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For a CE-transistor amplifier fig. The audio signal voltage across the collector resistance of 1.0kOmega is 1.0 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 1.0 V if the de base current has be 10 times the signal current, assuming V_(RE) = 0.6 V. Also calculate the voltage drop across the collector resistance. |
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Answer» Solution :Here `V_(0)=1.0V`, `I_(c)=(V_(0))/(R_(c))=(1.0)/(1000)=1.0xx10^(-3)A=1.0mA` `I_(c)=(V_(0))/(R_(c))=(1.0)/(1000)=1.0xx10^(-3)A=1.0mA` SIGNAL current through the base, `I_(B)=(I_(c))/(R_(c))=(1.0mA)/(100)=0.010mA` Now the base current has to be increased to be, `I_(B)=10I_(b)=10xx0.01=0.10mA` As, `V_(B B)=I_(B)R_(B)` `therefore R_(B)=(V_(B B)-V_(BE))/(I_(B))=((1.0xx0.6))/(0.10xx10^(-3)A)=4xx10^(3)Omega=4kOmega` dc Collector current, `I_(c)=betaxxI_(B)^(1)=100xx0.10=10mA` VOLTAGE drop across the collector resistance `I_(c)R_(c)=(10xx10^(-3)A)xx(100Omega)=10V` |
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