AC Current Source from a function generator

1. Jul 1, 2008

Jwink3101

First of all, i do not know if this is the ideal forum. If not, mods, please move it. Second of all, I know the title is redudant but whatever.

Anyway, I am doing some experimental work that requires us to be able to controll the current, but we want to have it be a sine wave. We had been using an AC current source made by Keithley (the only company that makes them) but we are having some issues and we are sending it back to be fixed.

So, we are left with using standard function generators and Ohms Law to be able to set the current. The problem is, the thing we are passing the current through changes with time and is not a constant resistance. Is there is relativly simple circuit I could build that would allow me to turn the equavalent of an ideal voltage source into an ideal current source? We are not talking very high currents. More on the order of 100mA max.

2. Jul 1, 2008

Staff: Mentor

What is the range of load impedance? What frequency range are you dealing with? There is probably a fairly simple way to do it with opamps, but the 100mA current is a bit high for standard opamps....

3. Jul 1, 2008

Jwink3101

I do not have a definite range of restistance but at least one test put it at about 1kohm. I have not tested it with the other load. The frequencies are actually very low. The maximum frequency we will use is 0.1Hz but ususlaly closer to 0.05Hz.

I have been trying to get it with a simple transistror and an volatge offset which we will account for, but, assuming my math and analysis are correct, is not very flexible. Ihave heard about current mirrors but they also suffer the same problem, i need negative currents as well.

For now, we are using the function generator and adjusting the current as needed as we go, but it is very inperscise.

Thanks for the help

4. Jul 1, 2008

f95toli

If the resistance of the load is low enough you can usually get away with just using a large value series resistor (e.g. if you are biasing a coil).
However, for currents as large as 100 mA you probably need an op-amp, just about any power op-amp will do the job so it shouldn't be too difficult.
But, as usual it is always best to measure both voltage and current (usually by measuring the voltage across a small resistor); the rule is NEVER to trust a source; at least not if you want accurate values.

5. Jul 1, 2008

Jwink3101

The problem is, we could use an op-amp but I need an AC source that will be the same regardless of the load. If that is easily doable with an op-amp, please let me know. I can only think of how to do it for a constant voltage.

6. Jul 1, 2008

Staff: Mentor

Well, 100mA into 1kOhm would take over a 100V AC source, which is not terribly safe. Can you say more about what you are trying to do? Why do you need a "constant AC current"?

7. Jul 1, 2008

Jwink3101

My 1kohm estimate must be off becuase come to think of it, when we use a current source (DC) we get 100mA for WELL UNDER the 40V max. Also, for AC, the 100mA is an upper bound

Without going into too much detail, we are passing current through a fluid (salt water sometimes, an acidic chemical reaction the others) with an array of strong magnets underneath. This gives us advection of the fluid.

While I am new to this stuff (only here for the summer) my advisor has been doing it for a long time now, We want to test something new now and need AC.

8. Jul 1, 2008

Staff: Mentor

I think the easiest way would be to use two current mirrors, one pullup and one pulldown, and combine their outputs to make your bipolar AC current supply. The simplest version would have zero-crossing distortion as one side shut off and the other turned on, but this could be minimized or eliminated, especially if you're running at room temperature for this circuit.

http://en.wikipedia.org/wiki/Current_mirror

That's the basics of a current mirror. I could try to sketch something up if I get some time tomorrow. What power supply would you use to power the circuit? Do you have something like +/-20V or +/-40V available?

9. Jul 1, 2008

Jwink3101

Thanks for the tip. That is also what I figured I would have to do...more or less. I read the article on Current Mirrors but I was not entirly sure i understood it (or the analysis). I will work through it again later tonight.

I will say from my preliminary analysis of it, it seems like there is a diode drop (in the transistor) to activate or deactivate the current mirror meaning there will be an around +/- .7V gap.

As far as DC supplies go, I more or less have avalible to me what I need. Being that it is the summer I can take bunch from teaching labs. hopefully, this is a temporary solution and we will have our Keithley one back soon.

thanks again

10. Jul 1, 2008

Staff: Mentor

In case it helps your work tonight, I'll try to describe in words what I was starting to sketch.

I sketched the current pullup half of the circuit... There is one NPN and two PNPs, to convert a + side voltage to a pullup current. (There would be a symmetric complemenary circuit to convert a - side voltage to a pulldown current.)

The NPN has an emitter resistor to ground, and its base is driven by Vin. When Vin is + (and over a diode drop), the collector current of the NPN is determined by the emitter resistor value. The collector of the NPN is connected to both bases of the PNPs, and the collector of the left PNP. The emitters of the PNPs are connected to the + power supply through emitter degeneration resistors. The collector of the right PNP sources the pullup current for the load (it connects to one side of the load, and the other side of the load is grounded).

Make a symmetric complementary version of this, and connect the right side NPN output transistor's collector to the output of the first circuit (the collector of the right PNP transistor). When the Vin is +, the first circuit sources Vin/Re, and when Vin is -, the 2nd circuit sinks Vin/Re. There will be crossover distortion with this simple implementation, and you also have to be careful not to break down the input transistors' BE junction with reverse bias when Vin is going the opposite direction. Like, when Vin is -, make sure that it doesn't go negative enough to cause Vbe reverse breakdown in the first circuit's NPN input transistor. Check the transistor datasheets.

You can mitigate the crossover distortion by making two Vin signals with an opamp stage, where the Vin+ is biased up by almost a diode drop, and Vin- is biased down by almost a diode drop. Come to think of it, if you make two Vin voltages like that, you can clip them so that you don't have to worry about the BE reverse breakdown issue. Vin+ doesn't have to be allowed to go negative, and Vin- doesn't have to be allowed to go positive. Just limit them with diodes in the opamp circuit.

EDIT -- added more to last paragraph.

Have fun. Post what you come up with.

11. Jul 1, 2008

Averagesupernova

A current source can be fashioned using an op-amp with the load as part of the feedback path with the op-amp configured in the inverting mode. All you have to do is find an op-amp that can source 100 mA.

12. Jul 1, 2008

Staff: Mentor

Yes, but those are traditionally DC current sources. The hard part of the OP request is for an AC current source. Do you know of an AC variation of the DC opamp current source circuit? I don't know one offhand, but it should be do-able, it would seem.

13. Jul 1, 2008

Averagesupernova

Ummm, yeah. Just feed the thing AC. Naturally a split supply is needed which is common with many op-amps. A push-pull current boost stage on the output of the op-amp could work if an op-amp cannot be found that can source the needed 100 mA. If the gain is kept low (large voltage swing on the input) offset voltages should be kept to a minimum.
-
Also, my previous post is incorrect in stating that the op-amp would be configured inverting. I meant non-inverting. Berkeman, I don't think what you quoted me as saying will change if I edit my post. The device being driven will not be able to be driven as single-ended either. There is an IC that is made to drive a speaker with up to 1 watt. I think that may be a good place to look for a high current output stage. For the life of me I cannot remember what the part number is. It has been around for ages.

Last edited: Jul 1, 2008
14. Jul 2, 2008

Jwink3101

Thanks everybody. I like the Op-Amp idea a lot more becuase I tend to prefer op-amps over transistors (for a variety of reasons the least of which is that they are slightly less fragile). Anyway, I am still trying to work out how to get it to split off the way I want to the negative input and then how to not make it short curcuit

To be honest though, while I am still interested for my own edification, our unit is getting replaced so I will be back to the real one.

I did some research and there are companies that make power op-amps that can handle 2A so that isn't a big problem. Considering I am making one one of them, if a compnent cost a lot more, it isn't that big of a deal.

Anyway, thanks for all the help. I will keep working.