GATE 2016 EE (SET 1) Q12

12. A transistor circuit is given below. The Zener diode breakdown voltage is 5.3V as shown. Take base to emitter voltage drop to be 0.6V. The value of the current gain \beta  is ________. 

Solution:

V_{B} = 5.3V

V_{E}= V_{B}-0.6= 4.7V

I_{E}= V_{E}\setminus 470=10mA

I_{B}=I_{1}-I_{2}=0.5mA

I_{E}\setminus I_{B}=\beta +1=20

\beta =19

 

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