# 3000W Battery Discharger Project



## Tesseract (Sep 27, 2008)

frodus said:


> ...I've finally gotten back to my discharger project. A while back I aquired 6 Darlington pair modules (400A/600V) and matching heatsink from a buddy of mine. I figured I had most of what I needed for a high-power battery discharger....


There are a few potential problems here. Actually, this circuit is pretty much a train wreck but I was trying to sugar coat things in that first sentence... 

Ok, the first problem is that the transistors in Darlington modules are optimized for switching and have very poor SOA in linear mode (it's even worse for IGBTs, btw). Much worse than the dissipation rating of the module would otherwise suggest. A reasonable estimate is to cut the current rating by 10x for up to 10V of Vce (eg... a 400A module might be good for 40A in linear mode). Terrible, huh? 

The MOSFET you are using for base drive is also not too keen on running in linear mode; that fabled positive tempco for Rds[on] that MOSFETs have is more than cancelled out by the (usually) negative tempco of Vgs[th], particularly in those MOSFETs optimized for switching (noticing a trend here? ).

I would also use a proper instrumentation amplifier for the shunt and heavy compensation to reduce the likelihood of oscillation. This is a minor quibble, however, as Darlingtons are extremely slow by nature so despite having a poor linear mode SOA are fairly docile creatures.

I would put the fuse above the shunt if you don't use a proper diff/instrumentation amp for the shunt otherwise the voltage drop across the fuse will add to the shunt signal. The placement of the fuse in the emitter connection is otherwise good as it provides a little bit of negative feedback (along with the shunt).

This is not an exhaustive list, it's just what I came up with while having a beer after driving the second leg of the trip back to Tampa, FL from Cape Girardeau, MO.


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## frodus (Apr 12, 2008)

> There are a few potential problems here. Actually, this circuit is pretty much a train wreck but I was trying to sugar coat things in that first sentence...


 thanks, analog electronics isn't my strong point



> Ok, the first problem is that the transistors in Darlington modules are optimized for switching and have very poor SOA in linear mode (it's even worse for IGBTs, btw). Much worse than the dissipation rating of the module would otherwise suggest. A reasonable estimate is to cut the current rating by 10x for up to 10V of Vce (eg... a 400A module might be good for 40A in linear mode). Terrible, huh?


An engineer friend gave me the parts when he learned I wanted to do a discharger project. He drew a schematic for me using the darlingtons as well as that FET to drive them. Most of the other help I've gotten was with respect to the op-amp circuit.

Here's the datasheet for the Darlington, is it really that bad?
http://www.evfr.net/uploads/mg400g1ul1.pdf

If it's that bad, are there other options? I'd like to salvage whatever I can with this project..... 




> The MOSFET you are using for base drive is also not too keen on running in linear mode; that fabled positive tempco for Rds[on] that MOSFETs have is more than cancelled out by the (usually) negative tempco of Vgs[th], particularly in those MOSFETs optimized for switching (noticing a trend here? ).


I could use a different MOSFET, again, this was chosen by my engineer friend. What would you recommend?



> I would also use a proper instrumentation amplifier for the shunt and heavy compensation to reduce the likelihood of oscillation. This is a minor quibble, however, as Darlingtons are extremely slow by nature so despite having a poor linear mode SOA are fairly docile creatures.


Which amplifier would you suggest? I am not set on this part yet. What kind of compensation are you referring to? We've filtered input and output as well as the supply.



> I would put the fuse above the shunt if you don't use a proper diff/instrumentation amp for the shunt otherwise the voltage drop across the fuse will add to the shunt signal. The placement of the fuse in the emitter connection is otherwise good as it provides a little bit of negative feedback (along with the shunt).


Unfortunately that would take a bit of rework.... I might just eliminate it altogether since I've got protection elsewhere. Might be able to fit 6 in there between shunt and Darlington.




> This is not an exhaustive list, it's just what I came up with while having a beer after driving the second leg of the trip back to Tampa, FL from Cape Girardeau, MO.


Thanks.... Just trying to finish this. Need to do a high discharge on a battery pack fairly soon.


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## PStechPaul (May 1, 2012)

I concur with Tesseract's critique. And there are other problems as well. So I will just point out the ones I see in addition to those.

