I think during the circuit analysis you should draw the internal schematic of the TL431, as it makes a lot more sense in this application. If you look closely on the circuit board, one of the voltage set resistors is bridged by a tiny PCB track, meaning that the REF pin of the TL431 is connected to the VCC of the capacitor. The TL431 will basically try to sink as much current as it can on its anode pin to keep the REF pin at 2.5V, which will ddraw current through the PNP transistor and therefore cause the NPN transistor to sink current. In this way, this circuit is very good (especially if they use a high quality TL431) as the current drawn will almost exactly match the current being put in to keep the capacitor at 2.5V. Since this is a fully linear circuit, technically those big power resistors are not needed, as the NPN transistor is designed to do all the dissipation work, however it will have a maximum current, which could be exceeded without those resistors. So the power resistors exist only to limit the current through the NPN transistor rather than to dissipate power like they were used in the switch on-off type protection circuits. Putting a LED with a small voltage drop should be possible in line with the base of the NPN, so long as there is enough voltage there, as this line is current driven, and shouldn't have too much current to overwhelm the LED (but just to be sure I would check the hfe value of that NPN).
Thanks - yeah I missed that tiny track to the TL431. I might try a red LED between the two transistors (I can easily cut that track on the reverse side).
Dear Dasrue Very good explanation! Took me a while.... Had to put it on a breadboard to see the effect of the Vref pin attached to Vcc. I agree. It is a very good circuit! Regards
Julian, Do you think you can do a video in the "SUPERCAP ACTIVE BALANCING" or SAB, using an SAB IC which does not waste capacitor energy with heat-emitting resistors--it distributes the energy among the series cap bank?
I think you misunderstood the circuit. This is in essence a linear regulator. The whole point is letting the transistor do the work, because you can't adjust resistors on the fly. If you want more power to be burned in the resistors, increase their value. This will limit the maximum current the protection circuit can "sink" though, because the voltage drop over the resistors will get to high at some point, even if you'd imagine replacing the NPN transistor with a short. Example: if you use a 1Ohm resistor above 2.5A the voltage drop will be above 2.5V. That's just Ohm's law. As a simple start you can remove one of the parallel resistors, but be aware that the two remaining resistor will now have to dissipate much more power (Ohm's law: higher voltage drop due to higher resistance at the same overall current; and the overall current now spread between two resistors instead of 3). And as someone mentioned, you missed the small trace that connects the ref pin to Vcc (you'd have to cut that, if you want to set you own ref voltage). The ref pin of a TL431 always has to be connected for it to work. It's not like it is an automatic 2.5V zener.
Thanks a lot for this video. I'm using similar 2.5V balancers in a hybrid system with a 4S li-ion pack in parallel and I have been vary of pushing these boards too hard and not going over 2.58V per capacitor being afraid I would roast them. Apparently I can safely go to 16V or 15.9V (2.67V per capacitor balancer board), which I need for my battery being charged without having the transistor desolder, and have much faster balancing, minutes instead of hours.
16:67 if you look at fig. 6 of www.alldatasheet.com/datasheet-pdf/pdf/54917/AIC/AIC431.html, the montage is almost the same and its main use would be to be a current limiter. Note that the reference pin of the TL431 (or AIC431, or other names too ) ***has to*** be connected even if almost no current pass through it, 4 uA. See section 9.3, second paragraph of the datasheet for the TL431, this time from Texas Instrument, that is the one that you used in the video.
I'm guessing that as the transistor is turned on harder by higher voltages, the current flow through the resistors will increase and so they'll start to dissipate more of the energy rather than the transistor having to dissipate it all.
I have 6 of these in series (12volt arrangement) and I noticed when the supercaps have discharged one of them was reading a negative voltage! I assume it was weak and discharged first but since the others are still discharging the weak one starts charging negative. It made me think the protection circuit should have reverse protection too.
The 620 ohm resistor has the simple function to limit the current through the TL431, standard design for zener regulated voltage supplies. So it doesn't try to clamp the entire capacitor to 2.5V itself which would be instant destruction as the currents involved. When the voltage reaches 2.5V the TL431 conducts which turns the PNP transistor on, as you have near 0V on Base, and 0.6V voltage drop from the NPN transistor on Collector.
