Filtering when designing a 4A linear power supply - Page 1
Forum section: electronics / projects

author: jtruc34
author_rank: Regular Contributor
timestamp: February 07, 2023, 04:11:50 pm
content: Hi,I'm designing a 4A linear power supply for a modular synthesizer. (This may seems a lot of current but modular synthezisers may have 20 modules at the same time.) It's 12V dual rail, using an LM317 and 337 (I think everyone goes by this route at one point in their life). There are two points of difficulty I'm facing.The first is I'm not sure how I should specify the allowable ripple. Modular syntheziser use a common +/- 12V rail and almost all signals are large signal. So ripple shouldn't be to much of a concern, at least I first thought. However, I'm worried, as their can be so much unshielded signal wires in a modular synthesizer that the ripple would get coupled somewhere. In particular there are control voltage following the standard of 1V per octave, and I've experimentally found out that my audible threshold would be about 2mVpp of ripple directly in the signal. However, I don't know how that would translate in the allowable ripple in the supply rail. There will also be a single small signal output for headphones (about 500mVrms), on which I estimated 50μVrms of ripple would be inaudible, which is much lower than the previous figure. However, I could specifically filter the supply of the headphone module, but again I'm not sure whether the unshielded 12V unshielded bus running everywhere in the synthesizer wouldn't couple by some way or on another.My second problem is how to choose the filter topology. I know this can be a complex topic, and I tried to apply my knowledge about linear systems, however, because of the bridge rectifier, this is a highly non linear system and this doesn't help much. The first obvious choice is a single capacitor (or capacitor bank), but I would need between 10 and 100mF, which I guess is fairly common for a power supply. But then, I wondered if I could make things better by using a pi filter, and that maybe I could use lower values of capacitors with the same ripple. However, I found it hard to design, and even simulations with constant current were not very intuitive, where sometimes increasing the inductance would have almost no effect on the ripple. What are the pros and cons of those topologies in my situation? Could I go even higher order? I'm aware I could have resonance problems, and I can't go higher than 100mOhms of series resistance without having too much voltage drop.I'm not very concerned about the price of an inductor since I don't plan on building ten thousand power supplies.Thank you for your help

author: Benta
author_rank: Super Contributor
timestamp: February 07, 2023, 07:58:22 pm
content: Normal term is "filtering capacitors", but that's misleading.The correct term is "storage capacitors", as they need to keep the DC voltage up during the 100 Hz voltage dips after the bridge rectifier.Rule-of-thumb is 2000 uF/A, but this can vary.So pi filter etc. won't really help unless the inductors are enormous, plus they will slow down the transient response of your PSU.I assume you're aware that LM317/337 types wil not deliver 4 A.Concerning ripple: I see very little risk there. The linear integrated regulators are very good at ripple suppression; on top of that modern audio opamps have excellent PSRR.I'd focus on potential signal ground loops, those can give you a lot of hum and can be hard to diagnose and fix.

author: floobydust
author_rank: Super Contributor
timestamp: February 07, 2023, 08:46:13 pm
content: Better recording consoles settled into using local voltage regulators fed by a higher regulated voltage i.e. I recall Neve used +/-19V main rails feeding 7815/7915 on critical strips but I could be wrong. Look at recording console designs.I tried using just a pair of master regulators but wiring/connector resistances came into play, as well as single-point (star) connections where required. You can do a bad PC board design and the high ripple currents from say 10-100mF filter caps creates ripple voltage due to trace/wiring resistances that the Vregs can amplify. I'm saying what you think is a good power supply design can actually perform terrible.You also don't want coupling between cards of audio frequency ripple, cross-talk. I think is more a concern than mains ripple.I found grounding is of more importance because even 5mV of hum/ripple (audible and easy to get at 1-4A) can get into many cards. You need a signal and power ground. Even then, RF shows up on the mixing bus node and is another challenge where single-ended audio runs between cards exist.Using LM317/LM337 powering a 24-track board near 1A +/-15V, I also found the recording engineer complaining about noise, I ended up adding huge 10,000uF caps on the regulator's outputs to filter that I think it was popcorn noise from the regulators. Then I got about 2mVpk and it worked fine.

