Need exspert advice for creating a load.

[mpilchfamily]

I'm trying to design a PSU Dummy Load. I've already worked out a design that uses the basic banks of load resistors and switches to give the aproximate load needed for testing. What i'm shooting for now is an adjustable load made from trasistors, that will give me a more accurate load. Here is a schematic of what i have worked out. Just wondering if this setup is correct or not and if those transistors will be OK on the 12v rails.

http://img229.imageshack.us/img229/7605/atachment2lr1.jpg
By mpilchfamily (http://profile.imageshack.us/user/mpilchfamily) at 2007-10-10

Thanks in advance.

[mpilchfamily]

I think the image wasn't showing correctly before but it should be fixed now.

[mpilchfamily]

I guess i'll just get a Sunmoon SM-268ATE+. Or at least the site i'll be reviewing for is talking with Sunmoon about getting one.

[Oklahoma Wolf]

I guess i'll just get a Sunmoon SM-268ATE+. Or at least the site i'll be reviewing for is talking with Sunmoon about getting one.

Probably the best way to go, unless Oleg over at Xbit will share his load tester plans... you'd need an electronics engineer to answer the above question.

[Super Nade]

I presume you are using Ib (base drive current) to turn the transistors on/off right? I don't see how you can do that individually in your config?

[mpilchfamily]

Your right. This design pretty much adjusts the current load on all the transistors at the same time. I guess that would not allow me to have the accuracy i would want. While the site I'm with is getting a sunmoon sooner or latter we are still continuing to design a DIY load tester for the fun of it and the education value in it.

[mpilchfamily]

My hope is to build my own loadtester and see how it performs when compaired with the Sunmoon. If it works out OK then i'll have an exstra load tester to loan out for someone else to help out on the PSU reviews and also for some verification of the number. Should help to get a better picture of the PSUs abilities running it on different test systems.

[cypherpunks]

That's actually a pretty good design. A basic voltage-controlled current source (well, load) using darlington NPN power transistors. The main thing I would like to know is the value of the load resistors ("10A" in the picture.)

Let's see... using the 1K pots (which will dissipate 144 mW > 1/8W on the 12V rail; are they rated for that?), you can adjust pull the emitter up to 2*Vbe below 12V. When pulling lots of amps, Vbe is a lot more than the usual 0.65 V; the data sheet suggests 2 V might be a more reasonable number. But that's in the power transistors; the darlington driver will be less. Let's call it 3V total, for a Ve of 9V.

At that value, the transistors can handle 30A of current each. Which is 90 W of heat generated in each, which is well under the power derating curve (150 W minus 0.855 W per degree C over 25 C means that you can do 120 W at 60 C case temperature and 90 W at 95 C case temeprature).

Note that at 30 A and 9 V Ve per transistor, you'll need 0.3 ohm emitter resistors and will be dumping 270 Watts into each, so you'll need some pretty honking emitter resistors.

With 6 transistors on each 12V rail, you can pull a total of 180 A or 2160 W, which should be plenty.

The other rails will need

You may have problems with the Darlington configuration on the 3.3V line; with two Vbe drops, there's not enough voltage headroom for a reasonable emitter resistor. You might just reference the 3.3V and 5V pots to a +12V line and make the problem go away; then you could drive the first Darlington transistor, at least, into saturation.

Oh, and I don't see a reason for such miniscule current-measuring shunts. Especially on the 12V lines, you can afford to drop some noticeable voltage there. Say a maximum of 1V at 100 A, that's still 0.01 ohms, which is 10x what you have. Or is it a power-handling limit? What's the power rating on those shunts?

Note that your current scheme is NOT temperature-compensated, so the current draw will drift a bit as the transistors warm up. If you set

If you want cheaper resistors and a lower maximum load, you can use smaller emitter resistors and adjust the transistor base circuit to limit Vb more

Ib computations... current gain is about 15 at 30 A, so you'll need 0.5 A base current on each of the three transistors, or 1.5A total. At that Ic, you can expect a beta of around 100 in the first-stage transistor, so you'll need 15 mA of base current. The 100 ohm base resistors aren't really necessary, but will drop 1.5V at that current level, which will cascade through the above computations to limit the current capacity more.

That limits Ve to 4.5V below the supply voltage, or 7.5V, so now you'd be dissipating 135 W in each transistor at 30A, which in turn means that you have to heat-sink those transistors to at most 42 C case temperature. That's more work, but doable. 30A at 7.5V means that your emitter resistors are now 1/4 ohm, 225 Watt.

You can also work though the computations limiting each transistor to 10A as shown. That means that you'd have a 720 W limit on the 12V rail, which isn't enough for the modern monsters, but would handle everything reasonably sane. (And you could add a fourth transistor, or just leave room to install one.)

Oh... don't forget that it is important that each transistor have its own emitter resistor, and in a TO-3 case, the emitter is the case, so you will have to insulate (well, at least 100 ohms, say) the heat sinks from each other.

