Please inform what ESD estimation software you choose . How to find voltage drop across 200pF capacitor ?? any formula.

Hallo J.

I did not use an estimation software for the analysis, only 'calculator Nr.1', means my brain.
Estimation of voltage drop is rather easy. Have a look at the schematic again:



At very first moment of ESD event, means exactly in that moment, when the contact is closed, C1 discharges into C2 and C3. The according current spike is extremely short lasting and extremely high. We can estimate it very roughly, when assuming, that contact time of a switch, or any two pieces of metal which are brought together, can be less than 1nsec. From well known C = (dQ/dt) / (dU/dt) we can estimate: I = dQ/dt = C x (dU/dt) = 1.98pF x 8kV/1nsec = 15.8A

Why 1.98pF? 8kV drops across series capacitance of C2 = 2pF and C3 = 200pF, which is 1 / (1/2pF + 1/200pF) = 1.98pF.

So, at very first moment more than 15.8A is flowing through C1, C2 and C3.

It's worth to note, that there's NO relevant resistance, which could limit this current flow! Current limiting is only caused by finite rise time of voltage at contact. Keeping in mind, that our estimated contact time of 1nsec is only a very rough approximation, we must assume even much higher currents flowing.

How long is 15.8A current flow lasting?
C1 does not totally discharge, when contact is closed. While C1 is discharged, C2 and C3 are charged, and very first current flow stops, when voltage drop across C1 equalls voltage drop across series capacitance of C2 and C3, means 1.98pF. Because charge Q is constant during discharge, voltage of equilibrum can easily be calculated: Uequi = Q / C = C1 x 8kV / (C1 + 1.98pF) = 7.922kV.
So, duration of current flow is about: dt = dQ / I = C x dU / I = 1.98pF x 7.922kV /15.8A = 990psec.

We can conclude: In our example, ESD event causes a current spike of more than 16A, which lasts less than 1nsec. (Estimation is very rough and current spike can be much bigger and shorter!)

Up to now, we assumed, that very first ESD current is not influenced by other components. Is this true?
After about 990psec C2 is charged to 7.843kV and C3 to 79V. 7.843kV / 11k = 0.7A and 79V / 100R = 0.79A, which means, that 15.8A current flow through C2 and C3 is highly dominating. So, it's true.

One word to parasitic inductivity: At very first moment, when voltage drop across C2 is still zero, it prohibits any current flow through 11k resistor, according to U = L x dI/dt = 0. When afterwards voltage drop across C2 increases, this voltage drop cannot totally equal voltage drop across 11k resistor, but a certain part must also drop across parasitic inductivity L. So, at very first moment of ESD event, current flow through 11k resistor is even smaller than estimated above.

After the first about 990psec, when 'equilibrum' has been reached, voltage drop across C2 and C3 begins to decrease again. Exact calculation of decay is a bit complicated, but fortunately not really necessary. We are just looking for the biggest time constant and will at least have an idea of total decaying time constant.
C3 discharges along 100R resistor and C2 along 11k resistor. Both pairs give a time constant of about 20nsec. C1 also discharges along 11k resistor and gives a time constant of about 2.2µsec. But this time constant is not really relevant, because after about 50nsec 4V7 zener diode has turned-on and limits current through opto-LED to less than about 35mA.

Bye,
Kai