What's up with this multiple capacitor stuff, anyhow? - Page 1
Forum section: electronics / beginners

author: Analog Kid
author_rank: Super Contributor
timestamp: November 16, 2024, 10:10:26 pm
content: Venturing into theoretical territory here: this is something that's bothered me for some time. I posted this some years ago in a different forum, but never got a really satisfactory answer. So here goes:Often we're advised to put more than one capacitor across supply rails like so:Reason being that the two caps shunt different frequencies of noise away from the load; HF by the small cap, LF by the larger one. (Pardon me if I didn't describe that exactly technically accurately: you all know what I mean here.) So let's say we do this:OK, fine: but my question is this: do we end up with anactualcircuit like this, because of the resistance of conductors?Or does it really end up being just this, where the small capacitance just gets absorbed into the much larger one? (Which is my suspicion.)After all, we're dealing with a vanishingly small resistance here on the order of milli- or micro-ohms, so basically negligible.How does this work? Color me puzzled.

author: TimFox
author_rank: Super Contributor
timestamp: November 16, 2024, 10:20:01 pm
content: If you consider only the ESR, you need to draw it at the capacitor top, not between the two capacitors.Electrolytic capacitors can have ESR in ohms, not milli-ohms.Worse than that:Each discrete capacitor has an internal parasitic inductance, which results in a series-resonant mode for that device.However, above the self-resonant frequency of the larger capacitance unit, its impedance is inductive (positive reactance).That inductance can parallel-resonate with the smaller capacitor, and you have a high impedance when you wanted a low impedance.Guess if I discovered that the easy way or the hard way?I had several JFETs in a 60 MHz amplifier system, where the active FETs were selected by applying negative voltage to the source bias resistor (resistor from JFET source to negative power supply).Each FET had a bypass capacitor, per the usual low-signal amplifier circuit.By chance, the two ceramic capacitors I chose produced a parallel-resonant mode at approximately that frequency.When the FET was active (source current flowing), the parallel resonance gave a high impedance at the frequency of interest, and a very low gain for the FET.When the FET was inactive (source resistor open), there was some coupling from the input through the gate-source capacitance to the parallel resonance, and some coupling thence to the next stage.Therefore, the gain in both states was very low, but the gain in the "inactive" state was actually higher than in the "active" state.Once we realized that, a suitable single capacitor at each FET source gave us the desired result.

author: SteveThackery
author_rank: Super Contributor
timestamp: November 16, 2024, 10:26:26 pm
content: In a normal circuit layout the resistance between the two capacitors is too small to be significant. The point is that the impedance of the large capacitor - especially if it's an electrolytic - can begin to rise at high frequencies, which is why it is common to bypass it with a low value capacitor of non-electrolytic construction, quite often a ceramic.So yes, it does end up effectively as a 1000.1uF capacitor, but one which has a significantly lower impedance at high frequencies than a 1000uF electrolytic on its own.That's the general idea.

author: Benta
author_rank: Super Contributor
timestamp: November 16, 2024, 10:41:38 pm
content: @Analog Kid,I suggest that you study some real capacitor data sheets.

author: Analog Kid
author_rank: Super Contributor
timestamp: November 16, 2024, 10:44:48 pm
content: Quote from: TimFox on November 16, 2024, 10:20:01 pmSo we're looking at something like this?Oy, what a mishegoss ...Is this true even at low-ish (i.e., audio) frequencies, or is this something only those RF guys have to worry about?

author: Analog Kid
author_rank: Super Contributor
timestamp: November 16, 2024, 10:48:41 pm
content: Quote from: SteveThackery on November 16, 2024, 10:26:26 pmWhat you write makes total sense.But I still don't see what prevents the small capacitance from just being "swallowed" by the larger one, thus preventing attenuation at higher frequencies. Wouldn't there have to be some kind of electrical separation between the two, other than an essentially zero-resistance conductor?I mean, we all know you can put two caps in parallel and you end up with a single cap of value C1 + C2.

author: David Hess
author_rank: Super Contributor
timestamp: November 16, 2024, 11:41:07 pm
content: The example you chose with the 7805 is correct but misleading. The bulk 10 microfarad electrolytic capacitor is suppose to be located near the 7805, however the 0.1 microfarad decoupling capacitor should be located near the load and not the 7805. If the schematic was drawn properly, then it would include the inductance of the relatively long inductive trace between the capacitors.

author: Analog Kid
author_rank: Super Contributor
timestamp: November 17, 2024, 12:57:58 am
content: Quote from: David Hess on November 16, 2024, 11:41:07 pmThus giving you a pi filter, right?