Connecting Q4 through a 10k resistor R7 to ground will not turn it off, since it's driven by U1 through a 470 ohm resistor R2. You could use a 10 ohm resistor for R7, or simply short the gate to GND.
The drain of Q4 is connected to the battery you intend to discharge, so the gate voltage of the darlington Q1 will never be higher than the battery voltage, and will vary during discharge. It should be connected to the 12V supply. Or, maybe this is being used to make sure the battery voltage never goes below the turn-on voltage of Q1, so you can't fully discharge the battery. If so, it's a bad way to implement such protection.
The best way to implement a battery discharger, IMHO, is to use PWM drive to the darlington transistor and use a separate resistive element to dissipate the power. I don't know the voltage of the battery pack or the current you intend to use, but for a 3000W discharger and a 400A darlington you will need to keep the current under 400A so the battery would need to be over 7.5 volts for 3000W.

Your resistor could be chosen so that it draws, say, 1000A at the lowest battery voltage expected (7.5V), or 7.5 milliohms, and use PWM of 40% for 400A. Then you can vary the PWM for higher battery voltage. The resistor could be made of strips of steel and possibly immersed in a bucket of water to keep the temperature in reasonable bounds. Plus you'd have a bucket of hot water which you could possibly use, rather than heating the air using fans (unless you want a room heater).

Or perhaps you meant 3000 amps which you might be able to do with six of the darlington modules in parallel (although that would be 500 amps each or 25% above their rating). And the 250A shunt would be grossly overloaded, so that is obviously not what you want.

The disadvantage of the PWM approach is that the current would not be constant. But you could add an inductor to smooth it out, although a 250A inductor is not a common item. However, you could make one with some heavy wire wrapped around a steel core or an old transformer with a gap to make it an inductor. But it's not exactly that simple. I'd have to redesign the circuit, and I'm not sure just how to do it at this point.

Actually your circuit is a constant current discharger, and the power will vary as the battery voltage drops. The adjustment is from 0 to 50 mV and thus output can be set from 0 to 250A, which is the shunt rating. You really need a resistor from the shunt to the inverting (-) input of the op-amp, because the input impedances should be balanced and it will allow the input protection of the op-amp to work with external current limiting. You really need to get rid of the fuse or move it elsewhere, because not only will it add resistance to the shunt circuit and cause errors, but if it opens, the full battery pack voltage may be applied to the op-amp which may destroy it.

Sorry to be so long-winded. I am considering the design of a similar device, using a PIC, and it would have the advantage of being fully programmable as well as performing a datalog to plot the entire discharge history of current and voltage as a function of time, and it could be set to stop at a certain time or depth of discharge. It could also discharge a single cell or a battery pack at a constant power if so desired. 

And, finally, if a space heater is not desired, perhaps the discharged energy could be pumped back into the local grid using something like a grid tie inverter as used for solar panels. You would just need to use a DC-DC converter as the battery discharge load. I guess this depends on the total Wh of the battery pack, too. A 10 kWh pack holds only about a dollar's worth of electricity, so it does not make economic sense to use a $100 inverter unless you are doing hundreds of discharge cycles.


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## frodus (Apr 12, 2008)

> The drain of Q4 is connected to the battery you intend to discharge, so the gate voltage of the darlington Q1 will never be higher than the battery voltage, and will vary during discharge. It should be connected to the 12V supply. Or, maybe this is being used to make sure the battery voltage never goes below the turn-on voltage of Q1, so you can't fully discharge the battery. If so, it's a bad way to implement such protection.The best way to implement a battery discharger, IMHO, is to use PWM drive to the darlington transistor and use a separate resistive element to dissipate the power. I don't know the voltage of the battery pack or the current you intend to use, but for a 3000W discharger and a 400A darlington you will need to keep the current under 400A so the battery would need to be over 7.5 volts for 3000W.


True, the battery is only ever going to 2V or higher. I considered using power resistors and may have to revisit that idea if these DP's don't work. This is a single cell tester, but could be used for packs up to the voltage of the darlington, and 3000W is what I can dissipate with the heatsinks and modules.

Also, there are 6 of these circuits. The schematic shows one circuit. 3000W = 500W each and under 170A.



> Or perhaps you meant 3000 amps which you might be able to do with six of the darlington modules in parallel (although that would be 500 amps each or 25% above their rating). And the 250A shunt would be grossly overloaded, so that is obviously not what you want.


 No, 3000W. 1000A max is what I'm designing for with all 6 modules in parallel. 500W each, and under 170A each, under the shunt rating.



