Electronics > Beginners

NiMH Battery Pack Substitution Effect on Charging Circuit

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6PTsocket:

--- Quote from: edy on October 30, 2018, 01:26:04 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?

--- End quote ---
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.

Sent from my SM-G900V using Tapatalk

edy:
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!

Rick Law:

--- Quote from: edy on November 03, 2018, 11:58:54 pm ---...
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!

--- End quote ---

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.

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