Electrical – lum – Calculating the leaking current for a tantalum capacitor

I would guess that \$22\mu F\$ is the minimum capacitance that the application can use.

Part characteristics can differ as their voltage rating increases, so be careful not to go 'too' high. (Don't put a 100V capacitor where there once was a 16V part). I would wager that if you can get something between \$22 \mu F\$ and \$47 \mu F\$ in 25V, you'll be fine.

You may also be able to use multilayer ceramic capacitors in parallel to get \$ 22\mu F\$ - just don't use electrolytic. Electrolytic capacitors are not 'high speed' due to their intrinsic parasitic characteristics.

The calculation is straightforward. The capacitor size is simply a question of how much voltage drop you can tolerate over the duration of the pulse. The average current from the battery is a function of the duty cycle.

ΔV = I × Δt / C

Solving for C gives:

C = I × Δt / ΔV

Let's assume you can allow ΔV = 0.1V. For your first example, this works out to:

C = 25 mA × 25 ms / 0.1 V = 6.25 mF

The average current draw is 25 mA * 25 ms / 2.5 s = 0.25 mA.

For the second example, the numbers work out to:

C = 50 mA × 100 ms / 0.1 V = 50 mF

Average current = 50 mA * 100 ms / 1.0 s = 5 mA.