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Question

Consider the circuit shown in the figure below:


Assuming all the transistors to be matched and working in active region with β=β1=β2=100,β399 and VBE(on)=0.7. If the value of I0=100 mA then the value of resistance R1 is equal to ___Ω.

  1. 86

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Solution

The correct option is A 86

Now, Ireff=ICl+IB3
and since IB1=IB2 and I0=k1=k2 ( transistors are matched)
IE3=2IB2
Also, IE3=(1β3)IB3
Thus, Ireff=IC1+IE3(1+β3)=IC1+2IB2(1+β3)
We can write is as, Ireff=IC2+2IC2β(1+β3)=I0[1+2β(1+β3)]

Ireff=100[1+2100×100]0.1 A
Now,
V=VBE5+VBe2=0.7+0.7=1.4 V

R1=101.40.1=86 Ω

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