author: golden_labels
author_rank: Super Contributor
timestamp: November 17, 2024, 02:30:27 am
content: See this Dave’s video, in particular starting at 07:10:

author: Analog Kid
author_rank: Super Contributor
timestamp: November 17, 2024, 03:01:03 am
content: Very good explanation.Question, though: yes, there's parasitic XLin every capacitor. But it's tiny, so isn't the effect he's showing (rising impedance w/frequency) a bit exaggerated? What is the actual (real-world) effect of this inductance on that curve?Seems as if the gross wiring on his test rig there is going to contribute waaaaay more inductance than any short PCB leads ever will.

author: TimFox
author_rank: Super Contributor
timestamp: November 17, 2024, 03:53:20 am
content: Quote from: Analog Kid on November 17, 2024, 03:01:03 amThe parasitic inductance gives the capacitor a self-resonant mode.Good datasheets specify typical self-resonant frequencies for capacitors.With normal capacitors, this resonance is not trivial!

author: Analog Kid
author_rank: Super Contributor
timestamp: November 17, 2024, 04:36:13 am
content: Quote from: TimFox on November 17, 2024, 03:53:20 amYou're sure about that?I just looked at a bunch of caps, a dozen or so, at Digikey. Looked at their datasheets.Not a single one even mentioned resonance.(Well, one had something on mechanical resonance and mounting technique.)Can someone show me a datasheet with this info?

author: golden_labels
author_rank: Super Contributor
timestamp: November 17, 2024, 04:51:51 am
content: Dave conducts actual tests towards the end. The charts from Red Pitaya show the impedance increases by an order to three orders of magnitude. And that is tested to 40 MHz only, with a quick and poor setup.(1)So the effect is certainly not exaggerated.Keep in mind that inductance may be small, but impedance it exhibits may go arbitrarily high with frequency high enough. It’s just a matter of where it becomes significant enough compared to XCand ESR.Two notes on charts. Whiteboard charts are using logarithmic scales for both impedance and frequency, while Red Pitaya only for frequency. Dave has drawn it the more typical way. If both scales are logarithmic, that V-shape actually remains the same and only its position changes depending on parameters. Second note: the V-shape assumes ideal capacitance and ideal parasitic inductance (sic!). In practice there is more to it, attenuating the curve and making it more bathtub-shaped. Most notably series resistance and capacitance itself being dependent on frequency. Blame molecules making up a capacitor, which — in the most mechanistic sense — can’t keep up with the tempo.(1)See the first chart produced by Red Pitaya. It shows that the test setup itself already introduces lower impedance at high frequencies.

author: Bud
author_rank: Super Contributor
timestamp: November 17, 2024, 07:33:33 am
content: Quote from: Analog Kid on November 17, 2024, 04:36:13 amSome Murata capacitors. Note Self resonant frequencies vs capacitor value, other manufacturers' products would be similar.

author: Andy Chee
author_rank: Super Contributor
timestamp: November 17, 2024, 07:50:08 am
content: Quote from: Analog Kid on November 17, 2024, 04:36:13 amhttps://www.vishay.com/docs/28108/mkt370.pdfPage 9, graph titled "Impedance as a function of frequency"Note the deep notches indicate resonant frequency.

author: SteveThackery
author_rank: Super Contributor
timestamp: November 17, 2024, 09:34:54 am
content: Quote from: Analog Kid on November 16, 2024, 10:48:41 pmAnd that is still true. Electrically, two capacitors in parallel are, effectively, one capacitor. But by combining them in the way described, the net result is a capacitor that combines the best characteristics of both - ie a large capacity (courtesy of the electrolytic part) with a low impedance at high frequencies (courtesy of the ceramic part).

author: sleemanj
author_rank: Super Contributor
timestamp: November 17, 2024, 09:49:52 am
content: Quote from: David Hess on November 16, 2024, 11:41:07 pmThat is not a correct statement.I quote from the TILM7805 datasheet....QuoteAll the characteristics for the 7805 are with 100nF output to ground.While there is no argument that 100nF is a common decoupling capacitor and logic devices should generally have on their power rail(s) physically close to the device (inadditionto LM7805 output capacitor, if any), the 7805 datasheet is pretty clear that if you are using an output capacitor, that 100nF as close as possible to the regulator is sufficient, indeed, their recommendation.

author: Vovk_Z
author_rank: Super Contributor
timestamp: November 17, 2024, 10:15:35 am
content: In real life you need small decoupling (like 0.1uF) near a load but not near 7805 or other regulator. That's just my observation from making low-noise power supplies for low-noise audio (I mean low-frequency) stuff.Those 78xx/317 IC are totally fine with just electrolytes near them.

