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I've run into quite the problem. I have a load that requires 12v@5A for under 2.5 seconds. However, my supply is a USB 5v@500mA. This is not an immediate load. The load will only need to be activated once per 5 minutes, so I have some time for charging of possibly capacitors or a lipo.

What would be the best way to go about this? I've could not find any boost converter for that much of a power increase, so I know I'm going to have to store that charge somehow.

I've researched adding a lipo and a charger circuit, however, I cannot seem to find a 3S lipo charger that is in a package with exposed leads. In previous exchange forums the BQ24115 was recommended, however I don't have the capability to solder that small of a device currently with my time restraints. Also, I don't see how that chip monitors 3 cells, as all the applications provided in the datasheet seem to be for only a 1 or 2 cell LiPo, which I still can't figure out how the chip balances the cells as I see no hookup for "cell 1 sense" or "cell 2 sense". There has to be an input somewhere for it to read the individual cell voltages, assuming that chip can do balance charge.

Capacitors. I don't know very much about them, however I do know it is a potential route. A colleague recommended I charge capacitors individually, then use a transistor to arrange them in series for the load dump. (because I need 12v out from the caps, with an input of 5v).

Any help would be much appreciated! Thank you.

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  • \$\begingroup\$ Rather than use the difficult-to-solder chip, try looking through the selection at digikey.com/products/en/integrated-circuits-ics/… \$\endgroup\$
    – Hearth
    Commented Oct 22, 2018 at 22:30
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    \$\begingroup\$ "I've could not find any boost converter for that much of a power increase", good, if you did find any then it would've been a scam. \$\endgroup\$
    – Harry Svensson
    Commented Oct 22, 2018 at 22:31
  • \$\begingroup\$ What exactly is that 12V@5A load? \$\endgroup\$
    – Harry Svensson
    Commented Oct 22, 2018 at 22:42
  • \$\begingroup\$ @HarrySvensson I have a device that requires 12v @ 5A to activate. \$\endgroup\$
    – AJ_Smoothie
    Commented Oct 22, 2018 at 22:55
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    \$\begingroup\$ RE "I've could not find any boost converter for that much of a power increase, " That's not surprising because a boost converter doesn't provide any power increase at all. The product of the output voltage and current, \$V_o I_o\$, will always be less than the product of input voltage and current \$V_i I_i\$. \$\endgroup\$
    – The Photon
    Commented Oct 23, 2018 at 4:59

1 Answer 1

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At 12v at 5 amp, for 2.5 seconds, you need to store 150 joules.

At 100 volts across a capacitor, the formula

Energy = 0.5 * C * V^2

may guide you.

A one farad cap at 100 volts holds 5,000 joules. Thus you need about 0.03 Farad cap.

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  • \$\begingroup\$ And for OP's case, he'll need a 12F cap for those 12V... \$\endgroup\$
    – a concerned citizen
    Commented Oct 23, 2018 at 8:15
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    \$\begingroup\$ @aconcernedcitizen That is assuming that he can get all the energy from the capacitor which is very unrealistic. Something more realistic would be to charge the capacitor to, say, 100 V and then use a buck converter to get it down to 12 V. In other words, \$J = \frac{C}{2}V^2 => J = \frac{C}{2}(V_{high}^2-V_{low}^2) => C = \frac{2J}{V_{high}^2-V_{low}^2} => C = 30\$ mF, assuming 100% energy conversion. - And yes, he could connect one end of the capacitors pins to the 5 V rail so the capacitor will be charged to 100-5 = 95 V which might be an easier capacitor to find. \$\endgroup\$
    – Harry Svensson
    Commented Oct 23, 2018 at 11:53
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    \$\begingroup\$ @HarrySvensson True, but then he'd also need a boost converter for the 100V rail. The plot thickens. :-) Note that I am not saying it's impossible, just a bit more complicated than one would need (which is also not saying OP is not willing). Also, I was wrong, it's 2.something F, I mistakenly (right word?) calculated for 5V not 12V (got confused with 5A). \$\endgroup\$
    – a concerned citizen
    Commented Oct 23, 2018 at 12:44
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    \$\begingroup\$ @HavocRC boosting up to 100V improves the storage efficiency of the capacitor. Note that voltage is the squared term in the capacitor energy equation. It allows you to store a larger amount of energy in a much smaller capacitor, lowering part size and cost. \$\endgroup\$
    – Chris Fernandez
    Commented Oct 23, 2018 at 19:01
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    \$\begingroup\$ It was an approximation, increasing the voltage 30 times has led to only a 2/3 reduction in physical size. \$\endgroup\$
    – Jasen Слава Україні
    Commented Oct 29, 2018 at 18:09

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