Constant current load for testing power supplies (guidelines)

[budm]

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

You are talking about power dissipation in two devices, one the current sensing which you can choose it to dissipate at much as you want it to be, then you have the power dissipation of the active device (MOSFET) that will dissipate most of the power, the same current flows through both devices.
So if you dissipating 1W on the current sensor, then you will be dissipating 99 W on the MOSFET so way getting around dissipating the power, cooling fan should be used..

[Dannyx]

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

Incandescent lamp has hot and cold resistance it is not the same as resistor, cold resistance can be 15 time less resistance than the hot resistance. just look up HOT/COLD resistance of the incandescent lamp.

So how much power dissipation is acceptable to your requirement?

To be fair, I'm not too sure what would be "useful" and what would be "overkill"....how does 100w sound ? It's probably still too much for your average ATX power supply or TV power supply, but might as well go bigger. From what I understand, there are two ways of going about this, neither of which is ideal: a low value resistor like the guy uses in the video makes controlling the thing rather difficult, since only minute voltages have to flow through the FET to reach very high amps instantly and also choosing and cooling FETs is a bit of a challenge which may become very expensive (going to gut some plasma TV boards hoping to find some nice high power components in there). On the plus side, the resistor can be of a relatively low power rating, since it's the FET that does the hard work.

Conversely, if the resistor has a high value, the control system becomes easier to implement (since the voltage has to be higher), but the power dissipation now happens in the resistor (assuming we want to reach the same current) so its power rating has to be really high, which is what I had in mind when I said it's not practical.

[budm]

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

I had another crazy idea: instead of a resistor, how about I use a lightbulb as a "shunt" ? You know, like a halogen lamp or something. If we think about it, a bulb is a resistive load, albeit it will change dramatically with temperature. That way, I can have a higher resistance which can withstand higher power dissipation. A resistor with a high impedance would need to have a VERY high power rating, which is both impossible and impractical to get....just a thought.

Incandescent lamp has hot and cold resistance it is not the same as resistor, cold resistance can be 15 time less resistance than the hot resistance. just look up HOT/COLD resistance of the incandescent lamp.

So how much power dissipation is acceptable to your requirement?

[Dannyx]

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

I had another crazy idea: instead of a resistor, how about I use a lightbulb as a "shunt" ? You know, like a halogen lamp or something. If we think about it, a bulb is a resistive load, albeit it will change dramatically with temperature. That way, I can have a higher resistance which can withstand higher power dissipation. A resistor with a high impedance would need to have a VERY high power rating, which is both impossible and impractical to get....just a thought.

[eccerr0r]

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

Now to do the same with a BJT instead of a mosfet. A different topology would be needed...

[redwire]

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

The shunt should be as small value as possible to not be a big hot resistor.
The shunt should be as large value as possible so the op-amp drift and offset voltage are minor compared to the current-feedback voltage.

So you have to choose a compromise. 0.01ohm dissipates 1W at 10A and 100mV of feedback voltage. This is 10mV/A. If you want to dial down to a 1A load, you have 10mV feedback signal which is getting pretty small but doable.


The hard part is cooling the mosfets. 12V 10A =120W of heat to dissipate.
You would need more than one mosfet for reliability.

[budm]

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

'0.015v/0.0015 Ohms=10a ' That is only 0.15 Watts (so at 50A it is still way below 1 watt). Why do you need it to be that low of the resistance, use higher Value so the Vdrops will be higher so it will above noise level for the Op amp to see. 15mV is very low signal.

[eccerr0r]

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

That's sort of the idea of the op amp, it should be doing all of the control feedback to ensure that you maintain a certain voltage across a resistor - which implies that a fixed current is going through that resistor.

Also as long as the resistor doesn't fry and can pass the current without the voltage going higher than the supply voltage, it doesn't really matter what the actual value of the resistor is. The low values are convenient as they're used for current measurement anyway, though in the CC load, what does not get dissipated in the resistor gets dissipated in the transistor.

As you're dealing with grounded resistors for voltage sensing, you need a op amp that can deal with voltages that low. LM324/LM358 and any RRIO op amp will do.