author: Benta
author_rank: Super Contributor
timestamp: February 07, 2023, 09:23:36 pm
content: All very good points from @floobydust, and the one about using POL (point-of-load) regulators is the best. Have local regulation on each module, and you're pretty much out of the woods.

author: dobsonr741
author_rank: Frequent Contributor
timestamp: February 08, 2023, 02:21:17 am
content: Seem counterintuitive first, but modular folks like a particular Meanwell switching supply:https://modularsynthlab.com/product/eurorack-power-supply-mean-well-rt65bThe key is per module filters, proper ground and POL as already pointed out.

author: srb1954
author_rank: Super Contributor
timestamp: February 08, 2023, 04:19:14 am
content: Quote from: floobydust on February 07, 2023, 08:46:13 pm19V sounds about right for the DC input voltage for a 7815.These are specified for RR with a 3.5V I/O differential and that RR gets significantly worse if the regulator headroom drops below that level. So, allowing for a little bit of ripple on the incoming supply, a minimum DC input voltage of 19V would required to comply with the manufacturer's recommended operating conditions. A little bit extra headroom on the incoming supply could possibly improve the RR even further.It should be noted that with the 78xx series the RR gets worse with the higher voltages as there is progressively less loop gain within the feedback loop to counter noise introduced on the supply rails.

author: pardo-bsso
author_rank: Frequent Contributor
timestamp: February 08, 2023, 11:36:22 am
content: Our studio console has every module fitted with a pair of 78 and 79 regulators fed from about ~16 volts and an extra wire and connector with the audio ground only.

author: jonpaul
author_rank: Super Contributor
timestamp: February 08, 2023, 04:21:57 pm
content: Euro Rack system is notorious for the poor power supply and grounds, creating noise and interference.Avoid SMPS, a normal linear regulator with electrolytic caps,and HF bypass should I fine, if,good grounds and wiring practice is usedJon

author: jtruc34
author_rank: Regular Contributor
timestamp: February 08, 2023, 05:47:48 pm
content: Quote from: Benta on February 07, 2023, 07:58:22 pmThis seems extremely low. This would mean about 5V of ripple at 100Hz. I know linear regulators can reject ripple but that seems way too much.Quote from: floobydust on February 07, 2023, 08:46:13 pmQuote from: Benta on February 07, 2023, 09:23:36 pmSo, in practice, I would have a global 15 or 16V regulator and many local regulators, each one with bypass caps?Quote from: dobsonr741 on February 08, 2023, 02:21:17 amBy filter, you mean simple bypass capacitors, or do you mean something more involved?Quote from: Benta on February 07, 2023, 07:58:22 pmQuote from: jonpaul on February 08, 2023, 04:21:57 pmQuote from: dobsonr741 on February 08, 2023, 02:21:17 amDo you have any specific advice on how to have a proper ground? What should I do now to avoid ground loops later?Quote from: floobydust on February 07, 2023, 08:46:13 pmSo, you would mean I would have two seperate bus tied down together at the PSU, one used for power and the other for signal? I don't exactly see how I would use that at the board level. Taking a simple example of an oscillator: all op-amps would be power by the +/- 12V. Where would I use the power ground? I'm not sure to see where I would use the power ground and where the audio ground.Thank you all very much for your feedback!

author: dobsonr741
author_rank: Frequent Contributor
timestamp: February 08, 2023, 07:53:29 pm
content: One common fault I found in my euro rack: the alu face plates are in contact with the signal in/out 3.5 mm jack’s GND. No noise if I keep the module “in the air”. Once screwed into the rack aweful noise appears. I’m planning to add input ground lift and differential output to the modules I’m about to build, so I control the GND reference point of the signal.Unfortunately not completely compatible with the basic euro rack design ideas, however a mixer output stage can reasonably separate the noise from the output of the whole rack.

