Bulk input caps and impedance - Page 1
Forum section: electronics / beginners

author: paulca
author_rank: Super Contributor
timestamp: April 26, 2024, 10:49:46 am
content: A split out question from my other project. Trying to pick out some of the basics I might be missing.If I put a cap directly across a DC barrel jack input, then when I connect the power to it, I will get a spark and the supply will basically see a 0 Ohm short for a fraction of a second and a load which seems to want to pull an ideally infinite amount of current.So, it seems to make sense to put at least a "token" value resistor in series before it. That or one of them magical to me, inductors.As it is a bulk input cap, it has to be able to pass at least the average current of the load.If I place an upper limit on the PSU of 200mA and choose a quarter watt resistor, then using ohms law....I want a resistor that will drop 0.25W maximum across it at 200mA under normal operating conditions.P / I = V0.25W / 0.200A = 1.25VR = V / IR = 1.25V / 0.2AR = 6.25OhmNow, for the "empty cap" short. Initially the resistor will be dropping all 36V of the input. So the "instantaneous" current through the resistor is limited toI = V/RI = 36V/6.25OI = 5.76AP = V*IP = 36 * 5.76P = WAY TOO MUCHHowever. It's transient. How long does it take for that volt drop to reduce as the cap charges?Well, we know that initially we would be putting 5.76A into the capacitor. This I believe can be input into the capacitor charge equation and produce a charge time / curve.I think this is the end of the line and as far as I can go, without dusting off the simultaneous equations I haven't explored in quite a while!As the cap charges the voltage drop across the resistor reduces... so the current passed with reduce as well.With 5.7A in and 0.2A out and, say, a 220uF cap. How does one calculate the total transient power dropped across the resistor and how do you determine if it's "sensible"?Can I assume that a quarter watt resistor will take a few micro seconds of 100W+?

author: paulca
author_rank: Super Contributor
timestamp: April 26, 2024, 10:55:29 am
content: If I was to use an inductor... as I am basically trying to "smooth" the current...Then am I correct in saying the ideal capacitor will "instantaneously" appear as 0Ohm, thus pull an ideally infinite current -ESR which will produce a "rising edge" with a VERY high frequency."Inductors effect voltage to resist a charge in current."So on the DC barrel jack connecting the current will "attempt" to spike but the inductor will instead charge it's magnetic field reducing the slope of the current into the classic inductor curve until it hits saturation, but which time the capacitor should hopefully be charged a good bit.Is that the "gist" of the idea?I assume then to actually calculate it you need all the values including the parasitics?

author: selcuk
author_rank: Frequent Contributor
timestamp: April 26, 2024, 11:02:35 am
content: You may use inrush current limiters for this purpose instead of a regular resistor. That is either a NTC or ferrite. They have ratings for inrush currents in their datasheets.If you want to use a resistor, there are "Pulse Withstanding" resistors on the market. These have inrush current ratings in their datasheets as well. An inductor will probably help but you need to check the DC resistance.

author: strawberry
author_rank: Super Contributor
timestamp: April 26, 2024, 12:20:21 pm
content: convert waveform to RMS power, that will produce heat within estimated timeframeneed to know resistors thermal resistance(long term dissipation) and thermal capacity(short term)then calculate close enough temperaturecan experimentally find hot spot temperature with IR cameraNTC wont work well when dissipated power is less than 1..2W

author: paulca
author_rank: Super Contributor
timestamp: April 26, 2024, 03:14:55 pm
content: I hooked a 220uF (cheap) cap to a 14.5V PSU. (Put a 1KOhm load on the cap).Put the scope on it. Connected the power with no resistor. 15us. 63kHz. With only 1/4 of that accounting for the majority of the rise.(The PSU does have a 5Amp limit, but it did not click with such a small cap.)With a 47Ohm resistor it takes 82 miliseconds.The linear (almost) portion of the initial rise, I can get a V/s for that. I should be able to calculate the current to produce that gradient? I should also be able to then compute the ESR of the full circuit? With or without resistor?Now I want to try what an inductor does.

author: paulca
author_rank: Super Contributor
timestamp: April 26, 2024, 03:29:57 pm
content: Quote from: paulca on April 26, 2024, 03:14:55 pm10uH torroid ... not what I expected.470uH - more what I expected, although it appears to only delay the steep portion of the rise.4.7mH and it's much better.... longer / smoother / lower gradient.What's the catch?

author: Siwastaja
author_rank: Super Contributor
timestamp: April 26, 2024, 03:56:25 pm
content: Inductor is possibly worse. Inductor is like a spring, it stores energy then releases it, which means a voltage peak.In fact, that's a problem with input capacitance to begin with: you are just concerned with the initial current peak, but missing the fact that any stray inductance (from wiring) stores significant energy because of that current peak, and if the current decays "too quickly", then a voltage peak is generated. L of wiring + input capacitance forms an oscillating circuit.Solution is a lossy element, just like a damper in car suspension: resistance. Do not put the resistor with series with the supply, that way you are generating DC voltage drop and losing power for no reason. Instead, put the resistor in series with the capacitor only.An electrolytic capacitor is a simple and cheap component which already incorporates enough internal resistance such that it's stable in itself with usual amounts of input wiring inductance.Only add explicit inductance when1) you have a real power quality or EMC issue you need to solve and which cannot be solved by capacitance and layout,2) you know what you are doing and can simulate or prototype/measure; you need to scan wide range of frequencies to prove that you did not create a resonance at some unlucky frequency making the circuit worse than what it was before adding the inductor.Seehttps://www.analog.com/media/en/technical-documentation/application-notes/an88f.pdf