> The disadvantage of the PWM approach is that the current would not be constant. But you could add an inductor to smooth it out, although a 250A inductor is not a common item. However, you could make one with some heavy wire wrapped around a steel core or an old transformer with a gap to make it an inductor. But it's not exactly that simple. I'd have to redesign the circuit, and I'm not sure just how to do it at this point.


 Yeah, that's what I was trying to avoid.



> Actually your circuit is a constant current discharger, and the power will vary as the battery voltage drops. The adjustment is from 0 to 50 mV and thus output can be set from 0 to 250A, which is the shunt rating. You really need a resistor from the shunt to the inverting (-) input of the op-amp, because the input impedances should be balanced and it will allow the input protection of the op-amp to work with external current limiting. You really need to get rid of the fuse or move it elsewhere, because not only will it add resistance to the shunt circuit and cause errors, but if it opens, the full battery pack voltage may be applied to the op-amp which may destroy it.


 Yup, it's a CC design with a 3000W max design (although flawed). I can move the fuse, or remove, it's not really required.


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## subcooledheatpump (Mar 5, 2012)

What about PWM with an (old) arc welder transformer secondary winding as the inductor? 

It's a fairly standard part if you can get your hands on an old arc welder.

Also, correct me if I'm wrong, but, isn't the size of needed inductor inverse to the switching frequency? At 20 KHz or more, how noticable is the current ripple?


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## Siwastaja (Aug 1, 2012)

subcooledheatpump said:


> Also, correct me if I'm wrong, but, isn't the size of needed inductor inverse to the switching frequency? At 20 KHz or more, how noticable is the current ripple?


True, 

and furthermore, the batteries tend to average out the result. The actual discharge resistor limits the current anyway (and may have some inductance, too). So, while I think adding an inductor to smooth out the current is a very good idea, it is not critical. I'd just wind a few dozen turns of thick wire over a large ferrite core and select such a discharge resistor that it can be used almost directly in the highest current case, and then have the PWM somewhere around 30-100 kHz, using high-current MOSFETs (don't forget proper drivers, gate protection, etc.).

I'd use any cheap microcontroller with an ADC to do the PWM drive. As said before, it can also do the datalogging, which probably is the main point in the project anyway, with no extra complexity.

Try it out and take a look at the current using a scope.


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## PStechPaul (May 1, 2012)

I found datasheets on the darlington modules here:
http://www.datasheetarchive.com/MG400G1UL1-datasheet.html

It seems that the lowest voltage drop for this module is 2.0V at 400A with a base current of 8 amps and Vbe of 2.7V. The voltage drop may be less at 250A, and the base current might be about 4 amps. So you won't be able to use a 12V 2A supply to drive them, and the battery will need to be more than 2.7V to use it for the base drive. I had thought these were IGBTs, so some of my previous comments are not valid for this design.

For a single cell tester the battery voltage will probably range from 4 volts to 2 volts. You might be able to make a voltage doubler DC-DC boost converter that has enough power and voltage to drive the base. You need about 15-20 watts. But that is not a trivial design. Maybe you can use a 6V or 12V battery for the power supply so you can get the current you need.

Really, it would be much better to use IGBTs or MOSFETs. You can get 150A 600V IGBTs such as 
http://pdf1.alldatasheet.com/datasheet-pdf/view/31020/TOSHIBA/MG150J1BS11.html
for about $10 each on eBay:
http://www.ebay.com/itm/TOSHIBA-FAN...123?pt=LH_DefaultDomain_0&hash=item43b159536b

They are much easier to drive, but the saturation voltage will still be a problem as it is specified as about 2.3V. So you really should look for MOSFETs. Because your voltage is very low, you can get 12V or 20V MOSFETs that are good for 100A or more, and just parallel a bunch of them. Or here are some 110V 300A MOSFET modules for $25 each:
http://www.ebay.com/itm/Semikron-po...825?pt=LH_DefaultDomain_0&hash=item337127e819


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## frodus (Apr 12, 2008)

PStechPaul said:


> I found datasheets on the darlington modules here:
> http://www.datasheetarchive.com/MG400G1UL1-datasheet.html


It's nice that you found that, but that's the same datasheet I posted in post #3



> It seems that the lowest voltage drop for this module is 2.0V at 400A with a base current of 8 amps and Vbe of 2.7V. The voltage drop may be less at 250A, and the base current might be about 4 amps. So you won't be able to use a 12V 2A supply to drive them, and the battery will need to be more than 2.7V to use it for the base drive. I had thought these were IGBTs, so some of my previous comments are not valid for this design.