author: Benta
author_rank: Super Contributor
timestamp: February 08, 2023, 08:17:17 pm
content: Quote from: jtruc34 on February 08, 2023, 05:47:48 pmActually. it's more like 4 V, but never mind.To cross the t's and dot the i's here, all suggestions go in the direction of a main pre-regulator capable of handling the 4...5 A for the whole rack, and having POL regulators on each module.The pre-regulator could even be a super-simple BJT/Zener combo circuit.Throwing big capacitors at it is a waste of space and money IMO.

author: mariush
author_rank: Super Contributor
timestamp: February 08, 2023, 08:41:23 pm
content: If you use a classic transformer, you can't depend on its output voltage to be a specific value.At very low output, the voltage may be up to 10-20% higher than the advertised value. So if your product is on stand-by or most modules are idle, the transformer may output a much higher voltage.Then, you have to account for much lower mains (AC) voltage - ex your 110v AC may be 100v AC, or may be 120v AC ... if the transformer ratio is 1:9 (ex 110v : 12v ac) then you could have almost 13.5v AC with 120v AC on the input, before we even get to that 10-20% extra at low current.You rectify that AC voltage with a bridge rectifier and then smooth out that rectified voltage with capacitors.You can sort of approximate how much capacitance you want to guarantee a minimum dc voltage with the formulaCapacitance = Current / [ 2 x ac frequency x ( Vdc peak - Vdc desired) ]So for example, let's say you have a 110v AC - 15v AC transformer, and you know you may have 14v AC if mains voltage is too low. At idle, and/or high mains AC voltage, you may have 16v AC or even moreRectified DC voltage will have a peak voltage of Vdc peak = [14 .. 16] x 1.414 - 2 x voltage drop on diode in bridge rectifier (around 0.7v..1v per diode) = 18.2v .. 21vSo you'd want to use at least 25v rated capacitors, but 35v or 50v rated would be better.So going back to the formula , you have Vdc peak of 18v , you have 4A current, let's say you want minimum 14v DC all the time so you could then use linear regulators to produce 12v....C = 4A / ( 2 x 50hz x (18v - 14v ) ) = 4 / 400 = 1 /100 = 10000uFI'd probably consider 3 x 3300uF or 2 x 4700-5600 uF , depending on how much space you can afford for them.You may want to think how much current you're gonna consume on negative voltages and how much on positive voltagesIt may be easier and cheaper to not have 2 identical secondary windings to have same current on positive and negative, maybe you could have custom transformer with less current on negative side. Or may even be cheaper to get just one secondary winding, and then use a switch mode regulator to produce some negative voltage, followed by a linear regulator like 337 to smooth that out even further.As for regulators, like others said ... you'll probably want to have multiple regulators, maybe let's say 5 of them, 1 for every 4 modules / slots.Regulators like LM317, 7812 , 7815 can typically do 1 - 1.5A of current, but there are other regulators like let's say LM1085 and *1085 (made by multiple companies) which can do 3AEven so, keep in mind linear regulators dissipate the difference as heat, so for example if you do 18v input, 15v output at 1A , that's (18-15)x1 = 3 watts of dissipated heat ... better to have a nice wide heatsink and attach multiple regulators on that heatsink to spread the heat instead of having one super hot regulator.There's also some linear regulators that are very good at filtering, rejection, maybe you'll find it better to go overboard and use a single 500mA... <1A linear regulator for each module, because such ICs have much better filtering and stuff.

