# Calculate capacitor to hold up current limited supply

I am designing a capacitor bank to hold up a current limited supply when a resistive load is switched by an NPN transistor to ground

The power supply is 2800V limited to 400mA and I require to draw 1.8A for a 15ms pulse.

How do I begin calculating the capacitor value I will need to hold the supply up with 150V maximum droop for 15ms (I'm looking for methods rather than answers, but it was easier to explain with the real numbers)

simulate this circuit – Schematic created using CircuitLab

• What's the power supplies reaction to an overcurrent situation? Oct 26, 2015 at 18:35
• when the current limit is reached, the voltage is reduced until the over-current ends - it is a bench top HV supply with the associated protection you would expect Oct 27, 2015 at 8:56
• What other behaviour would I expect rather than the fold back of the voltage - I wouldn't get any spikes of voltage unless there were reactive elements like inductors, is that right? Oct 27, 2015 at 11:15
• Well it might break down completely or switch on and off repeatedly, voltage spikes are unlikely. The thing why I mention this is, the supply might discharge your capacitor when the voltage breaks down, so that it has not the effect you want, in that case you'd need to add a diode to prevent that. Oct 27, 2015 at 12:10

For a capacitor, the charge $$\Q\$$ is proportional to the voltage $$\V\$$. $$\Q = C\times V\$$. Rearrange, $$\C=\frac{Q}{V}\$$.
Charge is current multiplied by time. $$\Q = 1.8\,\mathrm{A} \times 15\,\mathrm{ms} = 0.027\,\mathrm{C}\$$.
Voltage is $$\150\,\mathrm{V}\$$, from your question description.
So $$\C = \frac{0.027}{150} = 180\,\mu\mathrm{F}\$$.
Of course, this assumes the power supply contributes zero current during the pulse, and that the current of the pulse is a constant $$\1.8\,\mathrm{A}\$$. That doesn't match up with your circuit, which shows a fixed $$\2.8\,\mathrm{kV}\$$ source and a $$\2\,\mathrm{k}\Omega\$$ resistor, which will draw $$\1.4\,\mathrm{A}\$$, and the capacitor will do nothing. As Arsenal rightly points out in his comment, you should be concerned as to how your supply will actually react with that $$\2\,\mathrm{k}\Omega\$$ there.