NiMH Battery Pack Substitution Effect on Charging Circuit - Page 1
Forum section: electronics / beginners

author: edy
author_rank: Super Contributor
timestamp: October 30, 2018, 01:26:04 pm
content: Hey folks,I bought a bunch of Garmin Moto TC Rally cars for my whole family to race as they were on sale and I couldn't pass them up at $12.95CAN a piece ($9.95 US). They are fun little cars to race. I have some old iPhones which I have already downloaded the BlueTooth app for and so we are able to all race at the same time and do virtual tracks.Anyways, the issue is... one of the cars I bought had a dud battery pack. I know it's the battery pack because I already tested other battery packs (from the working cars) in it, and it worked fine. Also, the charging circuit was also working fine in that car as well. So it is definitely the battery pack, which simply gets hot but doesn't hold a charge. It could even be just one cell in the pack that is bad that is ruining the pack.I contacted Garmin to see if they have any support or warranty on the battery pack so they could send me a new one, but either way if I have to make my own or buy one and substitute it I will do so. They are just standard ones like this:Now I already have some separate AA NiMH batteries and a holder at home, and a connector which fits the car connector. I could just strap it in under the plastic shell as it won't fit in the battery compartment if I use AA batteries. I could use AAA but I was wondering if I could get more working time using AA instead:I can charge the batteries up externally and insert them in when I want to use the car. That shouldn't be an issue. I can't imagine it will cause any problems running the car except for a bit of extra weight with AA instead of AAA and I will get longer operational time as there is higher mAh.The question is....If I leave my newly made battery pack hooked up to the car and use the charger built into the car instead of taking them out to charge, will it cause any issues? Remember, the battery pack originally in the car (that never comes out) is 4xAAA in series giving 4.8V and 600mAh rating. I will instead have 4xAA in series, also 4.8V but probably in the 2000mAh range. Is it just a matter of taking longer to charge, or could it overload the charging circuit in the car?

author: drussell
author_rank: Super Contributor
timestamp: October 30, 2018, 01:39:11 pm
content: It will just take longer to charge.

author: edy
author_rank: Super Contributor
timestamp: October 30, 2018, 03:49:33 pm
content: Quote from: drussell on October 30, 2018, 01:39:11 pmOk thanks! I was worried about affecting the charging circuit somehow but it makes sense now since my wall-charger will work with both AA and AAA batteries (the bottom part slides up and down to provide the different lengths to meet the different battery sizes).... the actual electronics in the charger are "dumb" and don't know what size battery, as long as they are NiMH cells.In the case of the car, as long as I have NiMH cells and they are put in series to equal 4.8V, the charging circuit also shouldn't care whether I am charging 4 AAA or 4 AA. I'll try it and see how much longer the car runs! I will have to experiment with the charging time but the circuit does have a light that turns off when it is supposed to be fully charged, so it should still work the same and shut off appropriately.

author: mvs
author_rank: Frequent Contributor
timestamp: October 30, 2018, 07:48:06 pm
content: Quote from: drussell on October 30, 2018, 01:39:11 pmIt is not so easy... Lower relative charge current may affect NiMh charge termination (Negative Delta-Voltage detection). Some charge controllers also limit charging time.

author: ogden
author_rank: Super Contributor
timestamp: October 30, 2018, 08:06:31 pm
content: Quote from: mvs on October 30, 2018, 07:48:06 pmRight. Better use 1000mAh AAA. It's still 1.5x capacity and most likely current capability as well.

author: drussell
author_rank: Super Contributor
timestamp: October 31, 2018, 12:06:52 am
content: Quote from: mvs on October 30, 2018, 07:48:06 pmPlease elaborate. The voltage will still drop slightly as it peaks across the "full" charge level due to the Ni-MH chemistry regardless of what the total capacity of the battery pack is. have you ever actually seen a circuit that will charge a 600 mAh pack properly but refuse to charge a 2200 mAh pack?QuoteThat would potentially be a problem but I have a funny feeling that the charger built into a $10 RC car isn't going to be intelligent enough to stop charging a 2200 mAh pack because it times out thinking it should only be delivering 600 mAh. Most slow, simple Ni-MH chargers like that don't even terminate charge, they just trickle indefinitely.

author: drussell
author_rank: Super Contributor
timestamp: October 31, 2018, 12:08:21 am
content: Quote from: ogden on October 30, 2018, 08:06:31 pmWhy would 1000 mAh be okay but 2200 mAh not?Can you provide an example of somewhere that this is actually an issue?

author: Audioguru
author_rank: Super Contributor
timestamp: October 31, 2018, 12:58:33 am
content: Your photo shows Energizer AA Ni-MH cells with green at one end. They are cheap with a low capacity and are not made anymore. They are not listed on Energizer's website anymore. About 5 years ago I bought some by mistake. They are only 1400mAh and are still shown on Walmart's website at the price of normal ones. Normal ones are 2300mAh, come pre-charged and hold a charge for one year.

