An NPN transistor is used in common emmitter configuration as an amplifire with 1k\Omega load resistance. Signal voltage of 10\; mV is applied across the base-emitter. This produces a 3 \; mA change in the collector current and 15\: \mu A change in the base current of the amplifier. The input resistance and voltage gain are :

  • Option 1)

     0.33\: k\Omega ,1.5           Drift along river flow x= d\left ( \frac{u}{v} \right )

  • Option 2)

    0.67\: k\Omega ,300

  • Option 3)

     0.67\: k\Omega ,200

  • Option 4)

    0.33\: k\Omega ,300

 

Answers (1)

R_{L}=1K\Omega

\Delta I_{B}=15\mu A

\Delta I_{C}=3\: m A

V in =10 \; mv

\frac{Vin}{\Delta I_{B}}=r\; in

r\; in=\frac{10mV}{15\times 10^{-6}}

r in=0.67\; K\Omega

\Delta v=-\beta \frac{R_{L}}{r_{i}}

\beta=\frac{\Delta I_{C}}{\Delta I_{B}}

\Delta v=-\left ( \frac{3 mA}{15\mu A} \right )\left ( \frac{1K\Omega }{2000}\times 3 \right )

\Delta v=-300

 

 

 

 


Option 1)

 0.33\: k\Omega ,1.5           Drift along river flow x= d\left ( \frac{u}{v} \right )

Option 2)

0.67\: k\Omega ,300

Option 3)

 0.67\: k\Omega ,200

Option 4)

0.33\: k\Omega ,300

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