Boost converter explodes when connected to battery - Page 1
Forum section: electronics / beginners

author: fl3tching101
author_rank: Newbie
timestamp: August 11, 2021, 05:37:25 pm
content: So, sort of what the title says. I’m a bit at wits end. Let me give some context.I am working on a project that I’d like to be battery powered. It may draw a fair amount of power, so I got a fairly large battery (thishttps://www.amazon.com/dp/B07X7MD2JK/ref=cm_sw_r_cp_api_glt_fabc_RAWSCSN3V0NWM8XJTYFH?_encoding=UTF8&psc=1this205WH LiFePO4 pack to be precise). One of the components needs a voltage between 12-19v. So I wanted to use a step-up/boost converter to get a steady 15-17v or so no matter the battery’s current charge state. Easy enough.So for prototyping I have a bunch of these MT3608 boost converters lying around. I know they aren’t the best converters on the planet, and I’m open to suggestions of better ones with higher current ratings (preferably in the 5A region or more for a good safety buffer), but it’s what I have. So I get my bench supply out and hook one up and dial in 17v on the output, test varying the input voltage to the min and max of the battery and it stays rock solid at 17v. I even powered up the part that uses the voltage and it works great through the boost converter. So easy next step then, just hook up the battery to the input, it’s the same voltage as the bench supply, easy!Wrong. As soon as I connect the battery to the boost converter it pops and the smell of burnt electronics fills the room. I figured I shorted it out because I had some clips for my multimeter fairly clumsily attached, so I grab another and go through the same process, but am a bit more careful about not having anything near the converter. Same exact issue, though the pop is a bit less loud this time. So the next day I try one last time… this time I do the same thing essentially, and when I get to the battery, no pop! Except the output voltage now reads identical to the battery’s voltage instead of the 17V I had dialed in… odd… so I hook it back up to the bench supply to see what the deal is and… big pop, small flame or glowing red IC, magic smoke, burnt electronics smell…What in the living hell is going on? I understand that the battery can deliver more current than the bench supply, by a lot, but there’s not even a load connected to the boost converter! Only my multimeter! I triple, hell even quadruple checked the polarity. Why in the world does the battery seem to sabotage every boost converter it touches? For reference, this is basically an identical converter to the one I have:https://www.amazon.com/dp/B089Y7NDCR/ref=cm_sw_r_cp_api_glt_fabc_XGN56WVE6SC0DVT9Y1DS.From the internet, this appears to be a relatively correct schematic:https://images.app.goo.gl/3ACP78Td6vp2bXMq7Anyone have ideas?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 11, 2021, 08:17:47 pm
content: Inrush combined with a high-K ceramic capacitor. Put a TVS across the converter's input pins, say P6KE18A or so.Good thing they're pretty much disposable priced. They're made to be, too...Tim

author: Ground_Loop
author_rank: Frequent Contributor
timestamp: August 11, 2021, 08:34:59 pm
content: The Amazon comments on the part you reference indicate some similar issues.

author: fl3tching101
author_rank: Newbie
timestamp: August 11, 2021, 09:48:22 pm
content: @T3sl4co1lAh, that makes good sense! I have heard of TVS diodes before, but didn’t really understand their use. Sounds like something that might come up later too, so do you know of an assortment pack or something I could get that would have one I need now and other voltages for other projects? Would be really useful.And also, if you don’t mind, could you explain how a TVS solves the issue? My understanding is that the TVS will shunt away current when the voltage the TVS is designed for is exceeded. But I don’t follow how that helps in my case. Is the “high K” capacitor (might need a explanation for what that means as well!) causing the voltage to rise as the battery is connected beyond the voltage of the battery? Thanks!