author: Rick Law
author_rank: Super Contributor
timestamp: October 31, 2018, 01:45:38 am
content: Quote from: drussell on October 31, 2018, 12:08:21 amA 2200mAH battery may draw too much charge current - more than the charger circuitry can handle.Take an Eneloop, you can easily charge it at 0.5C, that is a whole 1A it can draw. If that little RC car charger circuit doesn't have appropriate current limit, it will try to push too much current through those tiny thin traces or through the tiny sub-sub-sub sot23 sized voltage/charge regulator. Poof goes the charging circuit.It may or may not happen, but it is certainly a risk.Tired of constant recharging, I was trying to replace a tiny 200mAH battery in a remote Andriod keyboard (with a new phone battery I have around)... When I saw the tiny IC and very very thin trace it had leading to the battery, I decided, if I wanted the battery to charge in place, I best just keep that 200mAH. That one was lithium 3.7V, but the same logic applies.(I understand that lithium has CC+CV phases and NiMH has entirely different charge-full determination, but cheap charge circuitry for lithium or for NiMH may skim on implementation... hence the risk.)

author: edy
author_rank: Super Contributor
timestamp: October 31, 2018, 03:11:43 am
content: Thanks for the input. Meanwhile I started putting together a substitute pack with stuff laying around the house. Here is what I did but no saying what will happen when I try to charge this thing. Here is the battery holder, same connector as battery pack:Batteries in place and rubber-banded together. These batteries are 2500mAh.Batteries installed, car turning on:It fits underneath the plastic shell, lots of room in there:For now, to test I just used rubber bands to hold it, in the future I will improve if it works out.Anyways, the car is $10 now, but originally it was priced around $99. I don't know if that necessarily reflects that the folks at Garmin well-engineered it, or anticipated that someone may try to screw around with the batteries. Nevertheless, I do have AAA's sitting around as well which I could try out but meanwhile I will try an over-night charge of my 4xAA pack and hopefully it won't burn out any of the circuit.

author: drussell
author_rank: Super Contributor
timestamp: October 31, 2018, 01:29:20 pm
content: Quote from: Rick Law on October 31, 2018, 01:45:38 amA completely discharged 600 mAh pack will look like a very low impedance load, just like a 2200 mAh pack will. It is thechargerthat limits the current, not the battery pack. Many simple chargers are just a resistor from a voltage slightly above the Ni-MH "full charge" voltage. Other cheap ways of doing it are just something like a LM317. Unlike Li-ion chargers, many Ni-MH non-rapid chargers don't even have any kind of management or sensing, they just leave the voltage on there but limited to a small enough current that they don't blow up the cell / pack. These won't care about the capacity. If it uses a real Ni-MH charger charger that senses charge termination, it won't care either unless they've intentionally put a timer on it or something, in which case you would need to start the charge cycle more than once to get it to full charge. It is still not going to burn up the charger.

author: ogden
author_rank: Super Contributor
timestamp: October 31, 2018, 02:58:32 pm
content: Quote from: drussell on October 31, 2018, 12:08:21 amSure. Issue can be way too long charging time for car with 4x bigger battery. 4x increased capacity will result in 4x longer charging time which for dumb 0.1C charger is... 40 hours

author: drussell
author_rank: Super Contributor
timestamp: October 31, 2018, 03:47:58 pm
content: Quote from: ogden on October 31, 2018, 02:58:32 pmObviously if you triple or quadruple the battery pack capacity, the charge time is going to triple or quadruple accordingly. That doesn't mean it is going to burst into flames and burn your house down.

author: ogden
author_rank: Super Contributor
timestamp: October 31, 2018, 06:32:23 pm
content: Quote from: drussell on October 31, 2018, 03:47:58 pmPerhaps for you only burned-down house is an issue, but for others quadrupled charging time could be as well.

author: edy
author_rank: Super Contributor
timestamp: October 31, 2018, 10:17:17 pm
content: Thanks for the help, yes I agree it could be an issue... I don't want to wait that long to charge the battery. I have another potential solution to bypass the need for using the internal charger which is designed for the 600mAh battery pack. That is, I could just charge the batteries by inserting them into an external charger and then putting them into a 4-battery holder that I use for the car. The external charger may be more rapid. But here is another solution where I don't have to remove and replace batteries each time.... Let me know what you think....Another option that I was thinking is to use one of those 5V Li-Ion cell-phone power packs. The car takes 4.8V but if I feed it a 5V output from a Li-Ion cell-phone charger, will it really care? I could attach the cell-phone power pack to the car permanently and charge it up by plugging it to 5V USB input. The output of the cell-phone power pack would be going to the battery input for the car. The car would see 5V and should be able to draw enough current out of the power pack to operate (most of these cell-phone power ups output 1-2A). Basically, this bypasses the need for using the built-in charger for the car, and in any case I am going to another battery chemistry, and I would still have the convenience of just "plugging in" quickly a charger cable for the car (in this case, the cell-phone power pack input instead of the car charging input), and not have to be removing and installing rechargeable batteries each time.Thoughts?

