Boost converter - can someone please explain the phenomenon - Page 1
Forum section: electronics / beginners

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 07, 2024, 06:08:34 am
content: I made a boost converter using the equations provided by Ti for calculating the power stage of a boost converter.My requirements are: 5Vin, 12V out with a power draw of 400mA from the load.My boost converter outputs 12.2V without a load but once I connect a 12VDC fan that takes 0.300mA, my output drops to about 6V. Input current also measures s 1.2Amp which is the max limited by my bench top power supply setting.If I bump the current limit on the 5V input to 1.4A, then my output voltage gets to 12V and the fan rotates at full speed. However, the output voltage continues to climb past 12V and things get awfully hot.What I don't get is why it's drawing so much power.From what I understand the formula to calculate the input current is In = (Iout * Vout)/Vin = 0.96A

author: Ian.M
author_rank: Super Contributor
timestamp: March 07, 2024, 06:21:41 am
content: Our crystal balls have gone cloudy, and its too much trouble to hack your CCTV so we've got no idea whatsoever what you are doing wrong!So far it appears you *may* be using a TI switcher chip, and have a problem with unstable output voltage. Without a full schematic, your PCB layout and closeup photos of the assembled board, we simply cant help significantly, though my first suspicion is you may have FUBARed the layout and noise is swamping the feedback signal.

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 07, 2024, 06:41:47 am
content: Quote from: Ian.M on March 07, 2024, 06:21:41 amAttached. I could not fit the 220uH inductor on the existing pad so I Frankenstein'ed itThe booster IC is an MC34063AD

author: Ian.M
author_rank: Super Contributor
timestamp: March 07, 2024, 07:20:08 am
content: It looks like you've got a massive loop area for the input cap => current sense => inductor => IC switching transistor => ground, and when the transistor cuts off, its extension from the inductor => diode => output cap, with the feedback divider in the middle of it, thus inductively coupled to it.The capacitors should be right next to R5 and the diode, with their ground legs towards the pin 1,8 end of the IC to minimise the loop area. If you've got a full ground plane, you may be able to fix it by drilling the board next to the cathode of the diode and the input side of R5 and bodging in more decoupling. Also try reducing the divider resistors by an order of magnitude to reduce the impedance at the feedback pin. It may also help to flip the inductor over so its leads cross to reduce the loop area.

author: pcprogrammer
author_rank: Super Contributor
timestamp: March 07, 2024, 07:45:56 am
content: Quote from: newtekuser on March 07, 2024, 06:08:34 amYou forgot to include efficiency. Only under ideal circumstances is the above true. The chip you choose was already discussedhere, as being not so good due to the use of BJT's for the switching.The efficiency found in the datasheet is at best 89% for a step up converter and at worst 58% for an inverting converter shown in the different test cases. So assume 80% which already brings up the input current to 1.2A.The datasheet also shows sample PCB layouts which are wise to follow.

author: Berni
author_rank: Super Contributor
timestamp: March 07, 2024, 07:55:12 am
content: There is so much wrong with the PCB layout indeed. That diode also looks undersized and doesn't have any copper area around it to aid in cooling.The chip used is also a old obsolete switchmode chip that has poor efficiency.For these kinds of things it is best to just buy ready made modules from Aliexpress. They cost less than what it would cost you to even just buy the components, let alone make a PCB and cost of your time to assemble it. The efficiency of those cheap modules is not quite world changing (since they are built down to a price), but it is good enough to give the power levels the listing says it does. You can still pimp it out for higher efficiency by giving it a better low mOhm inductor or better diode.

author: Ian.M
author_rank: Super Contributor
timestamp: March 07, 2024, 08:10:02 am
content: Quote from: Berni on March 07, 2024, 07:55:12 amThat very much depends on the module. Typically smaller Chinese DC-DC modules are good for a continuous load of under 50% of their nominal current or power ratings, and maybe 90% of their max. voltage rating. Try to push them anywhere near their ratings and they overheat within seconds.'LM2596' modules are a special case - most Chinese market LM2596 ICs are fakes, and seem to be clones of the older LM2576 IC. Unfortunately that operates at approx. 1/3 the switching frequency so if the design equations in the LM2596 datasheet are used (which is typically the case), the result is the inductor and decoupling caps are only one third of the value they should be. The current must thus be derated by a factor of three, (in addition to the 50% 'Chinese' derating) so a '3A' module is typically only good for 500mA.