author: Avelino Sampaio
author_rank: Frequent Contributor
timestamp: November 17, 2024, 10:26:02 am
content: This video from our friend Alan demonstrates a practical example of what David Hass said, about the importance of the 0.1 microfarad decoupling capacitor being located close to the load.

author: Analog Kid
author_rank: Super Contributor
timestamp: November 17, 2024, 10:27:05 am
content: Quote from: Vovk_Z on November 17, 2024, 10:15:35 amI think that's one of the things that Dave's video showed quite clearly. At least for LF work.

author: m k
author_rank: Super Contributor
timestamp: November 17, 2024, 05:00:12 pm
content: For the subject, first picture.Energy start coming out, first it fills 100nF and then 10uF.Capacitors have V/t curves of charging.If OUT voltage start dropping the time of that V/t coordinate system is reversed.So energy start coming out from both capacitors.But time is still constant, so voltage of smaller capacitor is changing faster.And so new V/t curves of discharging happen from overall voltage difference.When OUT voltage start rising V/t curves are renewed and time reversed again.So if time between dropping and rising is short enough the bigger capacitor's V/t curve is hardly dented, smaller capacitor's V/t curve has a bigger change.But amount of energy exchange in time is also limited, so position of smaller capacitors are not irrelevant.

author: David Hess
author_rank: Super Contributor
timestamp: November 17, 2024, 05:53:46 pm
content: Quote from: Analog Kid on November 17, 2024, 12:57:58 amIt does, but that is not the purpose. See below.Jim Williams discusses that in connection with low noise switching regulators. The inductance of even a short trace is sufficient to filter high frequency noise when combined with two capacitors in a Pi configuration. Most linear regulators are not fast enough for this to matter.Quote from: sleemanj on November 17, 2024, 09:49:52 amThe recommendations changed over time. The oldest application notes only mention using a small ceramic capacitor with the low power versions of these regulators to reduce high frequency noise. Sometimes a small ceramic capacitor is shown, and sometimes it is not, and it was practically never discussed. The bulk output capacitor was always optional.The bulk decoupling capacitance becomes part of the frequency compensation while its ESR adds a zero for phase lead to make up for the phase lag caused by its large capacitance combined with the regulator's output resistance. This is more important for low dropout regulators with common emitter outputs, especially at low current. Its ESR also acts as a rough termination for the PDN impedance quashing resonance in the PDN. A zero ESR decoupling capacitor cannot do any of this. High reliability circuits and low dropout regulators take other steps so that large ceramic or film capacitors with essentially zero ESR will work.The decoupling at the load requires the lowest ESR, and not so much capacitance. It is straightforward to calculate the needed decoupling capacitance for a given load based on charge transfer and allowable ripple, and 0.1 microfarads is usually higher than needed, but it is rarely worthwhile to go to this much effort when using cheap 0.01 or 0.1 microfarad capacitors everywhere almost always works just fine.So what might be going on here? There are three cases, and I think I can use one hypothesis to explain all of them. This ignores low dropout regulators that have special output requirements, and standard negative regulators which have a common emitter output making them related to low dropout positive regulators.1. Old application notes say to use small decoupling capacitors at the loads.2. Old application notes say to use small decoupling capacitors at the regulator output with low power regulators to reduce high frequency noise.3. New application notes say to use small decoupling capacitors at the regulator output.How designers use these regulators changed over time, and I think the application notes changed to reflect it. In the past the regulator was more likely to be located remotely, while now it is more likely to be used as a point-of-load regulator or in very close proximity to its load. Low power regulators have always been more likely to be used as point-of-load regulators.So if I take the old recommendation of bulk output capacitance optional, decouple at the load, and apply it were the regulator is used as a point-of-load regulator, then all three recommendations simplify to the same thing. The mandatory decoupling capacitor located at the load is now at the regulator's output because of proximity, and all three cases become the same thing.My own tests years ago found that the small ceramic decoupling capacitor had practically no effect on the regulator when the output bulk capacitance was present, and if it was the only output capacitance present, could make transient response worse. Jim Williams in Linear Technology application note 47 reported that some combinations of electrolytic and small decoupling capacitance would promote ringing but I do remember if this was ever explained.

author: Bud
author_rank: Super Contributor
timestamp: November 17, 2024, 07:19:24 pm
content: Quote from: David Hess on November 17, 2024, 05:53:46 pmThis is because self-inductance of capacitor A now resonates with capacitance of capacitor B, and vice versa.There is a paper i have somewhere with VNA measurements of parallel capacitors, which is the source of truth because stuff was actually measured, not just theoretically speculated. I'll post if i find it.