700F Super Capacitor Protection PCB - Reverse Engineer Update
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- เผยแพร่เมื่อ 1 เม.ย. 2018
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A final look at the protection PCB for the 630F and 700F 2.5V Nippon Chemi-Con supercapacitors. In the previous video, I missed a tiny track between the TL431 ref pin and Vcc (or CAP +). So this time I take a look at the internals of the TL431 and add a red indicator LED to the circuit. - วิทยาศาสตร์และเทคโนโลยี
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I watched the video about cell balancing and the detailed circuit analysis explanation (though I don't know).
Introducing the LED into the base circuit will cause an additional voltage drop across the LED and therefore the shunt transistor will be driven less hard, making it less effective. You might want to try putting the LED across (in parallel with) R400, with a suitable current limiting resistor instead.
I think he said he did that, but it didn't work as the voltage across R400 is too low.
Ah, I must have missed it. I reckon the voltage across R400 would be 2.5V minus the junction voltage of the shunt transistor when the transistor is conducting. It still looks like this might work. Basically, if the draining current is, let's say, 100mA, adding the LED will make it 115mA perhaps? I'd do some measurements across R400 while the transistor is active.
Transistors are current driven devices. Voltage drop on the base doesn't matter as long as it's above 0.6V
Oh, and this is hard to understand if you're used to look at a transistor as if it is a valve in a water pipe. :-)
Daniel, check Ohm's law and you know when a LED parallel to the R400 will work.
The whole point of the transistor is to "burn" power. It's not the resistors' job in this circuit. They prime reason they are here is to provide a current limit in case of very high over voltage, not to drop voltage.
That circuit is especially for series connected caps. when one capacitor charges before others, it gives a path for the current to flow to other caps.
if you want the led a little brighter you can add a resistor accross base-emitter of the D332. (1k would increase the led current to about 0.5 mA)
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I'm not sure if someone pointed that already, but it looks like this circuit pushes main current bit differently than you described (could be wrong, but maybe I'm not). When "variable Zener" saturates upper transistor (PNP), bulk of that current goes from + via E and C of upper one (same as you deciphered), but then through B and E of lower NPN. Classic path through C and E of NPN is limited via resistor. B-E path is not limited and acts as a diode. Probably that's why NPN one is much bigger than PNP element - because it is used in non-typical way and guys who designed the circuit weren't sure whether B-E junction survive this current in a small transistor.
Julian, you may need to pick up a couple of "diamond carbide scribing pens" (if you don't have any already) to do the kind of surgery you start at about timestamp 10:25 ...only a dollar-plus on ebay. I remember having them years ago--I am going to get more now.
Sir,
Its a nice video. I brought the balancing board. I tried testing the balancing board for verifying its regulation by connecting the supercapacitor and a variable power supply to the same slot and used a multimeter to measure the voltage of the supercapacitor. Not sure if the method is right because the multimeter would not be able to show the actual voltage of super cap as the source is also connected to same point. By this method, I find the voltage rising above 2.8V. Didnt check after that. Is there any way to ascertain that regulation is happening? What is the max applicable voltage to the board for the regulation to happen?
Julian do you know what's the maximum peak current for these?
I'm looking for some for li-on battery welder project.
8:23 *"when this transistor turns on fully..."* It can't turn on fully and therein lies the problem. If TL431 regulates its cathode to 2V as you found, then the S8550 PNP transistor is left with only 0.5V between emitter and base. The datasheet for these transistors says they require 0.66Vbe to turn on with a 1Vce. At 0.5Vbe you'd be getting less than 100uA. It makes sense that your red LED didn't light in series because the Vce would have been quite a bit larger than 1V at the 8550 BJT (forget about it being in saturation), including a 1.6V-1.7V drop of the red LED, and another 0.5-0.6V drop of D882. It's a surprise that the circuit functions at all! And it's not until you overstress the capacitor at 2.59V that you start getting some indication of balancing >1A, and that's really bad.
You know, Julian, I've built a very similar TL431 circuit for my capacitor charger (HV, of course!) and I've noticed something.
For both the transistors, there's no base resistor, so shouldn't the PNP base be shorted to ground through the TL431 and the NPN base to Vcc through the PNP? I should think a large current flows.
NNNI No, both transistors are operating in their linear region. The PNP transistor is only turned on a little bit, and so the current to the base of the NPN transistor is limited. That's why the LED only lights up with a dim glow. Because as the voltage increases between cap+ (Vcc) and cap- (GND), this circuit reacts by increasing the current through the NPN transistor's collector/emitter, so you get a negative feedback effect. During normal operation (that is when the supply current is limited), the PNP transistor should never get anywhere near to being fully turned on.
Mark 1024MAK Ah, I see, thanks!