author: Berni
author_rank: Super Contributor
timestamp: March 07, 2024, 08:16:06 am
content: I have bought a bunch of difference cheap Aliexpess DC/DC boards and they mostly seamed to have domestically designed Chinese chips on them.Maybe i was lucky but they did perform to what the listing says they do. That being said running a 3A rated module at 3A continuously will make it get ridiculously hot, the efficiency wont be great..etc but it does work. But when you run them at 50% load they run at a reasonable temperature. The PCB designs of the modules also looked fairly decent with minimized loop area and such.But yeah Aliexpress is a bit of a lottery pick at times, you do run into crappy fakes here and there. I just make sure to test stuff i get from there before using it.

author: Ian.M
author_rank: Super Contributor
timestamp: March 07, 2024, 08:20:30 am
content: I'd *rather* have a Chinese IC with a Chinese language datasheet, on any such module because it improves the odds that the design engineer understood the chip's design equations and ratings properly!If you've got the fake 'LM2596' modules and decent rework skills, it may well be worth fitting a genuine LM2596, which should get you up above 1A continuous, as you cant get the board and other components for the price of the module. However the diode is typically under-specced with insufficient trace area to handle its dissipation, and the inductor is likely to have a higher DC resistance and lower saturation current than desirable, so odds of getting anywhere near 3A are poor, even with forced air cooling.

author: pcprogrammer
author_rank: Super Contributor
timestamp: March 07, 2024, 11:06:50 am
content: A module likethisone is certainly enough for the OP's purpose. According to thedatasheetit can switch at max 4A.I have several of these modules but not used them with high loads in boost mode yet, but for buck purposes I have used the XL40xx versions and they are working fine on average load. I use one in my CV system for powering high torque brushless 6V servos from a 24V supply. Have not measured the currents but the module is lasting for several winter seasons without bringing up the temperature in the box up to high. ~32 degrees Celsius after a full day of operation.

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 07, 2024, 03:42:48 pm
content: Quote from: Ian.M on March 07, 2024, 07:20:08 amThanks for the suggestions! I am using a full ground plane on the bottom layer btw.

author: pcprogrammer
author_rank: Super Contributor
timestamp: March 07, 2024, 04:24:41 pm
content: It is not specified in the datasheet, but it might help to put a 100nF capacitor between pin 6 and ground as close as possible to pin 6. Might be that the ic is oscillating on the supply at high frequencies due to not being properly decoupled.I have seen this solve problems multiple times with different types of circuits.

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 08, 2024, 02:14:18 am
content: Quote from: pcprogrammer on March 07, 2024, 04:24:41 pmThanks, I'll try that! Btw, I put the board under a thermal camera and the only component that gets hot is the IC.I will re-route traces to put components closer to each other and also downsize the divider resistors as Ian.M suggested. If that doesn't work still, I'll try a different IC that is more efficient.My understanding of loops is that at least I should have no ground loops in my design because I'm using a solid ground plane and the return path for GND is "instantaneous" since the path between the component and the ground is direct.Based on this my options are either going four layer board and route power via a dedicated power plane, or as suggested, minimize the loop by re-arranging components closer together. Is this correct?The reason I'm not going with a device from Aliexpress is that I want to integrate this into a bigger project of mine that I'm designing from scratch and do not wish to use pre-made parts for it.

