Electronic – Calculate capacitance value and voltage rating in the event of ESD pulses

capacitancecapacitordischargeesd

I have a input power filter circuit.

It has ESD protection capacitors which help to protect the downstream circuit from ESD pulses.

ESD Capacitor

ESD Capacitors Specifications :

C0001, C0002, C0003, C0004 = 47nF, 100V, 10%, 0805

My ESD Pulse Specifications :

  • Contact discharge : +/-4kV & 150pF / 330Ohms
  • Contact discharge : +/-8kV & 150pF / 330Ohms
  • Air discharge : +/-15kV & 150pF / 330Ohms

Can someone please help me on how to do the calculations so as to justify that my capacitance value and voltage ratings of the capacitors are within the limits designed?

My understanding :

For example, I take the +/-4kV & 330pF spec

Q=CV ;

4kV * 150pF = 600nC

This charge will then get to the input ESD capacitors :

V=Q/C;

600nC / 47nF = 12.76V (And in this step, should I consider the capacitance value for the single cap 47nF or should I consider the equivalent series capacitance value of 23.5nF?)

So, my voltage rating of the capacitor 100V > 12.76V is appropriate for this pulse.

Is my approach correct?

My questions :

  1. The above calculation I have done for only positive 4kV. Can someone help me on how to understand when negative 4kV is applied and how to go about the calculation?

  2. I have justified only for the capacitance voltage rating. How to justify for the capacitance value? Can someone help me with the formula to justify the capacitance value?

Best Answer

4 kV pulse from 150 pF in series with 330 ohms

The charge is 0.6 uC (as you said) and, ignoring the 330 ohm resistor (which limits the peak current), that charge becomes (after applying the pulse) distributed between the 150 pF and 2 series 47 nF capacitors. Given that the 150 pF is really small compared to 23.5 nF (the series capacitance of two 47 nF caps), you can assume (without much error) that the charge is all "adopted" by the 23.5 nF.

This means that the voltage will be 25.53 volts. In other words twice the value you calculated because you need to regard the two 47 nF caps as being in series.

But, this doesn't accommodate the series diode and two more capacitors of 47 nF. In effect, the calculation above is valid for -4 kV. For +4 kV the capacitance is just 47 nF (ignoring the volt drop on the diode) because 2 series capacitors of 47 nF in parallel with an identical series connection of 2x 47 nF = 47 nF.

So, with a positive 4 kV pulse the voltage is 12.77 volts and, for -4 kV it is 25.53 volts.

Is my approach correct?

Broadly speaking yes. And, more importantly, it is conservative in that we have assumed a worst case of the 330 ohm resistor being shorted and the diode (D0002) being shorted.

Rating the capacitors at 25 volts or above would be OK for this example.

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