How a PSU With Too LOW Voltage Can Kill Your Electronics Devices! HOW A Buck Converter Works!
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- เผยแพร่เมื่อ 7 ธ.ค. 2022
- LER #336 I've seen this many times with items such as Android boxes and Set Top Box Streaming devices. Plug a 5V PSU into a 12V device and it goes Bang! Or at least it goes DEAD. Wanna find out why this happens and how to fix it? You just gotta press PLAY!
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Richard
If you renamed this video "How a buck converter works" it would probably be the best Buck Converter video on the internet.
I couldn't have put it better myself. I really enjoy his videos.
Renamed 😁
this is oversimplified and missing all the math (that is not complex too) that really explains why you would overload dc-dc components by undervolting them (proper ones have UVLO protections btw).
katkimshow did a great job on several dc-dc topologies here on YT and not overloaded with theory, just enough to really understand working principles and not the abstraction.
@@Mr.Leeroy Of course it is simplified, if you have to point that out then you probably miss the point of this channel. UVLO protection will not help you in this situation by the way 😉
@@LearnElectronicsRepair I understand that this channel is about repair only, and point this out in regard to "probably be the best Buck Converter video on the internet".
Why do you think UVLO won't work?
Basically the output power of a switch-mode power supply like this buck converter is fixed due to the voltages programmed by the feedback circuit and the current the load is drawing at this voltage. If the input voltage drops the converter needs more input current to produce the same output power. However, mosfet switches and inductors are designed for a certain maximum current. If this maximum current is exceeded the converter blows up.
*exactly* 😉
Yeah, depends on the load, but when converter dies, cause basically boils down to Ohm's law, less input voltage, more input current.
These circuits are designed not only with max currents in mind, but also highly optimized for input&output ranges of operation to decrease footprints and costs.
They may be configured for a more universal ranges, but this has to be accounted in design.
@@Mr.Leeroy Yes, good design can avoid this failure mode. The inductor needs sufficient current headroom for a big voltage input range. But this means more space on the PCB, higher component cost and probably less efficiency at the nominal input voltage. Additionally the controller chip must have a low voltage shut down in order to avoid even higher currents. The TPS562200 by the way has this feature. So it all boils down to the inductor selection.
@@UltraUltaderdritte at the same voltages efficiency would be better with bigger coil and beefier FETs, since this translates to lower ESR.
Efficiency usually lowers as you increase input to output voltage difference, but it depends on load percentage too.
Frequency is a big factor for safe operating range as well as inductance.
@@UltraUltaderdritte Exactly. You hit the nail on the head there!
Richard, you are the best teacher of electronics on TH-cam. I have watched a ton of your videos and I am never left wanting more info so I can understand the concept. Every video I learn key parts that fill in the missing pieces of the puzzle, for me to really understand certain concepts of electronics. Thank you my friend.
Yeah! I almost regret his channel was not the first I found before I got in to my retro HW hobby. It is absolutely the right place where to start if you are not graduated in this branch and you need to catch the important principes to go further.
there's also RSD Academy
Thank you for answering a question that has bugged me for ages 😃I couldn’t understand why a DAC board failed. I was certain on no overvoltage . Never considered UNDER voltage.
Many grateful thanks 🙏🏻
I have to admit that I've learned more about this type of electronics from you in a month than the last several years of watching and reading tons of other information. I suppose if i didnt have a good grounding from automotive and the other studies i've done I wouldnt learn as much, but you have an excellent teaching style and ability to communicate as well as a hell of a schematics drawing and explaining skills. Cheers and I hope to see the beauty of your island for myself some day.
My God, before I thought that it's not possible to kill it with lower Voltage, but now I see clearly, because the power supply for the lower voltage (5V) will be also designed to give more amperage than 12V PSU for the similar device. Probably 5V 1.5A vs 12V 1A :)
Thank You, you are awesome!
Such a great instructor. Always paints the whole picture in a very easy to follow explanation.
Thank you very much for this one! I never thought of that, that too low voltage can be dangerous.