author: Ian.M
author_rank: Super Contributor
timestamp: March 08, 2024, 02:40:58 am
content: Before giving up on this revision, drill the board (not on any track) next to the outer ends of R5 and D1 as I suggested above and next to pin 6 as Pcprogrammer suggested to get access to the ground plane and patch in 100nF thru-hole ceramic caps for more and better placed decoupling to see if it helps. Also, try to lower the divider resistances. If you don't have the exact values, do what you can and see if it then holds the correct output voltage for the new divider. With a 1K R2, R3 of 8K2 would be 0.5V low, 9K1 would be 0.63V high and putting the just removed 10K on top of the existing R3 in parallel would be 0.47V high, plenty close enough for a 12V fan.If my suggestions help, its then a good idea to respin the board to get the inductor on and improve the layout with fair confidence it will work as expected, though I'd still place pads for the extra decoupling so you can add them again if it doesn't behave with just the electrolytics.

author: ArdWar
author_rank: Frequent Contributor
timestamp: March 08, 2024, 03:33:25 am
content: The IC getting hot isn't unexpected, it isn't exactly efficient switcher and not particularly good at heat dissipation either (~90C/W is surprisingly bad tbh).The "loop problem" isn't (necessarily) about ground loop. Your switching loop is uncomfortably large and "loose". For boost converter, the critical loop is between switch transistor, output diode and output capacitor. at least try to make that loop as small and tight as possible. Granted your IC isn't exactly the fastest switcher so layout isn't as critical, but this is a good practice nonetheless.

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 08, 2024, 03:57:21 am
content: Quote from: Ian.M on March 08, 2024, 02:40:58 amYou were not kidding, I thought it was a figure of speech for actually adding the cap to the revised boardI’ll drill it, sure!As for the resistors I was thinking a 5.1K and a 44.2k for 12.08v Would these do?

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 08, 2024, 04:00:40 am
content: Quote from: ArdWar on March 08, 2024, 03:33:25 amThank you very much for the explanation on the switching loop! Next time I’ll respin the board I’ll correct that.

author: Ian.M
author_rank: Super Contributor
timestamp: March 08, 2024, 04:14:15 am
content: That's only a factor of two lower divider resistance, which may not be enough to swamp the noise we suspect is messing up the feedback. Go with my suggestion of ten times less, and if it improves stability, you can then get the quiescent current back down by removing R4, the 10K load.

author: MrAl
author_rank: Super Contributor
timestamp: March 08, 2024, 07:08:34 am
content: Hello to all,This has already been said, but I'd like to stress the efficiency issue a bit more.With an ideal converter, the input and output power relationship is simply:Pin=Poutand in terms of voltages and currents:Vin*Iin=Vout*IoutHowever, no converter is ideal which would mean 100 percent efficient. Instead, there is an efficiency factor that is always less than 1.For a real converter, the power relationship changes to (Eff is the efficiency factor less than 1):Pin=Pout/Effand since Eff is always less than 1, the power input is always greater than the power output.In terms of voltages and currents again this would be:Vin*Iin=Vout*Iout/EffUnfortunately, the efficiency Eff can be kind of low. To start, 80 percent, which means the efficiency factor Eff would be 0.80, and that means the input power will be a lot more than output power. This is easy to calculate too.Let's start with an input of 5 volts and an output of 10 volts at 1 amp.Since the input and output voltages are considered constant, and the output current also constant, the power relationship can then be written as:5*Iin=10*1 ( where again 5v input, 10v output at 1 amp output).That would make Iin equal to 2 amps in an ideal converter. Considering the efficiency factor Eff this would become:5*Iin=10*1/0.80or simply:5*Iin=12.5and so now that means Iin is higher than before by 25 percent:5*2.5=12.5so now the input current Iin is equal to 2.5 amps when in the ideal converter it was only 2.0 amps.Now let's drop the efficiency a little more, down to 70 percent. We then have:5*Iin=10/0.7or:5*Iin=14.29and this means the input current went up to about 2.86 amps.We can start to see how this messes up the original calculation with an ideal converter, and how much more current that we might see on the input than in the ideal case. Thus, the efficiency always has to be taken into account.With 5 volts input and 12 volts output and 1 amp output and 75 percent efficiency the input current would be:Iin=3.2 amps.With only 0.5 amps out the input current would be:Iin=1.6 amps.With only 0.3 amps out the input current would be:Iin=0.96 amps.All of these Iin figures are the average current levels though. The peak levels depend on the inductor value and the pulse width.If the inductor value is too low this will mean higher peak currents and that could mess things up also. If that is the case, the inductor value has to be increased. Because the inductor value is in the denominator of the current calculation, a larger inductor reduces the peak current. Because the pulse width is in the numerator, the shorter the pulse time the lower the peak current. Unfortunately, it is possible that this chip can skip a pulse which means the next pulse might be longer than usual. This would cause the inductor current to go up higher than usual, so the inductor value gets more critical.

