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The current gain of a common emitter transiter circuit shown in figureis. Drawthedc load line andmark the Q point on it . |
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Answer» SOLUTION :`BETA=120` Base CURRENT`I_B=(25V)/(1MOmega)=(25)/(1xx10^6)=25muA` `beta=(I_C)/(I_B),I_C=betaI_B` `I_C=120xx25muA` `I_C=3mA` `V_CE=V_CC-I_CR_C` `V_CE=25-3mAxx5k` `V_CE=10V`
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