I realized that I don't have much clue when it comes to bulk capacitance

[SiBurning]

All of the references you provided have the equations for minimum C, not one of which involves ESR.

You say C isn't important, but what happens if you don't have enough of it? Same for Ir? The voltage drop is going to be a lot worse than what you're getting from ESR.

You need to meet all of the criteria, ESR being only one of them.

[shadewind]

All of the references you provided have the equations for minimum C, not one of which involves ESR.

LM22674 lists the equation for the output voltage ripple which is a function of the inductance, ESR, frequency and capacitance.

LM2738 lists the exact same equation.

LM2675 doesn't list any technical details on choosing an output capacitor, it just has a table with recommended parts.

The TI and AVX app note I quote above supports what I'm saying as well.

You say C isn't important, but what happens if you don't have enough of it?

I never said C is not important. I was saying that you do not need a lot of capacitance for output filtering. By the time you've met all the other requirements, you usually have more than enough capacitance. What happens if you don't have enough depends on how much you're lacking.

Same for Ir?

Then you're going to have heat problems which breaks or reduces the lifetime of the capacitor.

The voltage drop is going to be a lot worse than what you're getting from ESR.

Do you mean the voltage drop from the lack in capacitance or current capability? Whether lack of capacitance is a bigger problem than ESR depends on how much you're lacking. Heat generation in the ESR is what mainly causes the limits on current.

You need to meet all of the criteria, ESR being only one of them.

Of course. My point is that voltage ripple being too high in a switcher design is rarely caused by lack of capacitance. ESR usually to blame.

[SiBurning]

All of the references you provided have the equations for minimum C, not one of which involves ESR.

LM22674 lists the equation for the output voltage ripple which is a function of the inductance, ESR, frequency and capacitance.

There are also equations for C in some of those app notes & data sheets.

[shadewind]

All of the references you provided have the equations for minimum C, not one of which involves ESR.

LM22674 lists the equation for the output voltage ripple which is a function of the inductance, ESR, frequency and capacitance.

There are also equations for C in some of those app notes & data sheets.

There are two different things that impose limits on capacitance. Many switchers have a fixed internal control loop compensation which assumes certain things about the external components. These assumptions place limits on the capacitance and you need to keep within these limits to ensure control loop stability, that is, to avoid oscillations.

The other thing is simply how much voltage ripple you can tolerate. If we assume an ideal capacitor with no ESR and ESL and only consider the effect of capacitance on the output voltage ripple, TI has a formula in their app note (as you say above) which gives the minimum capacitance for a given voltage ripple:

C = IL / (8 * fs * Vo)

Let's say our switcher has an average load current of 1A. A good tradeoff for component size vs. ripple current (IL) is to the IL at about 30% of the average load current. 500 kHz is a pretty normal frequency for switchers these days so let's use that. A maximum ripple voltage of 5 mV should be more than enough for most any purpose. Let's calculate the minimum capacitance based on these values:

C = 0.3 / (8 * 500e+3 * 0.005) = 1.5e-5 = 15 µF

That's not a lot of capacitance needed!

Let's pick a real electrolytic capacitor with a value as close to this as possible: http://industrial.panasonic.com/www-data/pdf/ABA0000/ABA0000CE114.pdf

We pick the 35 V, 22 µF one. The ESR is 0.36 ohms. We can calculate (or at least approximate) the voltage ripple using the following formula:

V = IL * ( ESR + 1 / (8 * f * C) )

We can rewrite this as:

V = IL*ESR + IL / (8 * f * C)

There are two linear terms, one which depends on the ESR and another which depends on the capacitance. The capacitance (or lack thereof depending on how you see it) term gives this much voltage ripple:

Vc = IL / (8 * f * C) = 0.3 / (8 * 500e+3 * 22e-6) = 3.4 mV

...and the ESR term gives this much voltage ripple:

Vesr = IL * ESR = 0.3 * 0.36 = 108 mV

The ESR term is greater than the capacitance term by a factor of approximately 32! It's certainly not what you would expect though.


Edit: It seems that the forum software cannot handle characters such as delta signs and ohm signs so you'll have to excuse the rather incorrect notation

[SiBurning]

That's exactly what I expect after you've reduced the ripple current due to capacitance. Capacitance is selected based on a tolerable ripple current. If you're not happy with the impedance loss in the capacitor, choose a lower... err... higher tolerable ripple voltage or find a cap or caps with lower impedance. (And don't forget to derate the cap for high frequency--something I have to leave to others to advise.)

Using capacitance is fine and good for the steady state. For dynamics, the Ir handling is also important. Just be aware that you want an Ir handling value for the capacitor based on what the circuit would do without the capacitor because this value specifies the limits to the smoothing & sourcing properties of the capacitor. It's not the same as the resulting in-circuit ripple current.

What's the question again?

[SiBurning]

It's kind of hard to figure out your perspective. A little glimmer of light shone through while cooking dinner, but not enough to formulate it clearly. It seems like you've fixed the 3 most important things (V, Ir, C)--or put another way, that those are known problems--and are now saying that the next one on the list (ESR) is the deciding factor.

That doesn't explain why you're still asking about bulk capacitance. You've more or less solved that, and the math shows that it's no longer an issue. Still, I only see this way of thinking as a viable position for the output capacitor, and only on the basis that the output of the regulator--and therefore, the input to the capacitor--is perfectly encompassed by that equation. It should be known, since it is a voltage regulator, after all. But you don't have this perfect information for the input side, and maybe not in the other circuits.

