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In the circuit, the value of beta is 100. find I_B,V_(CE),V_(BE) and V_(BC), when I_C=1.5mA. The transistor is in active, cut off or saturation state? |
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Answer» Solution :`beta=100,I_C=1.5mA=1.5xx10^(-3)A,V_CC=24V` `beta=(I_C)/(I_B)` `I_B=(1.5xx10^(-3))/(100)=15muA` To calculate `V_CE` we apply kirchhoff.s rule CEFDC, therefore `V_CC=I_Cxx4.7kOmega+V_CE` `24=1.5xx10^(-3)xx4.7xx10^(3)+V_(CE)` `V_CE=24-7.05=16.95V`  Again applying Kirchhoff.s rule to loop ABEFDCA, we get, `V_CC=I_Bxx220kOmega+V_BE` `V_BE=24-3.3` `V_BE=20.7V` GOING along loop ABCA,we get `I_Bxx220kOmega+V_BC=I_Cxx4.7kOmega` `15xx10^(-6)xx220xx10^(3)+V_BC=1.5xx10^(-3)xx4.7xx10^(3)` `V_BC=7.05-3.3=3.75V` As `V_CEltV_BE`, both the juction are forward biased.So, the transistor is in the saturation state. |
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