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


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.
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.
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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