Project: High power constant current programmable load

[Allen]

For my senior capstone design project for electrical engineering college, I will be designing a building a high power, constant current, programmable load.  Since so many people are posting or requesting dummy load DIY projects, I decided to start a thread documenting my project.  This project will go on for several months and is planned to be finished by mid December.  I also hope to pick up a few pointers from more experienced designers to help keep me from falling into "traps for young players".

Here are the ultimate specifications that I hope to achieve with this project:
Max power: 4.2 kW
Max current: 1,000 amps (must be able to draw full current at 2 volts)
Max voltage: Undetermined. Must be able to handle at least 4.2 volts, but will most likely be able to handle much more.
Max error: 1%

Required features:
Microcontroller controlled
LCD display (multiple if needed)
Cut-off voltage (load cuts off after source voltage drops below a certain threshold)
Powered from 120V outlet

Optional/desired features:
Display drawn charge (Amp-hours)
Display consumed power (Watts)
External thermocouple that can be used to monitor temperature of battery being tested and cut-off load if max temp is exceeded.
Ability to log voltage, current, drawn charge, and consumed power as well as some way to make this data available to a computer for analysis. Maybe log to a USB drive.

I will update this thread as my project progresses.

[EEVblog]

Required features:
Microcontroller controlled

But actual analog loop control?

Dave.

[Allen]

Yes, analog loop control.  I don't think the uC would be good for directly controlling something so non-linear.

[Allen]

For the main load section, I plan to use power MOSFETs with current sense resistors.

For the current sense resistors, I think the following would be a good choice: 100W, 10 mOhm, 1%, four terminal resistor (TGHGCR0100FE)
http://www.ohmite.com/cat/res_tghg.pdf
I plan to use 12 of these resistors (one for each MOSFET) which will give up to 83.3 amps and 69.4 watts power dissipation each at full load.

For the power MOSFETs, I think the following would be a good choice. ST Microelectronics, N-channel 100 V, 0.0045 ?, 220 A, ISOTOP
STripFET Power MOSFET (STE250NS10)
http://www.st.com/internet/com/TECHNICAL_RESOURCES/TECHNICAL_LITERATURE/DATASHEET/CD00002654.pdf
I plan to use 12 of these in parallel giving 83.3 amps at full load.  I like this MOSFET because it has a low junction to case thermal resistance of  0.25 K/W as well as a large base area of a little over 1 square inch.

I determined 12 of each would be needed for heat dissipation reasons.  I will post my thermal calculations soon.

[sonicj]

interesting!

why the individual current sense resistors vs a single high current shunt? the 12 resistors would run around $300usd while a single 1000A shunt is around $50usd.

4.2V really limits the usefulness of the product outside of lithium testing. 6V, 8V & 12V lead acid cells will be around for a long, long time...
my 2¢ fwiw
-sj

[amspire]

I would probably look at current sense resistors at around 0.0005 ohms to 0.001 ohms instead or 0.01 ohm.

The reason is that if the resistor can heat up 60degC at full power, you can only expect about 0.3% accuracy and that would be a pity if you look at the cost. Buying very low offset opamps costs very little by comparison, and you can end up with a much cheaper current sense resistor and a much higher accuracy.

The transistor looks good, but devices with that kind of packaging are expensive compared to several devices in a TO247 package.  It may be much easier to achieve your results with 42 100W modules, then your current 12 modules which will dissipate 280W on the mosfet and 70W on the current shunt.  280W means you have to keep those mosfets below 80 DegC and if the ambient can be up to 55 degC (as life gets hot near a 4200W heater), your heatsink has to be able to get rid of that 280 W per mosfet with an effective thermal resistance of less then 280/25 = 11 W/degC.
That is very ambitious.

At these sort of numbers, you can make your life much easier by just going for big numbers of much cheaper and more conservative lower power modules. A channel heatsink with a single fan and four mosfets can probably dissipate 200-300W. In a 19 inch rack width case, you make be able to fit 4 of these to get up to 1050W.  Four of these cases can add up to the 4200W.

