Constant current load for testing power supplies (guidelines)

[Dannyx]

Re: Constant current load for testing power supplies (guidelines)

Thanks for the input...my workplace is being reorganized at the moment (actually we're on the verge of disbanding, but nvm), so I won't have time/space to mess around with little projects for a while now, since work is where I spend most of my time and do stuff like this. I'm much too beat after a day at work to work on anything at home, so I only do the "design and research" here

So another important factor appears to be the choice of op-amp.... even a split-supply would not be too much of an issue, since I do have some center-tapped transformers lying around which could use a good project

[eccerr0r]

Re: Constant current load for testing power supplies (guidelines)

I don't think the LM741 won't work at all however. The key thing is that it requires the usual bipolar supply in order for it to work in this situation. You'll need the V- of the LM741 to go to at least -3V relative to GND, and the V+ supply probably +12V or so relative to GND. Yes the supplies are not equal to each other, but who's checking? Op amp won't care...

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If one can get away with just using a LM317, it's a lot easier to just implement it this way as everything but the sense resistor is inside the LM317 (op amp and dissipation transistor). Using the LM317 means you have to get all the way to 1.25V, so even in this prepackaged solution it's a fairly high sense voltage.

About the largest problem with the LM317 is that it's only 1.5A max. A LM350 will get you to 3A and LM338 to 5A, though each of these would require beefier and beefier sense resistors. A 10W resistor should handle all cases (5A * 1.25V = 6.25W, which is within 10W).

[momaka]

Re: Constant current load for testing power supplies (guidelines)

^ No, that is just for the V1 control voltage for the op-amp / comparator on the schematic you posted on the first page. Though I see now that they showed a LM741 op-amp used as a comparator - and that is a BAD idea, because LM741 is probably the lousiest op-amp you can use (believe me, I actually tried this circuit a long time ago, and until I switched it out for something better - in my case, an LM358 op-amp - the circuit would not work properly).

Basically the way this circuit works is the op-amp / comparator drives the Gate of the MOSFET based on the voltages it receives from its positive (+) and negative (-) input terminals. If the voltage on the (-) input is higher than the (+) input, the op-amp/comparator output goes low (i.e. whatever you used for Vss - the negative supply to the op-amp - in your case of that circuit, it is ground.) Conversely, if the voltage on the (-) input is lower than the (+) input, then the op-amp / comparator output swings high (whatever you use to feed Vdd to it).

So if you wonder how that circuit manages to work like that, think of it this way: the op-amp / comparator always wants to have the voltage between it's negative (-) and positive (+) inputs the same. Thus, if you set V1 to output 0.357V, then that will also be the voltage present across resistor R1. Why? Because if you set V1 to some positive voltage and there is nothing on the (-) input of the op-amp, it will output a high signal, which goes to the MOSFET gate. This turns the MOSFET ON and current starts going through R1. As the current through R1 rises, so does the voltage across R1. When the voltage across R1 becomes higher than that of V1, the op-amp switches its output back to low and the MOSFET turns OFF again. Then the voltage across R1 decreases again below that of V1, and the op-amp / comparator switches back the MOSFET ON. So in essense, this whole cycle repeats over and over.

However, it is worth noting here that this is NOT a switching circuit and the op-amp / comparator output never actually goes fully ON of OFF. Instead, after some time (called steady state), the op-amp / comparator outputs a fairly steady voltage that turns ON the MOSFET, but just only a little bit to drive enough current through resistor R1 to keep its (-) and (+) input terminals at balance. Thus, this circuit is actually LINEAR in nature, just like an LM317 regulator circuit. As such, at higher input voltages to the MOSFET, most of the power will be dissipated by the MOSFET rather than the resistor. So that FET will need a large heatsink. And even with a very oversized heatsink, you still need to make sure that you are not running your MOSFET past its safe operating area (look for an SOA graph in its datasheet).