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 08, 2024, 05:47:00 pm
content: dont expect anywhere near 100% efficiency for boost, think of current wasted to ground. my rule of thumb for boost is 50% efficiency. so to drive 12V 0.3A (4W) will need 8W from input ie with 5Vin equates about 1.6A thats already exceeding the top of IC spec, not really a good idea as well. next thing is check your inductor value, too small and it will into saturation quickly and IC will smoke. oscilloscope and a little bit experience to see switching node is golden at this stage. i tend to go with larger than recommended inductance, less stress to the IC, although ripple and noises may suffer, but its for later tuning as cooler switching IC is much more critical imho. this is what a lazy math stuff laymen guys like me will do fwiw...

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 08, 2024, 06:39:12 pm
content: Thank you all for taking the time to explain this! I’m not writing this IC just yet as I only need 12V briefly (less than 30 seconds) but are there any drop in replacements that are more efficient?

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 09, 2024, 02:54:26 am
content: Attached my revised design. I have resized the inductor footprint and diode and re-arranged components on the board. However, I can't come up with a layout where both traces from the inductor to the IC are short due how IC and inductor pins are placed, so I have to choose between making 5V or 12V traces long. In this case, I positioned components so that 12V traces are shorter.With my limited experience routing PCBs the only way I can see to make traces shorter is to go to a 4 layer board.

author: mariush
author_rank: Super Contributor
timestamp: March 09, 2024, 05:37:53 am
content: The maximum switch current of a MC34063 is 1.5A ... with 5v input, 12v output you're gonna exceed that with more than 250mA output current.You can use any online calculator and it will tell you that, for example tryhttp://www.nomad.ee/micros/mc34063a/There's a beefier version MC33167 / MC34167 that has a 5.5A switch current limit so it would handle higher output but it's more expensive and at this point, there's better cheaper switching regulators.

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 09, 2024, 06:03:26 am
content: Quote from: mariush on March 09, 2024, 05:37:53 amI looked at those but they have a min input of 7.5v and my supply voltage is 5v.Thanks for sharing the calculator btw!

author: mariush
author_rank: Super Contributor
timestamp: March 09, 2024, 06:29:42 am
content: Use something like MP3437 for example :~2.3$ SOT-23-8 / TSOT-23-8 :https://www.digikey.com/en/products/detail/monolithic-power-systems-inc/MP3437GJ-P/1809039310 QFN :https://www.digikey.com/en/products/detail/monolithic-power-systems-inc/MP3437GRP-P/18089144Cheaper but a bit less efficient :~1$ PAM2423 :https://www.digikey.com/en/products/detail/diodes-incorporated/PAM2423AECADJR/4033258

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 09, 2024, 06:41:18 am
content: Quote from: mariush on March 09, 2024, 06:29:42 amNice specs, many thanks!

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 09, 2024, 10:37:13 am
content: Quote from: newtekuser on March 09, 2024, 02:54:26 amimho, inductor/diode placement is less critical compared to feedback resistor as long as you dont place them unreasonably far away. but then feedback resistor layout will only affect transient/switching respond, this is a later stage to fine tune to get minimal output noise. but i guess your problem in OP has nothing to do with pcb layout, its overloaded / current limited issue ymmv.Quote from: mariush on March 09, 2024, 05:37:53 amthats why when i cant find suitable IC (those smps ICs that i found are pathetic when it comes to max current) i will go will external switching mosfet. with external mosfet we are free to choose for tens or hundreds of amperage output, but for smaller app and lower noise, built-in mosfet in smps IC is recommended. ymmv.