That would be a total of 64  65W mosfet modules. If each used a 0.001 ohm shunt resistor, a SMD 2W 100ppm shunt resistor costs about 50c, a IRPF250 is under $2, the opamp about $1 so you are looking at under $10 including PCB per module. The SMD shunt resistor would hardly warm up, so you will get about 10 times the precision of your current design.

Richard.

[Allen]

interesting!

why the individual current sense resistors vs a single high current shunt? the 12 resistors would run around $300usd while a single 1000A shunt is around $50usd.

4.2V really limits the usefulness of the product outside of lithium testing. 6V, 8V & 12V lead acid cells will be around for a long, long time...
my 2¢ fwiw
-sj

The reason I want to use a separate current shunt on each MOSFET is so that each one can be regulated individually.  I don't know how how consistent the characteristics of each MOSFET is.  I don't want a situation where one MOSFET is pulling 50 A and the one next to it is pulling 200A.

As far as the voltages, I plan for it to be able to handle higher voltages, but not while pulling 1,000A. I will probably have something in the software that will limit current based on voltage.

[Allen]

I would probably look at current sense resistors at around 0.0005 ohms to 0.001 ohms instead or 0.01 ohm.

The reason is that if the resistor can heat up 60degC at full power, you can only expect about 0.3% accuracy and that would be a pity if you look at the cost. Buying very low offset opamps costs very little by comparison, and you can end up with a much cheaper current sense resistor and a much higher accuracy.

The transistor looks good, but devices with that kind of packaging are expensive compared to several devices in a TO247 package.  It may be much easier to achieve your results with 42 100W modules, then your current 12 modules which will dissipate 280W on the mosfet and 70W on the current shunt.  280W means you have to keep those mosfets below 80 DegC and if the ambient can be up to 55 degC (as life gets hot near a 4200W heater), your heatsink has to be able to get rid of that 280 W per mosfet with an effective thermal resistance of less then 280/25 = 11 W/degC.
That is very ambitious.

At these sort of numbers, you can make your life much easier by just going for big numbers of much cheaper and more conservative lower power modules. A channel heatsink with a single fan and four mosfets can probably dissipate 200-300W. In a 19 inch rack width case, you make be able to fit 4 of these to get up to 1050W.  Four of these cases can add up to the 4200W.

That would be a total of 64  65W mosfet modules. If each used a 0.001 ohm shunt resistor, a SMD 2W 100ppm shunt resistor costs about 50c, a IRPF250 is under $2, the opamp about $1 so you are looking at under $10 including PCB per module. The SMD shunt resistor would hardly warm up, so you will get about 10 times the precision of your current design.

Richard.

I considered a larger quantity of TO247 type transistors, but then I would have to mount them to a PCB.  Would PCB conductors be able to handle that much current?

Or would I need to run wires from each module and consolidate them at the primary terminals?

[Scarionn]

Just remember of the TC (temperature coefficient) of all your parts. Because 1% at 1000A can be very hard to get.

[nctnico]

I'd go for regular transistors. These work like current sinks already so controlling them is much easier.

[Scarionn]

There is other things to think about instead of the transistors , like the 1000A connectors,PCB, the cables, and how compensate their resistance.Because you will need a cable with very low resistance less then 0,00005 to get 1% of precision.

[amspire]

I considered a larger quantity of TO247 type transistors, but then I would have to mount them to a PCB.  Would PCB conductors be able to handle that much current?

Or would I need to run wires from each module and consolidate them at the primary terminals?

No problem. If you had 64 mosfets, then each one is handling 18A. a 1cm wide track on a 2oz copper PCB can handle that with about a 10 degC temp rise.

There is other things to think about instead of the transistors , like the 1000A connectors,PCB, the cables, and how compensate their resistance.Because you will need a cable with very low resistance less then 0,00005 to get 1% of precision.

No - the cable has zero effect on the precision. The cabling resistance just has to be low enough to allow the sink to meet the 2V 1000A spec. Aiming at 0.0005 ohms would be a good target. If that was achieved with 4 rack width cases in parallel, each one would have to be less then 2 milliohms wiring resistance.

