> Q. For a CE transistor amplifier, the audio signal voltage across the collector resistance of $2.0 kOmega $ is 2.0 V. Suppose the current amplification factor of the transistor is 100, What should be the value of $R_B$ in series with $V_{BB}$ supply of 2.0V if the dc base current has to be 10 times the signal current? – LIVE ANSWER TODAY

Q. For a CE transistor amplifier, the audio signal voltage across the collector resistance of $2.0 kOmega $ is 2.0 V. Suppose the current amplification factor of the transistor is 100, What should be the value of $R_B$ in series with $V_{BB}$ supply of 2.0V if the dc base current has to be 10 times the signal current?

In This post you will get answer of Q.
For a CE transistor amplifier, the audio signal
voltage across the collector resistance of $2.0 kOmega $
is 2.0 V. Suppose the current amplification factor
of the transistor is 100, What should be the value
of $R_B$ in series with $V_{BB}$ supply of 2.0V if the dc base current has to be 10 times the signal current?

Solution:

The output ac voltage is 2.0 V. So, the ac
collector current $i_C$ = 2.0/2000 = 1.0 mA.
The signal current through the base is,
therefore given by $i_{B} =i_{C}backslashbeta =$ 1.0 mA/100 = 0.010 mA.
The dc base current has to be 10 × 0.010
= 0.10 mA.
$R_B = (V_{BB} – V_{BE} ) /I_B$.
Assuming $V_{BE} = 0.6 V, R_B = (2.0 . 0.6 )/0.10
= 14 kOmega $

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