+1 vote
in Linear Integrated Circuits by (37.1k points)
For the current repeater shown in the circuit, determine IC4 value, Where β = 75.

(a) 0.035mA

(b) 0.028mA

(c) 0.04mA

(d) 0.052mA

This question was posed to me during an online exam.

Question is from Operational Amplifier Internal Circuit topic in chapter Operational Amplifier of Linear Integrated Circuits

1 Answer

0 votes
by (39.8k points)
The accurate option is (a) 0.035mA

To make it clear: The reference current,  Iref = VCC-VBE/R1 =(15v-0.7v)/39kΩ = 0.366mA.

⇒ Iref=IC+4×IB


∴ IC= Iref×(1+1/β)

               =Iref×(1+1/β) = 0.366mA×(1+1/75) =0.347mA

⇒ IC1=IC2=IC3=0.347mA

To determine IC4,


⇒ 1.62kΩ =  25mv/(1+1/75)×IC4×ln(0.347mA/IC4)

⇒ IC4=0.035mA (find using trial and error method).

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