Increasing the PCb track width to about 32mm on a 2oz copper PCB would reduce the temp rise of the tracks to about 1 deg, and would probably make the total PCB resistance about 1 milliohms.

The IRFP250 I suggested may prove slightly marginal for on resistance - a 60A mosfet would make it easier to reach the targets.

Richard.

[Allen]

I considered a larger quantity of TO247 type transistors, but then I would have to mount them to a PCB.  Would PCB conductors be able to handle that much current?

Or would I need to run wires from each module and consolidate them at the primary terminals?

No problem. If you had 64 mosfets, then each one is handling 18A. a 1cm wide track on a 2oz copper PCB can handle that with about a 10 degC temp rise.

There is other things to think about instead of the transistors , like the 1000A connectors,PCB, the cables, and how compensate their resistance.Because you will need a cable with very low resistance less then 0,00005 to get 1% of precision.

No - the cable has zero effect on the precision. The cabling resistance just has to be low enough to allow the sink to meet the 2V 1000A spec. Aiming at 0.0005 ohms would be a good target. If that was achieved with 4 rack width cases in parallel, each one would have to be less then 2 milliohms wiring resistance.

Increasing the PCb track width to about 32mm on a 2oz copper PCB would reduce the temp rise of the tracks to about 1 deg, and would probably make the total PCB resistance about 1 milliohms.

The IRFP250 I suggested may prove slightly marginal for on resistance - a 60A mosfet would make it easier to reach the targets.

Richard.

What about where all the modules are connected together in parallel?  Would you do that with off board wiring, or could that somehow be done on the PCB?

[EEVblog]

What about where all the modules are connected together in parallel?  Would you do that with off board wiring, or could that somehow be done on the PCB?

On first thought I'd do it all with off-board wiring.
Build your XX number of individual constant current loads, each entirely separate, each with it's own heatsink and maybe fan and PCB etc.
and then simply wire them across your input terminals, maybe via some form of big arse bus bar arrangement.
So your bus bar is the only thing that has to carry all the current, and your input connectors screw directly into the bus bar.
Your individual modules then draw 1/XX amount of that current, which is much easier to handle with wiring and PCB etc.
Your digital control circuitry then knows what total current you need and simply divides that by the number of channels and then sends that command to each module.
Probably not the cheapest way, but the most reliable and foolproof in design I think. It would tolerate failure of several modules. And for yur report you could go into math about likelihood of failure and redundancy maybe?

Dave.

[Scarionn]

No - the cable has zero effect on the precision. The cabling resistance just has to be low enough to allow the sink to meet the 2V 1000A spec. Aiming at 0.0005 ohms would be a good target. If that was achieved with 4 rack width cases in parallel, each one would have to be less then 2 milliohms wiring resistance.

Richard.

I'm referring about the cables/tracks going for the resistors.The total resistance need to be at least 0,0005 ohms

[IanB]

I don't really get how people come up with these specs like 1000 A, 4.2 kW, 2 V?

Is it serious, or do people get their brain removed and replaced with a wet sponge when they go to college?

For instance at 1000 A in the system every milliohm of resistance is going to drop 1 V. Greater than 2 milliohms and your entire 2 V voltage drop is gone. (And 1000 A through 1 milliohm will dissipate 1 kW. Bye bye PCB tracks. They will vaporize.)

To have any practical device working will presumably need thick copper bus bars as fat as fingers, welded connections, giant cooling fans and ridiculously expensive parts.

Just to humour me, what is the industrial application of this device?

[amspire]

To have any practical device working will presumably need thick copper bus bars as fat as fingers, welded connections, giant cooling fans and ridiculously expensive parts.

Just to humour me, what is the industrial application of this device?

It can probably be built for under $2000 and I do not call that ridiculously expensive.

The need? It is not that hard to imagine needs.  Testing large stationary lead-acid cells. Testing 4KW DC welding supplies. Testing supplies that are going to power superconductor magnets. Who knows?
If Allen wants to build a 1000A load, that is fine with me.

