# 0-1A Bidirectional Current Source

1. Oct 29, 2013

I'm trying to design a bidirectional current source for the purpose of measuring very small (0.0001 Ohm) resistances. I've come across this circuit:

I can see that the transistors are there to 'boost' the current but I don't understand the purpose of the 1K resistors and the capacitor. I also don't follow how I can increase the current rating of this pump to 1A.

I've simulated the circuit in LTSpice and it works very well at 100mA but when I try to source or sink 1A I only get 850mA out.

Any suggestions on increasing the current sourcing/sinking ability of the above circuit to 1A or any other suggestions for such a pump?

2. Oct 29, 2013

### sophiecentaur

I don't have any opinion about that circuit, I'm afraid but I wonder whether you ever considered using an AC bridge circuit for measuring low resistance values? The Kelvin Double Bridge (Google it) has been used for ages to measure very low values of resistance. Old fashioned engineering is often a good way to go at the fringes of measurement.

3. Oct 29, 2013

### meBigGuy

I've read that circuit sucks (hard to keep stable), but have no experience with it. Maybe increasing the 1K resistors will increase the drive.

4. Nov 1, 2013

### George H

That's a bit of a weird circuit. It's using the power drawn by the opamp to turn on the transistors. (note the 1 k hom resistors are in the power supply leads.) (And I don't quite see how the feed back works.)

You might try a power opamp. Maybe the TCA0372. Can the load float or does it need to be grounded?
George H.

5. Nov 2, 2013

### Baluncore

The 1k resistors must be low enough to not turn on the power transistors when the op-amp quiescent current flows. When op-amp output current flows, it turns on the power transistors as it must also flow through the 1k resistors.

To increase output current you could use higher gain power transistors such as Darlington pairs.
An alternative would be to cascade another transistor output stage, but use 100 ohm resistors for the final stage for ten times the current.