Author Topic: NiMH Battery Pack Substitution Effect on Charging Circuit  (Read 6336 times)

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Offline Rick Law

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Re: NiMH Battery Pack Substitution Effect on Charging Circuit
« Reply #25 on: November 02, 2018, 05:56:47 am »
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.

Yes 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?
...
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
...
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!  |O

I 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.)
 

Offline edyTopic starter

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Re: NiMH Battery Pack Substitution Effect on Charging Circuit
« Reply #26 on: November 02, 2018, 12:27:33 pm »
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!  |O  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.  :palm:  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!
« Last Edit: November 02, 2018, 12:34:46 pm by edy »
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Offline edyTopic starter

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Re: NiMH Battery Pack Substitution Effect on Charging Circuit
« Reply #27 on: November 02, 2018, 02:19:55 pm »
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???  |O  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/3070
https://www.maximintegrated.com/en/app-notes/index.mvp/id/3256

In 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.  |O 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.  |O   Ok I'm done! I can't figure this out.  :horse:

[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
« Last Edit: November 02, 2018, 03:22:47 pm by edy »
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Offline Audioguru

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Re: NiMH Battery Pack Substitution Effect on Charging Circuit
« Reply #28 on: November 02, 2018, 07:57:12 pm »
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.
 

Offline Rick Law

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Re: NiMH Battery Pack Substitution Effect on Charging Circuit
« Reply #29 on: November 02, 2018, 08:08:01 pm »
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.


 

Offline 6PTsocket

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Re: NiMH Battery Pack Substitution Effect on Charging Circuit
« Reply #30 on: November 03, 2018, 10:50:13 pm »
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?
It 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.

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Offline edyTopic starter

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Re: NiMH Battery Pack Substitution Effect on Charging Circuit
« Reply #31 on: November 03, 2018, 11:58:54 pm »
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!
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Offline Rick Law

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Re: NiMH Battery Pack Substitution Effect on Charging Circuit
« Reply #32 on: November 04, 2018, 02:22:25 am »
...
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?
...
When I start swapping out the existing resistor I can do an actual current draw test and know better!

You 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 of BOTH resister 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=0.0227
 0.0227A is 22.7mA
So even if your LED drops 0 volt, the max current is 22.7mA.  You just can't get 200mA with that setup.

Leaving that behind since we know 220ohm is not right (or 200mA is not right)

Whatever current you really need (at minimum keep awake), let say it is 20mA for your lights+resister to keep-awake, since you have the current and 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 * R

So with current = 20mA  (0.02A) and resister=220ohm
power = 0.02*0.02*220
          = 0.0880 watt
A 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 = 37.5
Depending 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=43.75

Knowing 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.

« Last Edit: November 04, 2018, 03:40:46 am by Rick Law »
 


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