author: Vovk_Z
author_rank: Super Contributor
timestamp: February 08, 2023, 08:43:14 pm
content: Quote from: jtruc34 on February 08, 2023, 05:47:48 pm2000 uF/A is ok for a typical power amplifier. In your case you'll have a linear regulator after that, so you'll have 60-80 dB rejection (if you do it right).If 1.0-1.5 A is sufficient, then I would use LM317 instead of LM78xx, because LM317 may deliver higher rejection.LM338 works with 4 A fine and may give you somewhere near 0.1-0.2 mV RMS noise.It is easier and give much less space for mistakes if you'll do +- power channels as identical, with two isolated secondary windings.

author: jtruc34
author_rank: Regular Contributor
timestamp: February 09, 2023, 12:25:15 pm
content: Well, actually I've realized that I have much less headroom than anticipated. The transformer I have (and I can't justify buying a second one) is 12V, which means 17Vpp in normal conditions, but knowing mains can get as low as 210V and factoring the diode drops (480mV for the one I use), I only get 14.3V, which is way too low to have two LM338/317 in series, and with 1V of ripple it's only 13.3V.I'll have to opt for LDOs. Would you still recommend to regulate twice? Or maybe I could do something else at the PSU and only regulate at the point of load.Thanks

author: mariush
author_rank: Super Contributor
timestamp: February 09, 2023, 02:15:39 pm
content: 480mV sounds like schottky diodes... those aren't ideal to use to rectify AC voltage, especially in audio applications. Or maybe check the datasheet and see the voltage drop at higher currents, 480mV may be at 1A, not 4A. Also, you'd probably want a 10A or higher bridge rectifier, something much higher rated than 4A or whatever you think your device is gonna use.May even make more sense to trade low voltage drop for a package with bigger voltage drop, but which can be more easily attached to a heatsink, for example one of the GB* series which can be easily screwed down to a heatsink :https://www.digikey.com/short/j89wbrr3There are alternatives to using diodes to rectify AC voltage, but most are quite expensive or have some gotchas (for example the LT4320 in the video below is great if your product uses a lot of current, but it doesn't actually work while the ac output is below around 9v AC, so it's not great for low ac voltage output, and there's gonna be a very small period every cycle where there's no output to capacitors, which means you may need bigger capacitors to account for that if you insist on using it... for let's say a 48v rectified amplifer, it would be great)

author: jtruc34
author_rank: Regular Contributor
timestamp: February 09, 2023, 03:14:29 pm
content: Quote from: mariush on February 09, 2023, 02:15:39 pmWhy would I want to double the current rating? Why are Schottky diodes non ideal for this application? I'm working at about 15V, well below their low breakdown voltage. I've checked and the 480mV drop is at 5A.

author: Benta
author_rank: Super Contributor
timestamp: February 09, 2023, 09:50:02 pm
content: Quote from: jtruc34 on February 09, 2023, 12:25:15 pmConcerning transformer/storage cap selection, you're in even deeper trouble than what you describe.You can NOT say: 210VAC min. = 10.96 VAC output = 15.45 Vpeak - Rectifier drop = cap charge voltage.It's much worse.The issue is, that the caps are only charging at a fraction of the 100 Hz period time, meaning the charge current gets larger and larger when you increase the cap size, as the charging time gets shorter.This reduces your peak secondary voltage due to resistive losses in the transformer. And this drop can be several volts. At some point you'll reach the "point of diminishing returns" where increasing cap size will bring zilch, because the ultra-short charging time will drop the transformer peak voltage to a point where nothing more can be gained.Trying to calculate this is very, very complicated, Spice modelling quite easy if you can measure your transformer parameters.In your spot, I'd do the simulation in any case, as your headroom is severely limited, and you need to have an idea of where you are.317/337s et al are out, whether you have enough for LDOs is open.A last ditch attempt if all the above fails is adding windings to your secondary to increase the voltage, but this only really works with toroids.Let us know how you get along.Cheers.

author: Benta
author_rank: Super Contributor
timestamp: February 09, 2023, 09:51:47 pm
content: Quote from: mariush on February 09, 2023, 02:15:39 pmWhy? Please explain.Thanks.

