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inductor simulation circuit

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Sandeep Km

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Can u help me get the input impedence of the following circuit diagram? 11.png
 

It seems to me you exchanged inverting and non-inverting inputs of both op-amps. Supposing to reverse the polarity we have:

inverting input 1: V(1-)=Vo1/2
non-inverting input 1: V(1+)=V1

inverting input 2: V(2-)=Vo2+(Vo1-Vo2)*(1/SC)/[(1/SC)+R] = Vo2+(Vo1-Vo2)/(SCR+1)
non-inverting input 2: V(2+)=V1

From the first two equations: Vo1=2V1
From the last two: Vo2*[1-1/(SCR+1)]+Vo1/(SCR+1)=V1, since Vo1=2V1 we have:

Vo2*[1-1/(SCR+1)]+2V1/(SCR+1)=V1 then:
Vo2=[1-2/(SCR+1)]/[1-1/(SCR+1)]*V1 simplifying:

Vo2=V1*(SCR-1)/SCR

From the schematic I=(V1-Vo2)/R then:

I=V1*[1-(SCR-1)/SCR]/R that is:

I=V1/SCR² Then:

V1/I=SCR²

Since the laplace equation of an inductor is V/I=SL then the circuit will simulate an inductor of value L=CR².
Please check the math is correct.
 
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