Question

Explain with the help of labelled circuit diagram, the use of n-p-n Transisor as an amplifier in common base configuration. Why is common base amplifier preferred over common emitter amplifier.

Answer 1

Common base amplifier: The circuit diagram for common base n-p-n transistor is as shown in Fig. (n-p-n transistor is being used to explain the basic principle because of its simplicity). In this circuit the base lead is common to both the input (emitter-base) and output (Collector-base) circuit and is grounded. The emitter is forward biased by emitter base battery V_eb and collector is reversed biased by collector base battery V_ec, so output resistance is more input resistance i.e. R₀ > R_i. The input signal is applied across emitter-base and output voltage is obtained across a load R_L between collector and base in series with collector battery.

Let I_e, I_b and I_c be emitter, base and collector current respectively,

so I_e = I_b + I_c ............(1)

The potential drop across load R_L is I_cR_L

so V_c = V_cb - I_cR_L ..........(2)

When ac signal is fed to the input circuit, let the first half cycle of input voltage is +ve. It will make emitter less negative and hence emitter current decreases. Hence I_cR_L factor decreases. From eq. (2) we find that collector voltage V_c will increase. So corresponding to positive half cycle of a.c. input, positive output Half cycle will be obtained.

During second half cycle of a.c. input voltage will be negative and it will make emitter more negative, hence emitter current decreases. So I_cR_L factor also increases. From Fig.  we find that collector voltage V_c will decrease. So corresponding to negative half cycle of ac input negative output half cycle will be obained.

Thus, in common base amplifier, the output voltage is in phase with the input signal. Since input circuit has low resistance (30 to 50 Ω) and output circuit has high resistance (≅ 500 kΩ), so a large a.c. voltage will be developed across the load corresponding to a small change in the input voltage, so the signal gets amplified.

1. a.c. current gain (α_a.c)

the ratio of the change in the collector current to the change in emitter current at constant collector voltage is called ac current gain. Hence,

α_a.c. = \((\frac{\bigtriangleup I_e}{\bigtriangleup I_e})_{Ve}\) ...........(3)

2. d.c. current gain (α_d.c)

The ratio of collector current to the emitter current at constant collector, collector voltage is called dc current gain. Hence,

α_d.c. = \((\frac{\bigtriangleup I_e}{\bigtriangleup I_e})_{Ve}\) ............(4)

α_d.c. is very nearly equal to α_a.c., so we generally denote the current gain as α. Unless otherwise specified α means ac current gain. α is slightly less than 1 because collector current is slightly less than emitter current.

3. Voltage gain (Av)

The ratio of change of output voltage to the change in input voltage is called voltage gain. Hence,

A_v = \(\frac{\bigtriangleup V_0}{\bigtriangleup V_i} = \frac{I_cR_L}{I_eR_i} = \alpha\frac{R_L}{R_i}\)

So, Av = α (Resistance gain) ............(5)

Since R_L >> R_i, So A_v is quite high although α < 1.

4. Power gain (p)

The ratio of change of output power to the change in input power is called power gain. Hence,

Power gain = Voltage x Current gain

P = α. \((\frac{R_L}{R_i}). \alpha\)

So, Power gain P = a² (Resistance gain) ..........(6)

Power gain is moderate.