author: ogden
author_rank: Super Contributor
timestamp: October 31, 2018, 11:06:37 pm
content: Quote from: edy on October 31, 2018, 10:17:17 pmYou can use battery half-charged, but then what's the point to have such a big capacity and what's worse - it's (dead) weight that slows your RC car down?  What are charge time specs for that toy? - User manual shall tell about charging at least something.QuoteCar could be slower with 4.2V Li-Ion than fully charged 4.8V NiMH, but it will run indeed. On the other hand LiIon pack is lighter for given capacity, so car will benefit from it.

author: Rick Law
author_rank: Super Contributor
timestamp: October 31, 2018, 11:08:51 pm
content: Quote from: drussell on October 31, 2018, 01:29:20 pm[Emphasis added]Geeze, I thought the one I found (on a cheap hand-crank charging "emergency" radio) that was just a resistor (plus a diode) was unusual... Amazing how cheap can things get really.You are probably right that discharged battery packs would look like a low impedance load. But just the same, given all the other draw-backs, I won't risk it. You know, one missing piece of info is the charge current... May be that guy is just drip-charging it at 20mA or some such, and without termination at all. All the more reason not to use that to charge better batteries.[EDIT... adding this thought below:]While I said I wouldn't use that to charge better batteries, don't interpret that as suggesting one should charge the batteries individually.That is a 4xAAA pack, as a pack, the pack needs to be balanced. If the cells are charged individually, some would have more charge than others. During discharge, the lesser cells may "go negative" after their lesser charge is gone and the better ones are still pumping out juice. When a battery eventually "gone negative", that is the death for the battery.

author: Audioguru
author_rank: Super Contributor
timestamp: October 31, 2018, 11:48:48 pm
content: Edy, you did not lookup the charging voltage of a Ni-MH cell. The Chinese manufacturer printed 4.8V (1.2V per cell) because 4.8V is its discharging voltage.The charging voltage is way more than only 1.2V per cell.The charger circuit must be smart enough to limit the current and to disconnect charging when it detects a full charge. An overcharging battery gets hot and might burst.The charging voltage is shown to be more than 1.4V per cell (more than 5.6V for your 4-cells battery) on Energizer and Duracell datasheets:

author: Rick Law
author_rank: Super Contributor
timestamp: November 01, 2018, 04:42:15 am
content: Quote from: edy on October 31, 2018, 03:11:43 amThe way you have it rubber-banded to the topside made me think... if I may make a suggestion on anentirely different approach...How about those small USB power banks? The power bank odd to be able to rubber-band to the car the same way your picture shown.Yeah, USB power bank is going to output 5V+-5%'ish. Your original battery pack is 4.8 volt, that is 1.2v nominal for each NiMH cell. But NiMH cell is not exactly 1.2v always. On fresh charge, they are going to be 1.3v or 1.4v or perhaps even more. So your 4.8v pack is going to be 5.2v or 5.6v or more when freshly charge. That is just within the USB power-bank's expected output.My ASUS power-bank has 3x18650 in parallel (10050mAH total), boosted to output 5v, smaller than a pack of cigarette. You have all the charging solved for you along withover 7000mAH @ 5v. That is a good bit more than your 2000mAH 4xAA cells. With that, you can race all night with it...I got mine direct from ASUS, but B&H also has it:https://www.bhphotovideo.com/c/product/1173356-REG/asus_90ac00p0_bbt002_zenpower_10050mah_portable_battery.htmlIf you like this direction, check out this type of power banks (this types, not this one):https://www.aliexpress.com/item/Multicolor-New-USB-5V-1A-POWER-BANK-Suit-18650-BATTERY-External-DIY-Kit-Case-Box-Per/32760647698.htmlAs I said,this types, not this one. Look for a better quality oneas this one is crap. I actually got a couple of this exact one for experimentation. The boost board is crap, but it works. At 1A, it appears to run well but rather hot. This one is sized for unprotected cell. You can't fit a protected 18650 in it. While it works well for discharge, I wont charge the 18650 with it. I got two of those and I ripped apart one for experimentation. It has charge voltage that doesn't seem right. I have plans to test the other one. I have not had a chance to check the second one.

author: IanB
author_rank: Super Contributor
timestamp: November 01, 2018, 05:09:12 am
content: Quote from: edy on October 30, 2018, 01:26:04 pmHow long do the instructions say it should take to recharge the built-in battery pack? It is very likely that it is just a simple fixed current charger. A fixed current charger would probably take 6-10 hours to recharge. If the cars are supposed to recharge in 4 hours or less it may be a smart charger that is designed to stop charging when the batteries are full.Now then: the risk of overloading the charging circuit is close to nil. Not much risk at all.If it is a dumb fixed current charger then there is also no risk of damaging bigger batteries. However, be aware that if it takes 10 hours to charge a 600 mAh battery pack it will take over 30 hours to charge a 2000 mAh battery pack. You may not want to wait that long.If it is a smart charger it will also take 3x as long, but there is also a risk that the end of charge detection won't work properly with the bigger batteries and it doesn't stop charging when the batteries are full. On the other hand, since the batteries are so much bigger they will absorb the extra charge with less risk of harm.A faster option would be to remove the batteries and charge them externally, but you have to multiply this by the number of cars and the number of chargers you have.Using a lithium ion power pack is a possibility, but it might be overloaded by the cars and might perform less well than the NiMH batteries. You would have to try it and see.

