Boost converter - input USB 5V, output about 6V - Page 1
Forum section: electronics / beginners

author: Saimoun
author_rank: Frequent Contributor
timestamp: September 21, 2021, 11:41:15 am
content: HiI have not done much with SMPS before. I am trying to raise the voltage from a VBUS USB (5V'ish) to ideally 6V.The idea is to add a high-PSRR LDO just after to get a smooth 5V so I would say I am not too worried about the output ripple.I only need about 50mA of current.Because this is USB I cannot put I would like to use is a 4.7uF ceramic (inrush current limit and stuff).Is this doable? Is there a cheap DC-DC converter that could do the job?ThanksSimonPS/FYI: I cannot use the VBUS 5V directly because this is powering an audio circuit and my 5V is too noisy. I am trying this solution (i.e. raising the voltage and filtering with a high-PSRR LDO) after discussing it in this topic -https://www.eevblog.com/forum/beginners/adjustable-ldo-a-few-questions/

author: Zero999
author_rank: Super Contributor
timestamp: September 21, 2021, 12:42:05 pm
content: I'm pretty sure most boost converter ICs can do this, but what's the point?Use a decent filter. If inrush is a problem, then add a resistor and timer relay to bypass it, when it's charged enough.

author: Saimoun
author_rank: Frequent Contributor
timestamp: September 21, 2021, 03:06:30 pm
content: OK I like the idea, especially if it is simplerHow would you make the filter? The noise is about 100mV from a switching display (see attached the 5V rail noise, as well as the FFT - switching frequency is about 81Hz)Just FYI I tried a basic RC filter with 10 ohm and 4.7uF but that was not nearly enough.EDIT: Ok now I have also tried adding a 470uH inductor between the 10 ohm res and the capacitor - it made literally no difference. But that filters 3kHz and above, so it makes sense the low-freq noise (81Hz and resonnants) is still there.Please feel free to let me know if you have a better idea, otherwise I will look into my original idea of raising the voltage to add an LDO after.

author: David Hess
author_rank: Super Contributor
timestamp: September 21, 2021, 07:02:41 pm
content: If more input capacitance is required than the USB specification will allow, then a separate inrush current limiter may be required.Some switching regulators intended for direct connection to USB have provisions to limit input current, or this can be implemented with a common switching regulator and an operational amplifier.

author: Saimoun
author_rank: Frequent Contributor
timestamp: September 21, 2021, 07:16:40 pm
content: Thanks for the reply David.This one for example:https://datasheet.lcsc.com/lcsc/1810010322_MICRONE-Nanjing-Micro-One-Elec-ME2159AM6G_C84114.pdfsuggests a 22uF input cap. It has a "soft start" function, but I guess that is only to limit the current going through the converter and will not change anything about the inrush current created by the 22uF cap, correct?Will this regulator will work with a 3.3uF cap f.x.?Thanks

author: DavidAlfa
author_rank: Super Contributor
timestamp: September 21, 2021, 10:27:31 pm
content: For such low current, consider usign a charge pump voltage doubler like the LM2660, you only require a diode and a capacitor.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 21, 2021, 10:48:46 pm
content: I wouldn't worry too much about inrush capacitance; USB is generally supplied from a power controller at best (which will provide short-circuit current limiting and/or slew rate limiting, as well as fault protection), or polyfuses at worst (system probably browns out momentarily under fault conditions, until the fuse opens). Available current is supposed to be under 500mA, but is often much more, in part due to the loose tolerances of fuses, but also for un-negotiated 2A+ circuits that may be out there.Required filter capacitance can be calculated assuming the supply is a modest inductance (corresponding to maximum expected cable length, times an inductivity of say 0.6uH/m), while the load is a square-wave current (the current drawn by a buck converter, say) with average current equal the maximum total expected input current, and for this giving a reasonable value of voltage ripple at the input (say, 50mV or less?). Or for the boost/flyback case, you'll of course have lower (RMS and harmonics: triangular) input ripple current, and higher at the output (square/trapezoid wave); you can model with that too, just as well.Note that, since USB is shielded, it doesn't matter very much how well filtered your inlet is. Maybe knock off the highest harmonics, but the fundamental and low-harmonics ripple needn't be especially tight.In addition to input voltage ripple, you can use the same model to calculate expected input current ripple, assuming a stiff power supply at the other end of the cable (and perhaps assuming a short or even zero-length cable (directly plugged in device) for this case). Again, I wouldn't expect this to be very strict; I think I'd want to keep current ripple under 10% to avoid potentially confusing any power monitoring circuitry.So, all together, a CLC input filter should suffice, values as found via calculation and simulation, well damped (add a lossy R+C in parallel with one or the other, typically an electrolytic to save on space/cost), and with a unidirectional TVS (partly to shunt peak voltages during inrush / hot-plugging, partly for general immunity).Tim

author: DavidAlfa
author_rank: Super Contributor
timestamp: September 21, 2021, 11:35:10 pm
content: 4.7uF shouldn't cause any inrush current issue, it's very small.With much higher capacitances, like 100uF+, maybe...