The 620 Ohm resistor is there to supply the minimum required cathode current of 1mA that the TL423 needs to bring its voltage up to 2.5V. If it wasn't there then the TL423 would take a small current through the base of the PNP which would be amplified to a large current in the NPN long before the capacitor voltage reached 3V.
Yeah,from datasheet looks like D882 achieves best gain at 2+V/CE plus 8550 0.6v drop gives 2.6v roughly but it is apparently badly tuned with that zener ic. Also I wonder why they just don't use micro relays to switch all off...
Station240 you do not have 0V on the base of the PNP because on the 0.6V drop emitter to base which should give you 2.5-0.6 giving you 1.9V. But you have the 2.5V across the TL431, that is why Julian was not sure how this circuit works as in theory the PNP has no bias voltage. 2.5V on the emitter and 2.5V on the base, but it would have been nice for julian to measure the voltage on the base.
Since i'm not an electronics expert, but do like to tinker a little, i really like these drawn out schematics! You say you're not really happy with this protection board. If you had to design one yourself? Could you do a video of that? With schematic of course! :) Perhaps the led can be placed in series with the zener?
Great channel. So with all this talk about never exceed the rated voltage, just as with any warning, I'm curious. Yes I know when you whack an electrolytic up over the rating it explodes, BUT I'm dying to know what one of these baby's does. *Someone* should set up a proper bunker, notify the authorities, scan the surrounding quarter mile for any living object, don protective gear and remotely detonate one of these (for educational purposes of course) AHEM and to provide a visual reminder that won't easily be forgotten. I'll keep watching this and others for an answer but after a period of time I can no longer tolerate waiting I will do it myself. What kind of voltage/current do you think would provide the best result?
I think that in series with the base of the NPN transistor as you suggested would be an ideal place to put an indicator LED. You would probably want to use a red or green LED because they have a lower voltage drop and because a blue or white one would be very bright. The LED current would be roughly a hundredth of the input current when the capacitor is fully charged so about 10mA for 1A input current.
You neglected to measure the actual clamping voltage. My guess is 2.5V (LM431) + 0.7V to turn on the 8550 gives 3.2volts. You can test with a smaller, non-super cap.
With capacitors having relatively wide tolerances with regard to capacitance, I wonder what the tolerance for working voltage is? And I wonder what the tolerance is for that TL431 chip? You wouldn't use a 16V electrolytic capacitor in a circuit with a 15V power supply, you'd use one with a considerably higher voltage rating, say 25V. The numbers for this circuit are all way too close together! If you want to know if your base-connected LED would work, divide the collector current you want to see by the typical (or minimum) beta of the transistor. That will give you the base current at that point.
This circuit is basically a linear shunt regulator designed to prevent the capacitor voltage from rising above about 3V. The exact regulated voltage voltage depends on how much current it is being fed with. At high currents say 1A the voltage will be 2.5V (TL431) plus about 0.5V (Vbe pnp) = 3.0V. At very low currents say 1mA it will fall to just the 2.5V and if you leave it connected after charging it will discharge the capacitor to 2.5V. To test a protection circuit like this I would use the power supply in constant current mode with its voltage set above 3V without the capacitor connected and plot the voltage as a function of current. Bear in mind that the circuit can only handle a certain current limited by the power dissipated in the NPN transistor and the series resistor. The three parallel resistors don't appear to serve much purpose, they just reduce the dissipation in the transistor a bit. At 1A the resistors will be dissipating 0.4W and the transistor 2.6W. They look like 0.25W resistors so will be overstressed at any current greater than 1.4A. The junction to ambient thermal resistance of a TO252 power resistor mounted on a small pcb is somewhere in the region of 50 - 100 deg.C/W so at 20 deg.C ambient it can only handle about (150 - 20)/100 = 1.3W. So the rating of this circuit is only about half an amp. Bear in mind that during most of the testing you did you didn't know how much current the capacitor was taking. If you had left it running for long enough for the capacitor voltage to stabilise and all the charging current to transfer to the protection circuit then it would probably have become much hotter. Edit What I said above is not quite correct. Now that Julian has revealed that the reference pin of the TL431 is not connected to the Anode but to the positive rail we can see that it will attempt to regulate the capacitor voltage to 2.5V and not 3.0V. However this is not the intended configuration for the TL431 and it is only just barely capable of doing this because the anode voltage can only fall to 2.0V and so the capacitor voltage has to rise one Vbe worth above this to turn the PNP transistor on. So at low currents it will regulate at 2.5V but at higher currents it may rise to perhaps 2.6 -2.7V.
