Charging ultracapacitors with computer PSU

[drummerdimitri]

Hello people,

I am new to this forum and I hope to learn a lot about electronics during my stay here.

My electronics knowledge is limited since I am just a hobbyist but here's what I like to do.

I've moddified my 750W computer PSU by adding banana jacks to the case for 12, 5 and 3.3 V outputs to be used to either charge batteries directly or feed power to my battery charger (DC in).

Whenever I try charging my ultracapacitors, the PSU shorts out and shutsdown. I guess this is happening due to the very low internal resistance of the caps.

The PSU has 2 +12V rails with each capable of providing up to 32 A of current so that's 72 amps combined.

How would I be able to charge my ultracapacitors at around 60 amps without tripping the PSU?

I've tried adding resistors in series with no avail.

If anyone could help me out I'd be very grateful.

Thanks!

[tggzzz]

Details matter. Schematics help.

What is the voltage rating of the capacitors? Which PSU rail do you connect them to? What resistor values (and wattage) have you used? Do you connect them and the turn on the PSU or vice versa. What voltages and currents have you measured.

Get them wrong and smoke and noise will emanate from your room

[sacherjj]

Are you getting to 12V with ultra capacitors in series?  Have you looked at what is required as far as parallel resistors to those capacitors to do this without issues? 

[drummerdimitri]

Details matter. Schematics help.

What is the voltage rating of the capacitors? Which PSU rail do you connect them to? What resistor values (and wattage) have you used? Do you connect them and the turn on the PSU or vice versa. What voltages and currents have you measured.

Get them wrong and smoke and noise will emanate from your room

The capacitors are 2.7V and are being charged one by one since I am unsure of how to charge all 6 of them in series with proper balancing.

I tried charging on all rails with not luck. I used resistors from 0.8 to 3.8 ohms if i remember correctly. No idea about their wattage though.

I connected them and then turned on the psu and vice versa. Still no luck and shorts every time.

[drummerdimitri]

Are you getting to 12V with ultra capacitors in series?  Have you looked at what is required as far as parallel resistors to those capacitors to do this without issues?

6 caps in series would equal a max voltage of 16.2V, but since my PSU can only charge them up to 12V I will have to live with that.

So far I've been charging them one by one since I was not able to charge them at 12V in series without running into voltage balancing problems.

If you have any info regarding balancing in series please do share.

Thanks!

[tggzzz]

Details matter. Schematics help.

What is the voltage rating of the capacitors? Which PSU rail do you connect them to? What resistor values (and wattage) have you used? Do you connect them and the turn on the PSU or vice versa. What voltages and currents have you measured.

Get them wrong and smoke and noise will emanate from your room

The capacitors are 2.7V and are being charged one by one since I am unsure of how to charge all 6 of them in series with proper balancing.

I tried charging on all rails with not luck. I used resistors from 0.8 to 3.8 ohms if i remember correctly. No idea about their wattage though.

I connected them and then turned on the psu and vice versa. Still no luck and shorts every time.

And the other information I suggested you give?

[drummerdimitri]

You mean voltages and current? Like i said the psu would short immediately after being turned on so no values there.

[tggzzz]

You mean voltages and current? Like i said the psu would short immediately after being turned on so no values there.

Schematic?

What was the voltage on the capacitor afterwards?

What do you expect a 2.8V capacitor to do when you put 12V across it?

[drummerdimitri]

Where am I supposed to get you schematics from? It's a computer's power supply so there are none.

Input voltage shouldn't matter. It shouldn't matter what the input voltage is as long as the actual voltage of the cap remains at or below 2.7V correct?

[tggzzz]

Where am I supposed to get you schematics from? It's a computer's power supply so there are none.

Input voltage shouldn't matter. It shouldn't matter what the input voltage is as long as the actual voltage of the cap remains at or below 2.7V correct?

Oh good grief. The schematic of your circuit, not the PSU internals.

If you haven't bothered to make any measurements of voltage or current, and have only tried a very limited range of resistor values, then you really are stabbing around in the dark.

For example, what makes you think the voltage across the capacitor did remain below 2.7V?
What was the peak current flow?
What was the ripple current specification of the capacitor.
What was the capacitor's ESR?
What happens if you try a 1kohm resistor?
What wattage do you expect the resistor to have to dissipate?

[DanielS]

Schematic?

OP has already said he is simply trying to charge caps by hooking them up to a PSU and a series resistor... do you really need a schematic for a RC circuit?

[sacherjj]

If you try to connect a 2.8V capacitor across a 12V supply, it isn't going to act nice like a current limiting lab supply.  It is going to follow Ohms Law.  V = 12 - 2.8V = 9.2V

Not sure what the resistance of your super capacitor is, but I'm guessing hundreds of milli-ohms.  Let's just pick 0.5 ohm. 

V = IR

I = V/R

I = 9.2/0.5 = 18.4 A.

If it is 0.25 ohm, now we are at 36.8 A.   You power supply is most likely going into over current shutdown.   Put a big resistor in series (100K) and see what it does.  It will take a while to charge it.  Why not use the 5V rail?  It is closer to the voltage of the cap and generally has high amperage?

