# 0-5v controlled load I-V tester

Discussion in 'The Projects Forum' started by wayneh, Jul 25, 2011.

1. ### wayneh Thread Starter Expert

Sep 9, 2010
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I want to build a tool to place a computer-controlled load onto a DC generator such as a PV panel or rectified alternator. The only way I have to do this now is to use a bank of various resistors sharing a common ground and move the power lead from one to the next, manually reading and recording voltage drop across the various resistors. This works, but could be better!

Now I have a LabJack U3HV data acquisition device, which is very handy. Its analog outputs include a 0-4.5v output up to 45mA. It can sustain this current as a continuous short circuit of the output. I can set the output as I like from Excel (or many other languages).

I understand how I could use this output as a voltage reference to control a constant-current circuit such as the simple circuit attached. I'm pretty comfortable that I could build this and it would work fine within its limits. (The resistor values need refining first.)

But I'd like to think ahead to scaling it up to higher currents. Usually when I build something it works as intended (yay!) but then I realize a tweak here or there would have produced a more versatile design. So I'd like to reduce "builder's remorse" and anticipate future expansion before I build this time.

So I'm wondering how to parallel more of these circuits, or otherwise allow for expansion up to 30 or even 50A. I think I could just use more op-amps and literally build these circuits in parallel.

But I see problems I need help with:
1) Where can the base current come from? If I'm scaling a 30mA signal to 30A, that's 1000X current gain. Will the power transistors really require ~3A total at their bases to reach that current level? Could that current come from the source itself, instead of the control system?
2) How do I deal with DC ripple, in the case of loading a rectified alternator? I want this thing to act like a passive resistor, not an active filter. I don't want constant current throughout a ripple, just "constant" on average. Not a problem for PV testing.
3) What happens when I call for a base voltage of say 4v and the generator cannot supply enough current to meet that? Won't the base current max out as it tries? I guess that's where current-limiting to the base resistors comes in.
4) If I have 5 power resistors rated to 15W and 0.1 ohms, in series they provide 0.5Ω and 75W power dissipation. I'm not using fans or anything (yet), so I wouldn't expect to put more than 40W or so through them. Is that realistic though, or will a weak link develop in the chain?
5) I'm sure there will be more questions.

2. ### SgtWookie Expert

Jul 17, 2007
22,202
1,791
Why not use power MOSFETs instead of bjts?
Once a gate is charged or discharged, it requires virtually no current at all to maintain the state. You don't want changes of current to happen rapidly anyway, so the opamps' current source/sink ability to charge MOSFET gates doesn't matter very much.

Also, MOSFETs are pretty easy to parallel, as they have a positive thermal coefficient; as they heat up, their resistance increases which causes the flow of current to go through others in parallel with it.

You will of course need some way to carry all the heat away from the MOSFETs.

Put a low pass filter on the current sense.

Generator? Ahhh, did you mean alternators? You're testing alternators, correct?
[eta] OK, you said solar panels or alternators generating rectified DC. I'll refer to them in the following as generator/alternator.

If the generator/alternator can't supply enough current, it's output voltage would drop. Aren't you measuring the generator/alternator output voltage anyway to determine if it's holding the rated output voltage at the rated current? If not, you should be.

It's good you're being conservative, 47 Watts is the max you should have those resistors dissipate, and they will need plenty of air around them. 5*15W / 1.6 ~= 47W. 47 watts, .5 ohms is a 4.84V drop, 9.68A.

If 9.68A is the most you want to put through that part of the circuit, you would be OK. You would have a 4.84v drop across the sense resistors at that point.

Rather than use 0.1 Ohm resistors in series, consider using 2 Ohm, 20 Watt resistors in parallel.
5v/2 Ohms = 2.5 Amperes; 12.5 Watts, which is 20w/1.6 so it's a good match.
If you just keep your sense voltage at or below 5v, you won't exceed the resistor wattage+60%margin rating.
Two 2-Ohm 20W resistors in parallel = 5A @ 5v
4 " " in parallel = 10A @ 5v
8 " " in parallel = 20A @ 5v
12 " " in parallel = 30A @ 5v
etc.
It keeps the math easy.

