What kind of pass element should I use in a linear psu with switching pre reg? - Page 1
Forum section: electronics / beginners

author: Farrow
author_rank: Contributor
timestamp: January 08, 2021, 09:21:54 am
content: Hello, I've searched around the web for this answer but with no avail. After examining many designs from the web and old hp service manuals, the combination of tap switching and dc-dc buck pre-regulator as the primary power conditioning method seems like the way to go in terms of efficiency.Now the problem, PSRR. Which type pass element has the best PSRR? MOSFET, BJT or a Darlington?I don't even know if this is a valid question, this is my first time looking at rejection radios in a device.Any help would be very much appreciated!Thanks!

author: Whales
author_rank: Super Contributor
timestamp: January 08, 2021, 09:52:54 am
content: It depends how you control said MOSFET, BJT or darlington. With active feedback (eg an opamp) you can get really good PSRR until your feedback loop runs out of performance at higher frequencies.I can think of some specific non-feedback* circuits where a MOSFET pass regulator would outperform a BJT pass regulator, but it really depends on your controlling circuit.*technically an RC filter controlling an emitter follower might be perceived to have negative FB, it depends what level/method of modelling you're working at

author: CJay
author_rank: Super Contributor
timestamp: January 08, 2021, 09:56:44 am
content: I think it's less important than the overall circuit as long as the pass element is specified to be able to handle the current, voltage and is fast/slow enough (it needs to be fast enough but if it's too fast you might run the risk of it oscillating if the rest of the circuit isn't correct)So, there's no right answer to which is best, they can all be good or bad if you haven't designed the rest of the circuit well.

author: Farrow
author_rank: Contributor
timestamp: January 08, 2021, 10:20:20 am
content: Quote from: Whales on January 08, 2021, 09:52:54 amThank you for replying!I'm gonna be using op-amps in the feedback loop. Can you clarify the running out of performance part please?

author: Farrow
author_rank: Contributor
timestamp: January 08, 2021, 10:25:35 am
content: Quote from: CJay on January 08, 2021, 09:56:44 amYes you are right can't get far without a circuit. What do you mean by it needs to be fast enough? Let's say the switching frequency of the buck pre reg is 200Khz, my pass element has to be in the linear region at that frequency?

author: Zero999
author_rank: Super Contributor
timestamp: January 08, 2021, 10:27:50 am
content: Quote from: Farrow on January 08, 2021, 10:25:35 amAssuming it's a voltage follower, i.e. a gain of slighly under 1, then you want it to be as fast as possible, to avoid oscillation. The delay adds another phase shift between the error amplifier's input and output, which reduces the phase margin. The slower the output stage, the longer the delay, the higher the risk of oscillation.Quote from: Farrow on January 08, 2021, 10:20:20 amThe op-amp's open loop gain decreases, with increasing frequency, which will make it more hard for it to adjust its output voltage to the pass transistor, as the power supply voltage varies. The power supply rejection tends to go down, at increasing frequency. It's probably better to just use an LRC filter, rather than a low drop-out linear regulation, which can often be unstable.

author: Whales
author_rank: Super Contributor
timestamp: January 08, 2021, 11:50:07 am
content: Quote from: Farrow on January 08, 2021, 10:20:20 amLookup 'gain margin' and 'phase margin'. Quick and dirty summary: (in most feedback circuits) as you go higher in frequency you start to run out of both of these things. Without them your circuit is either useless at getting rid of x frequency (no PSRR) or actively does bad things at x frequency (like oscillate).Have a look at the datasheet for your opamp. It will have graphs of gain versus frequency and phase versus frequency. These are the absolute best case performance your opamp can provide, in practice your circuit may (will) make these much worse. Often you have to intentionally ruin your gain at higher frequencies (eg by adding more R & C into specific parts of your circuit) to stop your circuit from oscillating.QuoteSort of. "in the linear region" might not be the correct terminology here, that's not something that you typically consider to depend on frequency.Your switcher will leave some noise at 200KHz (andodd multiples of 200KHz, due to square-wave-ish switching). You will never fully get rid of this, but you can make it 'good enough' through several methods.Opamp-controlled linear regulators are not great at rejecting high frequencies, but simple R-C controlled linear regulators are very good at rejecting these.At the end of the day: make something, see how it performs and if it's good enough. If it's not: add more stages & try more ideas.

author: CJay
author_rank: Super Contributor
timestamp: January 08, 2021, 12:25:55 pm
content: Quote from: Zero999 on January 08, 2021, 10:27:50 amOscillation from power supply load fluctuation, I was thinking of it breaking into spurious oscillation at VHF because parasitic effects, my bad.

