf $V_{EB}$. The output signal is taken across load resistance $\left(R_L\right)$ connected across collector-Base circuit, which is reverse biased by a battery of emf $V_{CB}$.
The emitter current $\left(I_C\right)$ is given by
$I_e=I_b+I_c$ ...(i)
$I_b$ and $I_c$ are base and collector current respectively.
If no current flows through the load resistance, then the voltage across output circuit is the emf of the battery i.e. $V_{CB}$.
If current $I_c$ flows through $R_L$, then voltage across output circuit is: $V_o=V_{CB}+I_c{R}_L$ ...(ii)
Now, when the applied input signal increases on positive half cycle, the forward bias between emitter-base junction decreases i.e. becomes less negative. As a result, emitter current decreases and hence according to (i), collector current $I_C$ also decreases. Thus, potential drop across $R_L(=I_C{R}_L)$ decreases.
Hence, from equation (ii), we find that $V_o$ decreases.
When applied input decreases on negative half cycle, forward bias between emitter-base junction increases. Due to increased forward bias, emitter current $\left(I_e\right)$ increases and hence according to (i); $I_C$ also increases.
Thus potential drop across $R_L\left(I_C{R}_L\right)$ also increases. Therefore, from (ii) we find that $V_o$ decreases.
Thus, in common Base amplifier, collector voltage increases with increasing input signal and decreases with decreasing input signal.
This shows that input and amplified output signals are in phase.
Given, $R_L=600k\Omega$
$R_i$ = Input resistance = $150p\Omega$
Current amplification = 0.90
Voltage amplification = ?
Voltage Amplification= Current amplification x Resistance gain
$\Rightarrow =0.90\times \frac{R_L}{R_i}$

$\Rightarrow =0.90\times \frac{150p\Omega }{600k\Omega }$

$\Rightarrow =\frac{90}{100}\times \frac{1}{4}\times \frac{{10}^{-12}}{{10}^3}$

$\Rightarrow =22.5\times {10}^{-17}$