Electronic – the maximum voltage that the circuit can withstand

capacitoresdhigh voltagevoltage

Below is my circuit,

ESD Capacitors Specifications :

C0001, C0002, C0003, C0004 = 47nF, 100V, 10%, 0805
I have 2 basic questions :

  1. When a +4kV pulse & 330pF is applied, Does the 12.77V (Q=CV ; Q =4kV * 150pF = 600nC. Voltage at the top node = 600nC / 47nF = 12.77V) is shared by each capacitor separately (C0001 to C0004)?
    So each capacitor will hold 3.1925V?

  2. And what is the maximum voltage can this circuit section withstand in the case of positive pulse? Is it 4 x 100V = 400V ? Or only 2 x 100V = 200V (since the 2 sets of series capacitors are in parallel). 100V is the voltage rating of the capacitor.

Best Answer

There will be 6.38 volts across each capacitor (as previously answered here in comments): -

The 12.77 volts will appear where you have an arrow called "ESD pulse". If you ignore the small forward volt drop of the diode you can say that there is 6.38 volts across C0002 and C0004 and 6.38 volts across C0001 and C0003. If the rating is 100 volts then the capacitors can withstand 100 volts on each meaning 200 volts at the point marked "ESD pulse". This assumes the capacitors are perfectly matched. If they are mismatched by 10% i.e. one is 51.7 nF and the lower one is 42.7 nF then there will be proportionately more voltage developed across the capacitor with lower capacitance.

This is because the 12.77 volt seen during the pulse (as previously derived in my answer here) is shared equally between two series capacitors.

Given that the capacitors have a voltage rating of 100 volts, if they have the same value then the peak voltage withstand for two in series is 200 volts.

If one capacitor is low in value by 10% and one is high by 10% there will be 20% more voltage seen across the lower value capacitor hence, you can't really assume a 200 volt pulse withstand capability and it will be more like 160 volts.

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