As usual your hand-drawn schematics and info make it clear quickly. Thank you again!
Excellent video. I did A-Level electronics 30 years ago but never really got involved since, because I now work in IT. But I’m still interested and this video makes me realise how simple apparently complicated electronics can be!
Thank you
You are the best teacher I've ever seen. I'm from Brazil and I'm a Sgt of the Brazilian air force, but I LOVE electronics! You are REALLY helping me to understand how stuff works. God bless you Master! 🙏🙏🙏🙏🇧🇷🇧🇷🇧🇷🇧🇷
Thank you!! FinallY I understand the buck converter, voltage reference etc. After like 10 videos and tutorial read and seen. People like you we need at our tech. schools as well. I like your span from a cofee machine up to retro Hw (aka "old junk" for most of the people). 👍
I've seen dusins of videos about buck converter and understand how they work but this is the absolutely simples and easiest understandable video I've ever seen!
Yeah, the situation is just as bad when boost regulators are used below the required input voltage - they will start to try to pull a large amount of current from the source and either die or damage something up-stream. We had one that heated up the power cables a few weeks back as the bench supply used for testing was set too low (it uses an external MOSFet to generate higher current)...
Very informative, I'm not technician nor an Ece but I understand the concept and the logic the way he explained it. More power to your channel.
I killed a buck IC from a mini-pc, trying to use 6v instead of 12v. It convert 12v to 5v, and then the lower voltages are from the 5v. so i removed the IC/coil and put a buck converter board inside, and routed the output to the 5v rail. Worked fine. Took me a while to understand why a lower voltage burned the regulator, but eventually understand the limitations of tiny regulators.
you are awesome. no one could ever explain this better. now it all make sense why there is inductor in power supply circuits. i’m a software guy and electronics is abit hassle for me to learn. thanks for being such awesome teacher ❤
These short vids are great... you are a great teacher.
A great description. Thank you, I learnt loads ☺
Really clear and interesting description of the circuit and its working. Many thanks !
Excellent video Richard! such a pleasure tu watch your explanation.
Thank you for such a clear and easy to understand explanation. Good educational theory.
Another great video with an easy to follow explanation. Thank you Richard.
The best teacher!! You know exactly what is important to focus on.
Good video overall. I'd like to add that the upper right regulator here is a switching regulator which seems to be generating the 5V supply, not the 3.3V one. The 5V supply seems to be supplying the 3.3V and 1.0V regulators which seem to be of the linear regulator type. The triple rail IC generating the secondary 3.3/1.5/0.95V might also be fed off of the 5V rail as well (as the linear regulators suggest that the 3.3VSB and 1.0V loads are small).
That was a good video,you explain the circuitry very well I actually learned something new thanks .
Every day is a school day and today you taught me something, thank you :)
There is one more cause: at too low gate voltage the mosfet will not open completely, so it will have way higher RDSon and therefore dissipate a lot of heat. I had this case on a UPS with a dead battery. One of the mosfets exploded right before the protection circuit disconnected the battery.
What an eye opener. Thank you Sir, stay well.
Superbly explained!
Awesome video as usual Richard! Thank you!
Huh. you made so much sense of things. ive watched videos before but things didnt always click but i like your teaching style ^^
Thanks! Brilliantly explained. I'm only learning electronics and this type of well-explained example is very helpful.
Thanks Adam for the donation - it is much appreciated 🙂
The average current through the inductor (and by extension through the FET) should still be the same as when it was running on 12V though, it would just be the spikes that are higher. Which is still enough to cook a FET since P=RI^2. Increasing frequency might fix it though, something tells me a current-mode switching regulator IC might just handle it.
More importantly in my mind is that the 12V rail is likely being used for gate driving, so the R_DS ON would be significantly higher when running at a lower voltage. This is often what UVLO protection is meant to avoid, though ultimately it depends on the intended input range of the switching regulator.
Or you know, use a switching regulator with thermal shutdown.
Brilliant! Best explanation I’ve heard.