I can find an alternate 12V power supply, that is not an issue.



> For a single cell tester the battery voltage will probably range from 4 volts to 2 volts. You might be able to make a voltage doubler DC-DC boost converter that has enough power and voltage to drive the base. You need about 15-20 watts. But that is not a trivial design. Maybe you can use a 6V or 12V battery for the power supply so you can get the current you need.


I can find an alternate 12V power supply, that is not an issue.




> Really, it would be much better to use IGBTs or MOSFETs. You can get 150A 600V IGBTs such as
> http://pdf1.alldatasheet.com/datasheet-pdf/view/31020/TOSHIBA/MG150J1BS11.html
> for about $10 each on eBay:
> http://www.ebay.com/itm/TOSHIBA-FAN...123?pt=LH_DefaultDomain_0&hash=item43b159536b


I can afford some new modules if they're not too expensive, but I'd like them to be nearly the same footprint if possible to save from more fabrication/drilling/tapping.



> They are much easier to drive, but the saturation voltage will still be a problem as it is specified as about 2.3V. So you really should look for MOSFETs. Because your voltage is very low, you can get 12V or 20V MOSFETs that are good for 100A or more, and just parallel a bunch of them. Or here are some 110V 300A MOSFET modules for $25 each:
> http://www.ebay.com/itm/Semikron-po...825?pt=LH_DefaultDomain_0&hash=item337127e819


That might work. I don't need to really to larger packs, looking to do single cells or small 12V packs at most. I'd really like to build it for that 3000w though.....


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## Tesseract (Sep 27, 2008)

frodus said:


> ...
> An engineer friend gave me the parts when he learned I wanted to do a discharger project. He drew a schematic for me using the darlingtons as well as that FET to drive them. Most of the other help I've gotten was with respect to the op-amp circuit.
> 
> Here's the datasheet for the Darlington, is it really that bad?
> http://www.evfr.net/uploads/mg400g1ul1.pdf


Nah, you can still use the Darlington modules but they do need a lot of base drive (DC current gain is ~100) and you need to respect their often pathetic DC SOA curve. Unfortunately, there isn't an SOA curve in that datasheet, hence why I recommended you limit collector current to 1/10th of Ic[nom]. In any event, rarely will the DC SOA curve for a bipolar switching transistor allow more than half of rated Pd (ie - less than 700W total dissipation for this 1400W-rated transistor). Furthermore, rarely can you pass more than half of Ic[nom] with the transistor out of saturation (ie - in linear mode) because of hot-spotting which leads to thermal runaway and second breakdown.

These are the main reasons why bipolar transistors are used less and less these days, btw.



frodus said:


> I could use a different MOSFET, again, this was chosen by my engineer friend. What would you recommend?


Ditch the single MOSFET for a complementary bipolar buffer supplied by at least a 12V/10A power source (which can also supply power to the op-amps). A typical transistor pair used for this application are the ubiquitous D44 (NPN) and D45 (PNP). Tie both of their bases together and drive them with a resistor in series with the op-amp's output to allow up to, say, 40mA of drive current, then tie their emitters together and drive the Darlington's base with a resistor to allow up to, say, 2A of drive current. 



frodus said:


> Which amplifier would you suggest? I am not set on this part yet. What kind of compensation are you referring to? We've filtered input and output as well as the supply.


Any dual op-amp whose common mode input range includes V - (even if V- is ground) and which can deliver at least +/-40mA will work. I've had good results with STMicro op-amps optimized for audio applications such the TS922 (which can deliver 80mA of output current!). By compensation I mean loop compensation; ie - a lead-lag network. Roll the loop off at a fairly low frequency so the whole shebang doesn't oscillate. Since you are just trying to drain batteries at a constant current a loop bandwidth of 100-200Hz will be more than enough.


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## frodus (Apr 12, 2008)

Tesseract said:


> Nah, you can still use the Darlington modules but they do need a lot of base drive (DC current gain is ~100) and you need to respect their often pathetic DC SOA curve. Unfortunately, there isn't an SOA curve in that datasheet, hence why I recommended you limit collector current to 1/10th of Ic[nom]. In any event, rarely will the DC SOA curve for a bipolar switching transistor allow more than half of rated Pd (ie - less than 700W total dissipation for this 1400W-rated transistor). Furthermore, rarely can you pass more than half of Ic[nom] with the transistor out of saturation (ie - in linear mode) because of hot-spotting which leads to thermal runaway and second breakdown.
> 
> These are the main reasons why bipolar transistors are used less and less these days, btw.