[mpilchfamily]

Wow allot of that is going straight over my head. The idea was to try and limit each Transistor to only take a 10A load and not take the full 30A the Transistor is capable of. Those are 10A resistors on the emitters.

I've been considering a new design that uses mainly wire wound resistors to produce the load and then a simple setup of the trasistors to give myself a way to finetune the load. No matter how close in tolerance the resistors are the actual load produced will not be the same as what i calculate so i need a way to fine tune the load.

Here is the schematic i've come up with for that. It is missing the Shunt resistors that will be used for reading the amperage on the rail using a digital panel meter.

http://img147.imageshack.us/img147/9844/diagram3revision1kk3.jpg



As you mentioned i may need to have the pots receive voltage from the 12V rail but i worry about crossing rails like that and how it would effect readings.

[cypherpunks]

Go read The Art of Electronics, section 2.06: Transistor Current Source. It explains everything very well.

It's all a matter of what costs more. You can use transistors to run the resistors at less than their nominal power, saving you resistors and switches. But using resistors and switches avoids the need for a transistor circuitry and current-measuring shunts.

The problem is, there's no such thing made as a "10 A resistor", so it's not clear what you do want. To actually buy a resistor, you need a resistance and a wattage rating. I was talking about how to select that resistor. Basically, to get Imax = 10 A, you need to figure out what the maximum voltage (Vmax) across the resistor is, and use Ohm's law Vmax = Imax * R to pick the resistor R = Vmax/Imax.

Here's how you compute it:

The basic thing a transistor does is it looks at the voltage between the base and the emitter, and if that's higher than the "normal" Vbe (about 0.65 V for small-signal transistors, but significantly higher when handling mucho amps), it lets some current flow from the base, and a large multiple of that much current flow from the collector, to the emitter to try to pull the emitter voltage up.

Pulling the emitter voltage up brings it closer to the base voltage, and things restabilize with a fixed Vbe.

Likewise, if the base voltage drops, so Vbe would be in danger of dropping, and the transistor shuts off to let the emitter voltage fall, and the Vbe voltage is stabilized.

Now, Vbe actually does depend on the current, but not very much. So you can assume a reasonable current, figure out Vbe from that, figure everything else out for the fixed Vbe, and then if the current is within a factor of 2 of what you first guessed, the Vbe will be so close to what you first figured that it won't affect the computation in the slightest.

So if you feed the base of a transistor with a fixed voltage, it will act like a variable resistor between collector and emitter, continuously adjusting its value to produce an emitter voltage Ve that is Vbe below the supplied base voltage Vb.

If you then also put a fixed resistor R between that Ve = Vb - Vbe and ground, the current through that resistor will obey Ohm's law: I = Ve/R. So the current through the transistor to the emitter resistor to ground will

Vbe is not perfectly fixed - it depends slightly on the temperature, and the voltage Vce, and on the current flowing through the transistor, but it stays sufficiently still to base a good circuit on.


So let's take your example, and figure out what values are needed to get 10 A ththrough that resistor on the 12V rail.

At Ic = 10 A, the data sheet says that we can expect Vbe(on) to be around 2.2V. Also at that current, we expect the base current Ib to be around 1/40 of Ic, so it'll be 0.25 A. This is the Ic of the driving transistor in the Darlington configuration, and at this level, we can expect Vbe to be aound 1V, and its base current Ib to be around 1/100 of its Ic. Which is 250 mA, as mentioned, so the Ib of the driving transistor will be around 2.5 mA. Which, when the pot is dialed right to 12V, will drop 25 mV across the 100 ohm resistor.

So the full voltage drop from the +12V rail to the driven transistor's emitter is 25 mV + 1 V + 2.2 V = 3.225 V. (Now do you see why I think the circuit will have problems with the 3.3V rail?)

Which means that at 10 A, the emitter voltage will be at most Vmax = 12 - 3.225 = 8.775 V. We don't really have that much accuracy, so just call it 8.8V. So we need a 0.88 ohm resistor. That resistor will dissipate 8.8 V * 10 A = 88 W of power.

(Note also that you don't need a 2N3771 transistor, which costs over $2 each, just to drive the base of the main 2N3771. A lot of smaller and cheaper transistors can drive Ic = 250 mA.)

If you have a good heat sink on the transistor, but a lower-wattage *and* lower-valued resistor is more available, you use it, and limit the maximum range of the base drive potentiometer to ensure that its power rating never gets exceeded. (Actually, its voltage Vmax. V^2 = P * R, so if the resistance is R and the maximum power is P, the maximum voltage is sqrt(P*R). Then add the 3.225 V and add a dinky little resistor to the 1K trim pot to ensure that you can't set the base voltage above that.)

Anyway, that's the ballpark needed on the 12V rail. On the 5V rail the same math applies, but the voltage at the base is at most 5 V (unless you take my suggestion to use the +12V rail for all the base drive potentiometers), so the voltage at the emitter is at most 5V - 3.2V = 1.8V. To pill

Let's see what you have there right now.
On the 12V rail, 1.5 ohms will pull 8A and dissipate 96 W. Times two is 16A. 2 ohms will pull 6A and disspiate 72 W. Times three is 18 A. Then add 4A, 2A and 1A from the three smaller resistors, and you have a total of 41 A, 492 W.

