Adding capacitor banks inside conventional PC to compensate for temporary power line disruptions

Question

I've tried different ways to search the web for anyone who might have done this already, but can't find anything. So I'll ask here.

Would it be possible to add banks of large capacitors inside the case of a conventional PC to act as a sort of short-term UPS to hold up the various supply lines (ie 3.3 volt, 5 and 12 volt, etc) for perhaps several seconds to maybe a minute or so just to overcome the temporary loss or a glitch in the mains AC power? I know the effectiveness of this will depend on the current draw on any given line and the size of capacitors that are used. The one drawback I can see is that the inrush current into the capacitors on power-up might be more than the power supply can handle.

A slight tangent would be to use a 12-volt sealed lead acid battery (instead of a capacitor bank) and put it directly on the 12v supply.

Answer 1

Capacitors are sometimes used to provide emergency power to devices like SSDs so that the device can store any cached data to non-volatile memory and then shutdown. See for example: https://www.intel.com/content/dam/www/public/us/en/documents/technology-briefs/ssd-power-loss-imminent-technology-brief.pdf

Similarly, there are capacitors on your power supply that can provide charge during (very) brief dips in mains voltage. However, for back up power on a scale of minutes, uninterruptible power supplies use batteries, which store energy more cheaply than capacitors.

Answer 2

If you rectify the line voltage, maybe grabbing peaks at 160 volts, and you need to provide 100 watts for 100 seconds, and your downstream switching regulator has a 40 volt dropout, how big a capacitor will you need?

For 100 seconds, and 100 watts, you need to store 10,000 joules (watt * seconds) OR MORE. At 40/160 dropout, or 25%, that is close to 100% energy extraction, w'll ignore that 25% and compute what is needed.

Energy = 0.5 * C * V^2

and we manipulate

Energy / ( 0.5 * V * V) = C

10,000 / (0.5 * 160 * 160) = 10,000 / (0.5 * 25,600) = 10,00 / 12,800

C = 0.8 farads at 250 volt rating

Answer 3

Would it be possible to add banks of large capacitors inside the case of a conventional PC to act as a sort of short-term UPS to hold up the various supply lines (ie 3.3 volt, 5 and 12 volt, etc) for perhaps several seconds to maybe a minute or so just to overcome the temporary loss or a glitch in the mains AC power?

It would be possible but impractical, and your PC case would be the size of one wing of the Cray 1. That's because you want to extract most of the energy stored in the capacitors, and that means discharging them quite a bit - way outside of the tolerances of those supplies.

The only practical way of doing it with capacitors is by adding them to the high-voltage DC link of the ATX power supply, with proper safety and design precautions of course.

ATX power supplies are switching and typically will operate from 100-240V. So, if you're on 240VAC mains, the capacitors on the DC link can discharge from about 340V to 140V while the supply maintains good output voltages. This would extract >80% of the energy out of the capacitors. If you only got 120VAC mains, you'd need to step it up using an external transformer first or add a voltage doubler between the rectifier and the DC link.

As for how much capacitance you'd need? Say a 100W load, 100J per second of energy, 80% energy discharged from 340V - that's 2mF per second of hold-up time per each 100W of load. If you're gaming at 250W (let's say), you need 5mF per second of hold-up time. So, for example, 50,000uF for a 10s hold-up, and those have to be 450V capacitors. Not entirely impractical. Just remember that a 75% extra-fast stick of dynamite releases about 1MJ. 250W for 10s is 2.5kJ, so 1/4% of a stick of dynamite in energy terms. If you don't protect those capacitors and their circuit against short circuits, it'll literally blow in your face upon an inevitable short circuit. Each capacitor needs its own fuse.

Answer 4

Energy stored in a capacitor = \$ 1/2 C V^2 \$

To pull energy off a capacitor the voltage must decrease.

Therefore, if you want to store energy, the capacitor must not be on a fixed voltage rail, but before the regulator or power supply, at a place where voltage can go down a lot while the energy is pulled out of the capacitor.

A 470µF charged to 320V will store 24 Joules. Discharging it to 100V leaves 2.3 Joules in it, so there are, let's say 20 Joules per cap. If you want to power a PC from that, and the PC draws 100 watts, you'll need 5 capacitors per second.

Also a capacitor bank at that voltage is super lethal.

Therefore, it is a much better idea to just use a UPS, or a 12V battery and a PicoPSU.