Wow... finally I understand how buck converters work
The best teacher! Wonderfully described
That was brilliantly explained!
A very good, clear explanation. Thank you.
Great tutorial Rich
In the diagram, the freewheeling diode / low side MOSFET was missing, which is used to keep the current flowing into the output from the inductor as its field collapses.
The main issues I can see with a lower input voltage would be heat related, likely due to the MOSFETs not being fully switched on at the lower gate voltage, the longer on time of the high-side MOSFET, and maybe the higher average input current (the peak inductor current will be the same in both the 5V and 12V cases - typically less than twice the output current).
very interesting! I feel like there is an analogy to the classic 'greedy cup' water puzzle going on here. I have a lot to learn.
I brought this up in a forum over 10 years ago, said undervolting can kill a circuit and it wasn't taken too well and I didn't know enough to say why, just knew, from trial and error as well because i learn the hard way lol.
Brilliant analysis!
Excellent presentation!
Now I understand it in correct manner.
Before watching this video and without board examination, I just assumed, at lower voltage, more current would be draw by the load because of P=V*I. Therefore, circuitry traces, at least, would not handle it.
That was a very good explanation, thanks.
Essentially these are negative-resistance devices, and you often don't think of putting too much current into a device since normally the device can be depended on to limit current and the power supply just needs to limit voltage. I've always imagined this could be the case but didn't think it would really cause things to get damaged.
Awesome explanation. Thanks.
What a great factual video. Thank you.
Great vid Rich
Interesting. While I was working as a computer repair tech, one customer brought in his old Compaq laptop, saying for some reason it kept shutting itself off. So of course we tested it with the power supply he brought in with it to confirm the issue.
Sure enough, it would power on and start to POST, but it was like the moment it tried to spin up the hard drive it would shut down dead. When we looked at the power supply, we found that he was using a 9 volt power supply LOL!
For anyone not familiar, a laptop charger/power supply should be between 18 to 20 volts. I have no earthly idea how the thing even attempted to power on with a 9 volt power source, but it sure enough tried!
Honestly I wouldn't have been surprised if the laptop might have actually run if it had been connected directly to a car battery. I mean not like we'd dare do such a thing with a customer's computer, but it sure made me wonder.
I dunno, I guess the voltage regulator on the board was fairly flexible regarding input voltages, I would have never expected it to even attempt to turn on. At least it didn't damage the system. 👍
Thank you Richard!
thanks for the video im learning a lot
The current MUST flow through the inductor, but there is a voltage drop, and this is therefore a form of energy transfer which is what is stored as a magnetic field in the inductor. The current cannot rise instantly, it rises at di/dt = V/L, but any current entering the inductor must exist at the exact same quantity.
That was simply brilliant.
Good explanation thx for sharing with us.
Brilliant Richard.... took me right back to school 😉but this stuff was skipped 🤔
Well explained and simple thank u
Thank you, I also have a electronics background though wasn't never really my professional life (IT was), that said, my hobbies are in fact electronics+IT combined hehe, so this is quite handy knowledge :)
Good explanation
good video. thanks for sharing.
Would've loved a tiny ending section with common ways designers can incorporate protection in a scenario like this.
How do you mitigate it on a circuit level?
Thank you!
عمل رائع وممتاز احسنت
Good work
TY!!!
I think this is a design problem, if lower voltage kills your device, they intended to make it not last long/cheaps out too much. Many instances can make a voltage changes.
It's a dodgy devices, intended for it to be replaced often, to drive more sales.
So i think the title is a bit of misleading. A good device design will just stays off despite more current is being applied because of too low voltage.
Anyways, the explanation is good, thanks.
The feedback is compared with the ref voltage. According the difference the pwm is generated. The question is then when we have a lower voltage in, what do we have as vref. Along as we have the fixed vref, nothing will be destroyed. Is there anything in de datasheet that gives us a relation between vref and vin before stating dedtruction.
