1.

In the common emitter circuit as shown in Fig, the value of the current gain is 100. Find `I_(B),V_(CE),V_(BE),V_(BC)` when `I_(C)=1.5mA`. The transistor is in active, cutoff or saturation state.

Answer» Here, `beta=100,I_(C)=1.5xx10^(-3)A, V_(C C)=24V`,
`R_(B) = 220 xx 10^(3) Omega, R_(C) = 4.7 xx 10^(3) Omega`
`I_(B)=I_C/beta=(1.5xx10^(-3)A)/100=15xx10^(6)A`
In a close circuit GDCEFG
`V_(CE)=V_(C C)-I_CR_C`
`=24-(1.5xx10^(-3))xx(4.7xx10^(3))=16.95V`
In a close circuit GDABEFG
`V_(BE)=V_(C C)-I_B R_B`
`=24-(15xx10^(-6))xx(220xx10^(3))=20.7V`
In a closed circuit ABCDA
`V_(BC)=I_C R_C-I_B R_B`
`=(1.5xx10^(-3))xx(4.7xx10^(3))-(15xx10^(-6))xx(220xx10^(3))`
`=3.75V`
Since `V_(BE)gtV_(CE)`, so both the junction of transistor are forward biased. Therefore, the transistor is in a saturation state.


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