How to Power 13.2V 40A Battery-Operated Flash from AC Mains Safely and Reliably?

I own an electronic flash device that is run by battery only. The battery's specs are as follows:

Nominal / Charge voltage: 13.2V / 14.4V

Energy / Capacity: 66Wh / 5Ah

Max charge current: 4A

Continuous discharge current: 40A

Would it be possible to build a capactior that could "imitate" the battery yet draw power from A/C mains? What would the cost be of such a design?

Capacitors could provide the required 40A discharge current, but you need to know the amount of energy required to charge the photoflash capacitor inside the unit. This can usually be approximated using the Watt-Second rating the of flash.

The capacitor has to store enough energy to provide the require charge energy for a full power flash and maintain its potential above the required minimum potential to operate the DC/DC that is charging the photoflash capacitor. The required capacitor storage is likely to be 2-3x the charge requirement (since the voltage on the cap decays exponentially during the discharge).

You would also have to build a charge circuit for the capacitor that would charge the cap in less than the minimum cycle time between full power flashes.

This is all possible, but you might find that building a charger with a battery, sizing the charger to keep up with the full power flashes and staying with the OEM battery sizing is easier.

Thank you for the thoughtful reply. I believe the operating voltage of the unit is 12v 7a. Would it be possible to simply remove the battery cell from its cartridge and solder a 12v 7a power supply to the leads, or something to that effect? The unit is rated at 600Ws, and the battery lasts for 200 full power pops, if that info is at all useful.

The unit can be rapidly charged while in use. I'm just looking for a solution that eliminates a battery from the circuit, if not out of necessity than out of sheet curiosity. Thanks again!

600WS is a lot of energy for this type of system.

Capacitor energy = 1/2 * C * V^2

so C = (v^2) / (2U) = 12^2 / 600*2 = 0.12F or 120,000 micro-farads

My guess is that you are going to want about 3 times this or 360,000 uF

In order to get 40A, you will need to put lots of caps in parallel. Fortunately, this also adds the capacitance of the capacitors in parallel.

Assuming you want to do this on the cheap, a Cornell Dubilier - CDE SLPX333M025E9P3 is 33,000 uF, 25V, 5.3A ripple and an ESR of 18 milli-ohms. You will need 10 of them in parallel for a total capacitor cost of $35.

To charge them, you will need a power supply that can deliver 10A @ 12V for a 5 second cycle time. The power supply will have to be current limited because when discharged, the capacitor array will have an apparent resistance of 0.0018 ohms which would be >6000A charge current (assuming idealized zero resistance elsewhere) [I=V/R = 12/0.0018]. You can use a higher current power supply to reduce the cycle time, but pretty soon you will be at 40A and you might as well just skip the capacitors.