Active current load

I need a circuit which monitors current and will turn on a load to keep the current constant. The current will be about 400mA. It will be use to keep the power supply under a constant load and will shut down when the dut needs current but turn on when the dut current drops. Has anyone every built something like this?

Hi Brent,

This is a pretty interesting circuit. It looks like you are assuming that the DUT will be drawing more than 400mA if not and the DUT is drawing less then the circuit may short out the DUT. One other thought is that once the 400 mA load is enabled the DUT may have ~0V and be turned off.

I have a couple thoughts building on your circuit. Would it be acceptable if it was a linear circuit and it always drew at least 400 mA. for example if the DUT was drawing 200mA the load would draw an additional 200mA. And I think you probably want the same voltage on the DUT at all times, whether the load is on or not.

You didn't say anything about how much voltage you're willing to drop, power you're willing to dissipate, compliance range, etc.

Here's a real simple idea that may not suite your needs, but it's interesting to think about.

Yes I want to keep the same voltage on the DUT. It would be better if it were linaear circuit.

The DUT and the circuit would make the 400mA combined as you say. It would be great if it were a linear circuit since it would keep the current the same reguardless of the DUT current.

Voltage on the DUT is about 1.5 volts, the range would be 1.425 - 1.575V. There is no power limit. I can't put any parts between the DUT's ground.

To keep the voltage the same on the DUT. There has to be some compensation for the voltage drop across a sense resistor. What options are available for the power supply output. Are there any sense pins available? If not how much of a voltage drop can be allowed across the sense resistor? One other question what is the max current you could see the DUT requiring.

This is going to be on a big iron testers so yes there is sense lines. It will bepend on the part being tested. But It will be in the 200 mA range

Since the DUT needs to be connected to the common ground but you've got plenty of power supply headroom and power limit, something like my simple circuit flipped around to use a PNP might actually work. Instead of the magic external 5V source to drive the base, you can drive it with a zener from the main supply, resistor to ground, and cap to ground to kill power supply noise.

Will the DUT ever draw more than 400mA?

No the dut will won't draw more then 400mA

OK, so what's wrong with this then?

It is the right idea but I ran a simulation on the circuit and the voltage at the dut doesn't remain constant which it needs to, unless I am simulating this wrong. I pulsed the DUT current using a current source between 100mA and 200mA, measuring the voltage after the R1 resistor.

If the sense lines were connected the DUT then it might. I am not sure how to sim with sense lines, but I am sure it is possible. Let me think about it.

It should stay reasonably constant. The zener keeps the base of the transistor a fixed voltage below the power rail. That will hold the bottom of R1 at the power rail voltage minus the zener voltage plus the B-E drop of the resistor.

Of course transistor gain isn't infinite, the B-E drop not totally fixed, and the zener voltage not completely independent of current, but they are pretty good first approximations. If these were all perfect then the effective impedance of the supply presented to the DUT would be zero. In reality it should still be well under 1 Ohm. What did you use for the transistor gain, and what kind of supply impedance at the DUT did you see?

I just realized you attached the result of your simulation. Actually it shows the circuit working quite well. You got a 18mV variation when the current changed by 100mA. That's a effective impedance of 180mOhm, which is around the kind of numbers I was expecting.

You hadn't mentioned this need for "constant" voltage before. Actually "constant" is no spec at all since it implies exactly 0 impedance, which of course no circuit is going to achieve. You have to decide how much variation or how much impedance you can tolerate. If you need substantially less than 180mOhm, then this simple circuit isn't going to do it. Real specs would be helpful.

I just placed a 2N2907 transistor which has a gain of about 100 depending on the current. For the Dut load impedance, I am using a current load and I am not finding where the impedance is specified.

Cody, the sense lines wouldn't do any good since you would have to run the circuit higher then the Dut voltage so it would try to correct for the 11 ohm resistor.

I tried changing the resistor value to 25 ohms to get the voltage closer to the Dut voltage which I would need and I get a lot higher voltage swing about a volt. With the 11 ohm resistor I get about 60 mV swing.

I haven't played around with transistors for quite sometime so I am having to dust off some cobwebs.

I was talking about the impedance of the supply presented to the DUT. Since you say you want no variation in that supply voltage, you are saying that its impedance needs to be 0. Of course that's not possible, so we need to know what impedance, or what voltage variation, you can tolerate.

I don't know where you got 60mV variation from. Your plot shows 16mV. (I said 18mV before and just realized I picked off the wrong number from the graph. Sorry for the confusion). 16mV variation in response to 100mA load sounds pretty good. As I said, that's the range I was expecting this circuit to do. If you want better, then a different circuit topology is indicated.

Again, we need some clear specs. If 160mOhm impedance presented to the DUT isn't good enough, then we need to know what is.

The 60mV came from changing the transistor to an actual transistor on the simulation but it more like 50mV.

The impedance needs to be zero but I know we can't have that so it needs to be the lowest we can get.

The idea is to have the tester power supply in a saturated stated to eliminate the slow performance of the supply to the DUT. I don't know if this will even work.

Running this circuit at a faster switching speed I am getting some voltage spikes which wouldn't be acceptable.

So put a cap accross the DUT.

Don't get so hung up on simulations that you forget to think.

They don't want caps on the DUT because they are evaluating the DUT, but want a faster supply. This is the testing world they always want goes against the grain

I have a very simple and easy solution for you. Use an LM317T or K with a 3 ohm resistor between the output and the adj pin. You draw the current from the ADJ pin node. There will be a few volts of headroom required, but it is the simplest solution I know. BTW the 3 ohm resistor will need to carry the 400mA. I suggest 3 x 1ohm 1W resistors would be suitable for this.

Refer to the LM317 datasheets

The old classic technique:

http://www.eetimes.com/design/power-managemen...-rheostat-provides-decades-of-load-resistance

Maxim has used a circuit similar to this for decades to test their power supplies - load current is independent of applied voltage. I'm in the process of building one up right now - simultions show solid performance from 200 uA to 9A, 1V to 50V. An INA128 is a great high side monitor to look at the current.