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In the circuit shown in the figure, the input voltage V_i = +5 V, V_(BE) = + 0.8 V and V_(CE) = +0.12 V. Find the values of I_B I_C and beta. |
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Answer» SOLUTION :`I_B=V_i/R_B=(5V)/(80kOmega)` `=5/(80xx10^3)=5/8xx10^(-4)=0.625xx10^(-4)=62.5xx10^(-6)` `I_B=62.5muA` `I_C=V_"CC"/R_C =(12V)/(5kOmega)` `=12/(5xx10^3)=2.4xx10^(-3)` `I_C`=2.4 mA `beta=I_C/I_B=(2.4xx10^(-3))/(62.5xx10^(-6))=0.0384xx10^3 , beta=38` 
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