author: David Hess
author_rank: Super Contributor
timestamp: September 22, 2021, 12:14:33 am
content: Quote from: simonlasnier on September 21, 2021, 07:16:40 pmThat is right; the soft start function does not do anything to limit inrush current into the input capacitor.QuoteIt is difficult to say based on the datasheet, but see below about using a larger input capacitor. What would help is separately decoupling the regulator's Vin to remove ripple from the switch.I would look for a converter which allows setting the peak switch current to a lower value, which can be used to minimize the input capacitance requirement, or select a lower current converter.The USB requirements work out to about 10 microfarads of input capacitance, but it is hardly critical, and I doubt you could find any system which would work with 10 microfarads and fail with 22 microfarads.

author: Saimoun
author_rank: Frequent Contributor
timestamp: September 22, 2021, 10:39:52 am
content: Thanks all for the replies.@Tim: Understood about the capacitance, nice to know. Regarding the filtering I am a bit confused - maybe you understood this issue as being the noise on the 5V rail? The problem is the noise on the audio circuit which I want to power with 5V. Right now I am getting a noise of 5mVp-p on the audio output, which can be heard a lot. And I know this noise is the one coming from the power rail. AS I said I tried to filter it with a rather large inductor (I could even try with a 100uF cap as well - just to see) but it did not make any difference.EDIT: I think I get it, you thought the filter was to filter the output of the boost converter? That's not an issue since there will be an LDO right after. What Zero999 was suggesting was to drop the converter all together and simply filter the noise out of the 5V rail.@DavidAlfa: Good point - I actually have been considering it. Are you suggesting it only because the cost, or is it because it will be less noisy/more efficient?@David Hess:Quote from: David Hess on September 22, 2021, 12:14:33 amHow would you do that?Very interesting point about the peak switch current, I did not think about that. How to know its value though? It does not seem to be mentioned in datasheets... or should one assume a 1A converter has a 1A switch current?

author: DavidAlfa
author_rank: Super Contributor
timestamp: September 22, 2021, 02:16:54 pm
content: Because it's a lot more simple. You don't need specific coils. Depending on the frequency, you might use 2.2uF, 4.7uf, 10uF switching capacitor, end of story.However they drawback is they can only supply low currents.But for 50mA, it's ok.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 22, 2021, 02:30:50 pm
content: Quote from: simonlasnier on September 22, 2021, 10:39:52 amIt's both.You can filter the input to reduce noise towards the source, and you can filter the output to reduce noise towards the load. The input filter also serves to reduce noise from the source, of course.Note that the converter itself acts as something of a filter, but it's particularly good at that for high frequencies (where the switching inductor, bypass caps, and of course any additional I/O filtering, dominate), and low frequencies (where the control loop has high gain and thus is able to regulate over variations). It's the mid frequencies, where the filters do less and the control loop has lower gain, that you have the most susceptibility. LDOs suffer the same problem, rolling off at high frequencies; if you find you need more attenuation, you may need additional stages, or stages with greater immunity (a capacitor multiplier would be a fine application here, as you don't need the regulation of the LDO at all, the boost reg is already doing that), or lower-cutoff filters (gets bulky, but doable).Tim

author: Saimoun
author_rank: Frequent Contributor
timestamp: September 22, 2021, 02:47:25 pm
content: @DavidAlfa: Understood.@Tim: Ok I see - very smart! I guess what I hear (most) in the audio is like around 2kHz. That is filtered really well by the high-PSRR LDO. But I will experiment wil different boost converters and if they reduce the noise enough (without adding too much high frequency noise - which can be more easily filtered as you said) then maybe I could drop the LDO after the boost converter and run at like 5.5V.

