Reverse polarity protection for supercapacitor discharger

[markus.p]

Hi everyone,

I'm working on a handheld device (see attached picture) designed to safely discharge supercapacitor banks (24V, 33F). Functionally, it's a DC electronic load, and I'm trying to implement reverse polarity protection without compromising the ability to fully discharge the capacitors to ~0V.

Specifications

• Application: Handheld tool for discharging supercapacitor banks
• Capacitor Bank: 24V, 33F
• Target Discharge Voltage: ~0V (complete discharge)
• Available Supply Rails: 9V and 3.3V

Functional Requirements

• Reverse polarity protection is essential
• Must allow discharge down to near 0V (rules out Schottky diodes due to forward voltage drop)
• Ideal diode controller preferred, but most options require VIN > 3V, which is too high for this use case
• LM5050-1 was considered, but its VINmin > 1V limitation still prevents full discharge

Question Does anyone have recommendations for a reverse polarity protection solution that:

• Allows the capacitor bank to discharge fully to ~0V

Any insights or alternative approaches would be greatly appreciated!

[Stray Electron]

  Why are you going to all of the trouble to use an electronic load to discharge your super caps?  It would be easier and more reliable and therefore safer IMO, to simply discharge them via a set DC resistance.

   How many farads are you talking about here and is there a requirement about how fast they have to be discharged?  A real honest to God requirement and not simply wishful thinking?

  Oops ok now I see the Farads.

   If I had to absolutely do it they way that you're planning, i would take a look at some of the circuits that are sometimes used in automobile jumper cables that read the voltage and polarity and only if it's correct, does it then then engage a relay that closes the circuit between the charging battery and the discharged battery. I've looked at a few of those circuits and there is only a few parts in them.  The ones that I looked used a large standard automobile starter relay so they can take the current but I don't know if they can in your case. But the advantage is that the automobile relays are robust, cheap and readily available. 

[markus.p]

Thanks for the reply. A power resistor was, of course, my first idea. However, the difference in discharge time is major:
Discharge voltage: 50 mV
Max. discharge current: 20 A

• DC load: 50 sec
• Power resistor: 245 sec

The tool is intended for repetitive use in production, making the discharge time significant.
The "DC-load" circuit is basically just an op amp driving a linear NMOS from a current sense resistor, so it's not really that complex either.

I'll take a look at what you suggested

[Stray Electron]

Thanks for the reply. A power resistor was, of course, my first idea. However, the difference in discharge time is major:
Discharge voltage: 50 mV
Max. discharge current: 20 A

• DC load: 50 sec
• Power resistor: 245 sec

  If you use a power resistor then the discharge time can be anything that you want.   What is limiting you to a 20 amp discharge rate? 

   I haven't looked at the numbers yet but how much energy is 22 farads at 24 VDC going to generate and will a hand held device going to the able to handle dissipating that amount of heat?

   I don't understand this bit: "Discharge voltage: 50 mV"

[Terry Bites]

Test the polarity with a comparator and use the output to switch in the current sink or resistor.
The series resistor (R1) ensures that the comparator can accept negative inputs without bursting into flames.
Check the datsheet for the maximim current allowed in the ics protection diodes. LM393s are quite robust in this respect.
Dirt cheap too. How much current can your 9V supply provide, enough to pull in a chunky relay? Not much, use a mosfet.

The Eload you are describing has the topology as sketched out.
It seems simple but there's a catch.
The sink must be able to operate below or very near to 0V, that will be the hard part.
A controlled sink cannot go down to 0V without a negative supply.
It needs to have a voltage at least -(Vref+Vgs) to work.
Without it the sink will go high Z when there are still a few volts on the supercap.
Even with a very low vref and a bipolar npn or a Darlington the problem remains.
A negative supply capable of sinking the full current will be needed.

I'd stick with the car relay and a fat wirewound resistor.

You could switch in a series of smaller resistors as the discharge voltage drops to hurry things along. A staged discharge.
An quad comparator, a few mosfets and resistors might be just the job. Or do it in code if you feel the urge. LM3914 anyone?

Still, you've got 10kJ to dump overboard and thats going to push the thermal design a hand held unit to the limit. Get the oven gloves out.

[johansen]

is the -.3 to .4volts from a Schottky diode too much?

[markus.p]

That's true, but I am limited to max. 20 A due to the current rating of the connector used by the product holding the caps.

33F 24V is 9.5 kJ. I guess "portable" is a better word than "handheld". It's got a pretty beefy cooling solution.

With "discharge voltage", I meant the target voltage at which the discharge process is considered finished.

