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Find VE and IE for the circuit given below

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Mumbai University > Electronics and telecommunication engineering > Sem 3 > Analog electronics 1

Marks: 4M

Years: Dec 15, May 16

1 Answer
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Apply KVL at base loop,

10 - IERE-270k IBVBE = 0

10 = IERE+270k IB+VBE

10 = (1+β) IB x 1.5k +270k IB+VBE

10-0.7 = 101 x IB x 1.5k +270k IB

9.3 = IBx421.5x103

IB=0.022x10(3)A

IE=(1+β)IB= 101 x 0.022 x 10(3) A = 2.228 mA

IE = 2.228 mA

Voltage drop across RE=IExRE= 2.228 mA x 1.5k = 3.342 V

VE=VEE – Voltage drop across RE

= -5 + 3.342 = -1.658 V

VE= -1.658 V

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