Analog Electronics

Q1: A transistor with a = 0.9 and ICBO = 10 μA is biased so that IBQ = 90 μA. Then ICQ will be

A 910 mA

B 5 mA

C 5.25 mA

D 4.95 mA

ANS:A - 910 mA

ICEO = (β + 1)ICBO (9 + 1) 10 μA 100 μA ICQ = 9 x 90 x 10-6 + 100 x 10-6 = 910 μA.



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