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NiMH Battery Pack Substitution Effect on Charging Circuit

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edy:
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 under normal usage is 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.  ;)

Audioguru:
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.

edy:

--- Quote from: Audioguru on November 01, 2018, 01:24:52 pm ---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.

--- End quote ---

Yes, thank you for reminding me.... After a bit of magic smoke  :-DD  and 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 on
CAR FULL THROTTLE (without LED ARRAY) = enough to keep PowerBank on.
LED ARRAY ALONE (car sittle idle but on) = not enough for PowerBank to stay on
LED ARRAY + CAR NORMAL RACING = enough to keep PowerBank on   :-+

Here are some pictures....







Now I have the added benefit of seeing the car glowing nicely when racing at night. :P  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 all  :palm:   although 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?  :-DD)

[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.

Rick Law:

--- Quote from: edy on November 01, 2018, 12:49:11 pm ---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 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.
...
I would also have the added benefit of having ALWAYS-ON-HEADLIGHTS if I go with the LED example.  ;)

--- End quote ---

That 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.

edy:

--- Quote from: Rick Law on November 01, 2018, 06:59:12 pm ---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.

--- End quote ---

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? 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!  |O

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