Why do you always insist on using the capacitor in your tests to characterise the circuit? It works identically without any capacitor attached, you get instant voltage/current values (because of the cap you always have to wait to reach a steady state) and you can get rid of those backfeed diodes that further skew your results. And as a bonus it takes away your constant fear of damaging the cap ;) .
Good idea. Further, perhaps a small standard cap to ensure that the circuit does not un-necessarily discharge the cap.
I like the work that Julian does. Never boring. And I like the projects he is working on.
I wouldn't be complaining now that vids are regular again instead of a week apart - it would be the same vids further apart - , just be thankful for something that is essentially free. I'm not a fan of vids with no meat, like the lamp ones, but this one had enough. Either way you can turn it off if you're bored.
im glad he tested with the large cap, i think its good practice to test the whole completed circuit. takes longer and its a bit more work but your testing the intension and purpose of the circuit fully. working with electronics for many years will teach you this, ive learned the hard way by getting trolled by some electron gremlins myself
Hi Julian,
Is your beautiful Keysight oscilloscope broken ,or did you gave it back?
With that piece of instrument you could lose the " about ", " assume " and the " who knows " and get more precise numbers. ;)
But maybe i am missing the point here,if so ignore my comment.
I still enjoy watching your vids and learn from them .
Cherio.
Jacksat I don't think a 'scope would be that helpful here. But a multimeter testing the voltages around the circuit (across the resistors and the transistor junctions) would be.
Keysight is still working awesomely :)
Shouldn't the on voltage across the Base-Emitter junction of the first transistor be ~0.6 to 0.7 V? That would give you ~2V + ~0.6V = 2.6 V across the capacitor when the circuit is protecting the capacitor. Furthermore, forget splicing the LED into the circuit, it's reducing the protection circuit turn-on sharpness. Better to tag another transistor Base onto that point and drive the LED via its Collector, with a limiting resistor in series with the LED. Collector current comes from +Vcc.
Shouldn't it be as simple as removing the R400 load resistor, and replacing it with a LED+resistor combo? Or am I missing something? Do you anticipate the discharge current to be higher than the max LED current? Shouldn't matter with a CV supply, right?
This kind of protection need something like a fuse or polyswitch because its only dissipate the overvoltage in heat if you continu rising more and more current that will blow the protection and destroy the cap
Thank you
Can I use this circuit for charging a single cap using 5v usb?☺
Now it makes sense ;-)
I am curious why the engineers have those pads for a component across the ref trace. could you see any reason why they would want to have a component in that ref line? Admittedly I haven't read the data sheet, but I assume it is to raise the TL431 working voltage? What do you think the Voltage drop (and current) would be across (through) those pads if the trace was cut?
Would your LED work there? What about the other 2 pads. where are they in the circuit. Maybe a resistor and LED ? Humm Loving your ingenuity.
The 2 unused component positions are for resistors (a potential divider) to set different turn-on voltages.
Hello Julian, can you explain me how this circuit can regulate the voltage of the capacitors for an perfect balancing. Indeed it seems to miss the 2 resistors of 180k and 12k that are R4 and R5 and that should join the "leg" reference so that the transistor driven by the TL431 can prevent any voltage to rise above 2.7v on the super capacitor. They would be unfinished circuits.
I resume, the "reference" leg of TL431 is apparently not connected.
Conclusion: your TL431 can not regulate the voltage!
If you turn the board upside down you could use a surface mount on the back.
Also could you not just use two?
Would the leads of the red LED fit down the plated through holes for easier soldering?
No, I don't think they will fit. Those holes are very small but also he would have to remove the mask (10:40) to solder it. Another reason to use larger via holes when you have the room.
what about an optocoupler where that red led is. have one on each capacitor and the other sides in parallel so then that can offset your charge current from the charger.
can you put a link to shematic to your PWM5 controller
I could put it a better LED than that, try using a ULTRHIGH Bright LED or even HYPER one. I have order quite a few PCB boards made from JLCPCB. And I am using that TL431 for my secret projects.
It would be a *LOT EASIER and SAFER* to analyze the board's behaviour without the super capacitor connected. Use your current limiting PS and slowly step up the voltage while monitoring with your DMM. You could also monitor some other points in the circuit to take some of the guess work out of it.
Then it would be easy to compare the response of your LED modified board with the original. I think that you will find that adding the LED to the base of the NPN transistor will prevent it from turning on as quickly and "move" more of the heat from the resistors to the transistor. In the previous video you already said the transistor was getting hot while the resistors weren't so I don't think you want more heat in the transistor versus the resistors. This will also reduce the amount of current the protection circuit can handle which limits its usefulness.