You can use any voltage reference, a zener or something with a voltage divider will do.

If you want something quick and easy, you don't need to even work with discrete op amps. Just use an IC with an op amp in it with its own voltage reference: use a LM317 and a resistor to become a CC load. No other components are needed. Just use its internal reference of 1.25V to calculate a resistor based on the current you want to pass - for 1A, you just need a 1.25 ohm resistor (at least 2 watt, as it will be dissipating 1.25W). Remember the LM317 will be dissipating whatever the resistor does not dissipate to keep that current going, just like the transistor in the general case you schematic draws.

The LM317 circuit is just Input=your input. Output -> one end of resistor. Other end of resistor -> GND. ADJ-> GND. very simple constant current sink.

...

Oh me?
Do I do this to test power supplies?

hell no.

I just use a resistor or other resistive element (nichrome, light bulbs, etc.) and measure voltage across it. MUCH simpler and does not fry transistors. Not CC but if the PSU keeps the voltage constant (which it should else it's bad by definition!!!) we know exactly how much current and how many watts are going through it.
The only potential issue is that a resistor that can dissipate several watts can be expensive and a transistor that has your own heatsink could dissipate as much heat because of the semiconductor to metal contact could be better than ceramic resistors...

[Dannyx]

I can't help myself boys Looking around the internet, youtube to be exact, I came across an interesting little gizmo that could not only prove very useful to me in the long run as an electronics enthusiast, but could also serve as a good learning exercise: a constant current load. I thought "well what a cool project...I could use something that can give the power supplies I occasionally mess around with a good workout".

The vid that inspired me is here. What the chap achieved is VERY tempting to replicate (100A load), but ultimately way overkill for my purposes, especially since I haven't actually designed one of these before and need to start small, so I'd say.....10A ? That's probably too much still (1A would probably do it for a beginner, but hey....think big ) The device is rather simple, or so it seems at first, because when I really began looking into it, I was quickly faced with one or two quirks I'd like to discuss.

I DID do my research do a degree, so let's go over this. The basic schematic is below. The main ingredients are a series-pass FET and sense resistor. I like to think of the FET as essentially shorting the PSU under test to ground through the resistor. For this reason, the resistor has to have a very small value so it doesn't burn up. The shunts the chap used in his project are 0.0015 ohm 50A shunts (yes, I much prefer using the whole value than using "MILLI..." etc....bare with us ). How do I know the value? For once, I happen to have the same exact shunts on hand (came with some of those panel ammeters you see on ebay...anyway) and also the current and voltage are stamped on the side, so here's where good ol' Ohm's law comes in to work out the voltage: V/R=I so 0.075V/0.0015 Ohms=50A...plain and simple (triple checked my math to make sure I don't make a fool out of myself for not knowing Ohm's law...which will probably happen anyway )

So now that we have the shunts, what do we do with them ? Our chap uses two in parallel to drop the resistance even more, but since I don't need to go too crazy with the current for now, 0.0015 ohms should be enough to achieve a 10a load. I have 3 of those shunts available, so my idea is to build 3 individual loads so I can stress multi-rail supplies simultaneously. My math tells me I'd need to sink 0.015v (again, whole values please ) through that resistor to get 10a....I'll probably get corrected on this though, but 0.015v/0.0015 Ohms=10a ?

The part I'm most interested in is the control feedback: how on earth will I be able to call up just 0.015v ? He uses a fancy voltage reference which can go down to the nearest millivolt, but I don't have one and here's where I get stuck...I'd need to have the control voltage on the non-inverting input at a more manageable level (say 5v or thereabouts), so the feedback to the inverting input would need to be amplified. What would a good op-amp be for such project as a whole and also how could I go about controlling the FET at such low values ? An amplifier stage would need to go in the feedback loop between the sense resistor and the non-inverting pin of the "main" op-amp...the gain would need to be huge, so I'm not sure it's really possible. LOTS of issues here. Let's talk about FET selection a little later on...for now, THIS is my main challenge.