If the PNP is on, then it will short the base of the NPN directly to Vcc, so creating basically a dead short (except the BE junction of the NPN) across the supply. I assume there should really be a resistor between the PNP collector and NPN base. Then your LED should really be in parallel with the R400 shunt resistors so it only conducts when the NPN is on.
While watching the video I thought exactly the same. you measured the npn collector voltage which was close to the rail. The 1.5 amps your power supply was outputting had to come from somewhere. The only path is through the base. A resistor between the two transistors is not just desirable, I would say it is absolutely essential.
James Myatt yep, it's bad (but cheap!). In Julian's diagram he doesn't show where the supply or cap is connected, so I assume they both go between Vcc and ground. What's Vcc? This circuit tries to maintain it at 2.5v by shunting current, assuming that will cause a voltage drop from the supply. So the supply needs to be current limited or have a series resistor and the surplus turns to heat. I'm not familiar with the area, but I guess there are better designs for a coupla bucks more.
I agree with Superdau, simple linear reg. If you want an Led in there put a 2.5V Led where the TL431 is. Cant guarantee it would work but the led should draw enough current to pull the pnp on and force the npn to start draining power. Hope this helps. You might be able to just parallel it across the TL431 but it will draw much more power than the tl431. As minimum turn on is about 1ma.
The 620 resistor keeps the pnp transistor off until the reference voltage rises to let it conduct and then dropping it low turning it on and turns on the npn transistor through the 400 mill resistor shunting the voltage.
Normally with shunt type overvoltage protection circuits, the assumption is that there will be a fuse in series, so the high current drawn by the transistor in the event of the gross overvoltage will blow the fuse before the transistor or capacitor is damaged. Also it's the transistor you'd want the dissipation in rather than those resistors anyway, so it's not a problem that it does that in this case.
@2:24 those two diodes should have a voltage drop across them. So the Cap is not seeing the voltage the meter is saying. I have looked to see if anyone else has caught that, but have not found anyone posting about it. Voltage drop should be around ~.45 X 2.
It is a brilliant idea to demonstrate how a bridge rectifier works! You will have to slow down the cycle rate of the ac signal to 2 hz or so to make it visible and to visually follow the switching! You can use a signal generator. You'll have to limit the voltage of the input signal way down to 5V or so and limit the output to 20mA. But it will make a very good demonstration for kids to learn how this circuit works! Brilliant plan!
Julian, I don't think there can possibly be an efficient protection circuit that works purely IN PARALLEL to the device being protected. However it is designed, it has to take all the excess power from the power supply and dissipate it within itself. Its just a matter of choice which of the component is chosen to get hot, in this example it is the transistor, probably because it can withstand more heat than a small resistor. To have an efficient protection circuit it has to have some component(s) in series with the device under protection as well as in parallel.
Forget the super capacitor and just test the circuit with a PS and multimeter. This would make it easy (and fast) to *plot out a graph* of how fast it turns on, etc. You could also plot some of the other points in the circuit at the same time to help understand how it works.
Ray Kent - Na, just a brain surgeon... ;-) But the point is you don't always need to test circuits with final components when you can quickly and more accurately analyze them with a logical replacement. This also allows testing over a wider range than your final component would allow safely or without damage. ie. you can test your circuit past the 2.7 volts the capacitor is rated for without fear of the expensive component being damaged or exploding.