[Dannyx]

Re: Constant current load for testing power supplies (guidelines)

Yes, you can, and *without* a multi-turn pot. The trick: 100 to 270 Ohm resistor connected in series with a diode (to the diode's anode). Connect one side of the 100-270 Ohm resistor to your 3.3V/5V/12V rail and the diode's cathode to ground. The mid-point between the resistor and the diode will be between 0.6 to 0.8V, depending on the forward voltage of the diode. Now, even with a crappy pot (I suggest 1K or 10K), you will be able to dial anywhere between 0 to ~0.7V with much better precision. Need more precision? Add a second voltage divider connected between the diode and your 100-270 Ohm resistor. Two 470 Ohm resistor will drop that 0.6-0.8V forward voltage to 0.3-0.4V. If you do that, then use a 10K pot to get any range between 0 to ~0.4V with even more precision.

Your suggestion eliminates the FET and op-amp entirely then ?

[momaka]

Re: Constant current load for testing power supplies (guidelines)

Can I really get as low as 100mV with an op-amp and a control pot? Ok, it definitely has to be a multiturn pot...

Yes, you can, and *without* a multi-turn pot. The trick: 100 to 270 Ohm resistor connected in series with a diode (to the diode's anode). Connect one side of the 100-270 Ohm resistor to your 3.3V/5V/12V rail and the diode's cathode to ground. The mid-point between the resistor and the diode will be between 0.6 to 0.8V, depending on the forward voltage of the diode. Now, even with a crappy pot (I suggest 1K or 10K), you will be able to dial anywhere between 0 to ~0.7V with much better precision. Need more precision? Add a second voltage divider connected between the diode and your 100-270 Ohm resistor. Two 470 Ohm resistor will drop that 0.6-0.8V forward voltage to 0.3-0.4V. If you do that, then use a 10K pot to get any range between 0 to ~0.4V with even more precision.

[eccerr0r]

Re: Constant current load for testing power supplies (guidelines)

I'd say screw it, if I designed for 10A and it draws either 8A or 12A, it's close enough for a load tester...

But yes, though 100mV is cutting it close for a lot of op amps out there but not in the realm of impossible. I might go and try it and see what a few op amps will do...

Now I don't have a MOSFET I can burn (pun?) on this so I'll have to go with a BJT...

[Dannyx]

Re: Constant current load for testing power supplies (guidelines)

Then the other question, how accurate do you need to be...
If you don't care about 1% accuracy and just need to be in the ballpark, don't worry too much about offset voltage (most op amps are within 15mV, a lot are better than this) and just use around 100-200mV to keep most dissipation in the transistor. Just need to make sure you select op amps that can take inputs near the zero volt mark, so you don't need to bias.

Accuracy ? Well not laboratory-grade, that's for sure...I wouldn't be able to obtain such levels of precision with junk-box parts for one thing and secondly its usefulness would outweigh the effort. To draw 10a at 100mV would require a 0.01 resistor because 0.1v/0.01ohm=10a, correct ? (bare with me, I'm just getting the hang of things). The power dissipation would then be 0.1v*10a= 1w ? Can I really get as low as 100mV with an op-amp and a control pot? Ok, it definitely has to be a multiturn pot...if I were really ambitious, I'd try and to it PWM but let's get the basics down for now

[eccerr0r]

Re: Constant current load for testing power supplies (guidelines)

Then the other question, how accurate do you need to be...
If you don't care about 1% accuracy and just need to be in the ballpark, don't worry too much about offset voltage (most op amps are within 15mV, a lot are better than this) and just use around 100-200mV to keep most dissipation in the transistor. Just need to make sure you select op amps that can take inputs near the zero volt mark, so you don't need to bias.

[Dannyx]

Re: Constant current load for testing power supplies (guidelines)

The power handling and the resistance are not related, you can have 5 Ohm resistor that can only handling 1/8W (I.E. SMD resistor) of power dissipation and you can have 5 Ohms resistor that can handle 10W of power, same resistance but the physical size is not the same.

Yes, I believe that's what I said

[budm]

Re: Constant current load for testing power supplies (guidelines)

The power handling and the resistance are not related, you can have 5 Ohm resistor that can only handling 1/8W (I.E. SMD resistor) of power dissipation and you can have 5 Ohms resistor that can handle 10W of power, same resistance but the physical size is not the same.

[Dannyx]

Re: Constant current load for testing power supplies (guidelines)

" but a 5ohm 5w resistor is the same as a 5w 0.5w one, right ?" ????? How can they be the same?