author: Saimoun
author_rank: Frequent Contributor
timestamp: September 22, 2021, 09:30:55 pm
content: I have been spending the last few days reading tons about boost converters. I am trying to do the calculations for converting (min) 4.4V to 6V for about 20mA max current.The controller I want to use is this one:https://datasheet.lcsc.com/lcsc/2012102241_XI-AN-Aerosemi-Tech-MT3540-F25_C883495.pdfAnd for the calculations I have been following this:https://www.ti.com/lit/an/slva372c/slva372c.pdf(and a bit of this as well:https://www.powerelectronicsnews.com/the-dc-dc-boost-converter-power-supply-design-tutorial-section-5-1/)My problem is that for such a low output current (Iout=20mA), the inductor ripple current "should be" max 10mA (40% of Iout*Vout/Vin), and to get that at 1.2Mhz I need an inductor of at least 100uH!(see attached screenshot from the PDF from TI)The datasheet suggests 4.7 to 22uH for the inductor.Have I missed something, or maybe it's ok to have a higher inductor ripple current (like 300%) at such low output currents?

author: David Hess
author_rank: Super Contributor
timestamp: September 22, 2021, 10:45:07 pm
content: Quote from: simonlasnier on September 22, 2021, 10:39:52 amUse a separate LC or RC decoupling filter to the Vin pin. The regulator only draws like 3 milliamps (?) maximum, so the resistance can be relatively high, like 0.5 volts of drop at 3 millamps for 150 ohms. The idea here is to isolate the regulator's supply from the switching noise from the power switch so the input capacitance is more effective. Note that doing this means using *two* input capacitors with one across the USB input and one between Vin and ground.QuoteNo, the minimum switch current is higher than the output current. The datasheet mentions a 2 amp MOSFET and 2.1 amps maximum for the current at the LX pin.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 22, 2021, 11:35:36 pm
content: You're looking at a device which is very far from 20mA; note SW current limit 1.5A typical (with who knows what error bars..). So yes, 100uH would be rather high for it.But also, what's wrong with 100uH? You can get small chip inductors of much larger values, for such currents (10s mA). I'd be surprised if you need more than a 1206 (or up to 1212 size chip, inductors are oddball) for such a value, at high Q (reasonably low DCR / good efficiency).The important things to look for is current mode operation.Page 5 says "peak current mode". This agrees with the diagram, which shows sense current being added to oscillator ramp (slope compensation), then compared to error voltage. Control logic will be a flip-flop, and some gates to implement shutdown, soft start, blanking, etc.Peak current mode is usable down to a modest ripple fraction, typically over 30%. This is permitted by slope compensation, which comes with the downside of the peak current varying with pulse width. So instead of a reasonably sharp current limit, as voltage ratio goes up or supply voltage goes down, the current varies somewhat proportionally. The range of current limit corresponds to the allowable ripple fraction, which is why below 30% or so is unreasonable (the spread in current limiting would vary ~3:1, being essentially useless as the low end would be too little current, while the top end would be destructive).Without slope compensation, the minimum ripple fraction is 100%, i.e., the current rises to peak, then falls to zero, every cycle. This covers the span of DCM to BCM (discontinuous to boundary conduction mode, referring to time spent where the inductor current is above zero).Hah, they show compensation impedance back to FB, which if they truly are doing it that way, well that proves a point I considered in some other recent thread (I already forget which); just to say that, sometimes, you might find it useful to drive the FB pin in unconventional ways, well you'll need to consider how the pin responds. In this case, the block diagram is suggesting it will have a low impedance (virtual ground) response.Anyway, for self-contained regulators, you cannot control the value of the current sense resistor, or compensation R+C; you must use it at the recommended values, and that's that. Keep similar Thevenin equivalent values to R1 and R2 (i.e., around 13kohms, probably give or take ~2x), and you can probably use a little bigger inductor at the low voltage ratio you're looking at.You'll probably want to experiment with exact values (resistor divider, inductor, output cap), testing transient load response (go from 50 to 100% nominal load, say by switching a resistor with a 55 timer + MOSFET).Note they suggest mere 1uF caps, but 1uF at 1.5A for 300ns changes by 5 whole volts. You'll need larger capacitors than this. I think their interest is more to have a minimum of this much, as close to the regulator as possible; more, at up to a modest distance, will be required in total.Tim