A 0-10 A variable component on top of that when you already have 1A granuarity seems a bit silly. You could add a few more switched 100 W resistors, and then add a 1 ohm, 75 W resistor to the abse of a transistor to give you 0-8.8 A adjustable to fine-tune it between, say, 0 and 4*8 + 8.8 = 40.8 A.


On the 5V rail, a 0.5 ohm resistor will pull 10A, or 50 W total. Then 5A for the 1 ohm, and 2A for the 2.5 ohm. Um... are you sure you didn't mean for that to be 2 ohm (2.5A) instead?

Again, with 3A granularity (2.5A if the 2 ohm resistor gets fixed), the point behind a 0-10A adjustable component is unclear. Why not just do 10A, and build a 0-10A transistor circuit? Connecting the base drive pot and transistor to +12V, you can get the emitter of the transistor within Vce(sat), or about 0.5V, of the 5V rail. that will leave 4.5V to drop across a resistor. Get another 0.5 ohm 50W resistor, and parallel it with a 2 ohm, 12.5W resistor. That will make a 0.4 ohm combination, which will let the transistor circuit go from 0-11 amps.

The same basic idea applies to the 3.3V section, but the resistors get smaller.

Note that all of the transistor circuits have the downside that after adjusting them, you need to measure how much current they're pulling. You can do it once and make a calibrated dial, but that's work, too.

Switched resistors do have the advantage of instant "digital" readout. What kind of tolerance do you expect? When the resistors get small enough, you can switch to an R-2R ladder for the least significant bits to reduce the number of different *kinds* of resistors you need.

[mpilchfamily]

I haven't mentioned it yet but Thank You for your help. I will check out that site and learn more about Transistors. I think I'll just focus on a pure Resistor based load and worry about adding fine tuning much later.

OK, so we are ignoring the Transistors in the above schematic. I'm pretty sure i have all the resistor values i want to give the loads I'm wanting to have. All the power ratings on the resistors should be sufficient enough to handle the loads that will be exerted on them. Each resistor will have a tolerance of +/-1% though as higher loads are exerted voltages will fluctuate a bit also causing the Load to change. Overall the loads should stay within a +/-5% tolerance. I will still need the shunt resistors for the panel monitors to actively monitor the loads in order to maintain accuracy in my efficiency calculations.

Now if i wanted to get a 10A load on my design i would flip the switch for both the 1.5 Ohm and the 6 Ohm Resistors. Being that they are in parallel i get 8A from the 1.5 Ohm and 2A from the 6 Ohm. In total if i where to close all the switches on one of the 12v banks i would have a max load of 41A. Threw different combinations of those resistors i can achieve any load between 1A and 41A.

On the 5V bank the resistors give me a 2A, 5A and 10A load with a max load of about 17A. The 3.3v bank is also limited only offering me 5.89A and 10A loads with a max just under 16A. So they give me 3 basic load levels for those 2 rails. I may go back and add in more resistors to give me the kind of flexibility i have with the 12v banks. Of course the -12V gets a .6A load and the 5VSB receives a 2A load. All well withing tolerance to handle the amount of power that will be exerted on the resistors.

So all should be good to go with the resistors alone. Only thing i worry about is the margin of error here. When i set a 30A load on the 12v rail will it be > or < 30A. If its < 30A then some sort of arraignment should be made to be able to add the extra amperage that is missing. Better yet if I'm going to be basing my test loads on a percentage of the PSU's total output then I'll need to dial in the loads to at least the tenths of an A. There is allot more for me to work out before i add any transistors for fine tuning the load.

Hope that all makes since. In all just the resistors and without useing shunts or panel meters the parts will cost just under $500. When i am finally able to build this thing it will be a secondary test bed as i will be using a Sunmoon 268ATE+ once my site is able to pull the money together for it. The load tester is the last peice of the puzzle for starting PSU reviews. I just want to build this one for the exsperience, pluss i want to see how well it does against the Sunmoon. I know the Sunmoon is the best but if this unit does Ok i'd like to have another person doing there own review of the PSU and then we can compair numbers.

[cypherpunks]

It managed to eat a longer reply.

I just wanted to mention, it's easy to tweak the value of a power resistor downwards with some conventional higher-value resistors. Given a 1.5 ohm, 100 W resistor, you can add a 1.5K, 0.1 W resistor (dirt cheap) alongside it and thereby drop its value by 0.1%. The only annoying this is this only increases loads; adding resistance requires an ultra-low-resistance but high-current resistor, which is harder to come by.

Don't forget to measure and account for wiring resistance. Use star wiring; daisy-chains are a pain to compute. While you can do minor tweaks with copper, its temperature coefficient of resistance screws you up.