16 minutes lol but well worth it. Great explaining
Another device that could die from undervoltage is the Dragon from Atmel (now Microchip)
It has a boostconverter to make 12V from the 5V USB power. If the computer cannot supply enough current, the boost converter tries anyway and thus killing itself with overcurrent.
Cheers 👍🇳🇱
Heya, that was very clear to on school I didn't get that part lol
Hello, I have very little background in electronics and I wanted to ask a question. I have wirelless analog modulation UHF headphones. If I increase the source volume, then increases amplitude and logically should increase radio signal strength and coverage and vice versa? Or is it a misunderstanding and it works in different way? Thanks.
I'm getting mixed signals, that what you described works closely to the step up converter...
Interesting. I used a power tool battery to power my laptop via the power plug.The laptop would pull juice off the battery and run just fine and then switch to the internal battery a little early but I was getting like 80% of the useable power out of the power tool battery. However, this 18 volts nominal (instead of the 20V the laptop power adapter provided) must have damaged one of the power control components because the laptop died though it was when it wasn't powered by the tool battery at the time. But it's possible something else killed it. I have bought something like 8 of those laptops as parts machines and have only seen one other bad motherboard (It was really bad killed the guys ram and he had bought a replacement motherboard but didn't realize his ram had gone bad, probably killed the CPU too.)
Can't the chip which drive the mosfet limit current (so the amount of time it stay on) to stay within safe operation ?
I've got a 4 port Geffen HDMI switch that something has happened to and I can't figure out what actually went wrong as there is no visible damage to anything, as far as I can tell anyway. The difference with this unit however, is it actually took a 12v psu plug (i guess they never got the memo)... I still have the thing thinking I'll figure it out at some point. I've got a different unit in it's place so it hasn't been on my priority list.
What is the part number of the ic that provides 3 switched output voltages?
Great explanation! I learned a lot about cowent destroying chips 🙂
How many resistors, capacitors, and tracks got chopped in half with that metal pick swinging around?
Is it bad to use 5 volt adapter on 6 volt circuit, or does it depend?
I remember having a 12 volt tv box and i modified it to run with 5 volts usb, and i had another of the same of those 12 volt tv boxs and i tried to run it with 5 volts and then i heard some sparks and orange lights coming out of a 3 pinned smd looking thing.
This seems a bit counterintuitive to me. If you don't have enough voltage to start with, how does the chip even run in the first place to even supply the 5 volts instead of 12 to the mosfet. If the chip is insufficiently powered how can it do anything? Doesn't it need some kind of clock cycle with sufficient voltage on the supply side for it to switch on in the first place?
Nice
Yes my master!!!
Remember as the voltage drops the current goes up .Some component don't work properly and can't regulate the supply voltage
I experienced totally different situation. Swapped 12 with 5 v on molex to connect cooler and forgot to swap back. Connected DVD drive and it broken.
10:45 Your explanation makes it sound like the (1.5V) output is only supplied by the collapsing magnetic field in the coil. Really it is supplied through the coil while it's energized and by the coil, while it's not energized. Essentially it's just an LC low pass filter with a flyback diode, in order to create a path, for the coil to power the output. But you kinda omitted the diode in your example anyway. And in most buck converters, the diode is replaced with a MOSFET anyway, which is driven with the inverse of the PWM signal, to reduce power losses caused by the voltage drop over the diode. But I guess that kinda beyond the scope of what you were trying to explain.
I plugged my power supply into the USB connector on my android head unit, is this repairable or is there no protection on the USB circuit?
That's an unexcusable design flaw. An undervoltage protection for that circuit is as simple as a comparator (comparing the supply voltage to a voltage reference) connected to the enable pins of the buck converter ICs
There's no UVLO in this ic ? Thanks
Although the video was super interesting and I learned about how those voltage converters are working, I am still left wondering why it wasn't the chip that died, but rather the (25V rated) capacitor next to it. I am assuming it also couldn't handle the higher current, although it looks pretty big and beefy?
From reading research papers about ceramic surface mount capacitor failures, its mostly from localised thermal cycling of the board, bending the capactor enough to internally crack some of the layers and cause a short circuit.
