# Op~Amp Current Regulator?

Discussion in 'General Electronics Chat' started by ELECTRONERD, Jul 10, 2009.

1. ### ELECTRONERD Thread Starter Senior Member

May 26, 2009
1,146
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Hey Everyone,

I know that an Op~Amp can be a voltage regulator, but can it be manipulated to be a current regulator as well? Thanks in advance!

2. ### russ_hensel Distinguished Member

Jan 11, 2009
820
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Yes.

also Op amp current source and
Op amp constant current

3. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Sink or Source? Transistors do an excellent job too, so that is usually my first recommendation. In a standard inverting op amp, both feedback resistors have constant current going through them for a given voltage input.

4. ### millwood Guest

yes. there are many opamp based constant current sources out there.

5. ### steveb Senior Member

Jul 3, 2008
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eray likes this.
6. ### millwood Guest

or you can apply the input voltage on the non-inverting end and eliminate a couple resistors.

7. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
You mean to apply the input voltage directly to the + input of the opamp?

Which two resistors would you eliminate?

8. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
This is not a good implementation of an opamp current source (it's the Howland circuit;http://www.national.com/an/AN/AN-1515.pdf).

This circuit has the two resistors connected to the + input the same as the two resistors connected to the - input; both pairs are R1 and R2. This gives an output impedance of R1+R2, not very high.

By properly selecting the two resistors connected to the - input, the output impedance can (theoretically) be made infinite.

went into much detail.

9. ### millwood Guest

here is what I have. you can work out the math to get your desired current output, given the parameters.

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10. ### The Electrician AAC Fanatic!

Oct 9, 2007
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If this circuit is to be a current regulator, I would expect that for a fixed Vin and Vref, the current in R1 would be constant no matter what the value of R1, but that doesn't seem to be the case.

11. ### ELECTRONERD Thread Starter Senior Member

May 26, 2009
1,146
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Thanks for all the quick replies. I really want to understand the theory of op amp design, so your help is appreciated.

12. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
One way of looking at a op amp is the classic bridge, any bridge, be it Wheatstone or Wien. The op amp does what it has to to null the bridge.

13. ### millwood Guest

R1 really isn't the load: R1 is a parameter that helps set the current. for the circuit to work, you will need to put a transistor on the output so to isolate the current from R1.

for a current sink, the transistor's base will be to the opamp's output, and its emitter to the junction of R1/R2, and the collector will be tied to the positive rail via the load.

the transistor doesn't need to be perfect as any Vbe imperfection will be handled by the opamp's feedback network.

similarly you can set up a current source, going from ground to the negative rail.

14. ### millwood Guest

opamps are some of the easiest devices to understand, at least the ideal ones.

1) it has infinite open loop gain. this means that once you put an opamp within a negative feedback loop, its non-inverting pin and its inverting pin have exactly the same voltage.
2) it has infinite input impedance: no current going into / out of the inverting / non-inverting pins.

things get more complicated as you relax those two assumptions, and even more so when you factor in topology (vfb vs. cfb), ac performance, stability, etc.

but for what we do, and with a typical opamp, understanding DC performance, and ac stability will suffice 99% of the time.

15. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
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What you said, with respect to the original circuit, was "...or you can apply the input voltage on the non-inverting end and eliminate a couple resistors." You didn't say anything about adding a transistor.

It seems to me that the output current will be strongly influenced by the hfe of the transistor, rather than being determined by resistor ratios.

The original circuit can have an output impedance approaching infinity, and the current is determined by resistor ratios, and is therefore not much affected by temperature, or by parameter variations of a single transistor.

Do you have an expression for the output impedance for your circuit? Won't it be essentially just the ro of the transistor? That may not be very high.

16. ### millwood Guest

whether you need a transistor depends on if the load can be floated, ie. both ends off the ground.

no.

it will be affected by temperature, unless you go out of the way to take care of it.

I am not sure if i can give you one but I am imagine that it is pretty high, given that it is a ccs.

17. ### millwood Guest

really? what if it is?

18. ### millwood Guest

it is about 45kohm upto about 100hz and then down hill from there - poor CMRR.

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19. ### millwood Guest

here are two dc sweep charts, one for R5 (the load) at 22ohm, and another for R5 at 220ohm, or 10x bigger.

you can see for yourself. or you can work out the math for yourself as well.

BTW, I hope you know why the curve is a little bit flat on the right for the R5=220ohm chart.

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20. ### millwood Guest

let's see, .

1st chart: if I put one bjt there, and do a dc sweep.

you can compare the output current (Iout through the load resistor) vs. that we did earlier without the bjt.

they look identical, don't they? other than the regions where the opamp goes into clipping or the bjt is reverse biased.

you can also see the beta of the transistors used. it is in the 200+ territory. and there is one Vbe on the output.

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