Electrical – Measuring the peak discharge current of capacitor

In general voltage on the capacitor with respect to the current is governed by the equation:

\$v(t)= \frac{q(t)}{C} = \frac{1}{C}\int_{t_0}^t i(\tau) \mathrm{d}\tau+v(t_0)\$,

By the definition for CCS:

\$ i(\tau) = I \$,

from this we can derive that:

\$v(t)= \frac{1}{C}(I t - It_0) + v(t_0)\$

now assuming \$t_0 = 0\$ this simplifies to:

\$v(t)= \frac{1}{C}I t + v(0)\$.

What this means is simple! The voltage across capacitor will change linearly with time. The "rate" of change (or "slope") depends on the current magnitude and the capacitance:

• The bigger the capacitance the slower voltage changes.
• The bigger the current the faster voltage changes.
• The sign of the change (voltage rising or falling) depends on the sign or direction of the current. Obviously if current is flowing into capacitor voltagwe will rise if flowing out of capacitor voltage will fall.

This answer is not based on seeing a circuit diagram. A circuit would help a lot.

If the load is to mid-rail of the power supply and you are driving a sinewave at 145.6 Vp-p, then the current calculation is this.

1. Calculate the RMS voltage - 145.6 Vp-p is 51.5 Vrms
2. Calculate the load power - \$\frac{V^2}{R}\$ = 3.96 watts
3. Estimate input power - input power is going to be no-worse than twice output power so say 8 watts max.
4. Estimate dc current drawn from each 120V - each half cycle occupies 50% of the time of the sine period therefore each power supply is supplying 4 watts or 33 mA.

Sanity check, 240V x 33mA = 8 watts. OK I think I got it right.

Your power supplies can supply 50mA and this might be OK without adding capacitance (there'll be some inside the power supplies anyway) and the peak current taken by the load is

\$\sqrt2 \times 33mA\$ = 47mA.

I think you should be OK without adding capacitors and will run indefinitely as far as I can see.