author: shapirus
author_rank: Super Contributor
timestamp: March 09, 2024, 01:36:40 pm
content: Quote from: newtekuser on March 07, 2024, 06:08:34 amYou won't get more than ~250 mA on the output at 12 V when boosting from 5 V input with the MC34063A, unless you add an external switcher, and by then you'll be better off using a different chip altogether (take a look at e.g. MT3608, which requires an even simpler external circuit to operate).The MC34063A has a maximum allowed current of 1.5A for the internal switcher. It is also rather picky about the inductor you use. But once you get the external components right, it works quite well for what it is.Quote from: newtekuser on March 07, 2024, 06:08:34 amWhat is the value of Rsc? What is Isat of your inductor?You may find it helpful to refer to MC34063A calculators such ashttp://www.nomad.ee/micros/mc34063a/

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 09, 2024, 04:09:41 pm
content: Quote from: shapirus on March 09, 2024, 01:36:40 pmI'm using this 220uH inductor with Isat of 1.7A:https://www.digikey.com/en/products/detail/bourns-inc/SRR1280-221K/1970008Rsense value is 0.051 ohm, don't ask me how I calculated it bc I can't find my notes on it, or the formula in the IC data sheetI searched for MT3608 but Digikey and Mouser don't have it in stock.I actually purchased this part off eBay long time ago, but the number written on the IC I got is 7217P not MT3608 as in the auction listing. The package is also very small - will this be able to handle the heat dissipation?

author: shapirus
author_rank: Super Contributor
timestamp: March 09, 2024, 06:17:31 pm
content: Quote from: newtekuser on March 09, 2024, 04:09:41 pmWell I'd suggest using values provided by the online calculators as a starting point.Quote from: newtekuser on March 09, 2024, 04:09:41 pmNot sure what 7217P would be. The MT3608, per its datasheet, is supposed to be marked as "B628DC". I'm sure there are many alternatives that are similarly easy to use and can provide a decent output current at good efficiency, but I don't know them without searching. Maybe someone else can suggest something.Quote from: newtekuser on March 09, 2024, 04:09:41 pmSmall package is fine, if the resistance of the switching element is low enough, which is the case with MT3608. Besides it uses a MOSFET, unlike MC34063A which uses a BJT for switching. MT3608 is rated for 4A max switching current (not the same as continuous output current) and 0.6 W absolute maximum power dissipation. Let's say we want to run it at no higher than 0.4 W, which, at an efficiency of 90% (taken from the datasheet curves), would mean 4 W input = 0.4 W loss + 3.6 W output, and 3.6 W output at 12 V is 300 mA output. With 400 mA at 12 V on the output the power loss in the IC is going to be about 530 mW, still below the absolute maximum rating, but you do need to design the PCB with heat dissipation considerations in mind.

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 10, 2024, 03:34:17 am
content: Quote from: shapirus on March 09, 2024, 06:17:31 pmI got in touch with the seller and they confirmed the number on my MT3608 IC is the production code.

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 10, 2024, 06:40:24 pm
content: I took another stab at this design, this time using the more efficient MT3608 IC. When routing the PCB I am having trouble keeping the switching loop (IND, IC SW pin1, D) farther away from the feedback resistor network (R1, R2) to avoid coupling. I double checked the schematic and it looks in line with the example from the data sheet, but the way the rats nests are set up I must connect output of D as well as R2 to 12V output, so can't figure out how I can "isolate" them.Any suggestions?

author: Ian.M
author_rank: Super Contributor
timestamp: March 10, 2024, 07:10:50 pm
content: That's fairly easy - Flip C2 to bring its Gnd pin down the board and move its ground return current. Move R1 and R2 up next to R3, side by side to minimise the trace length to pin 3. The feedback must be tapped off at or after the capacitor pad, not at the diode, to reduce its coupling to the pulsing current in the switching loop, so connect +12V to the top end of the repositioned R1, from the C2 pad or output connector pad, running the trace up then parallel to the board edge, with a very well via stitched Gnd guard trace or fill right next to it to screen it from the switching node and diode.More via stitching for the MT308 Gnd pin would also help - if you cant fit a via under the IC (in addition to not instead of the one right of the pin), run a trace under the IC down towards the inductor with the via as close to the IC as possible.However you should be using wide traces, lands and fills for all the high current connections. Your other topichttps://www.eevblog.com/forum/beginners/mp3437-boost-converter-ic-vdd-pin-voltage/had a datasheet for the MP3437 with a suggested layout showing how to do this sort of stuff right. Study it and try to apply the style and concepts to this MT3608 layout!