Just because it's on your bench (table) doesn't make it a bench power supply! I expect it has terrible ripple voltage when compared to a proper bench power supply. Perhaps you'd be willing to put your oral assumptions to the test and measure the output stability with your Keysight DSO?
In principle, while the Vcc is less than 2.5v, there is no flow of current through the 620R or the R400 so the voltage of the base of the PNP is Vcc. Once Vcc goes higher than the 431 threshold voltage, 2.5v, it immediatelly shorts the base of the PNP to ground (but not exactly 0v as there is always some Ron) so it starts conducting and current flows to the base of the NPN and so the capacitor discharges somewhat through the R400. The 620R is there to avoid direct shorting of the cap when the 431 is activated. When it discharges enough for the volage of the 431 to fall bellow 2.5v again, it stops conducting. So the PNP stops and therfore the NPN too.
Dear Theodoros Kefalopoulos I don't think that the base of the pnp can go to ground because of the 431. The cathode will drop to about 0.7 below Vcc which is regulated to 2.5v The pnp operates in the linear region and acts as the current limiter for the npn. It does not turn on fully. The npn acts as a current amplifier and allows the big current to flow with a 2.5v voltage drop just like in a linear power supply. Kind regards Ps. The ref pin og the 431 must be connected to Vcc.
great video! but im confused looking at the TL431... if you pause the video at 5:40 you can see cap+ is connected to the bottom left pin of the TL431 [small trace thru R4]... in your schematic, cap+ is connected to the REF pin of the 431, but a look at the data sheet shows the bottom left pin of the 431 is not the ref pin, its the cathode. i thought at first it must just be the 432 [cathode and ref pin are flipped on 431 vs 432] but you can clearly see 431 stamped on the IC.. so i believe your schematic must be incorrect. In your schematic R1 [620ohm] is connected to the cathode pin of the 431, but on the PCB if you follow the trace out the top of R1 we can see it connects to the base of the PNP8550 and it looks like it also goes under the 431 connecting to the bottom right pin on [the reference pin of the 431]. So i believe you have your reference and cathode pins on the TL431 flipped in your schematic. let me know if anyone can confirm this! ive gone a bit crosseyed looking at this haha.. ill post another comment if i ever get around to building one of these.
Your review is great.. but the circuit is dainty.. It should OPEN the charge current, rather than dump it... I don't see using it in high current situations. Good Luck, and thanks! By the way, I don't understand how when fully charged, there was still 100ma flowing, minimum ? Just saying.
Hi Julian, thanks for this interesting video. I think that this circuit has to be understood in the following context : say you have several capacitors in series that you want to charge up. You can not be certain that the capacitance as well as the actual charge state of every capacitor in the battery is absolutely equal. There is a risk, even if you keep the overall charging voltage below the sum of the individual capacitors rated voltage, that the voltage across an individual cell exceeds the max spec. Therefore, the protection circuit kicks in, dumping the energy over the full cap, "into the next one" so to say. So say you have a 10 cap series bank, and charge it up to 24v; there is 2.4 volts per cell on average and once charged the current goes to zero and none of the transistors will actually dissipate anything. Some of them will (or may) only dissipate for a short period of time during the charging process. So for me, as long as the total voltage applied to the bank remain at least slightly below the overall max voltage of the bank (25v in my example) it is very unlikely that any of the transistors will get hot enough to desolder itself and fall off. Instead of seeing Vcc and GND think of it as "the previous cell" and "the next cell" which do not necessarily behave as a low impedance power supply, as you bench supply does.
Frédéric Bach that's interesting but maybe a bit exagerated? At end of charge you have 24v across a system that is in shunt conduction, it has no way of protectng each cell from overcharge other than by shunting. So at end of charge all of the energy will be dissipated as heat. It can neither disconnect a capacitor nor disconnect the supply, so it must sink all the current that the supply offers and dissipate it as heat.