I think it's just by chance this mag box had a shorted capacitor. When the customer brought it to me he brought a 5V PSU with it which he said he had connected, so my first assumption was that it would be chip that had blown as it is a common fault in these circumstances. It's possibly in this case that the capacitor failed for other reasons, and the customer could have previously been using the correct power supply and tried the 5V one when it would not work, but the issues explained in this video are still relevant.
So then why you even start with that one chip that needs 12 volts in the first place? Why not just set it up like the other box, starting with 5V? And what would be the problem with just using resistor-based voltage dividers?
Your explanation doesn't actually seem quite right...
It would make perfect sense if we were talking about making a simple buck converter from scratch, but these integrated ICs are actually much more sophisticated than that nowadays. In particular, I don't think I've ever seen one of these integrated chips that doesn't feature overcurrent protection as a standard feature, so it's pretty much impossible to blow the chip by just pulling too much current through its internal MOSFETs as you suggest. They're also generally designed to be fairly input-voltage-independent to specifically avoid this sort of issue.
What's more, looking at the datasheets for some of the specific models you listed, the TPS56x200 chips support an input voltage from 4.5V to 17V, and show reference designs and tables which list the component values required for different output voltages. None of these tables or implementation notes suggest that you need different inductor or capacitor values depending on your input voltage at all. They're just not designed to require that.
I could see potentially running into a problem generating the 5V supply on that board, given that the input is only 5V, so obviously the output can't go that high... In that case I'd expect the chip to either supply a lower voltage or just shutdown, though.
You are totally right. I am not sure, if the author really understands the principle of buck converters but omitting the flyback diode is always a bad omen. Note, that in normal conditions the inductor current does not depend on the input voltage, but it’s determined by the load. And it should not vary much. However, in case of boost converters, he would be right.
I recently had a 12v zvs driver die when the 12v battery dropped to 8 volts which killed both MOSFETs
Great diagram and tutorial as always but I'm not sure about it because if the input voltage drops then the PWM generator will have to apply the voltage for longer that makes sense but that's because the current needed to create the correct field and charge the cap will take longer but no extra current is needed, its same current out as comes in for the inductor and that's set by the load, the only way I see a problem is if the voltage drops to low and the reference voltage gets messed up then anything can happen. By dropping the input voltage there is less current and therefore less power available in the system. Well not sure I got my point across very well and I understand what your saying but I think there's more to it, maybe someone can tell me I'm talking rubbish down below....that's my 2 penneth worth !...cheers.
@andymouse123 Your 2 penneth worth is correct, but..... yeah there is always a but...
The circuit designer will know what the supply voltage will be and what the maximum load on the voltage rail is going to be, and calculate the pulse width required to supply that current.
They will then select the inductor such that, even at maximum load, it does not reach saturation (fully magnetised) and they will leave a bit of headroom as well (say 25% or something). Then you come along and stick 5V into the device.
The PWM obviously will try to maintain the correct output voltage, but it will have to use wider pulses to do it. So one of a few things are going to happen now.
The buck controller may still be able to produce the required output voltage (working hard to do it, therefore under more stress than usual).
The buck controller may reach maximum duty cycle and simply can not reach the correct output voltage.
The buck controller may drive the coil into saturation (or close to it) because the circuit is now running outside of the parameters the designer intended - and a saturated inductor is the same thing as a short circuit. Game Over!
Most likely the buck controller will have UVLO protection (under voltage lock out) to stop it running with too low an input voltage, but if the minimum permitted supply voltage is 4.5V for example, and the inductor saturates at 5V supply, for reasons explained above, UVLO will not save it either. I hope that helps clarify things. 🙂
@@LearnElectronicsRepair Thanks !
@@LearnElectronicsRepair lowering supply voltage will actually reduce peak current in the coil.
Load current is equal to average coil current, so making pulses wider will actually reduce peak current, making peak equal to average on Vin=Vout with 100% duty cycle.
