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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