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Question
In the circuit shown in figure, find the value of RC.
Solution
IE = IC + IB
IC = βIB
ICRC + VCE 6 IERE = VCC
RIB + VBE + IERE = VCC
From (3) Ie ≈ Ic = βIB
(R + βRE) = VCC – VBE,
IB = `(V_(C C) - V_(BE))/(R + βR_E)`
= `11.5/200` mA
From (2)
RC + RE = `(V_(C C) - V_(CE))/I_C`
= `(V_(C C) - V_(CE))/(βI_B)`
= `2/11.5 (12 - 3)` KΩ = 1.56 KΩ
RC = 1.56 – 1
= 0.56 KΩ
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