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Q.

For a CE transistor amplifier, the audio signal voltage across the collector resistance of $2 kΩ$ is 2V. Suppose the current amplification factor of the transistor is 100. The value of R_B in series with V_BB supply of 2V, if the DC base current has to be10 times the signal current is

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a

$4 kΩ$

b

$14 kΩ$

c

$28 kΩ$

d

$54 kΩ$

answer is B.

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Detailed Solution

The output AC voltage is 2V. So the AC collector current $i_{c} = \frac{2}{2000} = 1 mA$
The signal current through the base is therefore , given by
$i_{B} = \frac{i_{c}}{β}$
The D.C base current has to be 10 x i_B
$\begin{array}{l} ∴ R_{B} = \frac{(V_{BB} - V_{BE})}{I_{B}}, V_{BE} = 0 .6 V \\ ∴ R_{B} = \frac{(2 - 0 .6)}{0 .10} = 14 kΩ \end{array}$

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