[markus.p]

Thank you for your detailed answer, Terry!
I have a 9V battery (6x AA/LR6), so using a relay should be feasible.

My bad for not being detailed enough, here is the design of the actual load (there are two in parallel).
The "current sense resistor" is actually a hall effect sensor (ACS37012LLZATR-030B3) producing an output voltage proportional to the current.

Vref (LOAD1_VDAC) is supplied by the DAC of the MCU. The NMOS should (according to the datasheet) be able to sink 20A down to Vds=0.5V.

[Zero999]

is the -.3 to .4volts from a Schottky diode too much?

The Schottky diode would just blow up on its own, but might be a viable solution, with a fuse, so long as the diode is hugly over-rated.

EDIT: Unless you put the Schottky in series, but that would mean it'll take a long time for it to discharge below 200mV.

[Terry Bites]

Parallel up some cheap mosfets to get a lower Ron, better heat dissipation and lower device stress.
Fuse the drain to stop transistor melt down. MOSFETS often go short when they die.

Use the fault flag to trigger a latch and small signal MOSFET to pull Vref to ground (via a resistor in the Vref input of course).
Analog switches? Or maybe analog switches to ground the gates
That can be used to kill the current source even if the micro/ADC have brain farts.

Accuracy and speed don't matter I presume.
Add frequency compensation to the loop. Slug it as much as you like.

Good luck with your ingenious design. It looks rather good in my simulation.

[Zero999]

Parallel up some cheap mosfets to get a lower Ron, better heat dissipation and lower device stress.
Fuse the drain to stop transistor melt down. MOSFETS often go short when they die.

Use the fault flag to trigger a latch and small signal MOSFET to pull Vref to ground (via a resistor in the Vref input of course).
Analog switches? Or maybe analog switches to ground the gates
That can be used to kill the current source even if the micro/ADC have brain farts.

Accuracy and speed don't matter I presume.
Add frequency compensation to the loop. Slug it as much as you like.

Good luck with your ingenious design. It looks rather good in my simulation.

What does  H1 represent?

What happens if the polarity of V3 is reveresed?

You need to used back-to-back MOSFETs to prevent reverse current flow.

[Terry Bites]

Consider using a mosfet to do the switching. Here's why:

There is a big problem the DC relays which I forgot to mention in my response.
In an AC relay the arc is extinguished at the zero-crossing point(s). At higher currents the DC relay has no mechanism to stop the arc from continuing.
I learned this the hard way in a 24V lighting system. Dumb error! There are specialist high current relays that cost an arm and a leg. Avoid.
RC snubbers or TVS/ MOVs can help quench the arc.  Also see an ingenious solution here: https://sound-au.com/articles/arc-prevention.htm I think would have gone with a had if I'd known.

A relay capable of switching 20A or will likely need a fair bit of coil current. There are tricks that pull in the relay at full pull-in current and then drop it back to the holding current. A typical automotive 24V relay will have a coil resistance of ~350 ohms. So it will need 68mA to pull in but it could hold at 25% of this. Suck it and see.

[johansen]

is the -.3 to .4volts from a Schottky diode too much?

The Schottky diode would just blow up on its own, but might be a viable solution, with a fuse, so long as the diode is hugly over-rated.

The diode only has to handle the current sourced by the capacitor drainage device. 60 amp schott ky are like 1$

Consider using a mosfet to do the switching. Here's why:

Your mosfet resistor relay circuit is interesting, but you need a zener or tvs on the gate to prevent voltage spikes on the dc bus from blowing up the mosfet.

[NiHaoMike]

How about pairs of incandescent 12V bulbs in series as PTCs? They'll just work in either polarity.

[inse]

Why don’t you implement a polarized connector like XT60 instead of banana jacks on your device?
Does the cap bank have a polarized connection as well? (I‘d expect so)
In parallel you can implement the diode/fuse protection already proposed.
P.S. I am a little confused, does the rendering show the discharge device or the cap bank product?

[inse]

Now beat me for my unelegant primitive proposal. (Sorry, no microcontroller involved)
Cook off the charge via beefy smoking hot resistors - may also be halogen lamps as already proposed.
Use a relay to parallel an additional string once the voltage has dropped to speed up discharge while limiting max current.
The switch shall avoid arcing/surge while connecting the bank.
An analog meter will indicate when the bank is sufficiently discharged.
If you have one with center position, it will work regardless of polarity

[inse]

Markus, please don’t abandon your discussion.

[MariuszD]

Just for fun, why make a protection circuit if you can make a circuit that works regardless of polarity?

In practical implementation, protection against amplifier saturation, when the circuit is disconnected, needs to be added. I marked the current surges caused by this.