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I have an electronic component with following characteristics:

Operating voltage: 4.5V
Peak operating current: 2000mA for 600uS every 4000uS.

The device can tolerate voltage drop of upto 400mV during the peak current burst.

I want to use a capacitor (placed close to the component) to supply the entire peak operating current requirements so that the burst current does not have to flow all the way from main source to the component. Also this will give me the option to use a source of lower output current capacity instead of one that can handle peak current requirements.

Assuming the capacitor's ESR to be <= 0.1 ohm. How do I come up with an optimal capacitance and voltage rating of the capacitor?

[I am seeking a theoretical answer, I will add some headroom while actually selecting the component.]

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  • \$\begingroup\$ @Brian Drummond The problem definition is not non sense unless you assume things. The device operates in different modes. Most of the time the device is in idle mode. At that time it consumes 200mA of average current. During moments of activity, the consumption shoots up to 2A 600uS pluse every 4000us. The average current during this stage is different and I have not specified it here (but it can be calculated if needed from the given data). Anyway average current consumption during idle stage is not required here. So I'll remove it to avoid confusion. \$\endgroup\$
    – Dojo
    Mar 24 '14 at 14:13
  • \$\begingroup\$ That makes it clearer : it is just the peaks you need to cater for. Will remove earlier comment. \$\endgroup\$ Mar 24 '14 at 14:20
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Capacitor Discharge through Constant Current Source

From the link, you can get the equation for constant current through a capacitor. This allows you to assume a worst case spec of 2 Amps for the entire 600uS.

Assuming you can virtually fully charge the capacitor in the off time (15% duty cycle).

\$v(t)= \frac{1}{C}I t + v(0)\$.

\$ C = \frac{It}{v(t) - v(t_0) }\$

Plugging your numbers in, without taking into account the 0.1 ohm ESR you should end up with 0.003F. Keep in mind current is negative since you're discharging it. With 0.1 ohm ESR taking out 200mV of voltage and assuming the supply voltage is only 4.5V, you'll only have 200mV left to play with. Plugging in 4.5V-4.3V you'll end up with needing 0.006F.

If you don't want the worst case for a resistive load and instead want to go with something more accurate to save on cap costs you could use the RC equation: \$V(t) = V_0(e^{-t/ RC}) \$ With that, you would have to integrate it and then solve for C. For R, you would use 4.5V/2Amps which you sort of implied was your maximum current.

Also, the original question asks what voltage the capacitor should be. Good design should be rated 25-50% above what's expected. So it should be rated at ~5.6V to 6.75V.

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    \$\begingroup\$ What about the ESR? 2A through 0.1 ohm is 200 mV, which uses up half of your allowed delta-V. \$\endgroup\$
    – Dave Tweed
    Mar 24 '14 at 15:12
  • \$\begingroup\$ Good point, I'll update the answer to adjust for this. \$\endgroup\$
    – horta
    Mar 24 '14 at 15:14
  • \$\begingroup\$ Using exponential RC decay to calculate the capacitor value won't buy much headroom in this case when compared to just assuming a constant 2A, since at 4.1V the current would still be about 91% of what it was at 4.5V. In cases where the voltage dip is small compared to the working voltage one can usually assume constant current and come very close to the true answer with much less math. In this case the extra math will only buy about 5% smaller capacitor. \$\endgroup\$
    – user4574
    Apr 25 '16 at 20:05

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