author: mariush
author_rank: Super Contributor
timestamp: February 10, 2023, 04:13:01 am
content: Maybe I shouldn't have said. It was based on various "audiophile" advice I've heard in the past, but didn't think about it thoroughly before writing it and now that I think about it I can't find a good reason why they shouldn't be used.As for my statement about why double the current, I was simply thinking of how to keep that rectifiers diodes cool - passing 5A through a single diode, even with forward voltage of 0.5v, will cause a couple watts to be dissipated on it, which means the diode will heat up significantly.One of those bigger packages will be easier to mount to a heatsink, and there will be more headroom, you may be able to use a smaller heatsink or maybe even no heatsink, because the package helps a tiny bit to spread the heat.

author: srb1954
author_rank: Super Contributor
timestamp: February 10, 2023, 06:15:02 am
content: Quote from: jtruc34 on February 09, 2023, 03:14:29 pmUsing a capacitor input filter the diode current waveforms are very spiky i.e. they have a high peak current to average ratio. Due to this high peak current the diode power dissipation can be higher than a simple measurement based on the DC rating would suggest.It is instructive to do s SPICE simulation to see this effect but you will need to include the transformer parameters to get an accurate simulation of the current waveform and the diode power dissipation.

author: jtruc34
author_rank: Regular Contributor
timestamp: February 10, 2023, 10:32:58 am
content: So, I've realized there is no way my 230:12 transformer is gonna do it, even with LDOs (without buying LDOs so expensives that I could buy a second transformer), and unfortunately my transformer is toroidal but its hole is filled with some hard polymer which I couldn't remove, so adding turns is out of question.The two options I have is 230:15 or 230:18, and I decided against the latter because I would need some massive heat management. The only way I could be safe with 15V is still with LDOs (taking a margin of 10% on mains power), and I've opted for mariush's suggestion of GB's bridge rectifier.To chose the 13.5V, there will be a maximum of 1V drop on U2/U4 taking a bit of margin gives 13.5. So 15*sqrt(2)*0.9 - 2.2 is 16.9V. There will be at maximum a 1V drop on Q3/Q6 and a 0.5V drop on U3/U5, it thus leaves a margin of 16.9 - 13.5 - 1.5V = 1.9V. If I can squeeze the ripple below 1.9V I'm good. I will have to do some testing to be sure.Do you think my solution is sensible?Thanks

author: Benta
author_rank: Super Contributor
timestamp: February 10, 2023, 11:02:32 am
content: You're ignoring the issue I outlined in reply #16 completely. You really need to take it into account.

author: mariush
author_rank: Super Contributor
timestamp: February 10, 2023, 11:54:30 am
content: I think it's over-engineered actually.You calculated 2.2v for the bridge rectifiers when it will be more like 0.8v per diode, and it would be cheaper to parallel a couple bridge rectifiers to get lower voltage drop, than complicate your life with those transistors.For example here's this $1.3 30A bridge rectifier :https://www.digikey.com/en/products/detail/micro-commercial-co/GBU3010-BP/12177159Check datasheet .. the 1v drop typical for 15A is at 25c, while in use the rectifier's gonna be more like 50-70c on a heatsink ... and at 4A and 75c you're looking at around 0.8v drop40 mF may be a bit too much, it may blow the fuses and cause a big current spike on the bridge rectifier at startup - again a simple solution could be just paralleling two bridge rectifiers.If you're gonna have regulators for each set of 4 why even bother with pre-regulators... lose those and just use a couple bus bars or thick traces to bring whatever voltage to the local regulators

author: Benta
author_rank: Super Contributor
timestamp: February 10, 2023, 07:47:42 pm
content: Quote from: mariush on February 10, 2023, 11:54:30 amYou still don't get it.Peak current is going to be 10..,.20 times higher than average at the voltage peak. Ever heard of "crest factor"?