Thanks!

Datasheet for the IRFZ44 for reference
http://www.vishay.com/docs/91291/91291.pdf

1/2 Ic(nom) of 400A is just fine, that's ~200A per module. I'm aiming for 3000W total for the entire assembly, so 500W each.... At 3V (I realize voltage will fall fairly fast at 1000A on a single battery).... but 500W, 3V would be under 170A per module.... so that's just fine. The guy that gave them to me actually used single units for ~200A constant curent loads, which is why he gave them to me when I said I wanted to build one.... I'm designing so each module dissipates ~500W max, and most times much much less. There's 6 of these 400A Dpairs. So if I ran under 700W at say 500W, 6 modules would be 3000W total. My schematic shows only one module driver circuit, sorry if it was unclear.

And as the voltage falls, I don't care what power does, I just need to keep current constant.... so it may start off as a 3kW load, but at 2.5V, I may only be at 2.5kW, and 2V at 2kw, which is fine as long as I keep it cool.




> Ditch the single MOSFET for a complementary bipolar buffer supplied by at least a 12V/10A power source (which can also supply power to the op-amps). A typical transistor pair used for this application are the ubiquitous D44 (NPN) and D45 (PNP). Tie both of their bases together and drive them with a resistor in series with the op-amp's output to allow up to, say, 40mA of drive current, then tie their emitters together and drive the Darlington's base with a resistor to allow up to, say, 2A of drive current.


I'll consider those, thanks.



> Any dual op-amp whose common mode input range includes V - (even if V- is ground) and which can deliver at least +/-40mA will work. I've had good results with STMicro op-amps optimized for audio applications such the TS922 (which can deliver 80mA of output current!). By compensation I mean loop compensation; ie - a lead-lag network. Roll the loop off at a fairly low frequency so the whole shebang doesn't oscillate. Since you are just trying to drain batteries at a constant current a loop bandwidth of 100-200Hz will be more than enough.


Thanks. I'll look into those. As long as they can be used with a single sided supply.


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## frodus (Apr 12, 2008)

I'm changing the direction on my project. I just don't have any spare time anymore with my full time job + running my own business + finishing my EV.

I've got all datasheets, circuits and the heatsink calculations done for a discharger if you want to build one, or use it for something else. Each module should be good for 500W. 

6 9" 5x5 bonded fin aluminum heatsink (drilled and tapped for the Darlingtons and Shunts)
6 MG400ULGL1 Darglington Transistors 400V 600A
6 250A/50mV shunts
some IRFZ44 MOSFETS + matching heatsinks
some TS922 Op Amps
various resistors and small caps
Mounting standoffs for the shunt


If anyone is interested, I can sell parts to make a a single module, I've got enough to build 6, just no time to finish. Add a fan, finish the driver circuit and it should be good. I've been doing 50A load tests without much issue at all. 

$100 each module, $500 for everything.
OBO


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## Sunking (Aug 10, 2009)

Perhaps I am missing something here but why does does the load need to be active vs passive?

I work in the utility and telecom sector and we use load banks which are just special resistors then look like air filters with a fan blowing through them. 

The older units you can get with a song that use simple toggle switches to obtain about any current/resistance you want.


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## frodus (Apr 12, 2008)

I need constant current discharge, like the West Mountain Radio CBA, but higher current. The current with a resistive load changes as voltage drops. I want a dead-nuts-on current for battery curves. 

Lets say you start with 3.2V/100A. The resistor would need to be 0.032ohms (lets assume it's sized appropriately for wattage). The problem is, down around 2.5V when the current becomes ~78A.

That's no-good for battery discharge curves. This isn't for my motorcycle, or testing my own stuff, I have a side business and battery testing is one of the things I do for clients.


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## Sunking (Aug 10, 2009)

frodus said:


> I need constant current discharge.


OK I understand, I do battery capacity testing for a living in the telecom and utility sector..



frodus said:


> That's no-good for battery discharge curves. This isn't for my motorcycle, or testing my own stuff, I have a side business and battery testing is one of the things I do for clients.


Then use industry accepted Professional Equipment like Albers Battery Capacity Test System.


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## frodus (Apr 12, 2008)

I've found what I need (lab grade electronic load), thanks though. I'm only testing single cells and have my own logging equipment.


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