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I would like a formula to be able to calculate how long it would take 6 2.7v 350f ultracapacitors connected in series to degrade from 14.4 volts to 11.5 volts given a parasitic draw of a specific load.

For example: A bank of capacitors replacing the lead acid battery of an automobile, charged to 14.4 volts by the alternator. How long can the car sit with the engine off before the voltage dropped to 11.5 volts? 11.5v it's a minimum voltage before the starter will not turn over the engine.

6 2.7v 350f ultracapacitors at 16.2v 58.33f Balancing board

Q. Do the balancing board resistors cause the capacitors to drain faster? Q. How fast does a ultracapacitor drain on its own not connected to anything?

12v system Parasitic current draw of 0.03Amps 58.33f Starting voltage 14.4v Minimum voltage 11.5

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Let's look at a simple analysis first. The basic equation relating current and voltage is $$\frac{dV}{dt} = \frac{i}{C} $$ where i is in amps and C is in farads. For a current of .03 amps and 58.33 farads $$\frac{dV}{dt} = 5.14 x 10^{-4} $$ so in one hour $$\Delta V = 1.85\text{ volts}$$ Since 14.4 minus 11.5 equals 2.9 volts, your ultra caps will last about $$\Delta t = \frac{2.9}{1.85} = 1.57\text{ hours}$$

This is so short a period that you simply don't need to worry about self-discharge currents.

EDIT - Along the same lines, you need to take a look at the load on a car battery. The big-ticket item is starting. Starting loads are typically 100 to 200 amps.

For a current of 100 amps (and this is best-case, remember. Starting in cold temperatures takes rather more.) we find $$\frac{dV}{dt} = 1.71 $$ and for a 2.9 volt drop, $$\Delta t = \frac{2.9}{1.71} = 1.69\text{ seconds}$$ For 200 amps, of course, the time is half that, or about 0.85 seconds.

So you might want to reconsider using caps (ultra or otherwise) for a car battery replacement. Also, just for fun, you need to check the spec sheet for the maximum current which can be drawn from an ultracap without damaging it.

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  • \$\begingroup\$ Perfect that is exactly what I needed to determine the size of the capacitors I need to use. I may need to come up with a way to isolate the parasitic load. Thanks \$\endgroup\$ – Rob C. Dec 1 '18 at 8:18
  • \$\begingroup\$ Even with the load isolated, check out Edgar's answer and Solar Mike's comment. Driving a starter, these caps are likely to last about a second or less.. \$\endgroup\$ – Brian Drummond Dec 1 '18 at 13:31
  • \$\begingroup\$ @RobC. - See edit. It occurs to me I really should have addressed your larger question, rather than focussing on what you asked. \$\endgroup\$ – WhatRoughBeast Dec 1 '18 at 17:09
  • \$\begingroup\$ @RobC. - Don't ask these sort of questions in comments - start new questions. Also, you should be aware that the CCA rating for car batteries assumes a minimum voltage of 7.2 volts. Run that through the calculations and see what sort of operation times you get. \$\endgroup\$ – WhatRoughBeast Dec 1 '18 at 22:25
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You will need to carefully study what the ESR is for these ultra capacitors. Some of these types of capacitors may have an ESR that is higher than a standard lead acid battery making them less than suitable for providing high amperage currents for running a starter.

Other manufacturers make ultra capacitor types that are specifically designed for providing starting assist.

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A battery holds up its voltage during a discharge until it is nearly dead. A capacitor voltage drops quickly in the beginning of a discharge.

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Car batteries are designed for the high currents that a starter requires. We could be talking of 100A or even more.

But the calculations are straight-forward.

6 * 350F capacitors in series gives you a capacitance of 350F/6 = 58F. 100A for a drop of 3V gives you a duration of 1.75s. SO that better be a very fast starter, if you give it twice the drop or half the current that's 3.5s.

30mA drain (which is 100 times higher than the self-discharge current for these size capacitors) for a drop of 3V would be 97 minutes.

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  • \$\begingroup\$ 300A and around 500A or 600A for larger petrol engines... \$\endgroup\$ – Solar Mike Dec 1 '18 at 7:10

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