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For a BJT circuit shown , assume that the 'beta' of the transistor is very large and V_(BE)=0.7V. The mode of operation. |
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Answer» Solution :`V_(BE)=0.7V` Input junction is a forward BIASED . SINCE , `V_(BE)=0.7V` `V_(CE)=V_(BE)+V_(CB)` `V_(CB)=V_(CE)-V_(BE)` To determine `V_(CB)` we find `I_(C)` `I_(C) -=I_(C)=(2-V_(BE))/(R_(2))=(2-0.7)/(1kOmega)` `I_(C)=1.3mA` `V_(CE)=V_("CC")-I_(C)(R_(1)+R_(2))` `=10-1.3mA (10K+1K)` `V_(CE)=-4.3V` `V_(CE)=-4.3V-0.7` `V_(CB)=-5V` |
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