author: jtruc34
author_rank: Regular Contributor
timestamp: February 10, 2023, 08:35:30 pm
content: Quote from: Benta on February 10, 2023, 11:02:32 amFair enough, I ignored it because I don't know the characteristics of my transformer yet but maybe I'm too optimistic of actually being able to get 2V of ripple with 4-5A. What values are relevant for the problem you outlined? Is the series resistance of the secondary enough to model it in SPICE or should I get the inductance as well as the coupling factor? I'm not sure. The inductance is not very hard to measure but I guess it will strongly depend on the saturation of the core which is a mess to model.I'll measure the secondary resistance of the one I've got and hope the new one will have similar characteristics.Quote from: mariush on February 10, 2023, 11:54:30 amThey are not there to reduce drop, they are there because the linear regulator is not capable of delivering 4-5A. (LDOs with such a high current are really expensive)Quote from: mariush on February 10, 2023, 11:54:30 amnThat's a good point, for which I blindly assumed without thinking that it was better to do so, as many people directly suggested to use preregulation. I'm not sure if it is strictly necessary, but in my case it serves the purpose of dissapating most of the heat in the PSU and not on each per-module regulator, which is good because I won't have to mount heatsink on each of them but only on the one in the PSU.

author: srb1954
author_rank: Super Contributor
timestamp: February 10, 2023, 08:46:54 pm
content: Quote from: jtruc34 on February 10, 2023, 08:35:30 pmTo model a transformer correctly you need to include the reflected primary resistance as well as the secondary resistance: Rtot= Rsec+ Rpri/N2where N is the primary/secondary turns ratio.On larger transformers you also need to include the effects of leakage inductance but on small transformers the winding resistance dominates the transformer regulation and the leakage inductance can generally be ignored.

author: jtruc34
author_rank: Regular Contributor
timestamp: February 10, 2023, 08:48:41 pm
content: Well, I've just measure between 140 and 150 mOhms on both secondaries. With a bit of luck it will be similar on the new transformer. Now that I think about it, it's really the leakage inductance that's important here, which I could assume is negligeable for a toroidal transformer, the secondary impedance is probably dominated by the DC resistance. Do you agree?EDIT: Well, I didn't see srb1954's answer, which answers my problem. Thanks!

author: Benta
author_rank: Super Contributor
timestamp: February 10, 2023, 09:02:48 pm
content: Which at anoptimisticpeak current of 15 A results in around 2.1 V peak voltage drop. This is consistent with my own experience. Plus, the primary resistance is not included yet. It'll be around the same.

author: jtruc34
author_rank: Regular Contributor
timestamp: February 10, 2023, 09:15:43 pm
content: Quote from: Benta on February 10, 2023, 09:02:48 pmTaking into account the primary resistance of my 12V transformer, I simulated with Schottky diodes (I think there is no real reason not to use them, and it's also easier for me for thermal design in this particular case), I got the following result (image linked below). As you can see, the current spikes doesn't really seems to depend on the value of the capacitor, and increasing it really does seem to reduce ripple.I don't fully understand this result, maybe I've done something wrong.

author: Benta
author_rank: Super Contributor
timestamp: February 10, 2023, 09:29:34 pm
content: I'm not familiar with your simulation program, but isn't I1 a current source?

author: jtruc34
author_rank: Regular Contributor
timestamp: February 10, 2023, 10:16:08 pm
content: It's LTspice. Yes, it's a current source, but here I would call it a load.

author: Benta
author_rank: Super Contributor
timestamp: February 10, 2023, 10:38:18 pm
content: Right.Just for fun, I threw your circuit together quickly, but had to scal the current to 10 % (I didn't have any 20 A schottkys in my Spice library), so it's operating at 0.4 A.I use just a resistive load, this makes things simpler.the schematic is super-simple.Plot 1 is cap voltage and R1 current. You see the crest factor in the current plot, and that the peak current is in this case three times higher than average. Plus, the DC voltage is disappointing.Plot2 is the voltage at the AC input of the rectifier bridge. The peak voltage "capping" is typical of linear power supplies.