author: edy
author_rank: Super Contributor
timestamp: November 01, 2018, 12:49:11 pm
content: Ok, so I made a small adapter for USB to the car-battery-input so I could hook up a Li-Ion Powerbank to the car so I can run it off the powerbank. I no longer use the charging circuit in the car, I would just plug a microUSB to the powerbank when I need to charge up the powerbank. Here are the pictures, followed by explaining what my initial testing reveals:So as you can see, I have also the nice added feature of seeing the percentage of battery used on the display of the powerbank!I've played around with the car, it works great. Good power, lasts quite a long time compared to the previous 600mA NiMH battery, and the weight added is not much more. I'm happy with the results so far, except for one thing.... The power drawn from the car undernormal usageis NOT enough to keep the PowerBank on always and it will time-out and shut off after a specified number of minutes. I can always turn it on again by pressing the button, it takes 2 seconds. But still, it can be annoying.I've tested the car by just running it full-speed THROTTLE ON MAX forward, upside-down on the table.... And the powerbank stays ON! Therefore there must be some current-sense that the PowerBank uses to determine whether it is being utilized or not, or whether the phone it is supposed to be charging has cut out of charging. I must be BELOW that threshold during normal car racing, unless of course I am in FULL THROTTLE ALL THE TIME.So one thing I may need to do is create a PARASITIC current draw that is just enough to bump the power bank above the threshold. Maybe add a string of LED's or something in parallel to the car-battery-input. The current drawn by the LED's would not be enough on it's own, and neither the car itself under normal use, but together maybe the both of them may be just enough to get over the threshold when racing normally. I would not know how much extra I need of course unless I start measuring, but if I experiment and just start off with 1 LED, then 2 LED, and keep adding them, I will get a better idea. Yes the powerbank will not last as long but it is unlikely to make a huge impact. I would also have the added benefit of having ALWAYS-ON-HEADLIGHTS if I go with the LED example.

author: Audioguru
author_rank: Super Contributor
timestamp: November 01, 2018, 01:24:52 pm
content: Of course you must know that the resistor in series with an LED, the actual forward voltage of the LED and the supply voltage determines the current in the LED.Also you must know that most 5mm diameter LEDs are rated at 20mA and about 30mA is their absolute maximum.

author: edy
author_rank: Super Contributor
timestamp: November 01, 2018, 01:40:06 pm
content: Quote from: Audioguru on November 01, 2018, 01:24:52 pmYes, thank you for reminding me.... After a bit of magic smokeand experimentation I ended up with the following solution. It is one of those LED flashlight arrays that I pulled out of some dollar-store item, added a 220 ohm resistor (to keep the magic smoke from releasing) which was sufficient to power the LED's to a reasonable brightness, and enough to allow to keep the powerbank running under NORMAL CAR USE (without activating the auto shutoff)..... Yet not enough to keep the powerbank running when the car is completely idle.That is....CAR NORMAL USE (without LED ARRAY) = not enough for PowerBank to stay onCAR FULL THROTTLE (without LED ARRAY) = enough to keep PowerBank on.LED ARRAY ALONE (car sittle idle but on) = not enough for PowerBank to stay onLED ARRAY + CAR NORMAL RACING = enough to keep PowerBank onHere are some pictures....Now I have the added benefit of seeing the car glowing nicely when racing at night.I also have a very obvious indicator of whether the powerbank is still on or not, since the LED's activate as soon as the lights are on. I have still to do more testing, I didn't have enough time to really use the car for extended periods of time.... But based on previous experience at least, the PowerBank never shut off and I was racing I believe longer than previous attempts (when the PowerBank did shut off). No more smoke so far (like my previous LED experiment when I did not use a resistor at allalthough nothing got damaged on the LED board where it was coming from, I unplugged in time.... although that may be another cool realism effect for the car, don't you think?)[EDIT: Added below content before any further posting on this thread]One more thing... I have a 220 Ohm resistor now but I can increase the resistance and get dimmer LED output to prolong the Powerbank. I will have to experiment and keep increasing resistor value and see when normal car usage will shut down the powerbank. For example, if I bump it up to 1K ohm and my powerbank stays on, I'll do that. If I bump it up to 22k ohm and it still works, even better. I will be limiting the current through the LED's to the point where they barely are visible, and more current will go to the car. But no matter what the sum of the two currents drawn when the car is in normal use will have to exceed the Powerbank "shut off" threshold.