author: shapirus
author_rank: Super Contributor
timestamp: March 10, 2024, 08:26:25 pm
content: BTW, while you're at it, take a look athttps://embeetle.com/#blog/return-currents. Useful!

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 11, 2024, 01:07:41 am
content: Quote from: Ian.M on March 10, 2024, 07:10:50 pmThank you very much for the suggestions! Re-attaching the revised layout. The only DRC I get now are about the new unconnected vias. I have concerns about thermal imbalances from using those fills but I'm not able to use thermal reliefs because some of the spaces are too narrow and copper won't fill through.Hope I didn't misunderstand anything.

author: Ian.M
author_rank: Super Contributor
timestamp: March 11, 2024, 01:17:44 am
content: Looks a lot better but there are a few small improvements left. Move C1 left a bit and put its Gnd via just down from it to make space to widen the trace to the inductor. Also widen the trace from the inductor to the diode. The 12V trace to R1 can (and probably should) be narrow as it carries negligible current.N.B. the new vias should all be connected topside by a ground track or fill, which you might as well join to the IC's Gnd pin for better heat dissipation.

author: shapirus
author_rank: Super Contributor
timestamp: March 11, 2024, 01:24:27 am
content: Why not rotate the inductor 90 degrees and run the traces alongside each other to reduce the loop area?

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 11, 2024, 02:35:43 am
content: Quote from: shapirus on March 11, 2024, 01:24:27 amCouldn't see the forrest for the treesIs this what you had in mind?

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 11, 2024, 02:36:58 am
content: Quote from: Ian.M on March 11, 2024, 01:17:44 amThank you very much!!! I have updated the layout in my previous post. The DRC errors are also gone after connecting the vias topside and hooking up to the IC GND pin.

author: shapirus
author_rank: Super Contributor
timestamp: March 11, 2024, 10:35:51 am
content: Quote from: newtekuser on March 11, 2024, 02:35:43 amYes, more like in the attached picture. Cut corners where you can. I also see no reason (there may be!) why not to increase the width of high-current tracks where you practically can.I'd also turn J1 and C1 180 degrees to avoid that narrowing of the power carrying track between C1 and the IC, and dragged C1 a little lower to put its ground via in line with GND of J1 and the IC.Then, I'd drag J2 a bit lower as well, again to improve the current return path, and, I think, rotate it 180 degrees too, so that its terminals are placed very close to the respective ones of C2.Draw your return current traces *before* pouring copper on the ground plane :).C2 and J2 could be moved closer to D1.Also, routing using copper filled polygons is inconvenient, as it makes it difficult to rework when you move parts. It's easier to use traces (of whatever width is suitable), as they are dragged along with the respective footprints, but mainly because the layout looks much tidier with them. Polygons are fine, but they I think should be added when everything else is ready and only where you really need them.

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 18, 2024, 10:44:33 pm
content: Another failureMy output is only 4V something and it draws 200mA just idling and getting very hot. I wonder if the IC I got is indeed a MT3608.I chose a 6uH inductor (per calculations, I should have used a 3.3uH but the IC data sheet says 4.7uH is minimum), R1 = 27K, R2 = 1.4K (0.6 x (1+27K/1.4K)) = 12.1VDuty cycle was calculated based off 92% efficiency (1 - (5x0.92)/12) = 0.61I checked for shorts and there are none.Inductor -https://www.mouser.com/ProductDetail/652-SRP1265A-6R0MDiode -https://www.mouser.com/ProductDetail/757-CMS04TE12L%2cQ%2cM27K resistor -https://www.mouser.com/ProductDetail/71-RCS080527K0FKEA1.4K resistor -https://www.mouser.com/ProductDetail/667-ERJ-P06F1401V10K resistor -https://www.mouser.com/ProductDetail/710-56011212000422uF cap -https://www.mouser.com/ProductDetail/963-MSASE21GBB5226MT

