In a common-emitter transistor amplifier, signal voltage across the collector resistance of 2 kΩ is 2 volt. What should be the value of resistance R_B in series with V_BB voltage supply of 2 volt in the input circuit, if the DC base current is 10 times the signal current in R_B ?
(Given : Base emitter voltage, V_BE= 0.6 voit; and current amplification factor $β = 100 .$

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$10 kΩ$

$7 kΩ$

$14 kΩ$

$26 kΩ$

answer is D.

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Detailed Solution

$\begin{array}{l} {(i_{c})}_{AC} = \frac{{(v_{m})}_{AC}}{R_{C}} = \frac{2 wolt}{2 kΩ} = 1 mA \\ {(i_{B})}_{AC} = \frac{(i)_{AC}}{β_{s}} = \frac{1 mA}{100} = 0 .01 mA \\ {(i_{B})}_{DC} = 10 {(i_{B})}_{AC} = 0 .1 mA = 10^{- 4} \\ A \\ R_{B} = \frac{V}{I_{B}} = \frac{2 - 0 .6}{10^{- 4}} = 14 kΩ \end{array}$

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