author: Rick Law
author_rank: Super Contributor
timestamp: November 01, 2018, 06:59:12 pm
content: Quote from: edy on November 01, 2018, 12:49:11 pmThat is great! Looks great too...Interesting "car normal use" is not enough to keep your power bank on and needs a whole array or LED to keep it awake. My ASUS power-bank would stay on with a meager 20mA.Why not just use a resistor(short prevention)+VR and a DMM in-line to check for the exact "stay awake" current, then size your ballast resistor(s) to it. You can save the max juice for driving.If a couple of LED would do it, that would be just the right number for your car's head-light! If that turns out to be the solution, can you post a photo of it? Your car should look very cute with a pair of "eyes". I'd love to see that. But if more is needed, it would look fun if you make it into a red+blue light bar on top like a little police chaser... That would be cute too.

author: edy
author_rank: Super Contributor
timestamp: November 02, 2018, 03:46:34 am
content: Quote from: Rick Law on November 01, 2018, 06:59:12 pmYes it is strange but the whole array of LED with a 220 ohm resistor on one end (and some value of SMD resistor on the other) is not enough to keep my PowerBank awake. I could test the current draw on that and see what it is, but we could just calculate it. I am a bit confused as to what is going on with my calculations, as I seem to be missing something as the results are ending up strange. I will explain...It looks like my parasitic circuit has 24 LED's, standard white regular size it seems, would assume 3.2V 25 mA. So if we have 24 LED's each taking 25 mA each, then 24 in parallel x 25 mA = 600 mA being used in total across all of them? Is this correct? I'm supplying 5 V, the drop across the parallel LED's is still 3.2V for each, as well as the parallel array, so 5 - 3.2 = 1.8 V which is what will need to "drive" the LED's... and that has to be at 600mA current. To figure out the resistor, using V=IR, V/I=R, 1.8V/0.6A=3 Ohm resistor? Therefore, if I was powering my LED's with a 5 V supply, I would need a 3 Ohm resistor? Seems like it is too low.... Therefore, I would expect that that SMD resistor on there (which I can't read the numbers on) is somewhere around 3 Ohm because it needs to make sure only 600 mA total is getting shared (25 mA each) across those 24 LED's from a 5V supply. I hope my calculations are ok so far?I think I am screwing up somewhere here in my thinking... If I add more and more LED's, and the current drawn goes up for each LED added, then I need to use a smaller and smaller resistor. Intuitively it seems to make sense but if I carry out the calculations for 1, 2, 5, 10, 20, 50, 100 LED's in parallel I seem to get smaller and smaller resistors to ensure I am not limiting my current, as long as the supply can deliver it. I must be missing the "internal resistance" of the battery or some other factor that is limiting current, since I am not using an ideal power supply? Is this where I am going wrong? Let me continue then with my messed up calculations...However, when I powered the LED array by the 5V PowerBank I noticed it was very bright and some "magic smoke" was being released from one of the components (maybe from insulation near one of the supply wires soldered to the board) as they were thin and could be heating up? That leads me to think that this LED array was *NOT* originally designed to be powered by a 5V supply but could have been in a flashlight using I would guess 3 AA batteries (2 AA would be under the 3.2V forward drive voltage, but 4.5V would be enough). In that case, if that SMD resistor was designed expecting a 4.5V supply it would be needing a 2.17 ohm resistor to provide 25 mA current across each LED (600 mA total), and running it at 5V would have provided 830 mA total current (or 34.5 mA per LED). BUt I don't think it was the LED issue as it would have boiled off a few and smoke would have been coming from them. Instead smoke came from part of the board, either a solder joint, the SMD resistor (which would have burned open I guess, not shorted) or the insulation from my thin wires. Either way, it was running HOT and too much current was being drawn.... yet it survived even after 10 seconds of light smoke being produced (a "whisp") and I continued to use it after once I added the 220 Ohm resistor.Having said the above.... I can't understand how the addition of a 220 Ohm resistor would still allow the LED to turn on. Based on calculating a single resistor for 24 parallel LED's, that would have yielded 5-3.2=1.8V through 220 Ohm, V=IR or V/R=I, 1.8V/220Ohm=8.1mA?For the entire parallel array? Then that current gets divided across 24 LED, it would be a fraction of a mA? How is that possible? I know my math is wrong here somewhere, I just can't figure it out!

