Question

A transistor is connected in CE configuration with V_CC = 10 V. The voltage drop across the 600 Ω resistor in the collector circuit is 0.6 V. If α = 0.98, the base current is nearly
1. 6.12 mA
2. 2.08 mA
3. 0.98 mA
4. 0.02 mA

Answer 1

Correct Answer - Option 4 : 0.02 mA

Concept:

The common base DC current gain is a ratio of the value of the transistor's collector current to the value of the transistor's emitter current, i.e.

\(α = \frac{{{I_c}}}{{{I_e}}}\)   ---(1)

Also, the common-emitter current gain is the ratio of the value of the transistor's collector current to the value of the transistor's base current in a transistor, i.e.

\(β = \frac{{{I_c}}}{{{I_b}}}\)   ---(2)

Using Equation (1) and (2), we get:

\(β =\frac{α}{1-α}\)

Analysis:

The given CE transistor can be drawn as:

DIAGRAM

With α = 0.98, we get:

\(β =\frac{0.98}{1-0.98}\)

\(β =\frac{0.98}{1-0.02}=49\)

From the above circuit, the collector current will be:

\(I_c=\frac{0.6}{600}=1~mA\)

V_CE is obtained by applying KVL from Vcc to emitter ground as:

10 - 0.6 - V_CE = 0

V_CE = 9.4 V

Since V_CE > V_CE(sat), the transistor is working in the active region and we can write:

I_C = β I_B

The base current from the above will be:

\(I_B=\frac{I_C}{\beta}\)

\(I_B=\frac{1m}{49} = 0.02~mA\)