Question

For a CE transistor amplifier, the aurdio 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_(B B)` supply of 2.0 V if the dc base current has to be 10 times the signal current. Also calculate the dc drop across the collector resistance (Refer to Figure).

Answer 1

The output ac voltage is2.0 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 ) //beta = 1.0 mA// 100 = 0.010 mA`. The dc base current has to be `10 xx 0.010 = 0.10 mA`
From `V_(B B ) = V_(BE ) + I_(B) R_(B) R_(B) = (V_( B B) - V_(B E) )//I_(B) `. Assuming `V_(BE) = 0.6 V` .
` R_(B) = ( 2.0 - 0.6 ) //0.10 = 14kOmega `