author: Rick Law
author_rank: Super Contributor
timestamp: November 02, 2018, 05:56:47 am
content: Quote from: edy on November 02, 2018, 03:46:34 amI am a hobbyist (ie: not a pro, could be wrong), but I think your estimate for 600mA total may be off.If you are connecting it right or unless your power bank is quirky, the "magic smoke" should show whether the led array is connected alone or connected together with the car. I suspect you mean the "magic smoke" would appear when the car (while also connected) is not running.If the car and the LED array are indeed in connected in parallel and magic smoke shows when car is not running, that would indicate to me that the power-bank is not holding its voltage regulation too well under load - high enough to "magic smoke" to appear when array is alone, but drop to below that when car is running.LED's varies a lot particularly when you say you got it at a dollar store or some such. I suspect the whole array of LED's are just LEDs. If you have say just two in the array, due to variation, one LED may have voltage drop 3.1v and while the other has voltage drop at 3.2v. In this scenario, at least one of them would not be running at its natural voltage drop. With your array, who knows exactly what the combined system's voltage drop would be? I think it would be a better bet to grab a DMM to measure the current instead of calculating it, or to measure the actual voltage across the LED array, and the voltage drop across the resister. Now you have all the info to do a calculation of the current.I suspect 100mA is plenty for an array like to light up well. I would add a VR POT in series with the ballast resister to find the max current without smoke, then replace the ballast resister with that (10%-20% larger resister to reduce the current to below smoke-point to give yourself some safety margin.)

author: edy
author_rank: Super Contributor
timestamp: November 02, 2018, 12:27:33 pm
content: That makes sense, somehow 600 mA didn't sound right to me either and a 3 Ohm resistor seemed too small. Then again, the LEDs aren't full brightness but with so many of them together even at 30-40% brightness they still look fairly bright. If I am using 100 mA for the whole array (with the car off, not being utilized) then V=IR, or V/I=R would indicate a 5-3.2V/0.1A=1.8V/0.1=18 Ohm resistor should be enough. Nevertheless, I am using a 220 Ohm and whatever SMD is on the board, and still getting them to light up, which according to my calculations would have limited the current to the LED array to V/R=I, 5-3.2V/220=8.1mA, for the ENTIRE LED array. Something is definitely wrong here with my thinking or my assumptions!I need to back to basics and understand the behaviour of LED's in series and parallel, how voltage and current changes along the circuit... that may be the error in my thinking.Even if the forward voltage drop on these LEDs was say 2V (not 3.2V), and they were all matched theoretically so the current would divide equally between all of the LED's.... AND we then limit current to each LED to say 10mA (perhaps visible but not bright and about half the usual), we would still consume 240mA total shared across all 24 LED's. So wouldn't we need to supply the array 240mA of current? Is that correct? So in this case, V=IR, V/I=R, 5-2/0.24=12.5 Ohm resistor needed. Again, something is way off here in my assumptions.I will have to put my multimeter in series with the resistor and see what current goes across the LED array. I can't wrap my head around the calculations with this, and I don't have enough real-world experience and measurements to get the right idea of what is actually happening! Seems like it should be such an easy beginner problem yet puzzling.Nevertheless, this is how you learn which is why this is now bugging the heck out of me until I understand! Funny how one problem (dead NiMH battery back on a $10 clearance toy needing a fix) leads down the rabbit hole into so many other learning opportunities.PS - By the way, the PowerBank powered car with the LED array and 220 Ohm resistor is doing awesome! No magic smoke, I race it for hours! The car does NOT shut off, the PowerBank has only dropped from 100% to 88% after more than an hour of use! If I do not drive the car for say 2 minutes, it automatically shuts off anyways (as the LED array alone can't keep it on). So I have the benefit of a well-lit car for night driving, plus I can just park it from wherever I am and leave it alone and it turns off by itself. So cool!