ElmerFuddGun the LED on the base does not affect the cathode current at all! Nor does it affect how hard the transistor turns on. Remember, a BJT is a current amplifier. It has a gain and that gain is not set by a LED. It could be limited by the LED as Julian mentioned, but the gain is probably greater than 100. 20mA x 100 = 2A (and he never got that far)
Jeff - Yes, I understand how transistors work but look at the circuit again. 2.5 V is your limit and then you have the NPN base emitter voltage, the red LED voltage and the drop over the PNP. I'm saying it might have an effect on how hard the NPN turns on and at what voltage. Consider what would happen if you used a blue or white LED instead?
But the real point is that Julian should be analyzing the circuits (both normal and with LED) without the super capacitor to fully see the effects. Like I said in the last video, this is a simple circuit that he could measure multiple voltages at different Vcc voltages and plot out the results by hand or in a spreadsheet. Testing such a circuit and hoping the capacitor doesn't go bang when you go over the specs isn't the best idea in my opinion.
ElmerFuddGun - A red LED has a voltage drop of about 1.8V at 20mA. It'll probably start to glow dim with as low as 1.5V leaving 0.7 to 1V to spare, more than enough for the transistor to turn on. VBE(SAT) at Tj=25C starts at 0.65V with IC=10mA.
The 8550 will most likely saturate (acts as a switch) and have neglectable emitter-collector volgate drop. Julian suggested 0.2V but I think that is a bit high. After all the current is tiny and the 8550 would be on as hard as it can (in order to maintain its gain).
The D882 will try its best to amplify the base current hFE times. It looks borderline and that is exactly what Julian said at 8:18 :-) Oh, and VBE(SAT) drops with Tj by the way.
ElmerFuddGun
Perhaps you could show us on your channel? Oh wait you don't have one.
Very good explanation, I am from sri lanka,❤😂
what is type of PNP 8550 and NPN D882 is? I gonna make DIY protection board by myself can y'all tell me?
I still think there's a base resistor missing to limit the base current through the NPN transistor. There's basically a dead short in the unmodified circuit when the transistor is on, which is why there's little current through the shunt resistors.
James Myatt No, both transistors are operating in their linear region. The PNP transistor is only turned on a little bit, so is acting like a resistor and so the current to the base of the NPN transistor is limited. That's why the LED only lights up with a dim glow. Because as the voltage increases between cap+ (Vcc) and cap- (GND/0V), this circuit reacts by increasing the current through the NPN transistor's collector/emitter, so you get a negative feedback effect.
Can i use this method to discharge 18650 Battery to 4.1volts?
Yep! Youd just have to populate the missing resistor divider he mentioned in the previous video ^_^
iv seen caps like these on ebay for a fiver. i want to make a portable jump starter for my little boat on deptford creek. the seller gives this little schpeel, as follows.. "Please charge to 16V then discharge to 2-4V and recharge again before DIY, and to do this for about 6-10 times,then connect it to the battery in parallel directly"
any ideas on what's going on here please?
it mean u must clear curent in cap..u can do it with led and R 1K. Resistor and Led must be series and paralel with cap.it will be clear curent for 2 hour.if led is off or curent cleared u can use your cap to your electronic.
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Just out of interest, I don't know the answer. How much voltage does the voltage meter steal ( if any) when it's checking the voltage?. If someone out there could help, that would be great
Neil Tonks A typical quality digital multimeter will have an impedance of 10M ohms. A cheap digital multimeter will have an impedance of 1M ohms. So at 2.5V a cheap meter will draw 2.5uA or less. This is such a small amount that it will have next to no effect on the circuit.
I wonder why those boards don't just fit a couple of suitably crafted zeners to discharge the overvoltage. Maybe the turn-on is not sharp enough.
Pentti Kantanen
The current through the transistor can be much higher than through a zener.
there a 1.2v infrared diod led that can work great than the 2.5v red led than consider great than the 3v green or 3.3v blue led
That's a really big cap to be pushing past it's voltage limits. I know it should be ok at 2.6v but I would not want to near it if it failed.
Nah. +0.1v or even +0.5v may get it a little bit hot and maybe shorten it's lifespan, but that wont make it pop. You have to pump 50v in reverse or something like that to make it dangerous to stay close if it explodes. It ain't about the size, some small caps can shoot their aluminum cover directly to your eye if you mess things up
I am thinking You need to treat that RED LED and the 8550 and the D882 like they are all in series == VoltLED + Vdrop 8550 + Vdrop across the gate of D882 >>> to get a R { in series with the LED} = (Vs - VLedTotal ) / I . I know R is going to be really small, but it might be important to prevent your magic smoke from escaping from the LED.
At 18:30 Perhaps due to ("zero hysteresis" + "non-zero noise"}?