author: Zero999
author_rank: Super Contributor
timestamp: April 26, 2024, 04:22:41 pm
content: Quote from: paulca on April 26, 2024, 03:29:57 pmWhat's the ESR of those inductors?How do they compare to a resistor of the same value?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 26, 2024, 05:39:36 pm
content: Quote from: paulca on April 26, 2024, 03:29:57 pmAn apparent delay is a likely result of saturation: as current goes up, incremental inductance decreases, so dI/dt accelerates, and peak current moves from a gentler to more aggressive trajectory.This can be exacerbated by ceramic capacitors, which saturate in an analogous way (ferromagnetic and ferroelectric materials have many similarities), with the difference that, since capacitance is reducing, the peak voltage Vpk ~ sqrt(L/C) * Ipk can be many times higher. (Which, since for an LC resonant circuit, normally this applies for the initial step, and you'd expect Ipk = Vin / sqrt(L/C), and Vpk = 2Vin.)Wiring itself has inductance, within a geometric factor of the permeability of free space, 1.257 uH/m. Usually the geometric factor is less than 1, say 0.3-0.6. So 0.5 to 1uH per meter of cable length. Typical cabling is fractional to a couple uH, so you get time constants with ~10s uF in the low us range.Damping is best introduced with capacitor ESR, and a good way to provide it is to put excess capacitance in parallel with the existing capacitor, and put the ESR there. Electrolytics are a good source of this.Tim

author: xvr
author_rank: Frequent Contributor
timestamp: April 26, 2024, 06:49:26 pm
content: QuoteThere is should not be with simple capacitor. Especially with 220uF. Something definitely wrong. May be capacitor was plugged in reverse, may be rated voltage of capacitor is much less that PSU output.Normal PSU should not even notice capacitor of such value.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 26, 2024, 08:00:42 pm
content: Quote from: xvr on April 26, 2024, 06:49:26 pmAre you sure?What are typical values for output capacitors on a PSU?What happens in the instant such a supply is hot-plugged to a discharged capacitor?What happens to a mechanical contact (such as a hot-plugged connector) in the instant it becomes conductive? Could sparking occur?Tim

author: xvr
author_rank: Frequent Contributor
timestamp: April 26, 2024, 08:10:53 pm
content: I'm sure. To give sparks 220uF is not enough (if voltage is moderate). Normal PSU can withtand up to 1000uF. May be not hot plug, those.

author: tooki
author_rank: Super Contributor
timestamp: April 26, 2024, 08:15:20 pm
content: Quote from: xvr on April 26, 2024, 08:10:53 pmThe OP’s PSU is running at 36V. That is more than enough to cause visible sparking even at moderate currents. Never mind the quasi-short-circuit current of a ceramic input capacitor, especially if there’s also a ceramic output capacitor.

author: xvr
author_rank: Frequent Contributor
timestamp: April 26, 2024, 08:16:03 pm
content: Double check. Connect 300uF to 5V PSU - no sparks. Discharge of capacitor afterwards give a spark

author: xvr
author_rank: Frequent Contributor
timestamp: April 26, 2024, 08:17:35 pm
content: Yes, 36V could give a spark. If OP needs a hot plug some circuit could be required

author: Zero999
author_rank: Super Contributor
timestamp: April 26, 2024, 09:04:52 pm
content: Most components can handle power spikes without a problem.A MOSFET and zener diode can be used to make a crude current limiter.

author: Jwillis
author_rank: Super Contributor
timestamp: April 27, 2024, 07:52:46 pm
content: Quote from: paulca on April 26, 2024, 10:49:46 amTrue an ideal capacitor will "Draw" infinite current provided their is an infinite supply. But if restricted to 5A then the maximum is 5A. Now even 5A seems to be a lot, so Lets look at the mechanical structure of the circuits leads and internal plates of the capacitor and compare that to the fusing current. Typically you would see that 20AWG (capacitor lead gauge for a 100uF capacitor) is capable of 5A. But the fusing value of 20AWG is 158 A for 1second or 882 A for 32 mili seconds.https://en.wikipedia.org/wiki/American_wire_gaugeAs you can see the amount of current that a simple piece of 20AWG can handle is incredibly high for short periods of time.The Time Constant RC is variable as this is dependent on the series resistance and capacitance of the circuit.https://en.wikipedia.org/wiki/RC_time_constantBut typically the charge time of 1RC (63.2% total charge or peak supplied voltage) is only a few micro seconds. By that time the amount of current across the capacitor has also dropped by near 63.2%.Of course you can lengthen the time constant by adding a resistor in series with the capacitor there by increasing the ESR, but keep in mind that that resistor will also slow the discharge of the capacitor. So in a PSU filter this becomes undesirable because slowing the capacitors charge and discharge will increases output ripple voltage. Also depending on the value of the resistor and the desirable RC time constant that resistor begins to heat up over repetitive charges and discharges of the capacitor. Typically you don't want the capacitor to discharge lower than 63.2% per cycle. You can work out the dissipation factor of the capacitor if you like.https://en.wikipedia.org/wiki/Dissipation_factorSo for filter capacitors, in your PSU, you want low ESR with a low dissipation factor. A capacitors operating properties are listed in the data sheet and you would look at that to see if the capacitor your using is adequate for the application.

author: soldar
author_rank: Super Contributor
timestamp: April 27, 2024, 08:37:07 pm
content: Quote from: paulca on April 26, 2024, 10:49:46 amhttps://en.wikipedia.org/wiki/RC_time_constant