author: edy
author_rank: Super Contributor
timestamp: November 02, 2018, 02:19:55 pm
content: Ok, I got a LED ring for some test purposes just to get some measurements to help me understand better what is going on. I've included some photos below and results of my test:I measured the resistance of that single SMD resistor mounted on the board (white in color, no other markings) which was found to be 10.9 Ohm. I plugged in the 18 LED ring to a 5V 1.8A-rated adaptor meant for charging a BlackBerry phone. I then measured the current across the RED wire (which you see cut in the photo) by connecting my DMM in series and got 0.143 A or 143 mA.Now for the calculations. If there is 5V applied, and the forward voltage of white LED is assuming 3.2V, then 5-3.2V/10.9=1.8V/10.9=0.165A or 165 mA. That seems to be close to 143mA that I measured. So perhaps the LED has a higher forward voltage, say if we try 3.4V then we get 5-3.4/10.9=1.6/10.9=0.146A or 146mA, which is very close.Ok, so the entire ring of 18 parallel-connected LED's is using 146mA, which means if we divide the current equally between all of them (assuming the have identical characteristics) we would get 146mA/18=8.1mA per LED???I am thinking that these LED's are working at low current and the 20mA assumption doesn't apply here, or the forward voltage is wrong for these LED's. So I had a look here and it seems to give some clues as to what is going on:https://www.maximintegrated.com/en/app-notes/index.mvp/id/3070https://www.maximintegrated.com/en/app-notes/index.mvp/id/3256In particular, this graphic:and this...I guess I am still trying to understand what is meant by applying a voltage to the LED and the forward-voltage of the LED in the articles above. For example, take this equation:So if I apply 3.4V to a white LED whose Vf is also 3.4V, this equation doesn't yield any useful results at all because I will have R=0/I, or RxI=0. Now if we are to start increasing the V beyond the Vf of the LED, say 3.6V applied across the leads of a 3.4Vf LED, then I have R=0.2/I or RxI=0.2. Now if I increase my R I will be decreasing my I, and I want to pick an I that is ~20mA or whatever the particular safe operating range is for that LED. So let us say we want I=0.020A, we have RxI=0.2 or R=0.2/0.02, so R=10 ohm?In these charts above, and in the diagram below, white keeps showing up as having some of the highest Vf and also still needing 20mA to operate (or I think 20 mA is the MAXIMUM and they should be well below):So what gives? How can we be powering so many white LED with such little current? I don't think we are anywhere close to 20mA for most of these LED arrays, it seems that they are running in the 10mA or below range. In the case above for my 18 LED ring light it is going to be close to 8.1mA per LED. But in the case of the car light I have, with that 220 Ohm resistor being used I cannot see how it is drawing enough current to power all those LED. The calculations show the entire array would draw 5-3.4/220=0.0073A or 7.3mA.... divided by 24 is 0.3mA per LED.Then again, if that was resistor was actually RED RED BLACK GOLD and not RED RED BROWN GOLD (I will have to measure it, I can't see the damn colors anyways) then it would be 22 ohm (not 220) and we would be up to 3mA per LED. Still very low though but more reasonable.[EDIT:]Ok.... I measured it, it is 0.217kOhm or 217ohm for sure. The SMD on that 24LED board is 1.4ohm for sure. So yes the SMD is negligible as I assumed, but that is a 217ohm resistor. There are 5V being supplied by PowerBank, if the average of the Vf of all the LED are around 3.4V, then the amount actually passing through each LED once past the breakthrough Vf would be 5-3.4V=1.6V which all the LED's "see" across them, but now they have to share V=IR, V/R=I or 5V/217ohm=0.023A or 23mA among all of them, so current through each individual LED is 23mA/24mA or approx 1mA.Ok I'm done! I can't figure this out.[EDIT 2:]I just raced the car again for I'd say 15-20 minutes and the PowerBank dropped from 88% to 78% used. Which begs the question, why didn't they use a Li-Ion based system from the start? This car is so efficient it is able to power the LEDs and run for hours on a single charge, and the PowerBank doesn't even take that long to charge. This is a relatively small one also, at most it may have 2 or 3x18650's in them (assuming each is about 2600mAh capacity, then you have 2x2600=5200mAh, on the lower end you may have 3x1800=5400mAh, I can't imagine more cells). Here is what I'm using:https://www.amazon.com/Justin-Power-Bank-Display-Black/dp/B00UMVVH4U

author: Audioguru
author_rank: Super Contributor
timestamp: November 02, 2018, 07:57:12 pm
content: Since you have a 220 ohms resistor (actually 217 ohms) in series with the LEDs then measure the voltage across the resistor then calculate the current I= V/R.If you connect a current meter in series then the resistance of the meter adds resistance which reduces the actual current.Yes, 24 LEDs each with a few mA is VERY bright. 24 LEDs each with 1mA is reasonable, like 1 LED with 24mA.

author: Rick Law
author_rank: Super Contributor
timestamp: November 02, 2018, 08:08:01 pm
content: I am not very experienced, so I will share with what knowledge I have... (assuming I am right, which at my experience level... ahem... shall we say less than perfect...)Your last point first:Looks like $$$ to me...My ASUS power bank is 3x18650 in parallel and boosted, that was about $20. Yours at your guessed 2x18650 is under $20. It could be in serial and bucked down. Bucking would give it better efficiency particularly in higher current draw.At perfect efficiency, yours would be: 5200mAh*3.7v/5v=3848mAh at 5v.3848mAh is boosted/buck at perfect 100% efficiency which is impossible, so say boosted at 80% efficiency, you are looking at about 3000mAh.Yeah, 3000mAh@5v is a hell of a lot better than original 600mAh@ 4.8v. But they could get close with just 2500mAh with 4xAA instead of 4xAAA. Heck, they could have use 1000mAh AAA, that is almost double the capacity at the same physical dimensions. Cost... A penny here, a penny there...Back to your Ring of 18 LED;Your 143mA is more inline with what I would expect. Those LEDs look like run of the mill 1/4 watt to 1/2 watt LEDs. I think 20mA each (20x18=360mA) will make them look blindingly bright. Human eye is not that sensitive. You would not be able to quickly tell the difference between say 20mA and 30mA if you don't have them in side-by-side to compare. It one compare by view one first then view the other and compare from memory of how bright the first one was, it is hard to judge.It looks like they are all paralleled with a single ballast. The smaller white/cream-color SMD does have R2 label, so that looks like the ballast. I have no idea what the other bigger white color thing is. My experience level is not deep enough for me to ID it.Assuming it is flat DC - With 18 LEDs running in parallel and so easy to access,I would ignore the graph and just use a dmm to measure what the combo LED's voltage drop is. After all, there is no assurance that all 18 LEDs are from the same batch. Even if it is, how do I know that the specs I got are for the one soldered into the darn thing.You can see the two rings that the LEDs are soldered to. Just measure the voltage there and you'll know the Vdrop.

