Capacitor sizing for 120VAC blackout / brownout power backup

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UPDATED (more clarification)
Essentially what I am trying to do could be called "SuperCap Based UPS".
In other words, to replace the battery in a DC to AC Inverter circuit with a Super Capacitor to provide temporary 120V AC power to an appliance that draws about 0.08 amps or about 10 watts @ 120VAC.

Will I get more runtime by charging up a large 1 Farad Cap at 12V or would it be better to store 120V in a Cap that is capable of this voltage, and instead have the inverter convert from 120VDC to 120VAC?

I'm trying to provide as much run time to the end device (which obv. uses 120VAC).

So what's the most efficient way to get the most runtime at 120VAC from a capacitor powered inverter solution?

Best Answer

Let's say your "inverter" works just like a battery backed up UPS, and that it has a pretty wide margin on its storage voltage, going from 14V down to 7V before it taps out.

That means your main capacitor in case of the low voltage option can only drop 7V.

The voltage drop across a capacitor is an integral over time of the current signal taken from it, which becomes a very nice fiddly bit of magic concerning systems of differential equations once you add in a 10W constant load and a variable converter efficiency, so I'm going to roughly ball-park it, because I'm lazy and it's evening and I have a million things to do.

(( Ref: Wikipedia page jumped to the spot where the voltage current relation of a capacitor is given ))

How will I do this?

  1. Coursely take 80% efficiency for a converter going from 12VDC to 120VAC (which may be seriously overestimating it in a DIY scenario, to be honest).
  2. Estimate the current draw to be constant, calculated at a capacitor voltage of 9.5V, rather than the exact average, which voltage I drew from my large hat of "that'll probably do". If you want to do other use cases, you can take the average, since a constant current assumption will be off any ways.
  3. Simplify the integral for constant current, which then becomes a simple linear equation: V = (I*t) / C.

So, the current from the capacitor will be:

I = (10W / 0.8 [=efficiency]) / 9.5V =~ 1.32A

Which then can be put into the simplified linear equation for the assumption of constant current (be aware, this is a very broad and lazy assumption):

V =~ (1.32A * t) / C

Let's say you want only ten seconds of power, with the known voltage drop of 7V across the capacitor, that becomes:

7V ~= (1.32A * 10s) / C

Which becomes:

C =~ 13.2As / 7V =~ 1.88F

Let's quickly do that for 120VDC as well:

Same assumptions, but the voltage range will be 80V to 120V, probably, so a drop of 40V is allowable, estimating the constance of current at the 90V point:

I = 10W / 0.8 / 90V =~ 139mA

with t=10s:

40V =~ 1.39As / C

C =~ 1.39As / 40V =~ 35mF --> Charged up to 120V = very, very lethal.

So, you see, I've already used a lot of assumptions about all the stuff you're not giving us about your project, and how you will personally be able to complete the electronics and it's still a lot of calculation, even though I made a very bad and broad assumption of constant current

The final choice will depend on fixing all the parameters and some will intertwine. There's no solution to that and that's what makes electronics design a difficult field.

This is just your very first, very broad ball park. But to be honest, re: "very, very lethal", if you are asking this question I don't really think you should be considering anything above 30VDC to store energy the likes of this.

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