Sanity check on capacitor value for current spikes - Page 1
Forum section: electronics / beginners

author: alexthe5th
author_rank: Contributor
timestamp: July 12, 2020, 10:26:22 pm
content: Hey folks - looking for a quick sanity check on a few calculations I've made.I've designed a circuit that intentionally trips circuit breakers (this is for a flight simulator) with the use of a solid-state relay to short the breaker on command. The breakers are connected to a 28V supply, have an integrated 153Ω resistor, and the worst-case current draw for each breaker is 200 mA for no more than 25 ms before the breaker is fully tripped (open circuit).There's a total of 64 circuit breakers in the simulator, and if all of them trip simultaneously, that's a total of 12.8 A which is a lot more than a typical wall wart is able to handle. To take the load off the power supply, I'd like to put a capacitor on each breaker to handle the worst-case 25 ms @ 200mA current spike. (Maybe "spike" isn't the best term - it's basically a step function.)This is the math I've used:C = Q / V = I * t / V = 200 mA * 0.025 s / 28 V = ~178 uF. Therefore, closest standard size up would be a 220 uF electrolytic.Since there's no periodic charge/discharge cycle (this is a one-shot affair - the breakers have to be manually pushed back in after a trip), I don't think there's any concern here about the time for the capacitors to charge back up after they discharge.So with that said - if I place a 220uF electrolytic between the power and ground rails for each breaker, should that handle the current spikes in the way I'm hoping it will? Is there anything I missed here?Thanks in advance!

author: Manul
author_rank: Super Contributor
timestamp: July 12, 2020, 10:49:47 pm
content: Unless I'm missing something, it seems to me that you want the power supply to stay at ~28V, but calculated capacitor charge for 28V drop.So 0.025s at 0.2A is 0.005C. If you want for eg. max 1V drop, it will be 0.005C / 1V = 5,000uF.

author: alexthe5th
author_rank: Contributor
timestamp: July 12, 2020, 11:17:40 pm
content: Quote from: Manul on July 12, 2020, 10:49:47 pmGot it, thanks! Totally forgot to factor in the drop. Oof - 5000 uF, that's a massive capacitor, and 64 of them will take up a ton of space on the board.Not sure what other options I have here, though.

author: Manul
author_rank: Super Contributor
timestamp: July 12, 2020, 11:41:52 pm
content: Well, it will work with less probably. 1V drop is small. If you allow a bit more, like 4-5V worst case drop, 1000uF may be enough. But total 64,000uF is huge. Not sure how your typical wall adapter will like it at startup.

author: Manul
author_rank: Super Contributor
timestamp: July 12, 2020, 11:48:14 pm
content: I guess other options is a proper power supply or short the brakers in fast succession, not all at once.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 13, 2020, 02:29:10 am
content: May consider going the other way -- your voltage is convenient for a lead-acid battery. Good for heavy momentary loads. Two 12V motorcycle batteries in series would do.Tim

author: alexthe5th
author_rank: Contributor
timestamp: July 13, 2020, 08:03:33 pm
content: Got it. I'm leaning towards just getting a large (400W+) 28V power supply that can handle the current and be done with it - something like this:https://www.antekinc.com/ps-4n28-400w-28v-power-supply/The lead-acid batteries aren't something I'd really considered, but you're right in that they're well suited for this application (makes sense, given they're used to drive starter motors). With that said, this is a fixed setup so portability isn't really a factor, and more importantly, I just can't be bothered having to keep them charged. Thanks for the suggestion though!And as for the capacitors, loading up my board with 64 large (or 1 monster) electrolytic caps doesn't make a ton of sense - they'd add so much cost and board space, not to mention that the impact on the power supply from charging them is another can of worms to deal with.Thanks for all your help - and if anyone has other ideas how for to power a 25ms blast of 12.8A @ 28V, let me know...

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 14, 2020, 02:45:09 am
content: The batteries will stay charged as long as you keep them at the float voltage -- very close to nominal operating already. You'd just use a small (1A or less?) supply to do that.It may also be of interest, to get more bang from your buck with the capacitors -- you could use a 24V supply and a boost module capable of the specified load current. As the input drops, 20V, 16V, 12V -- the boost continues to produce power, indeed drawing more current as it goes, so the curve drops sharply, but this utilizes the capacitors much more than brute force bypass. And may be cheaper than a brute force offline supply.https://www.seventransistorlabs.com/Calc/PSHoldUp.html#cpFor a 330W load, 24 down to 12V, for 25ms, I get 19.1mF. Beefy, but affordable.Heh, I even know what I would do, as I have a 6 x 4700uF 25V array on hand (for similar sorts of testing), and I've made one-offs like this 12-24 to 18V 20A Cuk converter,https://www.seventransistorlabs.com/Images/CukHack5.jpgwhich I'd only need to make a slightly beefier boost variant of, using the same controller board (and fewer mods, because the controller is actually made for boost in the first place).But an off-the-shelf module will do. Uh, beware of noise from the usual eBay offerings, may want some additional filtering or shielding to keep the test environment clean.Tim