Once you decide to design a monster like this, you may as well push the specs as far as you can. If Allen needs 1000A at 4.2V, it makes perfect sense to spec the design at 1000A at 2V, and if possible, 1000A at 1V would be even better.

Richard.

[Scarionn]

For instance at 1000 A in the system every milliohm of resistance is going to drop 1 V. Greater than 2 milliohms and your entire 2 V voltage drop is gone. (And 1000 A through 1 milliohm will dissipate 1 kW. Bye bye PCB tracks. They will vaporize.)

You are right. At last post I said he will need a 0.0005 ohms wire to maintain it on spec, but he will need a 0.00005 at least to maintain it on spec. 50W of dissipation.

[amspire]

For instance at 1000 A in the system every milliohm of resistance is going to drop 1 V. Greater than 2 milliohms and your entire 2 V voltage drop is gone. (And 1000 A through 1 milliohm will dissipate 1 kW. Bye bye PCB tracks. They will vaporize.)

You are right. At last post I said he will need a 0.0005 ohms wire to maintain it on spec, but he will need a 0.00005 at least to maintain it on spec. 50W of dissipation.

I see what you mean.  At 1000A current, a wattmeter measuring the internal terminal voltage is accurate for the load's dissipation but inaccurate for measuring the DUT's dissipation. Good point.

To check the real total load dissipation, an external precision current shunt or DC clamp plus a voltmeter across the leads of the DUT are needed.

Richard

[Scarionn]

DUT?

[amspire]

DUT = Device Under Test

[EEVblog]

For instance at 1000 A in the system every milliohm of resistance is going to drop 1 V. Greater than 2 milliohms and your entire 2 V voltage drop is gone. (And 1000 A through 1 milliohm will dissipate 1 kW. Bye bye PCB tracks. They will vaporize.)

True. But 1mohm resistance is pretty high when you are talking huge conductors and massive bolt interconnects etc.

To have any practical device working will presumably need thick copper bus bars as fat as fingers, welded connections, giant cooling fans and ridiculously expensive parts.

I don't know about "ridiculously" expensive, but certainly a bus bar approach with the terminals directly bolt (or weld) connected as I mentioned would be the key.
Once you have that bus bar and connection that can carry your 1000A without much drop, then it's almost trivially easy to connect on as many smaller loads as required in parallel, and each one can be measured accurately, so you will know the total load current.
If you wanted to take into account the drop on the bus bar and/or connectors then you'd need an additional diff amp to measure the drop, wired directly onto the bar at appropriate points. That's likely necessary with the original requirement of 1% accuracy.

Here is an online calculator:
http://www.bdbatteries.com/buss-bars.php
For example, a 12" long copper buss bar 1" wide and 0.1" thick is 0.00008126 (0.081mohms) and drops only 81mV at 1000A.

You'd probably have to worry more about your connecting leads and connections to do with the DUT more so than the design of the load itself.

Dave.

[Bored@Work]

You'd probably have to worry more about your connecting leads and connections to do with the DUT more so than the design of the load itself.

My biggest worry would be the heat. 4 kW is enough to operate a 1000 C furnace, or heat your house.

My second worry would be the electro magnetic induction. Turn 1000A on, and ... Or maybe even just the ripple of the regulation. I would run the numbers here first.

And I have to agree, this is not something I would recommend as a final year project. To many things to juggle in the air for a beginner. One of those things "if you have to ask how to do it you are not qualified to do it".

[Allen]

I have another question for you guys.  The datasheet for one of the MOSFETs I am considering using says that the maximum operating junction temperature is 150 degrees C.  Do you guys think that is a safe number, or do you think I should shoot for a lower limit?

[T4P]

I have another question for you guys.  The datasheet for one of the MOSFETs I am considering using says that the maximum operating junction temperature is 150 degrees C.  Do you guys think that is a safe number, or do you think I should shoot for a lower limit?

Of course not. that would shorten the life of everything.
maybe 80C would be better but it's still scalding