# What are the considerations for designing a capacitor bank?

I am trying to understand capacitor banks and capacitor discharge, and how to work out what I need to generate a short pulsed current.

If I would like to use capacitor discharge to produce a sudden burst of current in a circuit. This would be triggered by a MOSFET acting a switch connecting the load. What do I need to consider when selecting the capacitor, and what determines whether I need a capacitor bank, and what configuration it is, series or parallel?

I am looking at pulsing a 40 A current through an inductive coil (2 uH) and would like a rise time for the current of less than 1 us, which would require 80 V to drive it according to my calculations for the coil.

When considering the design of the capacitor load, I think the first thing is to make sure it can handle the voltage supply. The capacitance will determine the charge and discharge time in combination with the resistance of the circuit.

What other factors do I need to consider? What dictates whether one needs to create a bank of capacitors, or if a single capacitor would be sufficient?

And can the voltage source be left connected to the capacitor load, while the system is operating (switching)? Or would be a one-shot type of thing, charge the capacitor, disconnect, press the switch for the capacitor to discharge into the inductive load, reset?

• You need to consider the capacitance. How much energy do your capacitors need to dump into the coil? Work out the equations to tell what the capacitance is that you need to store that much energy at your desired voltage. Be careful about resonance, make sure your circuit doesn't end up oscillating. – Hearth Oct 23 '18 at 0:40
• Also if you are -stopping- the pulse just as suddenly (you don't mention a fall time) you need to consider polarity reversal on the capacitor bank. Polarized capacitors (aluminum, tantalum) dielectric films degrade when reverse biased. If it's just a charge dump, you'll probably be OK. – isdi Oct 23 '18 at 0:58

Vary basic formula: $$u_L=\dfrac{di_L}{dt}$$ will give you 80V. But this is just inductor voltage not the supply and the rest of the circuit.