author: David Hess
author_rank: Super Contributor
timestamp: January 08, 2021, 05:54:48 pm
content: I do not recommend relying on high frequency line rejection to reduce switching noise; that is where passive LC filtering are better. But as a practical matter, bipolar transistors will have higher performance because of lower capacitance and higher transconductance.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 08, 2021, 08:52:54 pm
content: What noise level are you after?A tracking post-reg is an interesting project, I won't begrudge you for that; but I think the number of people building or using such designs, is out of proportion to those who actually need the performance -- or even have the tools and procedures to measure it.I find it's easier to just filter the SMPS. You can't avoid issues of ground loops and EMI from a functional standpoint, you might as well just make it better in the first place. There's only so much an active circuit can cover for. Without adding shielding, wideband noise levels around 30dBuV are reasonable.A tip is to choose higher switching frequency, so that the filters are smaller and more effective. Filtering say 50kHz ripple takes more effort, and can in fact be a good application for an active filter or regulator.Tim

author: julian1
author_rank: Frequent Contributor
timestamp: January 08, 2021, 11:11:43 pm
content: I wonder if is it practical to cascade LC filters?The frequency response/characteristics don't compose in a 2 + 2 = 4 way.But perhaps that doesn't matter for low-pass, for filtering switch noise, and when there is no signal fidelity to care about?. The drop-off still looks steep for the naive approach - "cascaded 2 pole" chart here,https://www.analogictips.com/design-analog-filters/Or interpose cap-multipliers (or regulator/op-amps steering pass-transistors) in-between stages to reduce passive LC interaction.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 09, 2021, 01:15:33 am
content: The characteristics do compose; however, the values do not. They need to be adjusted to account for the impedance from each section. This is... rather messy (in fact, the underlying problem is polynomial factorization), so we tend not to do it that way.The practical solution is to simply consult a table of the values required for a given filter order -- all the hard factoring has been done for us, we need only apply it.One last catch: we need to maintain the conditions assumed in that factoring. Termination impedance is probably the most common thing to run afoul of. The usual way to supply termination, when a full R is unsuitable, is to use an R+C, with a large enough C that the filter sees that resistance around the critical frequency range.Tim

author: julian1
author_rank: Frequent Contributor
timestamp: January 09, 2021, 03:02:07 am
content: Quote from: T3sl4co1l on January 09, 2021, 01:15:33 amTo create an example. For a 4-pole low-pass filter at 5kHz (a decade down/40dB from switcher freq of 50kHz),And with filter specified at 50ohm.Then to impedance match the output as seen by the filter - we need a RC notch-like filter - creating some 50Ohm impedance at 5kHz (and above). And with R chosen low enough to allow desired DC current.And for the input? Also the same 50Ohm at the filter freq of 5kHz?For termination matching, the key point is to avoid purely resistive termination (eg. output mosfet voltage regulator). ie. because the point of a power supply is a low-impedance DC output able to handle load variation.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 09, 2021, 05:17:05 am
content: The impedance, by the way, is given by desired AC regulation -- that is, output impedance.So, if we're making a 3.3V 1A regulator and we want it within 10% for all loads (including full-step, or peak-amplitude sine (at any frequency), load currents), we need a variation of < 0.33V, or an output impedance under 0.33V/1A = 0.33Ω.So we would choose a filter impedance of this order.When the filter is poorly damped, there will be an impedance peak, probably in the transition band (i.e., around cutoff), which gives a peak resistance Q times higher than this -- thus we want a low Q, i.e., well damped filter. (There will probably be valleys corresponding to Zo / Q as well.)As for the filter prototype, a singly-terminated type is often the way to go. If we have a typical power supply with modest size switching choke, and large filter caps, then the added filter can treat those large caps as shorted terminations. The branch facing the low impedance will always be a series inductor. Then we need to apply a resistive termination on the other end of the filter (which can be L or C depending on order; we'll probably choose even order, because capacitors are usually cheaper or smaller than inductors), and also ensure that our load impedance is never lower than that termination (else we'd short it out, and screw up the filter's response).So with that covered --Quote from: julian1 on January 09, 2021, 03:02:07 amNot a notch, though an RLC series resonant tank can be used in some cases (like, if you need a constant resistance filter:http://jeroen.web.cern.ch/jeroen/reports/crfilter.pdf).Typically, R = Zo, and C >= 2.5 Co, where Co is the total capacitance used in the filter. This gives an RC cutoff a fair factor below the filter's cutoff frequency Fc, ensuring resistance is dominant in the transition band (near Fc).If we didn't have to worry about DC, R alone would suffice, but of course we wouldn't want to burn the supply's entire load in one dumb termination resistor (or actually, quite a lot more than full load, given the earlier suggestion!), so we use a coupling capacitor to block DC to it.QuoteIdeally yes, but since there are one-port-terminated filter prototypes, we can get away with just the one.(And just to flesh that out -- the capacitor would be on the order of C >= 2.5 / (2 pi F Zo) = 1.6uF. Which, for a 50 ohm filter, yeah, that sounds about right; for a real power supply, obviously just sub in whatever impedance is needed.)Also, if we placed termination between the regulator's big filter cap, and the filter, we wouldn't be able to use a shunt R+C, it would be shorted out by the big cap. Instead we would need a series R||L.Which, in case we have a load that may have very low impedance -- this is exactly what we would use at the output as well. Or if we can't make assumptions either way, use both! (R+C and R||L networks are often seen in audio amps, for this reason -- who knows what kind of whacky impedances a variously-resonant speaker might present, nevermind piezo tweeters!)Tim