Ray Kent, in my opinion, at the end of the charge the system reaches equilibrium: 24 v at the supply and 24v at the capacitor bank, no more potential difference. Current would go down to zero. By the way I watched the video done by Andreas Spiess, mentionned by "MrToM" a few comments below, it explains the stuff very well (much clearer than I did I think)
Dear Julian, I'm sorry I'm not getting the point. The charge current is limited by your bench supply, isn't it ? Why would a capacitor mismatch influence the actual current ?
Frédéric Bach hi Frédéric, I agree with what you just said 99.9%. But (if you're happy to play along), this equilibrium is only (very nearly) achieved if you use, say, a 12v source to charge a cap to 12v. That goes well okay up to about half charge but keeps slowing down following an exponential law. It soon becomes a trickle charger. If you want to speed it up by connecting a fully charged car battery at about 14v then zero charge transfer between the car battery and the cap can only be achieved here by dumping current through the power transistor (and relying on external resistance to drop the voltage) until the car battery gets tired. Then you get equilibrium. Except you still have a small curent drain through the circuit, but that's nothing to write home about. Amusing anecdote: I had a lead acid power pack that wanted to keep me informed that it was okay by keeping a small led lit. Guess the outcome, yep, left alone, it destroys itself. Bad design.
Julian, use a laser temp gun (at any hardware store) to get pin-point temps on components! Also... in the context of the PNP transistor: don't think in terms of voltage to trigger... think current (they are current devices!). Once the Zener sharply conducts it will provide base to emitter current to the PNP which in turn provides emitter to base current in the NPN causing it to conduct and dissipate the excess voltage as heat!
Hey, i'am in dispair. I got 10x 2,7V 500F Caps, welded in series. With a quite similar Board as yours, but the transistor gets hot at 1,6V, and stop discharge at 1,2V. i tried to weld different resistors because of 2,7V instead the stock 2,5V (Voltage divider: R1: 1,25k R2: 18k on stock => 2,667V on the formula in the Datasheet 431) nothing changes. i have no clue why, EVERY board has the same behavior :/ greets from Bavaria :) The raccoon 🐾🦝
First useful comment! The ref input should be connected to the cathode of lm431 to get a 2.5V zener behaviour. This design is quite bad and let the voltage raise to 3.1V before real activation. There are certainly some leakage current and because of the dual stage transistor system it's amplified so it may start to let a tiny current flow under 3.1V but nothing enough to protect the super capacitor if you charge it worth a decent current.
All of the collector current of the PNP goes into the base of the NPN so it appears to be dependent on the current gain of the NPN for correct operation. There maybe sufficient current with a Vbe of, say 0.4V, across the PNP to start turning on the NPN. Probably a 2V zener may be more appropriate. It's probably better to charge the cap to 90% than to overcharge it.
Because it starts clamping at 2.5V, I'd think the reference pin of TL431 must be connected to +VCC. If it was not, it would not work properly. Are we sure there wasn't something connecting it that was missed? (it would clamp at around 3.1V or so otherwise)
Pete Allum maybe to do with age. I'm old English and I much prefer the zigzag line to a box. I also prefer the use of a blob to indicate that two wires are connected and the use of a little arch to indicate that they're not.
Pete Allum Both the zig-zag resistor symbol and the rectangle resistor symbol are acceptable in the U.K. I personally am not bothered either way, but if drawing a schematic myself, I use the rectangle resistor symbol as I'm lazy and it's quicker and easier to get right.
I am aware that both symbols are acceptable here although the zigzag one is standard in the US. It's also traditionally associated with wire wound resistors, rather than modern metal film and carbon varieties.
I don't like this circuit. I'll stick my neck out and hope someone will explain why I'm wrong. For the health of the cap we need to limit the max charging current and the final voltage, these are to do with different physical properties. To get the fastest charge time we must treat them separately. So.... current limited power supply at higher voltage and op-amp comparator measuring voltage across cap and turning off series fet when it gets to 2.5 volts, or whatever. Disconnect rather than use a transistor as a heater.
That's okay if you're just charging a single capacitor, but this kind of protection circuit is designed for use on multiple capacitors in series. When one capacitor is full, you want to shunt current around it so the rest of them continue to charge. When all of them are charged, then you can turn off the supply, but not before.
