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How much capacitance do I need on a 3.3 V rail with loading of ~500 mA?

I have 22 µF capacitors that I can use.

In general, what is a good rule of thumb on how much capacitance is required based on a voltage rail and its current value?

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    \$\begingroup\$ We need a lot more information on your application. What are you trying to accomplish? \$\endgroup\$ – HikeOnPast Dec 12 '12 at 5:44
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    \$\begingroup\$ It depends on the load and switching frequency. In some cases, the layout and place of caps becomes more important than the quantity. As HikeOnPast said, need more details. \$\endgroup\$ – jbarlow Dec 12 '12 at 5:55
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    \$\begingroup\$ @user1406716 - Your question as it is now is basically "I'm driving somewhere. How much gas do I need?" You will need to give a lot more information about exactly what is on the 3.3V rail, what magnitude spikes you're seeing , what the layout is like, etc... before anyone can really answer your question. \$\endgroup\$ – Connor Wolf Dec 12 '12 at 8:11
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    \$\begingroup\$ The rules of thumb you are after depend mostly on other factors than the voltage and total current. Most important is the rate and duartion of current change(s), the response of your power supply to such changes, and the effect the resulting voltage variations will have on your components. 'Microcontroller' is some indication (but why not give the type?), but 'FET' could mean anything from switching a 2mA LED to PWMming a 100A motor. But as a 0-knowledge starting point I would say: 100nF near the micro-controller, and 22u somewhere on the board. Better answers require better info. \$\endgroup\$ – Wouter van Ooijen Dec 12 '12 at 10:25
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    \$\begingroup\$ How many caps? All of them! \$\endgroup\$ – user3624 Dec 12 '12 at 15:10
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There is no "rule of thumb". Cutsey shortcuts will get you into trouble more often than not anyway. There is no substitute for actually knowing what you are doing. When you do, you can easily derive the answer.

You have to ask yourself what the purpose of the capacitance is. Since you said "3.3V rail", it seems like this power supply is being driven by a voltage regulator of some sort. Look at its datasheet and put whatever capacitance on the output it says.

Some older low dropout (LDO) regulators assumed the best caps you could get were tantalum. These have some minimum equivalent series resistance (ESR), and the designers of the regulator took this into account when compensating the control loop. As a result, some of these regulators require a minimum ESR to be stable. Nowadays you generally use a ceramic cap on the output, which have very little ESR, and usually none of it guaranteed. More recent regulators have been designed with this in mind and are "0 ESR output stable".

Anyway, do what the datasheet says, including paying attention to the type of cap or its ESR and its physical placement relative to the regulator. That should be fine for the basic power supply. Then at each digital chip that uses this power supply, put a small ceramic cap (100nF - 1µF) directly accross its power and ground pins as close as possible to the chip. This is called a "decoupling" capacitor, and is for a different purpose than the larger capacitance at the output of the power supply.

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Capacitance for what?

If your answer is ripple filtering then, yes, there is a equation for that. No matter how large a ripple capacitor you use, you will have an unavoidable level of ripple. So we have to talk about maximum allowed ripple voltage here.

Here is the equation:

\$ C_{min} = {Iout \times dc \times (1-dc) \times 1000 \over f_{sw} \times V_{P(max)} } \$

Source: http://www.ti.com/lit/an/slta055/slta055.pdf

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In its current form, your question has no real, good answer. There is no magical formula that gives you a good amount of capacitance to have on a board given a voltage and a load current.

You could be bypassing the output of a voltage regulator or power module... a linear regulator might specify 10uF at the output to be stable while an integrated power module might want 100uF at its outputs. An accelorometer IC I've worked with draws less than a milliamp at operating current but specifies a 4.7uF decoupling cap to better handle noise on the supply rail. As you can see, there isn't always a relationship between load current of an IC and desired capacitance for bypassing it. You have to do a little reading about the things you're bypassing.

So... the first thing you should do is check the datasheet. If they don't specify anything, and they usually do, then I'd put 1uF cap for every power pin on the IC. Put as small a package as you can reasonably assemble (0603s aren't hard to hand solder with a good iron and some patience) and put it close to the pin. Connect it to ground with a short via. This will get you in the ballpark of where you want to be... in terms of proper decoupling.

Don't forget, though... there's no substitute for understanding the powers at work here and learning why and how much bypass capacitance you should have. Our best resource on the subject thus far as another question about decoupling capacitors: here

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