If you connect capacitors in series, include a resistor in parallel with the capacitor that has 10 times the leakage current of the capacitor when charging.  This helps balance things out.  Voltage is initially a function of capacitance in a series combination, but eventually becomes a function of leakage resistance.  You should only series caps with the same capacity. 

[tggzzz]

Schematic?

OP has already said he is simply trying to charge caps by hooking them up to a PSU and a series resistor... do you really need a schematic for a RC circuit?

He has indicated his equipment in his possession, but hasn't provided useful information such as series/parallel connection of multiple capacitors, nor which PSU rails, nor what he has done with two rails of (nominally) the same voltage, etc.

I have found that getting myself or others to accurately define the situation is half the battle - and often the act of defining it enables I/they to realise the error and correct it.

But I was somewhat intemperate, and for that I apologise.

[tggzzz]

If you try to connect a 2.8V capacitor across a 12V supply, it isn't going to act nice like a current limiting lab supply.  It is going to follow Ohms Law.  V = 12 - 2.8V = 9.2V

Not sure what the resistance of your super capacitor is, but I'm guessing hundreds of milli-ohms.  Let's just pick 0.5 ohm. 

V = IR

I = V/R

I = 9.2/0.5 = 18.4 A.

If it is 0.25 ohm, now we are at 36.8 A.   You power supply is most likely going into over current shutdown.   Put a big resistor in series (100K) and see what it does.  It will take a while to charge it.  Why not use the 5V rail?  It is closer to the voltage of the cap and generally has high amperage?

If you connect capacitors in series, include a resistor in parallel with the capacitor that has 10 times the leakage current of the capacitor when charging.  This helps balance things out.  Voltage is initially a function of capacitance in a series combination, but eventually becomes a function of leakage resistance.  You should only series caps with the same capacity.

Initially, when the capacitor is discharged, the current will be 12/0.5=24A (or 12/0.25=48A).

As current goes into the the capacitor and it charges up, the voltage across the capacitor will rise asymptotically to 12V - unless of course something nasty happens when the capacitor's voltage rating is exceeded Or something nasty occurs if the capacitor's ripple current rating is exceeded

[drummerdimitri]

If you try to connect a 2.8V capacitor across a 12V supply, it isn't going to act nice like a current limiting lab supply.  It is going to follow Ohms Law.  V = 12 - 2.8V = 9.2V

Not sure what the resistance of your super capacitor is, but I'm guessing hundreds of milli-ohms.  Let's just pick 0.5 ohm. 

V = IR

I = V/R

I = 9.2/0.5 = 18.4 A.

If it is 0.25 ohm, now we are at 36.8 A.   You power supply is most likely going into over current shutdown.   Put a big resistor in series (100K) and see what it does.  It will take a while to charge it.  Why not use the 5V rail?  It is closer to the voltage of the cap and generally has high amperage?

If you connect capacitors in series, include a resistor in parallel with the capacitor that has 10 times the leakage current of the capacitor when charging.  This helps balance things out.  Voltage is initially a function of capacitance in a series combination, but eventually becomes a function of leakage resistance.  You should only series caps with the same capacity.

I see what you mean. In that case wouldnt it be better to use the 3.3V rail instead of the 5V?

All I would have to do is calculate the resistance value based on the current I am looking for and the 3.3V input correct?

How can I find the leakage current of the capacitors?

[tggzzz]

All I would have to do is calculate the resistance value based on the current I am looking for and the 3.3V input correct?

And stop the voltage exceeding 2.8V.

How can I find the leakage current of the capacitors?

Leakage current is usually very loosely specified, and is highly dependent on temperature, voltage, capacitor age. In short don't rely on any value other than those stated in the data sheet.

You have read the capacitor data sheets, haven't you?

[drummerdimitri]

I have read the data sheet and it says that the maximum leakage current for my capacitors is 4.2 mA. What resistance will I need in that case?

[sacherjj]

Initially, when the capacitor is discharged, the current will be 12/0.5=24A (or 12/0.25=48A).

As current goes into the the capacitor and it charges up, the voltage across the capacitor will rise asymptotically to 12V - unless of course something nasty happens when the capacitor's voltage rating is exceeded Or something nasty occurs if the capacitor's ripple current rating is exceeded

Oh, yeah.  Been working with batteries that never go to 0V too long.  As I have said many times lately.  It's not as bad as I thought.  But WORSE. 

[tggzzz]

I have read the data sheet and it says that the maximum leakage current for my capacitors is 4.2 mA. What resistance will I need in that case?

That will depend on the rest of your circuit - hence the question about a schematic. It is unlikely to be relevant for a single capacitor plus series resistor - and as I have indicated, you are likely to have other failure modes with that topology and a 3.3/5/12V supply.

When doing your calculations, bear in mind that the leakage current could easily be 4.2uA and still be within specification.