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 18, 2024, 11:19:18 pm
content: Quote from: Mechatrommer on March 18, 2024, 11:12:55 pmThe inductor value was chosen based on calculations. However, I swapped it with a 220uH inductor and it behaves the same.

author: shapirus
author_rank: Super Contributor
timestamp: March 19, 2024, 09:34:51 am
content: Do you have an oscilloscope? Check if there's any switching activity going on at all.

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 20, 2024, 01:43:13 am
content: Quote from: shapirus on March 19, 2024, 09:34:51 amI probed the switching node at pin 1 with the scope and when I measure frequency it only reads around 60Hz. Datasheet says it should oscillate at 1.2MHz. Did I get a dud?

author: shapirus
author_rank: Super Contributor
timestamp: March 20, 2024, 02:50:21 am
content: Quote from: newtekuser on March 20, 2024, 01:43:13 amQuite possibly, as long as everything is wired correctly, input voltage is correct etc. I think we can rule out the board layout, as it should work, albeit poorly, even on a solderless breadboard.So if everything else is done correctly, then it must be the IC that's wrong. But before this can be concluded, you need to make damn sure that everything else is done right and you're not seeing some normal operation under certain conditions.I'd also suggest looking for switch mode controller ICs at a reputable distributor that's comfortable for you to buy from, whatever it might be. Narrow the search down to what you can obtain, then look at the datasheets to narrow it further down to what will be comfortable to use. Maybe get a few different ICs. Try them on a solderless breadboard, or maybe on a solderful (meh) breadboard to get at least some results that'll tell you that you're on the right path.

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 20, 2024, 05:34:12 am
content: Quote from: shapirus on March 20, 2024, 02:50:21 amI did notice one issue in that I routed the connection to the EN pin incorrectly. Should have been tied it to supply voltage for startup but I tied it to ground via 10K resistor (R3). Maybe the 10K resistor wasn't necessary anyways.However, even after removing the resistor, there is no change in behavior, maybe I had damaged the IC at this point.My other suspicion was improper diode orientation, but I measured it with a multimeter in diode mode and it only conducts to the output and not backwards to the switching node.Unfortunately suppliers like Digikey and Mouser don't have this IC in stock until late April.

author: MrAl
author_rank: Super Contributor
timestamp: March 20, 2024, 05:37:12 am
content: Quote from: newtekuser on March 18, 2024, 11:19:18 pmOne of the things about a switching converter that is a little different than other stuff is that the inductor has to have the right inductance and also the proper operating current specification. Some inductors just won't work in some applications because they saturate with DC current too easily.This could be the problem, and this ties in with the inductance and the switching frequency.

author: pcprogrammer
author_rank: Super Contributor
timestamp: March 20, 2024, 06:48:37 am
content: And that is why I like to use Aliexpress modules. For myremote controlproject I'm usingsmall buck converterboards, that I take the components of and solder them to my own board. Can also be cheaper if it is only for a small series. These complete boards are so cheap that buying loose components costs more.I'm using one ofthesein a satellite receiver to power the VFD from 12V to 35V. Not a lot of current draw for this, but these modules can do about 2A on the output.Edit: Fixed a link

author: radiolistener
author_rank: Super Contributor
timestamp: March 20, 2024, 07:49:15 am
content: Quote from: newtekuser on March 07, 2024, 06:08:34 am300mA at 6V means that the fan has resistance about R = 6 V / 0.3 A = 20Ω.If you want to keep 12V at 20Ω load, then you're needs to provide P = 12^2 / 20 Ω =7.2 Wpower for your fan.DC/DC has efficiency about 80%, so you're needs to provide at least 7.2 W / 0.80 = 9 W on your DC/DC input.Since your DC/DC input is 5V, you're needs to provide at least minimum 9 W / 5 V = 1.8 Amps.But as you said,your lab PSU max current is limited with 1.2 Amps which is just P = 1.2 A * 5 V = 6 Wpower.While your load wants to get 7.2 W power and your DC/DC needs some power loss.DC/DC cannot produce additional 3 W power from nothing and this is the reason why output voltage drops down.As you can see your lab PSU just unable to provide required 9 W power.