There is a circuit sim online, it's just that you might have to breakdown that 3 pin diode thingy because it's probably not in the library. www.falstad.com/circuit/
protection voltage has to be Vref+Vbe=2.5V+0.7V=3.3V. It is higher than 2.7V which is cap max charge voltage. The design is not suitable for 2.7v cap.
Question regarding its charging procedure and working.
Question -1) Is For super capacitor the charging voltage should be always10 percent less than the capacitor rated voltage.To increase the life of capacitor.
Question-2) regarding charging current to these capacitor for example 3 volt capacitor of 500F connected in series of 27 pcs.so total voltage will be 81 volts and capacitance will be 18.5 farads.Hence what will be the safer charging current.if we connect it to 79 volts,45 amps power supply.(transformer type power supply). Is there any problem to capacitors.
Question-3) after capacitors fully charged it means voltage across the series capacitance network will be 79 volts is equal to power supply voltage. hence in these condition if the capacitor network remains in the circuit parallelly with the power supply with no load connected. Is there any problem of heating capacitors.
Question-4) Is there any need for protection circuit.i seen protection boards available in markets.
Question-5) What is the leakage current of these capacitors.500F,3V
Question-6) terminal resistance of these capacitors.500F,3V
waiting for your reply
Using super capacitors to supply power as an alternative to Lithium ion batteries is doomed to failure. Just because something is technically possible and sort of works, does not make it a good idea. Using super capacitors to absorb huge but brief load transients in a power supply might have some merit. Using these super capacitors to supply brief peaks of current at the end of a long relatively thin power cable could make sense if average power demands were relatively modest, but with extreme bursts of power required at intervals, i.e. boom or bass box amp.
Several people have told Julian that, but he ignores them studiously. Julian is spinning out a theme as a means of making more money-making videos while he comes up with new ideas. Can't really blame him; he's trying to earn money from TH-cam. So long as enough people keep watching, and patrons keep paying, he'll be happy. And besides, making videos keeps him away from other mischief! ;-D
I'm offgrid on solar. If you crack the supercapacitor pack that'll make my day ... To run my washing machine
Am I missing something,.. why not just disconnect the super capacitor,.. get rid of the diodes,.. ( they just skew results,... ) and characterise the protection circuit by controlling the input voltage ( with a current limit),.. and see if the protection circuit does what is expected of it,.. and if the newly inserted diode is a benefit or degradation on the performance/function of the protection board.
now we need a '431' with user variable REF voltage :)
Huh why? You have to option to set a voltage with the resistor devider (ok it must be grater than 2.5V but everything above that works)
I forget what authority I got this from, but it basically goes like this "a super cap is a way of taking up more space, storing less energy, and paying more money to replace a lithium battery". I know that a ceramic cap can be cycled zillions of times, but what of this technology?
Ray Kent All electrolytic capacitors (including super capacitors) have a lifetime that is determined by operating temperature (both from the environment and self heating), ripple current (causes self heating), over voltage abuse, and of course the quality of materials and manufacture. If used within the manufacturers recommendations, they can be charged and discharged an almost unlimited number of times over their rated lifespan. For DC applications, the losses (in the capacitor only) are much lower than the losses in a rechargeable battery system. The big disadvantages are low energy density compared to batteries and an exponential voltage drop as they discharge.
Authority? I think not.
Dave Rhodes I live near the magical forest of Broceliande, where Merlin is forever entrapped. Occasionally a bird perches on my shoulder. That's my citation. If there is a benefit to using supercaps instead of lithium cells for a portable radio I'd like some facts and figures, else others may be listening to magical birds. Julian has not given them.
Rapid charge (if I can make it work) and almost infinite lifespan. Ironically, I don't listen to music much :)
Julian Ilett haha, ta for that. Maybe you should do a walkie talkie instead....
You might be interested in a paper of mine which talks about the voltage sag and rebound of a supercapacitor. Explains why you need to hold a SC at the charged voltage for a significant amount of time to ensure internally it is actually charged to the state you expect: www.sciencedirect.com/science/article/pii/S0378775317301684
But you cannot see it unless you use camera
This is not a protection circuit. It's a balancer.
pro tip: Just drill the track next time.
close the blinds we cant tell with the sun
its not working, 400ohms load even if you get the main transistor full shunt, that 400ohm wis not going to stop that 1amp, imagine all that energy blowing up with that 0.75volt above 2.5volts hehehe, that circuit is good for 3.7 volt super caps. 0.5volts on the frst transistor is not going to turn it on
Those 2 thick diodes cause like 1.2V drop. That is bad. Must be removed or add more volts
but bottom line the protection cct is rubbish
the 431 actually is a zener-wannabe
YAWN!! Oh no not Super Caps Again!!!!!!....Boring
First...
':D
Why is your comment sixth?