I am working on a powerboard that may need to switch up to 2kA and while working on some simulations the question occurred to me. How much current can flow in/out of a ceramic smd capacitor when constrained to a very small time frame? For example; the capacitor is 10uF and 100V and the pulse is 0.1-1us, which equates to a frequency of 1-10MHz, and in the datasheet for the capacitor this gives an ESR and characteristic impedance of about 0.01 Ohms. Staying within the 100V rating, this gives a peak current of up to 10kA. Will capacitors actually allow current at that level, even for such a short duration, or is there some other factor that comes into play?

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    \$\begingroup\$ ...What about wiring inductance? 10kA at 100V is a megawatt, it is questionable if any SMT circuitry will suffer that for even a microsecond or turn into .... google "exploding bridgewire detonator". \$\endgroup\$ Commented Jul 24, 2017 at 22:42
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    \$\begingroup\$ Let's assume that the capacitor and the wiring to & from the capacitor got a total of 1mΩ real resistance (no imaginary part). 10kA => \$10×10^3\$, and \$P=I^2×R\$, let's plug in the values and see how unrealistic it is. \$(10×10^3)^2×10^{-3}=100kW\$ Yeah... I don't think you want to do this. With 10mΩ you get 1MW, same as rackandboneman. A detonator. \$\endgroup\$ Commented Jul 24, 2017 at 23:33
  • \$\begingroup\$ @rackandboneman that is exactly why I was skeptical. I believe the factor I forgot to take into account was the inductance of the pads and other PCB features, which would have a very strong limiting effect with pulses this large and fast. \$\endgroup\$
    – Redja
    Commented Jul 25, 2017 at 12:48
  • \$\begingroup\$ Imaginary part of resistance will not dissipate - nor consume - the energy... that will happen in real resistance generating real heat, delayed as it may. If it is sufficiently delayed/stretched to not cause such a heat pulse, your circuitry doesn't work - if it isn't, your circuitry doesn't exist (afterwards). \$\endgroup\$ Commented Jul 25, 2017 at 15:56

1 Answer 1


Assume 10nH total inductance. Given V = L * dI/dT, or conversely, I = 1/L * integral (V * dT),

the peak I = 1/10nH * integral (100V * 100nS) = 10^+8 * 100 * 1e-7 = 1,000 amps.

If your total inductance (caps plus solder pads plus GND inductance of vias + vias to a shared high current bus) is 100nH, the peak current is only 100 amps.


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