My bad, I misspelled it, I meant the value (the resistance in ohms) is the same even though the power rating is higher, so there exist 5ohm resistors (as an example) of both 1w and 5w or even 0.5w...does this make sense ?

[budm]

Re: Constant current load for testing power supplies (guidelines)

" but a 5ohm 5w resistor is the same as a 5w 0.5w one, right ?" ????? How can they be the same?

[Dannyx]

Re: Constant current load for testing power supplies (guidelines)

Well let's say the voltage range will be the usual rails of a PSU: 3.3, 5, 12 and maybe throw 24v in there for good measure, so yeah, a pretty wide range...might have to build multiple loads appropriate for each voltage.

[Dannyx]

Re: Constant current load for testing power supplies (guidelines)

But why? Total power dissipation of ALL the resistors will be 2W at 1V Vdrops, so each one of the 10 resistors will only dissipate 0.20W. So how did you come up with 5 Ohms at 5W resistors? 2A of current is distributed through each one of the 10 resistors, SO ONLY 0.20A of current will flow through each resistor, simple Ohm's law.
5 Ohms x 0.20A = 1V

I said 5w only to have more headroom in case I need to push things and I gave 1v as an example - I can go over that if I want more current draw, right ? It's not like the power rating of the resistor affects anything other than the size....ok, so maybe it does...tolerance and price and stuff, but a 5ohm 5w resistor is the same as a 5w 0.5w one, right ? No really, am I correct ?

[eccerr0r]

Re: Constant current load for testing power supplies (guidelines)

Hmm, maybe they weren't resistors after inspecting the pictures again. I thought I saw ceramic but due to the focus they may be heat sinks after all. It's a Seasonic and posted in the "Could a PSU tester cause a power supply to start smoking?" thread.

I've seen PSU testers with many parallel resistors before so it's not uncommon to do this. Of course you want to run the resistors close to their limits to reduce costs. If using common 10W resistors on the 5V line, you want to use 10W=5V^2/R R=2.5 (use 2.7) ohms and multiply that out. About 2A per leg here. For 12V, 10W=12V^2/R R=14.4 ohms, use a 15 ohm resistor per 10W. This time less than 1A per leg here due to the higher voltage.

Using the transistor as the dissipation device, you can just get the exact value you need, don't bother with paralleling resistors as the resistors will not be the major component.

Keep in mind regardless of transistor or resistor load, the dissipation will be voltage dependent. At least with a CC load, the power will go up linearly with voltage whereas pure resistive, square of voltage. Either way, the expected voltage range must be taken into account.

[budm]

Re: Constant current load for testing power supplies (guidelines)

Had another brainfart: how about some resistors in parallel ? Like 10 5ohm 5w resistors in parallel (could get expensive)... Would that give me 0.5 ohms and 50w max dissipation ? If I put 1v across this thing I'd get 1v/0.5ohm=2a. Then, 2a*1v=2w ?

But why? Total power dissipation of ALL the resistors will be 2W at 1V Vdrops, so each one of the 10 resistors will only dissipate 0.20W. So how did you come up with 5 Ohms at 5W resistors? 2A of current is distributed through each one of the 10 resistors, SO ONLY 0.20A of current will flow through each resistor, simple Ohm's law.
5 Ohms x 0.20A = 1V

[Dannyx]

Re: Constant current load for testing power supplies (guidelines)

What other thread, pardon my asking

[eccerr0r]

Re: Constant current load for testing power supplies (guidelines)

That's also a very common way of making load testers, you can see that Asus load test on that other thread, it's full of parallel resistors instead of using transistor loads...

[Dannyx]

Re: Constant current load for testing power supplies (guidelines)

Had another brainfart: how about some resistors in parallel ? Like 10 5ohm 5w resistors in parallel (could get expensive)... Would that give me 0.5 ohms and 50w max dissipation ? If I put 1v across this thing I'd get 1v/0.5ohm=2a. Then, 2a*1v=2w ?

[Dannyx]

Re: Constant current load for testing power supplies (guidelines)

Cooling the thing is my last concern at the moment, since I do have some nice fat heatsinks around which should do the job nicely. Implementing that op-amp control is the main challenge