author: Saimoun
author_rank: Frequent Contributor
timestamp: September 23, 2021, 09:31:25 am
content: Thank you both very much for the replies. I can see I have a lot more to learn about SMPS before I can design something decent - funI have now ordered a bunch of parts, both LDOs, boost converters and charge pumps, with the associated passives. So I will try a few things, experiment, and probably write back here what I foundIn particular I found this IC: LM2775 -https://www.ti.com/lit/ds/symlink/lm2775.pdfIt's basically a charge pump but which outputs a regulated 5V - this might be perfect to create the audio 5V rail from my digital 3.3V rail (which has much less noise than the USB 5V rail). And the noise created by the charge pump is very high frequency so should be handle by the audio filter anyways.And this part seem to be very common since I found plenty of chinese equivalents (fx:https://datasheet.lcsc.com/lcsc/1808081634_SHOUDING-SD6210A_C250809.pdf).@DavidHess: Ok I understand how to do it now. So that would reduce the noise (from switching but also the original noise of the 5V supply) on the Vin pin of the switching IC. I am not sure why this would help though?@Tim: I understand, basically to design a proper low current converter I would need one where I'm in control of more aspects (so probably a controller instead of a converter). Anyways that would be too complicated for me. I've ordered some 100uH which I will experiment with.

author: David Hess
author_rank: Super Contributor
timestamp: September 23, 2021, 07:10:49 pm
content: Quote from: simonlasnier on September 23, 2021, 09:31:25 amExcessive ripple at the Vin pin caused by high peak switch current and low input capacitance could cause all kinds of weird behavior. I suggested this as an alternative to using a larger input capacitor, but it is not the ideal solution. Better would be to lower the peak switch current with a different switching regulator.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 24, 2021, 01:18:14 am
content: Note that charge pumps tend to be very dirty by themselves: they work by hammering one capacitor into another. And as with a hammer, large peak currents can be generated in the process. Basically this is used to double the supply (or halve, or invert, or various other small rational numbers), then for the regulating types, it's LDO'd down to the desired output. Some may be controlled so that it's only switching enough to maintain the output within bounds (without a separate dropping stage); in this case, effectively the LDO is the switch resistance itself. They're good when not much power is needed (up to low 100s mA), at a convenient voltage ratio, and the noise is tolerable (or there's room to filter it).Anyway, you will soon be able to measure all this yourself.Tim

author: Saimoun
author_rank: Frequent Contributor
timestamp: September 27, 2021, 07:15:11 am
content: @David: Oh ok I see. I will definitely try that@Tim: are boost converter not doing the same, "hammering" current in and out of an inductor? Or do you think charge pumps are generally more noisy?

author: Siwastaja
author_rank: Super Contributor
timestamp: September 27, 2021, 07:36:40 am
content: Inductor by definition smooths the current. Current follows a nice triangle ramp in CCM, or in DCM, a "clipped" triangle, but there are no current peaks, except from small amounts of stray capacitance.By connecting capacitors, current is only limited by resistance (implicit or explicit). Efficiency is crap, and peak currents are high. Resistance can be increased to help with EMI, but then at some point the pump doesn't perform the desired voltage doubling (or whatever) because the time window runs out and not enough charge gets transferred.In boost, input current is this smooth triangle while output current is more like square wave. As a result, this output is what causes more noise issues. The peaks are still just what the ratio of power to duty cycle needs, not kind of "infinite" as seen in charge pumps. So charge pumps are most noisy, then comes the "hammered" side of a switcher (like output of boost, or input of buck), and least noisy is the inductor side of said switchers (input of boost, output of buck).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 27, 2021, 10:00:21 am
content: When I say "hammer", I literally mean hammer -- similar dynamics at work, a moving mass impacting a stationary one, delivering huge peak force, with some kind of RLC (or more complicated) ringdown waveform. Velocity and voltage, force and current, mass and capacitance, and spring rate / bulk modulus and stray inductance, are the analogous quantities here.Likely the switches aren't that strong, and it's more akin to bopping something with a rubber mallet -- maybe there's a little rebound, or it just kind of goes thud, and the peak force isn't all that high, like, not nearly enough to cause damage. (Not to cause concern -- this isn't a wear mechanism or anything.) Electrically speaking, the worst that can happen is the switch draws whatever short-circuit current it does, a few amperes or something, and that'll only happen under momentary startup conditions, or short circuit. (Which, if lacking protection, will lead to overheating and destruction, which will probably take some cycles to happen. It's not like it'll smash and fracture like hammering things sometimes does.)So it's not like the magnitude is unlimited. Worst case, it's set by the size of the switches; in normal operation, it will be less. How much less, is hard to say though. For example, the classic IC7660 does a good, I forget how much exactly, 10s of mA I think on switching edges, just sitting there. Partly shoot-through (poorly timed switching), partly actual load current.In any case, the peak current is more than enough to cause bounce or ringing in the supplies. If your main supply rail has a lot of capacitance, you might compare that to hammering an anvil. Which obviously doesn't move much when struck, but it's not zero. Indeed anvils ring quite loudly when struck (at least without any damping).More exactly, the amount doesn't matter, it's a proportional, dynamical process -- the same thing happens regardless, just a matter of scale. But you might not mind below some threshold e.g. 50mVp-p or 1% of the total (say 5V nominal), etc. If you need a lower noise voltage, or higher current, you need a lower supply impedance: more C, less L, less ESR.You do get a similar effect at the input side of a buck, or the output of a boost (or both of a flyback), where the current changes relatively suddenly (at the rise/fall time of the switch). Here, it's more of a step change, as the level goes between zero and whatever flat-ish-topped waveform is drawn (including tilted all the way to a sawtooth). And obviously, this can be some amperes easily, but so is the load current in that case.Tim