author: jtruc34
author_rank: Regular Contributor
timestamp: February 10, 2023, 10:55:13 pm
content: OK, but I still don't really understand the result of my simulation. You said increasing the capacitance would increase the current spike which does not seem to be the case in my simulation. The DC voltages does become better by increasing the capacitance.

author: Benta
author_rank: Super Contributor
timestamp: February 10, 2023, 11:56:38 pm
content: Try replacing I1 with a suitable resistor and let's see if we get similar results. That might be interesting.

author: jtruc34
author_rank: Regular Contributor
timestamp: February 11, 2023, 12:37:28 am
content: Quote from: Benta on February 10, 2023, 11:56:38 pmI've just tried and the difference is insignificant, but by the way, I do have similar results to you. I have a pretty significant drop in voltage (about 2 to 3V), but the only thing is that you said there is a diminishing return in picking increasing values of capacitance, and that the current spikes would get higher and higher, which it doesn't, or at least not in the values of C I've tested.

author: Benta
author_rank: Super Contributor
timestamp: February 11, 2023, 11:32:57 am
content: Quote from: jtruc34 on February 11, 2023, 12:37:28 amYes, that statement is only true up to a certain point. The transformer internal resistance saves the situation here.

author: Kleinstein
author_rank: Super Contributor
timestamp: February 11, 2023, 03:13:03 pm
content: Schottky diodes tend to have relatively high series resistance. This can cause the relatively identical and not that short current peaks.In addition to the transformer resistance one could consider adding a fuse (or 2). The added resistance is not that bad.

author: jtruc34
author_rank: Regular Contributor
timestamp: February 12, 2023, 06:06:39 pm
content: Well, thank you for all your support, I think you have given me pretty much everything I need to know. One small thing, still: do you think it's a problem that I exceed the ripple current for the first cycles when charging the capacitor? Except from blowing the secondary fuse (which I can just choose smartly to avoid that problem).

author: Kleinstein
author_rank: Super Contributor
timestamp: February 12, 2023, 08:09:46 pm
content: There is usually not problem with a higher paek current for the first few cycles. The worst case loss for the series resistance on charging a capacitor is equal to the energy stored in the capacitor. Normal electrolytic capacitors have no problem absorbing that much as heat without getting really hot.A large size (e.g. more than some 100 VA) toroidal transformer may need some kind of soft start anyway. There can be a quite large current spike from driving the core to saturation due to remanence and random start phase.

author: srb1954
author_rank: Super Contributor
timestamp: February 12, 2023, 11:49:24 pm
content: Quote from: jtruc34 on February 12, 2023, 06:06:39 pmElectrolytic capacitors are generally OK for absorbing the high initial peak current, which will be limited mainly by the transformer resistance anyway.However, you need to pay attention to the surge currents through the diodes. The diode will be specified with a maximum surge current IFSMwhich should never be exceeded even on a single cycle. The diode datasheet may also have a derating curve for that IFSMfigure that shows the maximum surge current that can be tolerated over durations longer than a single cycle. You should always ensure that the peak diode currents come under the allowable limits of that curve at all times during the startup period.

author: jtruc34
author_rank: Regular Contributor
timestamp: March 15, 2023, 12:46:53 pm
content: I'm resuscitating this post because someone questionned the usage of POL regulation. From what I understand, the goal is to minimize the effect of stray wire resistance and inductance when the load is far away from the source, but I don't see how POL is achieving more there than simply putting bypass capacitors near the loads.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 15, 2023, 02:05:45 pm
content: Bypass capasitors are for the higher frequencies / shorter time scale. The local regulator is maily for the longer time scale / lower frequency part, e.g. regulate after the resistive drop on traces. With relatively low voltage parts (e.g. 1.8 V or 3 V) the tolerance in absolute voltage got smaller.