Super Capacitor bank Design for Battery Swap

Question

We need to maintain approximately 4-5 A at a max DC voltage of 84 V for 15 s or less.

Is this possible and how many super caps or banks of super caps will it take?

We have an electric motor and controller running off a 20S14P Lithium battery at 84 V max and low voltage after discharge of approx. 65 V.

We are doing a hot swap battery exchange, and with the current capacitors inside the controller we only have about 4 seconds before the controller shuts down and we have to do a restart.

We would like to extend the shutdown time by at least 10 seconds so the controller never shuts off before having the old battery out and a new battery connected.

Plugging another battery into the system before removing the discharged on will not work because we never know exactly what the voltage will be. To much of a spike or discharge for a lithium battery.

Answer 1

Rough calculation

5 Amps x 20 Sec = 100 Coulombs of charge

80 Volts / 5 Amps = 16 Ohms DC resistance of the load

100 Coulombs / 80 Volts = 1.25 Farads

16 Ohms x 1.25 Farads = 20 seconds time constant

Allow 10% voltage drop in 20 seconds

Capacitor sizing 1.25 x 20/2 = 12.5 Farads minimum at 150 Volts minimum

This will require a whole helluva lot of low voltage supercapacitors in series to meet the minimum voltage, then paralleled to make the required minimum capacitance.

For example, you would need about 2300 of Amazon's 4 Farad 5.5 Volt units at an approximate cost of $10,000.

Answer 2

As pointed out by Phil's answer, the amount of super-capacitors for such high current for so long is not very realistic.

One option might be to have a small (fixed) lithium battery that handles the transition. You don't need much capacity (5 A for 15 s = 5*15/60/60= 0.02 Ah), but you require a somewhat big current (compared to the capacity).

You can probably use the smallest capacity you can find in 20S (just check the current rating).

An alternative, that might be cheaper, is to use a low(er) voltage battery for the swapping, and a DC-DC converter to maintain the correct voltage.

If you can, I think the best solution is to follow Neil's route: connect the 2nd battery before disconnecting the 1st.

If your only issue is that the full battery will try to charge the empty one with high currents, then you just need to add a diode on the + connection of each battery to block reverse currents (size it properly for heat dissipation). I would recommend you use a Schottky diode (lower voltage drop), or if you do your own PCB, then even better, an "ideal diode" made of a MOSFET + ideal diode driver (voltage drop an order of magnitude smaller, so far less power losses/heat to dissipate).

If you fear to have a big inrush current because your system itself will react this way, then there are special hot swap ICs with integrated inrush current limitation (some integrate the MOSFETs, some drive external MOSFETs).

Answer 3

Plugging another battery into the system before removing the discharged on will not work because we never know exactly what the voltage will be. To much of a spike or discharge for a lithium battery.

... unless you treat the unknown voltage of the battery as you would have to treat the variable voltage of the supercaps, and use it to drive a DC-DC converter

... or given the high voltages involved, >65 V, a diode drop would be little loss of efficiency, and you could simply diode-OR the batteries.

If you're worried about the sudden rise of bus voltage from 65 V to 84 V, then a series pass FET per battery supply could be used to make the transition as smooth as you like. Two FETs would be a lot cheaper than a bank of supercaps.