1
\$\begingroup\$

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?

\$\endgroup\$
  • 3
    \$\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\$ – rackandboneman Jul 24 '17 at 22:42
  • 2
    \$\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\$ – Harry Svensson Jul 24 '17 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 Jul 25 '17 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\$ – rackandboneman Jul 25 '17 at 15:56
1
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.