author: David Hess
author_rank: Super Contributor
timestamp: January 09, 2021, 01:37:49 pm
content: Quote from: julian1 on January 08, 2021, 11:11:43 pmAbsolutely, and that is commonly done because the larger inductance required for the lower cutoff frequency filter will have a lower self resonant frequency allowing higher frequencies to bypass it.

author: julian1
author_rank: Frequent Contributor
timestamp: January 09, 2021, 10:38:57 pm
content: Quote from: David Hess on January 09, 2021, 01:37:49 pmTo restate to confirm my understanding - lower resonant frequency, due to higher parasitic capacitance, that allows high-freq noise to 'blast' through the network.I was thinking the approaches to reduce coupling capacitance in switching transformers for lower noise (eg conductive screens between prim/secondary, physical distance between windings etc), would also apply to LC filter design.Does this mean, that if one were optimizing for low-noise, one would favor an output filter with lower characteristic impedance?So, for the same corner/cut off frequency, one would go for a larger capacitor(s) and smaller inductor(s) value - to benefit from the lower parasitic capacitance (of the inductor)?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 10, 2021, 12:35:49 am
content: I don't know that you would do that, for that specific reason? I've already given reason to do that, and the resulting value is typically quite low (in relation to system-level impedances i.e. 50-150 ohms).Might be worth discussing high voltage supplies, where you typically don't have the option to filter anywhere near the load resistance: consider 1kV 1mA, a 1MΩ load. You can hardly find >1H chokes, and you definitely can't find any that are any good at some MHz! A big consequence is, HV supplies tend to have stupidly high peak fault currents, in relation to nominal output. Lots of capacitive energy storage. It can even be preferable to use RC filtering -- the DC drop is hardly any matter (1kΩ costs only 0.1% of regulation here), and the resistance works at all frequencies (very little stray capacitance) and helps limit surge current (assuming you use a very good pulse rated resistor, of course!).The other thing you can do, is stagger cutoffs. Add a small LC (higher cutoff) at the end of a large (low frequency) filter. Fill in for the zeroes in the rest. Have done that plenty of times, to help EMI. With smaller values, damping can be easier to obtain -- i.e. use a ferrite bead, assuming DC current doesn't saturate it.(A note: series inductors and shunt capacitors contribute poles to the transfer function. Poles are lowpass. Zeroes are highpass, and caused by ESR and ESL in shunt elements, and EPR and EPC in series elements. This can be beneficial when positioned carefully -- an Elliptic type filter uses complementary poles and zeroes to give a very steep filter response with maximum compromise: passband and stopband ripple of specified magnitude. When incidental, it tends to limit the stopband attenuation of filters that are otherwise supposed to be asymptotic; really bad cases can have whole passbands at random frequencies. Or, y'know, if that's something you want -- a multibandpass filter -- that's also something that can be constructed.)Tim

author: Vovk_Z
author_rank: Super Contributor
timestamp: January 10, 2021, 01:29:06 am
content: Quote from: Farrow on January 08, 2021, 09:21:54 amI used neither bipolar transistors nor Mosfets but integral regulators (LD1084/LM350) with a success. Of cause with a small LC filter after DC-DC (XL4016). Differential noise was not the main problem but PCB design and selecting right L and C parts was. I am talking about 1 mV (+-50%) RMS noise value in a 10 Hz --5 MHz range.