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I'm a DC guy but I think my calculations could work in AC too. I just need confirmation before start putting things together

Ok, I have a 3D printer it draws 5amps, 220 volts its internal power supply can't hold up for 10ms ( average relay operation time )

So I'm thinking about using AC power.

based on my calculation:

Energy = Power x time ( in sec)

Power = Voltage x current

so my total Jouls in second is

5x220x1 = 226 j/s ok I need to feed 226 jouls for 1 second to keep my printer active for relay to switch

now, to be on the safe side I want double my uptime.

I believe a 50uf is more than enough Am I right?

I wan't to put the capcitor in paraller with the printer and the power source

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  • \$\begingroup\$ what does this say about the power supply? ... Ok, I have a 3D printer it draws 5amps, 220 volts its internal power supply can't hold up for 10ms ( average relay operation time ) \$\endgroup\$ – jsotola Jun 23 at 17:26
  • \$\begingroup\$ "Energy = Power x time ( in sec)". Energy = power x time. When using SI units then energy (J) = power (W) x time (s). SI units named after a person have their symbols capitalised but are lowercase when spelled out. 'V' for volt, 'A' for ampere, 'K' for kelvin, 'Ω' (capital omega) for ohm, 'J' for joule, etc. \$\endgroup\$ – Transistor Jun 23 at 18:07
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A capacitor on the AC side won't do what you seem to think it will do.

A capacitor will pass AC.

If you put a large capacitor across your AC source, it will in effect be a (near) short circuit - cue sparks and bangs and tripped circuit breakers.

If you put a large capacitor in series with your AC source, then still won't help. It will act like a series resistor and reduce the voltage to your device.

If you want to use a capacitor to power the machine for a little while, then you will have to do it on the DC side.

You're going to need a much larger capacitor than you think, though.

Your formula assumes that you can get all of the energy back out of the capacitor that you put in. That's only the case, though, if you can discharge the capacitor to 0V.

You can't do that. Your machine needs some minimum voltage to operate. You have the normal operating voltage, and the minimum voltage. You can only use that voltage difference when calculating the energy you can use from the capacitor.

Better to use a UPS that will maintain the AC output regardless of the input.

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  • \$\begingroup\$ So, Capacitors work differently in AC? also, printers power supply is a switching type, maybe I can do it wit large DC caps? because I read somewhere that you can connect DC to AC-DC switching power supplies \$\endgroup\$ – far2005 Jun 23 at 17:30
  • \$\begingroup\$ No. They act the same whether it is AC or DC. You don't seem to understand the implications of what alternating current means. In the case of AC and a capacitor, the AC charges the capacitor to a particular polarity and voltage, then discharges it and charges it to the opposite polarity. It does that 50 times a second. \$\endgroup\$ – JRE Jun 23 at 18:01
  • \$\begingroup\$ What is don't understand is, it ac caps pass the current, then what is their use?? \$\endgroup\$ – far2005 Jun 28 at 12:47
  • \$\begingroup\$ There are no "AC capacitors." There are "capacitors." Period. No AC, no DC. \$\endgroup\$ – JRE Jun 28 at 13:05
  • \$\begingroup\$ @far2005: Take a look at the wikipedia capacitor page. \$\endgroup\$ – JRE Jun 28 at 13:06

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