You have been given hints by several people about how to go about experimenting to find the cause of your problems (e.g. use a 1k or 100k series resistor). If you cannot be bothered to take the hints, there is nothing we can do.

[drummerdimitri]

Ok so here's a little update on what happened today so far.

I used a lead pencil's core to limit the current being drawn from the PSU using the 3.3V rail.

I was able to get one capacitor charging at around 0.2 amps but that was far from ideal since my hobby charger can charge them at 5 amps.

Then, I decided to use the 12V rail for charging using the pencil as resistance (around 2.5 Ohms) and using my clamp meter, i registered an inrush current of 13 amps and that is when the PSU shorted.

What I don't understand is why would it stop working if the 12V rails are rated at 32 amps each?!

So what I did was use a 14 year old PSU and to my surprise, it was able to charge the cap at 25 amps initially with a direct connection from 0 to 2.5 V while the current slowly decreased to 8 amps eventually and poof went the PSU.

I don't know what happened, but when i opened the case, I found a 250V 6.3A fuse that was blown so I replaced it and it blew immediately upon power up.

Anyway, I've salvaged some of the old PSU's parts and threw the rest in the garbage.

Now the question to ask is why can a 10 year old 350W PSU charge a cap at 25 amps but a new 750 W PSU can't?

There must be something wrong with it correct?

Let me know what you think.

[timb]

Computer PSUs aren't designed for this sort of thing.

The new one is going into under voltage, over temperature or over current mode and shutting down so it doesn't die like the older unit just did. Which means the old one likely doesn't have that sort of protection in it.

Computer PSUs aren't designed to drive high current capacitive loads like this. They're meant for high current resistive and medium current inductive loads.

Most likely the capacitor is pulling the PSU into constant current mode, which means the voltage is dropping to supply the required current. The PSU sees this as a short and turns off to stop the component and itself from being damaged.

So this isn't going to work.

Go on eBay and look for a 500W AC to DC supply for LED lighting. These are designed to run in a CC mode.

Alternatively you could get a Deep Cycle Marine Lead Acid Battery from the auto store or Wal-Mart and run that in parallel with the 12V output of your computer PSU. The battery will act as a buffer to supply the needed current and the PSU will keep it charged. (This is a huge hack and may kill the battery eventually, but it will work.)


Sent from my Smartphone

[DanielS]

PC power supplies have over-current, over-power and short-circuit protections that monitor the rate of change of their output current to detect abnormal fast rise.

If you do not want your PSU's over-current protections to get false-positive, you need to rate-limit your current rise to something like 3-4A/µs. When you stop charging, you also need to ramp the current down at a similar rate or the PSU's output voltages may overshoot and cause it to shutdown due to tripping over-voltage protections.

Haswell-ready computer PSUs are supposed to be able to cope with load transients up to 8A/µs.

[drummerdimitri]

Computer PSUs aren't designed for this sort of thing.

The new one is going into under voltage, over temperature or over current mode and shutting down so it doesn't die like the older unit just did. Which means the old one likely doesn't have that sort of protection in it.

Computer PSUs aren't designed to drive high current capacitive loads like this. They're meant for high current resistive and medium current inductive loads.

Most likely the capacitor is pulling the PSU into constant current mode, which means the voltage is dropping to supply the required current. The PSU sees this as a short and turns off to stop the component and itself from being damaged.

So this isn't going to work.

Go on eBay and look for a 500W AC to DC supply for LED lighting. These are designed to run in a CC mode.

Alternatively you could get a Deep Cycle Marine Lead Acid Battery from the auto store or Wal-Mart and run that in parallel with the 12V output of your computer PSU. The battery will act as a buffer to supply the needed current and the PSU will keep it charged. (This is a huge hack and may kill the battery eventually, but it will work.)


Sent from my Smartphone

Thanks for the in depth reply.

I was thinking of getting a high powered hobby charger such as the icharger 308DUO for all my charging needs be it capacitors or batteries. Only problem is the cost plus I would need to by a high powered DC power supply.

[drummerdimitri]

PC power supplies have over-current, over-power and short-circuit protections that monitor the rate of change of their output current to detect abnormal fast rise.

If you do not want your PSU's over-current protections to get false-positive, you need to rate-limit your current rise to something like 3-4A/µs. When you stop charging, you also need to ramp the current down at a similar rate or the PSU's output voltages may overshoot and cause it to shutdown due to tripping over-voltage protections.

Haswell-ready computer PSUs are supposed to be able to cope with load transients up to 8A/µs.

I see your point. Any idea on how to do this though?

[DanielS]

I see your point. Any idea on how to do this though?

The most efficient option would be to build a simple DC-DC switching power supply and set its soft-start ramp to provide the 3-4A/µs load slew rate.

A much less efficient option would be to use a pair of op-amps: one integrates a fixed current to generate a ramp signal while the other is wired as a a voltage-controlled current source following that ramp.

In both cases, you need to shut down the regulator (short the soft-start cap on the PWM chip or the integrator cap on the op-amp) so current starts at 0A instead of max when you hook up the next cap.