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For a CE transistoramplifier, the audio signal voltage across the collector resistance of 4.0kOmega is 4.0V. suppose the current amplification factor of the transistor is 100. What should be the value of R_(B) in series with supply of 2.0 V if the de base current has to be 10 times the signal current. Also calculate the dc drop across the V_(BB) collector resistance. |
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Answer» Solution :The output as voltage is 4.0V. So the ac collector currnet `I_(C)=4.0//4000=1.0mA`. The SIGNAL current through the BASE is therefore given by `i_(B)=i_(C)//beta=1.0mA//100=0.010mA`. The dc base current has to be `10xx0.010=0.10mA`. ASLO `R_(B)=(V_(BB)-V_(BE))//I_(B)`, Assuming `V_(BE)=0.6V` `R_(B)=(2.0-0.6)//0.10=14kOmega` The dc collector current `I_(C)=100xx0.10=10mA` |
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