Working of the below Op-amp circuit

[FreshmanNewbie]

I have seen this circuit at my workplace and I'm trying to understand the working and the purpose of each component in the below schematic.


I also happen to note that this was a constant current circuit where R5 - 3ohms happens to be the load resistance, whose value can be from 3-50ohms.

To understand about the constant current Op-amp circuit working, I read how it works. But in those circuits, the Op-amp would have a negative feedback configuration and the output of the Op-amp would drive the base of an NPN transistor which would drive a constant current connected to the emitter of the NPN transistor. The Op-amp will try to maintain the voltage between the inverting and non-inverting terminal and try to regulate the load current through the load resistor.

But in my above circuit shown above, can someone help me how, the constant current through the load resistance is achieved? I am not sure why there is a shunt regulator placed in the negative feedback path?

If possible, please also help me understand why there is a capacitor of 100nF between the inverting and non-inverting terminals of the Op-amp? I have not seen this type of a design. Would like to know the role that each component plays so that I can understand the design better.

 

[KlausST]

Hi,

let´s reduce it to this:

Generally it regulates the R7 voltage to be the same as R3 voltage.
R3 and R1 form a voltage divider. The voltage across R3 is 0.512V.
So the constant current it 0.512V/2.34 Ohms = 0.219A.

C1 is for stabilisation.

Klaus

 

[FreshmanNewbie]

Hi,

let´s reduce it to this:
View attachment 164921
Generally it regulates the R7 voltage to be the same as R3 voltage.
R3 and R1 form a voltage divider. The voltage across R3 is 0.512V.
So the constant current it 0.512V/2.34 Ohms = 0.219A.

C1 is for stabilisation.

Klaus

Could you please simplify more for me? I am confused as to how the voltage across R3 will be the same as R7? Could you please explain a little more with an image on how the voltage across R3 and R7 would be made the same through this circuit? And the values of R2 and R5 are different in your circuit.

And what is the purpose of 47k in the circuit and R5?

 

[KlausST]

Hi,

one key point of a regulating OPAMP is that the (difference) voltage at it´s inputs can be seen as zero.

Now see the upper connection of R7. It is connected to the upper connection of R3. Thus both have the same "upper" voltage.
Now see the lower connection of R7. It goes to OPAMP input.
If the difference voltage between the OPAMP inputs is zero, then the other input has the same voltage. The same as R7_lower.

This signal goes to the lower connetion of R3. Thus both (r7 and R3) have the same "lower" voltage.

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The values of R2 and R5 are not that important, thus i made no input. It´s just the default view.
If you modify R2 or R5 from 50% to 200% of their initial value it won´t change anything significantly.

*********
R1 = 47k .. as already mentiones forms a voltage divider. If you disconnect R1 then the output current will be close to zero.

*********
R5 is the load.

from your post#1:

I also happen to note that this was a constant current circuit where R5 - 3ohms happens to be the load resistance, whose value can be from 3-50ohms.

Klaus