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 :


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. And 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?

  • \$\begingroup\$ I would certainly use a simulation tool for this (as I would for your later question on inrush current). \$\endgroup\$ – Andy aka Apr 24 at 8:51
  • \$\begingroup\$ Yes I would use a simulation tool too. But I would just want to understand / calculate it theoretically before \$\endgroup\$ – Newbie Apr 24 at 9:01
  • \$\begingroup\$ Why are you placing these 47N caps in series? You should place them all parallel. \$\endgroup\$ – Huisman Apr 25 at 20:11
  • \$\begingroup\$ With respect to what do you apply the ESD pulse? \$\endgroup\$ – Huisman Apr 27 at 13:14
  • \$\begingroup\$ With respect to ground \$\endgroup\$ – Newbie Apr 27 at 14:20

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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  • \$\begingroup\$ Thank you for the answer. But 2 questions. 1. At -4kV, the diode blocks and only the first set of capacitors get impacted. So, 25.53V get accumulated. But the bottom capacitor is connected to ground right. So, my understand is that, the 25.53V will equally split into 12.76V across each cap, with the bottom side being +ve with respect to the top side. So, the charge will immediately get a path to ground through the bottom resistor. Am I correct? And my second question, could you please help me to check value of the capacitance whether the chosen is appropriate? Please help me with that formula. \$\endgroup\$ – Newbie Apr 24 at 9:56
  • \$\begingroup\$ Bottom resistor??? What formula??? \$\endgroup\$ – Andy aka Apr 24 at 9:58
  • \$\begingroup\$ Sorry, What I meant by Bottom resistor (It was actually bottom capacitor, Let me edit that) - "How is the voltage accumulated in each capacitor when -4kV is applied (I mean the polarity of the voltage across each capacitor after the pulse is applied) and how will that 25.53V get discharged to the ground? And the second question is, similar to the steps above to validate the voltage rating of the capacitor, is there a way / formula / calculation to validate my selected capacitance value? \$\endgroup\$ – Newbie Apr 24 at 10:15
  • \$\begingroup\$ The polarity of each capacitor follows the polarity of the applied voltage. The -25.53 volts will remain -25.53 volts until something discharges it. It can't be discharged through the diode so, unless you have a bleed resistor it remains. The formula is the one you (and I used) and that is that charge is (always) conserved. \$\endgroup\$ – Andy aka Apr 24 at 10:22
  • 1
    \$\begingroup\$ No, I said charge conservation; energy is NOT conserved. There is no final capacitance other than what you have in your diagram and that is a bunch of 47 nF capacitors - they share the 0.6 uC charge with the 150 pF capacitor when a connection is made. \$\endgroup\$ – Andy aka Apr 24 at 11:09

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