author: shapirus
author_rank: Super Contributor
timestamp: March 20, 2024, 10:19:39 am
content: Quote from: newtekuser on March 20, 2024, 05:34:12 amBut is the EN pin connected to VCC now? "EN" means enable, active high, and the datasheet says:A high input at EN turns on the converter, and a low input turns it off.

author: shapirus
author_rank: Super Contributor
timestamp: March 20, 2024, 10:24:08 am
content: Quote from: pcprogrammer on March 20, 2024, 06:48:37 amYeah true. Sometimes, however, the level of switching noise they emit into the air is excessive and could be reduced with a proper layout. Sometimes the dimensions of those complete boards aren't what you need, and sometimes you just want to make your own for the hell of it, or for learning.But yes, if your end goal is to have a dc/dc converter and having it made yourself isn't a requirement, then aliexpress is full of ready made solutions.

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 20, 2024, 04:12:17 pm
content: Quote from: shapirus on March 20, 2024, 10:19:39 amYes, now with R3 removed, it is only connected to VCC and IN pin of the IC.

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 23, 2024, 12:42:27 am
content: Update: I purchased a set of cheap boost converter boards from Amazon so I can replace the IC on mine and lo and behold my board now works fine!It does not overheat anymore and I get steady 12.2V when running the fan.I also swapped my bad IC onto the donor board to prove it's bad and it behaves the same, only get 4V out of it.Thank you all for the comments/suggestions!

author: shapirus
author_rank: Super Contributor
timestamp: March 23, 2024, 12:50:22 am
content: Quote from: newtekuser on March 23, 2024, 12:42:27 amGreat. What are the markings on those ICs that do work?

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 23, 2024, 01:01:55 am
content: Quote from: shapirus on March 23, 2024, 12:50:22 amB628 and two dots following that.

author: shapirus
author_rank: Super Contributor
timestamp: March 23, 2024, 01:15:45 am
content: Quote from: newtekuser on March 23, 2024, 01:01:55 amSo the datasheet was rightQuote from: shapirus on March 09, 2024, 06:17:31 pm...at least, to an extent.the two dots in place of the "DC" can be packaging type or whatever.

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 23, 2024, 02:46:12 am
content: One thing still eludes me, the IC data sheet says the chip switches at 1.2MHz but when I probe the pin with the scope it measures 2.1MHz, albeit with an exclamation mark so it doesn't like something. I increased sampling to max 5GS/sec with no change.

author: pcprogrammer
author_rank: Super Contributor
timestamp: March 23, 2024, 07:07:17 am
content: Quote from: newtekuser on March 23, 2024, 02:46:12 amLook at the waveform to see if there are glitches or ringing that screw up the scopes frequency determination.Post a picture of it here so we can see for ourselves.

author: newtekuser
author_rank: Frequent Contributor
timestamp: March 23, 2024, 06:01:15 pm
content: Attaching waveforms. The first two are mine, last one is from the amazon purchase which is very stable at 1.2MHz.Besides layout difference, some component values are different: I'm using 22uF caps (https://www.mouser.com/ProductDetail/963-MSASE21GBB5226MT) and 6uH inductor (https://www.mouser.com/ProductDetail/652-SRP1265A-6R0M) whereas they use 10uF caps (measured with DER EE LCR meter) and 22uH inductor.

author: pcprogrammer
author_rank: Super Contributor
timestamp: March 23, 2024, 07:29:55 pm
content: Wow, that are some waveforms.On your second picture, looking at the signal it seems to be switching at little over 1MHz. This based on 200ns per division if I read your screen right, and the switching moment being the sharp cutoff with the high frequency ringing. About 4.5 divisions so 900ns > 1.111MHz.The rest of the waveform might be resonance in the coil.But I'm no expert on boost converter signals.