author: Saimoun
author_rank: Frequent Contributor
timestamp: September 27, 2021, 11:36:49 am
content: Ok!Thank you both for this very clear explanation - it makes sense. My only two concerns are the noise on the output - which hopefully will be sufficiently low (measuring will show) - and EMI but hopefully this is also contained as the product's case is entirely made of metal.I'll update this thread once I have tried things

author: Saimoun
author_rank: Frequent Contributor
timestamp: October 06, 2021, 07:11:57 pm
content: HeySo I received everything today, I have not tested the boost converter option (yet) but I tried the 5V regulated charge pump and it works beautifully.On the audio output I got a little bit of white noise compared to when I was using the 3.3V High-PSRR LDO, but I cannot hear at all the 300Hz noise from the 5V rail (makes sense since I put the charge pump on the "clean" digital 3.3V rail).I am thinking in the real PCB the noise should be quieter - now it is just soldered over the current PCB with cables, so not ideal.I also added a 2.2nF bypass cap (on top of the 100nF) to filter the high frequency noise from the charge pump in front of each audio IC.Do you think the white noise comes from the switching of the charge pump? Or could it be only from the fact that the charge pump is connected using cables and not PCB traces?I tried to see on the oscilloscope but it's just general noise of about 50mVpp at high frequencies, the 1.2MHz switching freq is not at all predominant.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 06, 2021, 09:23:43 pm
content: Is that just by ear or is it measured? 50mV at high frequencies, probably about typical, this is... bad layout just a lashup kind of thing?Tim

author: Saimoun
author_rank: Frequent Contributor
timestamp: October 07, 2021, 07:58:41 am
content: By ear I can clearly hear the difference between using the original design with the 3.3V LDO (which is dead silent) and using the charge pump (which has a bit of a white noise).And just to be clear I do not have a PCB with the charge pump (yet), so for now the charge pump IC is simply soldered on a small adapter board and then attached to the original PCB (i.e. the one with the 3.3V LDO) using short cables.EDIT:Nevermind, this noise is clearly unrelated to the charge pump as the exact same can be seen on both version (the quiet 3.3V LDO version and the charge pump version). I guess this could come from the USB power supply (external) which for sure has a switcher inside.See next post for audio range noise floor measurements.Quote from: simonlasnier on October 06, 2021, 07:11:57 pmThis was measured. It is very strange, the noise I can see is basically:* The noise floor around 10mVpp* A "clump" of 60mVpp noise at 10-11Mhz, of about 500ns, coming every 21us (=47kHz)I've attached a few of the noise, the associated FFT, and a zoomed view of the "clumps".

author: Saimoun
author_rank: Frequent Contributor
timestamp: October 07, 2021, 08:45:05 am
content: So I tried to measure the noise on the audio range, since this is what matters to me. First I set the audio opamp in the circuit to add +30dB on the audio output.There I can clearly see (and hear) the noise is louder, and that it is just white noise, there are no particular frequencies sticking out.PS: I am reaching the limitation of my cheap digital oscilloscope since the values are really low (about 2mVpp) so I am not sure how accurate this is.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 07, 2021, 09:17:04 am
content: Could well be due to high frequencies being detected, resulting in variable offset levels at low frequencies. But without a tight differential measurement I guess there's not much to say just yet?Tim

author: Saimoun
author_rank: Frequent Contributor
timestamp: October 07, 2021, 11:38:07 am
content: Ok I understand - thanks for the help anyways TimIronically it is a lot cheaper for me to order a new PCB and try the design properly rather than buying a proper oscilloscope so I'll do that