In this context, everything you've been saying suddenly makes sense. Hoping that helps clear up some confusion.

The chart jahonen posted seems very relevant to me at this point.

[shadewind]

The key here is to differentiate between switcher output filtering and general bulk capacitance. Though you only need 15 uF to filter the ripple voltage, you still need more capacitance to handle load transients due the fact that when the load suddenly changes, the voltage will dip due to the inductance along power supply chain - the current cannot change that fast. That's why you add more bulk capacitance, to handle load transients. The more capacitance, the lesser dip in voltage.

But I don't know how much to add! There has to be some approximation formula or rule if thumb based on expected load transients?

So as you say, there's been some misunderstanding. I have my output filtering figured out but I am a bit fuzzy when it comes to figuring out how to handle load transients. So far, it's been mostly guessing.

[SiBurning]

I'd refer to the difference as light vs heavy load current. At least in terms of what the capacitor's doing. For transient purposes, I certainly don't think of the filter cap as part of the power supply in the heavy load case, but rather as a secondary source of current. I always model this in terms of current demands, i.e. a capacitor keeps a stable voltage by supplying needed current when (the rest of) the supply can't. (Maybe my reply #23 will make more sense in that context.) You need to figure what the worst case current demand will be, or rather, the worst case you care to deal with.

There's no simple formula or rule of thumb. Mostly guessing sums up the process pretty well. If you want to do better, You need to figure out what expected load transients you want to handle, and pick a set of caps that can handle the current burst. Basically, what everyone's been saying all along.

Ask some questions... When the load changes to what? When you power on a computer, all the devices start up at the same time, causing a burst in current demand. A speaker, for physical reasons, might suddenly present a 0.5-0.25 ohm load at some frequencies. (The main current demand in audio comes from the signal, not from impedance changes, but it makes little difference except that the relative stability of load impedance--a false simplification--explains why power supplies are generally unregulated.) A blinking light might require a short spike of current every time it turns on. Do you have some similar situations?

The difference between high load and low load can be thought of as an additional ripple current on the cap since it may have to supply the current deficit. This is why the Ir specification for a capacitor is so important, possibly more important than the capacitance.

I generally make 3 main decisions in specifying capacitors.

• ripple current-- You can choose capacitors where the peak instantaneous current is less than the rated ripple current handling, or figure out the rms (of any recurrent peaks and the average of random peaks) and select accordingly. Which to choose, or where in between, depends more on voodoo than reality, but boils down to a typical case of tight tolerance vs over-speccing, but any heat will cause some... let's just call it nonlinear behavior.
  
• capacitance-- Select capacitance based on how far along the discharge curve you want to be after a transient, and how fast you need to charge & discharge. You obviously can't flatten the cap at any point, and probably want to stay in the linear portion of the discharge curve, say 5% for some applications and more for others? Another, admittedly controversial, point is just how fast can any particular cap source current, and do you need various sizes to supply current at different speeds? There's a (indefinite?) relation between this speed and the selection of ripple current handling.
  
• ESR & leakage current--An audio power amp might suddenly face a 0.5-0.25 ohm load, so it might be wise to have a low ESR so as not to dissipate too much power in the caps. There's not much sense in constantly wasting power & losing current through the cap. On the other hand, higher ESR and some leakage can actually stabilize things in other places.
  

If you want to think of load as purely inductive--I tend to think of load more in terms of current demands--and worry about how a drop in load causes more of a voltage drop internal to the supply (with the filter cap modeled as internal), then the simple answer is to make sure the supply impedance is much lower than the heaviest load. Of course, that's really the whole job of the chip or electronics in the supply.

[Conrad Hoffman]

I believe one test commonly done with switchers is to attach an AC coupled signal generator, often with an added power transistor and limit resistor, to the output, and hit it with a square wave or pulse. How the switcher handles the transient current draw is indicative of how well the feedback and compensation has been set up. This same test should tell you if the bulk capacitance is sufficient, since they all work together. Other than that, IMHO, there are too many unknowns to bother calculating it.

[shadewind]

I should probably add that all my calculations and discussions regarding switchers have been about buck switching regulators. Much of it does not apply to, for example, boost converters where the output of the inductor is switched as opposed to the input as is the case in a buck regulator.

The difference between high load and low load can be thought of as an additional ripple current on the cap since it may have to supply the current deficit. This is why the Ir specification for a capacitor is so important, possibly more important than the capacitance.

For a buck regulator in continuous mode, the ripple current doesn't change with the load, it is only dependent on the inductance (assuming fixed duty cycle). At low loads, you enter discontinuous mode but that's another topic. Because of this, it doesn't matter if your load current is 0.5 A or 1A, the ripple current the output capacitance has to handle is constant. What does matter is when the suddenly changes, that's when you have voltage dips or voltage spikes.

I believe one test commonly done with switchers is to attach an AC coupled signal generator, often with an added power transistor and limit resistor, to the output, and hit it with a square wave or pulse. How the switcher handles the transient current draw is indicative of how well the feedback and compensation has been set up. This same test should tell you if the bulk capacitance is sufficient, since they all work together. Other than that, IMHO, there are too many unknowns to bother calculating it.

Yes, this sounds like a very good idea, simply measure it. But when you're making something as a hobbyist, it's a bit of a pain to first make a board and find out that the capacitance didn't cut it and then have to make another board. Of course, if you're using parts that have standard footprints (mainly ceramics and leaded electrolytics) you can always use parts with more capacitance unless you have to actually add more of them.

I suppose the only thing you can do is add what you believe to be enough and then a little more and measure until you develop a feel for it?