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ESE Electronics 2018: Official Paper

Option 4 : 0.02 mA

CT 3: Building Materials

2962

10 Questions
20 Marks
12 Mins

__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:

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

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