# Deciding capacitor value to use in my circuit

I have a power source which supplies 1A at 5V. I am using a switching regulator which supplies 1.3A at 3.3V. The load requires a burst of 2A without reducing the voltage across it than 3.2 V for 477 usec after every 5 minutes. The normal operating point is 500 mA at 3.3 V. I want to use a capacitor to provide that extra current for the required duration without reducing the voltage. How to choose proper capacitor value with low ESR?

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What are the line and load regulation on the 3.3V regulator? – Ignacio Vazquez-Abrams Jul 1 '14 at 14:01

You start with $$Q = CV$$ and differentiate both sides, $$\dfrac{dQ}{dt}= C\dfrac{dV}{dt}$$, which can be represented by $$I = C \dfrac{dV}{dt}$$, rearranging variables you get $$\dfrac{I}{\frac{\Delta{V}}{\Delta{t}}} = C$$

Using your values from above, $$\dfrac{2 [A]}{\frac{(3.3 - 3.2)[V]}{477 [us]}} = 9.54 [mF]$$

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Thanks for the reply. Can you please give the Resistor calculations also (there would be some small resistor in series). I want the capacitor to provide extra current only. The regulator will keep supplying normal operating current. – Ankit Jul 1 '14 at 14:15
You lost track of your decimal point there. The time is 477 us (not 0.477 us), making the answer 9.54 mF (not uF). – Dave Tweed Jul 1 '14 at 15:06
@DaveTweed thanks – placeholder Jul 1 '14 at 15:10
@Ankit the ESR is mainly important for heating calculations, but you are pulsing every 5 minutes so that should be OK. If the ESR is large then it eats into your 100 mV droop (30 mOhm * 2 A = 60) so choose caps with ESR in the single digit mOhm range and then scale the capacitor upwards a bit. Since there is tolerance issues on the caps, I'd say size it to be 2X the value above or better. – placeholder Jul 1 '14 at 15:15

The large capacitor values are a suggestion that this may not be the best way to solve the problem.

If you can rate the 3.3V regulator to supply 2A for short pulses, you can put the reservoir capacitors upstream of the regulator, where voltage stability is less important.

So another approach is to use similar calculations for a reservoir capacitor on the 5V rail, allowing the voltage to sag, perhaps by 0.5V, during the current pulse. This will allow a smaller capacitor. The 5V supply will briefly current-limit until the end of the pulse, then charge the capacitor back up to 5V.

But take care that this doesn't cause trouble for any other sensitive circuitry on the 5V supply.

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And make sure that the load and line regulation on the 3.3V regulator will keep the voltage high enough while the load current increases and line voltage decreases. – Ignacio Vazquez-Abrams Jul 1 '14 at 18:09
Thanks. I will go with this approach only. The capacitor values comes out to be close to 500 uF. – Ankit Jul 2 '14 at 6:18

If the regulator is too slow to supply significant current in the 477usec interval (seems unlikely) and you needed to supply the entire 2A additional current with less than 100mV droop you can easily calculate that the cap has to be around 12,000uF allowing 20mV for the cap impedance (10m ohms @2A).

Two 8200uF caps such as EEU-FC1A822L in parallel would give you 16,400uF with 8m ohm impedance.

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The regulator can supply the required current only if I use the input source which supplies more current than about 1.5A at 5V but my source supplies 1A at 5V. Can you please provide the calculations. – Ankit Jul 1 '14 at 14:23
What does the 1.3A mean then? I assume that it can't supply 2.5A at the output. – Spehro Pefhany Jul 1 '14 at 14:25
It can supply. The output power can't be greater than the input power. Since the input power is (5V*1A), the output power has to be less than that. With around 86% efficiency, the output current is 1.3 A at 3.3V – Ankit Jul 1 '14 at 14:26
It means you need a capacitor before the regulator to supply enough current to it for the half a millisecond (I'm pretty sure the 3.3V regulator can handle 2A for that long). – Ignacio Vazquez-Abrams Jul 1 '14 at 14:27
@Ankit: It looks like you're going to have problems with inrush current. The large 10-12 mF cap will look like a short circuit when it's discharged. When you power up the system, you'll get considerably more than 1.3 A into the capacitor, for short amount of time. How does your regulator handle an overcurrent condition? Hmm... Perhaps if it can handle the inrush it can also handle the output pulse? – bitsmack Jul 1 '14 at 16:48