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For a CE-transistor amplifier Fig.the audio signal voltage across thecollector resistance of 1.0kOmega is 1.0V. Suppose the current amplification factor of the transistor is 100, what should be the value of R_(B) in series with V_(BB) =0.6V. 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=1.0 mA` Signal current through the base, `I_B=I_c/beta=(1.0mA)/100=0.010 mA` Now the base current has to be INCREASED to be, `I_B^(')=10I_B=10xx0.01=0.10mA` As, `V_(BB)=V_(BE)+I_B^(')R_B` `:. R_(B)=(V_(BB)-V_(BE))/I_B^(')=((1.0-0.6)V)/(0.10xx10^(-3)A)` `=4XX10^(3)Omega=4kOmega` DC collector current, `I_C^(')=betaxxI_B^(') =100xx0.10= 10mA` Voltage drop across the collector RESISTANCE `I_C^(')= (10xx10^(-3)A)xx(1000Omega)=10V` |
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