author: 6PTsocket
author_rank: Regular Contributor
timestamp: November 03, 2018, 10:50:13 pm
content: Quote from: edy on October 30, 2018, 01:26:04 pmIt will just take longer to charge. With a common chemistry, the bigger the cells the more heat they can take and the faster they can be charged. The only drawback is by charging at a AAA rate it will take longer that it would at the rate AA's can handle. It will take longer than the other cars but you will get more run time.Sent from my SM-G900V using Tapatalk

author: edy
author_rank: Super Contributor
timestamp: November 03, 2018, 11:58:54 pm
content: Thanks. So I've settled on using the PowerBank to power the car as I get much longer running time and I can charge the PowerBank directly from a USB 5V wall-wart, completely bypassing that internal NiMH charger in the car. As far as the PowerBank turning off, I can manage to keep the car going using the parasitic current draw of the resistor and LED flashlight, but when my son was driving the car he was less aggressive and it turned off at some point and I had to activate again.To remedy this situation I will hook up a lower value resistor (now I'm using 220 ohm) which should increase the brightness of the LED array but not exceed the mA too far and I will need to calculate that I am still within the power rating of the resistor (1/4 W?). I mean if I am using 5V x 200mA that is already 1 Watt, right? I may have to parallel a few resistors so they can share the current load so as not to exceed power rating.When I start swapping out the existing resistor I can do an actual current draw test and know better!

author: Rick Law
author_rank: Super Contributor
timestamp: November 04, 2018, 02:22:25 am
content: Quote from: edy on November 03, 2018, 11:58:54 pmYou are right arithmetic-wise, but wrong application-wise.5V is the voltage across your resister AND the LED in serial. So 5v at 220 mA is the power ofBOTHresister and LED together.To get the wattage dissipated by the resister, use the Vdrop across the resister only, that would be the right way of calculating the resister's wattage.But your problem started sooner than that,at 5v, you can't get 200mA current with a 220ohm resister.With 220ohm at 5v, the most current you can get is:5/220=0.02270.0227A is 22.7mASo even if your LED drops 0 volt,the max current is 22.7mA. You just can't get 200mA with that setup.Leaving that behindsince we know 220ohm is not right (or 200mA is not right)Whatevercurrentyou really need (at minimum keep awake), let say it is 20mA for your lights+resister to keep-awake, since you have thecurrentand the resister value, you can calculate the power with that two known value:Power = I * V (V is unknown, but we know V=IR, so replace V with IR)Power = I * I * RSo withcurrent= 20mA (0.02A) and resister=220ohmpower = 0.02*0.02*220= 0.0880 wattA 1/8 watt in decimal is 0.125watt. You have a good margin left over. A 1/4 watt is 0.250 and that is 2.84 times the wattage needed (for 20mA and 220ohm).As to determining the minimum keep-awake current:Try using a Variable Resister (or trim-pot) if you have one. Use a 500ohm pot or a 1K pot in series with 10ohm-ish. The 10ohm in series is there just to make sure you don't have a dead-shot when you lower the pot too much.When measuring the required resistance for the resister, just don't forget to measure the total (ie: trim-pot plus the 10ohm) instead of just the trim-pot.By the way, I am quite sure your "keep-awake" current for the power bank should be below 200mA and probably at or below 100mA. It would be a rather bad design for the power bank to go to sleep when it is still charging a battery at 200mA+. At 100mA, it is early but is arguable. One of my two 18650 charger typical cut out at just below 100mA, another charger is at (around) 50mA.EDIT: adding this:By the way, don't waste time increasing/decreasing a bit at a time. Newton's method works here too...Start with say 400mA, if it stays awake, cut it in half to 200mA, then half to 100, then half to 50, then half to 25... keep halving until you hit when it goes asleep.If it stays awake at 50 but goes to sleep at 25, now you know the right number is between 25 to 50. Go 1/2 way between 25 and 50 for the next cut: (50+25)/2 = 37.5Depending on if 37.5 is awake or asleep, choose it as high or low.Say it is 37.5 is awake and 25 asleep, next 1/2 way is (37.5+25)/2=31.25.But say at 37.5 it is asleep (but 50 is awake), you know it is between 37.5 and 50, (37.5+50)/2=43.75Knowing which half to use, you can cut it again. Here, you have in between 37 to 50 is 13 integers. Next cut will be 13/2=7 rounding up, then 4, then 2, then 1 single number (if you can indeed be that precise to a single mA). I would be quite please with +-5mA there (ie: a 10mA range).Cutting it in half and choose the right half is a much faster way to narrow in on the target.You don't really need the exact decimal each cut, you could have use 31 or even 30 for 31.25, or 40 for 43.75. Since you are determining which next half to use, you don't need it to an exact 50/50 half.