Mouser stocks these:
http://www.mouser.com/ProductDetail...=sGAEpiMZZMtlubZbdhIBIP2z8nsdzoA2POj4rqUbb3U=

You say you'd like a 30A capability; so 30A/2.5A per resistor = 12 resistors; cost is \$40.68 + shipping for those resistors - plus whatever you are planning on connecting them up with.

Your MOSFETs are going to need pretty good-sized heat sinks on them.

If your gen output is 14v and 30A, then 12 resistors dissipating 150w total will have 5v dropped across them, leaving 9v across the MOSFETs; 30a*9v = 270 Watts. You'll need to divide that up among enough MOSFETs to keep them from melting. Let's say you didn't want any MOSFET to dissipate over 25 Watts; you would then need 11 MOSFETs with heat sinks.

If the alternator regulator had a fault that caused more than 14v output, power dissipation in the MOSFETs would rise very quickly. Also, some newer alternators are generating in the range of 15v to take advantage of the newer AGM batteries that can take being charged more quickly. I don't know if you want that capability or not.
The most capable GM alternator that I know of is rated for 130A output.

You might find this helpful: http://mcmanis.com/chuck/robotics/projects/esc2/FET-power.html
Lots of stuff about MOSFET current capacity and power dissipation - even though in your application you're deliberately wasting power, you still have to keep the MOSFETs from melting down.

Last edited: Jul 26, 2011
3. ### wayneh Thread Starter Expert

Sep 9, 2010
14,787
5,277
D'OH!! (Slapping forehead) Great solution to the base current problem. The answer is, because I happen to have some power bjts I was thinking of using. Not a very good reason, since I have power MOSFETs too.
Simple enough.
Different situation. I'm working with experimental alternator designs and configurations. There are no ratings to go by. I'm mapping the current-voltage curves.

The ones I have on hand are IRF540 in the TO-220 package, versus the 2N3055 and 2N3771 in the TO-3 cans. I haven't studied the relative heat dissipation issues, but I suppose the cans are a bit easier.
That is a great, straightforward and practical approach to the problem. Should be required reading for anyone wanting to post a MOSFET question.

4. ### ErnieM AAC Fanatic!

Apr 24, 2011
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Well... there is one problem when you go to test your alternator output, or any AC source with this load: It simply is not a resistive load: it is a constant current load.

These loads always try to pull the same current from your source. If your source voltage doubles, the current stays the same. In a resistive load that current would double.

There are ways to make a power resistive load. Basically, instead of using the fixed DC output of the LabJack you need a voltage that is a proportion of the source voltage. The simplest way to do this is to put a pot there.

But you can't control a pot from LabJack. You need something else to convert the DC and the source voltage to the drive voltage, something like a multiplier circuit.

(Nope, I ain't got one in my back pocket of I would share it.)

5. ### wayneh Thread Starter Expert

Sep 9, 2010
14,787
5,277
That's the problem I asked about, sort of. I'm concerned about the AC component - the ripple in the rectified alternator output. Is that what you're referring too? Because otherwise holding constant current is exactly what I want, so I can see voltage at a given current load.

Wouldn't the suggestion of a low-pass filter on the sense voltage fix this problem? The op-amp would only see the time-averaged sense voltage and would be calling for a time-averaged current. An increased voltage from the source would cause increased current into the shunt resistor(s), but this wouldn't be seen by the op-amp unless it was sustained. I'd use a time constant to give me a few cycles to average before the control can respond.

UPDATE Now that I think about it, the op-amp might see an average voltage after a filter but the main current transistor sees the rise at the shunt, and limits current as the alternator's voltage goes up. So yes, current at peak voltage would be less than through a purely resistive load. Hmmm...