Electronic Devices - Special-Purpose Op-Amp Circuits - Discussion

Discussion Forum : Special-Purpose Op-Amp Circuits - General Questions (Q.No. 9)
9.
Refer to the given figure. This circuit is a setup for

an antilog amplifier.
a constant-current source.
an instrumentation amplifier.
an isolation amplifier.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
3 comments Page 1 of 1.

Claude said:   4 years ago
Using KCL in the node : Iin = IL + I.
Iin = input current =(V-0)/ RL,
I = current going to the op-amp,
IL = load current.

Since we know that op-amps have very high input impedance then I would be equal to 0
therefore, Iin=IL + 0 = Vin/Ri, and since both the Vin and Ri are constant, the current in the load will always be constantly creating a constant current source regardless of the value of the load resistance.

Claude said:   4 years ago
Using KCL in the node : Iin = IL + I.
Iin = input current =(V-0)/ RL,
I = current going to the op-amp,
IL = load current.

Since we know that op-amps have very high input impedance then I would be equal to 0
therefore, Iin=IL + 0 = Vin/Ri, and since both the Vin and Ri are constant, the current in the load will always be constantly creating a constant current source regardless of the value of the load resistance.

Gautham said:   8 years ago
Could someone explain this?

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