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 12, 2021, 12:49:20 am
content: Sure. What's happening is, you have -- I assume some length of hookup wire, maybe a few feet, which has inductance, on the order of mu_0 = 1.257 uH per m of length. It's usually lower than this, by a geometry factor -- wires paired closer together, or broad facing sheets, have less inductance than loose wires, or loops or coils -- but it's in the ballpark.So, say you have 1uH and you connect the battery to the RLC series circuit. L = 1uH, R is small (capacitor ESR, battery ESR, wire resistance, and C is whatever that capacitor is, apparently something like 22uF?This is a transient problem, where the capacitor charges from initial zero, to battery voltage, through the L and R. R << sqrt(L/C) so it resonates and rings down. Voltage reaches a peak of 2 * Vbat in 1/2 wave or pi sqrt(L C) = 14.7µs, and the peak inrush current is Vbat / sqrt(L/C) = 56A or so, reached in 1/4 wave or 7.3µs, and every 14.7 thereafter (but decaying, as it rings down of course). (It'll be less than this due to resistance, though it's not clear how much; R is probably 10s of mΩ here, so it will ring for maybe a dozen cycles, give or take.)Now, 24V would still be just within the rating of the device, though it's higher than you might've expected.But we're missing one key point. The capacitor is nonlinear. You don't get 22uF in a chip that tiny, and have it work for any voltage.This is where "high K" comes in. High-value ceramic capacitors are made with some blend of barium titanate and friends, which has an extremely high dielectric constant (~10k), but saturates easily (above a modest electric field, k drops precipitously). This isn't well covered in datasheets, you have to go looking for the characteristics. Example:https://www.digikey.com/en/products/detail/murata-electronics/GRM31CC71C226ME11L/8323503Datasheet doesn't say anything about it. Gotta click through the SimSurfing link to see characteristics. It's -66% at 12V!The consequence for our little RLC circuit is, at first the capacitance is large, which takes time to charge up, putting excess current into the inductor (~50A). Once voltage gets going, C drops like a brick and dV/dt goes way up. Consequently, peak voltage shoots up -- it's not going to be double the input anymore, more like 3, 4 times, or even more.The regulator can't handle anywhere near that much, so it breaks down, and some amperes flow into it the wrong way. Needless to say it doesn't last long, and out comes the magic smoke.So, by putting a TVS on there, the peak is simply clamped. The TVS is a beefy zener diode, which just holds the voltage ~constant when pushed above its breakdown voltage (not really constant, it has some internal resistance -- but a 5 or 10V range is a far sight better than without). It can handle lots of amps, so the peak current isn't a problem. The inductance just discharges into it a bit slower (over say 20 or 30µs) then the voltage settles down.Another option is using "too big" of a capacitor, with some ESR -- an electrolytic, say 100uF 25V, with ESR ~ 0.2Ω. This will charge more slowly, limited by its own resistance (which is substantial compared to other resistance in this circuit), and while the peak current is higher (on the order of Vin / ESR) the voltage doesn't overshoot or ring. At least up to some maximum supply inductance (a few meters, because sqrt(4uH / 100uF) = 0.2Ω).With illustrations:https://www.analog.com/media/en/technical-documentation/application-notes/an88f.pdfTim

author: thm_w
author_rank: Super Contributor
timestamp: August 12, 2021, 01:14:04 am
content: Specifically what was the battery output voltage?I've not personally ever had one of these modules die, but they are not quite the same as the Amazon photo. Mine use 1206 ceramic caps, of about 20uF. The amazon photo shows 0805 caps which I would doubt are 20uF. So that would make me want to follow teslacoils suggestion of the electrolytic capacitor addition.Quote

author: fl3tching101
author_rank: Newbie
timestamp: August 12, 2021, 03:05:55 am
content: @ T3sl4co1lWow! That’s a damn good explanation and makes perfect sense! There is one question still though, why does the battery do it but the bench supply doesn’t? Is it because it is capable of delivering the higher amperage that this scenario incurs? I know the bench supply can limit the current, but I figured the ringing wouldn’t have cared about that current limit. And the electrolytic capacitor solution is a great quick fix since I have plenty lying around! It’ll be good enough of a hack to last until I can get something a bit beefier (and more reliable) for the full project. Thanks so much!@thm_wIt is a 12.8v 4s8ishp LiFePO4 battery pack, it’s an off the shelf pack so not 100% sure how many parallel cells are there. But it has a built in BMS that triggers at something like 48A. The battery reads more like ~13.4V with full charge though. I believe I got these boost converter straight from AliExpress a couple years ago, so it wouldn’t surprise me if they didn’t have the right values or used cheaper caps that aren’t up to par. Like I said above, I think swapping on an electrolytic cap is good temporary solution. I don’t need it to charge up quickly by any means, so all is well. If I was going to use them in the final design I’d definitely wait around for the TVS to ship, but since I’m just trying to get a proof of concept going it doesn’t much matter. Thanks for the note though! Sounds like it’s a bit of a roll of the dice on these boards. Which I’d say it usually is from these no-name brands, but for as cheap as they are it’s a gamble worth taking usually.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 12, 2021, 09:31:45 am
content: Quote from: fl3tching101 on August 12, 2021, 03:05:55 amIt's actually a good question, though the answer may also now be in front of us.A bench supply doesn't need to deliver pulse or fault current (and, I would say, shouldn't!), but most of them do -- a common sight inside them is a big fat electrolytic at the output.But therein lies the answer. If its ESR is comparable to the resonant circuit, then that will limit peak current more than the inductance, and inrush will be relatively mild.It's hard to say, because electrolytics from 50mΩ to 1Ω are available in those values; who knows, you'd have to look inside to see what it is.But it's certainly possible; and it seems unlikely that the battery would have nearly as high an ESR.QuoteBTW, that's adding a capacitor, not swapping! It's better with both.Tim