Yes, high-side MOSFET RMS current is most likely what will get it fried, though not due to saturation.
Low-side MOSFET/diode will have it easier, though)
Im not quite sure i understand why all this circuitry is required just to step 12v down to 1.5, i can only assume the main reason is to maintain the 1.5v at a steady voltage , which stays steady despite differenced in load . Is this why these converters are used ?
No. it's mostly due to energy efficiency. If the supply is 12V and the output is 1V, and the load is 1 amp then the power dissipated is:
Load 1 Volt X 1 Amp = 1Watt
Linear Voltage Regulator 11 Volts x 1 Amp = 11 WATTS
Get it now?
Given I know that this particular device is built to a price and/or designer doesn't expect user to plug in incorrect supply, but if a designer so wished, what could be implemented into the design that would prevent the device from starting up unless it sees the correct voltage, supervisor chip of some kind? something that generates a power good if voltage conditions are met? what would be the solution? or is there more than one way to solve that?
Low voltage cutoff circuit would work.
Those boxes are tremendously cheap thus very low quality design. I have seen blown units because of too small heatsinks, burning AC/DC adapters etc. Sad story - we are fighting e-waste and still flooding world with crap. Thanks to people as Richard it mihht be a bit better.
@@user-dx8dc1be5k UVLO (under volts lock out) will not help you here if, for example, the UVLO cut off voltage is 4.5V but the inductor already saturates when the supply voltage is 5V, due to the longer on-time. Once an inductor is saturated (fully magnetised) it is basically a short circuit and the controller/mosfet will be destroyed, Even driving an inductor close to saturation is not going to be very good for the buck controller.
@@LearnElectronicsRepair the uvlo , you mention here is of that built into the chip...
The circuit designer has to provide an external u.v. shutdown based on input & output requirements....
From the looks of it , this seems a voltage mode buck. i.c. so some external overcurrent
protection would help... As seen in data sheet, there's an EN ( enable ) pin for such external add-on... protection..
Another point to consider in buck switch..., they are " high " side switches...so its max. g-s drive is the difference between input & output voltages...( assuming an N ch.mosfet ) , with source lifted to v- out.
Also the mosfet gate driver requires at least 8-10 volts ( bootstrapped...see i.c datasheet... there's a .C-bst between pins 2& 6) ...for reliable switching...else turn on is not proper & device tends to operate in linear region (can cause thermal runaway failure ).....good gate driver design always have a gate driver u.v.l.o. too.( latch mode) for extended switch life..
Needs a duty cycle limiter in circuit
so where does the reference voltage come from
I don't understand one thing if instead of 12 v put 5 v the buck converter should be more safe because deal with less voltage... And the pwm shouldn't have so hard work because the 3v the time for open and close the MOSFET, to generate the 1.5 v is not so hard like one 12 v spikes... And it doesn't need so much current to generate the 1.5 v ... Right? So I don't understand the title of video... Greatings from Portugal 🇵🇹 🌟
@Henrique Ferreira That may seem to be the obvious case, but there are other considerations here. If the mosfet is in the on-state for long enough to full magnetise the inductor (drive it into saturation) then the inductor becomes just an ordinary bit of wire and effectively shorts the mosfet through the load to ground, killing the mosfet, and quite possibly the load which is suddenly exposed to 12V or 5V instead of 1V or whatever it prefers to consume.
When you apply too low a supply voltage (wrong PSU) then the PWM must turn on the mosfet for longer on each pulse to get the correct output voltage, and it will try to do that.
Now the circuit designer expected 12V supply and knew the maximum possible load current of the device/voltage rail so he selected an inductor of a suitable value to handle the expected pulse width, plus a bit more for headroom/tolerance ,and never get close to saturation. When you come along with a 5V PSU and the PWM is now driving the inductor with pulses 100% or a bit more wider than the expected maximum, it is quite possible the inductor will be driven into saturation (instant death for the mosfet) or driven close to saturation (slightly slower death for the mosfet) I hope that helps clear it up a bit