Protection board, battery holder, and converting the ride-on car to a battery pack

In the previous post I showed the lead-acid battery that “couldn’t cut it.” I said there would be a second part, so today I’ll tell you what I ended up having to do to get something that works.

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The lead-acid battery just wasn’t working out, so I decided to convert the ride-on car to a lithium battery pack.
At first I wanted to use a pair of 32700 cells, but then I thought that, first, it didn’t need that much capacity, and second, I felt like giving the thing a little more power. In the end, I decided to use two Samsung INR21700-33J 3270mAh cells. Besides, they were originally designed for “Tesla,” so why not put them to their intended use—in electric transport, even if it is a small one :)

The cells had enough capacity, and the voltage was a little higher, which was actually a good thing. The only snag was the aluminum positive terminal, so I bought a holder for a pair of 21700 cells. The ride-on car has room for two 6 V, 4 Ah batteries. My neighbor even thought about putting a 12 V battery in there, but I talked him out of it, since the car (or, more precisely, its gearbox) probably wouldn’t have handled it.


We bought the holder locally. As a reminder, aluminum battery terminals are a real headache: welding them properly with a home welder is quite difficult. If you’re not in a hurry, you can buy a similar holder on AliExpress - link.


I liked the holder overall. As is often the case, the leads are pretty thin, but I liked the springs: they’re not too stiff, and not so soft that the battery falls out.


There was just one problem: according to the markings and the design, the cells are meant to be installed in parallel, with the positive terminal on one side and the negative on the other. I needed to install them end-to-end, with the terminals facing the same way, but the markings got in the way. If this ends up in someone else’s hands, there’s almost a 100% chance they’ll install the cells incorrectly.
So I printed out a couple of small labels and taped them over the original markings.


Next came the choice of protection board. First, I ordered a couple from AliExpress: one standard board and one for LiFePO4, since I hadn’t decided which cells to use yet. Both boards are available at the same link
But I didn’t feel like waiting, so I rummaged through my stash and found three types of boards:
1. For “standard” cells, without a balancer - link
2. For “standard” cells, with a balancer - link
3. For LiFePO4, with a balancer - link.

The second and third links are to the exact stores I bought from. For the first one, I just found a similar board, but I think this is actually the most interesting listing - link.



Anyway, these boards had been sitting around for about 5 years, and now I finally found a use for one of them. I chose the version with a balancer for “standard” lithium cells. I could have added a separate balancer, but that would mean another board, wires, and so on—not an option at all.


The specs say the board is rated for 10 A, 7.4 V, with a pulse current of up to 20 A.
Model: HX - 2S - JH20 overcharging range: 4.25-4.35 v plus or minus 0.05 v
Size: 46.7 * 23 * 3.15mm over voltage range: 2.5-3.0 v plus or minus 0.05 v
Maximum working current: 10 a working temperature: - 40 - + 50 °
Maximum instantaneous current: 20 a storage conditions: - 40 - + 80 °
Static current: less than 10uA effective life: more than 50,000 hours
Resistance: less than 300 m Ω short circuit protection: to protect and restore must charge
Charging voltage: 8.4 v-9v
The protection board size and device description;
The protective plate size is 46.7 * 23 * 3.15mm



A closer look at the components. The transistors AOD514 are arranged in a 2S2P configuration: 30 V, 46 A, 5.9 mΩ at 10 V gate voltage and 11.9 mΩ at 4.5 V.


The controller is marked 20CB, so its specs can be looked up. But honestly, I think the last option would have been a better fit here: it has a lower threshold of 2.8 and an upper threshold of 4.25, and, well, I'll get to the third “and” later.


Next, the assembly steps.
First, drill a couple of holes in the bottom of the holder and a couple in the inner plates.
To mark one of the holes, you'll need to position the protection board against it. You'll see why as we go through the build.


Next, solder a thin wire to the center contact (the balance lead), degrease the surfaces, and stick on a couple of strips of double-sided tape.


I feed the soldered wire through the drilled hole and stick the board in place. That hole was drilled with the board positioned against it.
I bend the holder contacts over and solder them to the board. I don't think I need to point out that they must have the same polarity as marked on the back of the holder, right?


Then I route the balance lead through the prepared holes and bring it out through the hole on the other side of the holder.
There's a detail worth noting: if you look closely at the photo, you'll see that the wire doesn't just run to the point where the batteries connect. It runs along the side of the battery whose negative terminal faces that way. This is for safety: if the wire's insulation and the battery insulation are accidentally damaged, the wire will simply connect to the matching pole, and nothing will happen.




The balance lead, which now comes out near the opposite contacts, is used to connect the batteries.
I didn't cut the wire; I just slid the insulation back as needed and wrapped the exposed part around the terminals, then soldered it and clipped off the excess terminal length.


And that's pretty much it. All that's left is to solder on the output wires. I think it turned out quite neat.


One big advantage of the board is that it has auto-restart. That's not always useful—in some cases, you'd actually want it not to restart automatically. But for this application, auto-restart is more convenient, since if the protection trips, you won't have to carry the little car home. Trips can be false alarms, too.


Out of curiosity, I checked the voltage drop at 10 A.
1, 2. Voltage drop across the transistors, about 0.06 V with the battery fully charged, and rising
3. Total voltage drop, including the shunt resistors, about 0.09 V
4. As things warm up and the battery voltage drops, the voltage drop across the transistors increases. After a while it was already 0.0626 V; I didn't bother waiting any longer.


I tried loading it. The rated maximum current is 10 A, so I set the upper limit to 15 A. The current reached that level, but the board didn't shut off.


The datasheet shows a typical application circuit, but the board has a shunt, so my first thought was to reduce the shunt resistance.


I desoldered one resistor, which should halve the protection trip current.


I connected it to the load and raised the current to 16 A. Nothing happened; the board didn't shut off.


All right, let's keep going. I connected the two shunt resistors in series, so the protection trip current should now be four times lower than it was originally. I even started wondering whether the shunt and transistors were connected in series, which would explain why changing the shunt didn't affect the current proportionally.


15 A, and still no shutdown. This is ridiculous...


And then I take a closer look at the datasheet and spot the third “and” I mentioned at the very beginning.
The thing is, the specified overcurrent protection threshold is 200 mV. With a 2.5 mΩ shunt (two 5 mΩ resistors in parallel), that gives a trip current of... 80 A!
What's more, even after changing the value to 10 mΩ (two 5 mΩ resistors in series), I got a trip current of 20 A, which is still too high for me because the batteries aren't high-current cells at all.
And that's exactly why there was a third “and”: the last chip has a trip threshold that's twice as low, which would be better.


Originally, the circuit looked like this: the voltage drop was measured only across the shunt.


Realizing that this was complete nonsense, I decided to change the circuit. Yes, the shunt's resistance doesn't drift with temperature or supply voltage, unlike the transistors, but losing voltage across both the shunt and the transistors is hardly a great solution.
So I changed the circuit as shown below. In this case, the total resistance is the shunt resistance plus the transistor resistance, or about 10 + 6 = 16 mΩ here. The calculated protection trip current is therefore 0.2/0.016 = 12.5 A.


At first I just wanted to move the original resistor and add a jumper, but I accidentally broke it, so I installed a through-hole resistor instead. It makes no difference to how the circuit works.


I ran a load test and got a protection trip current of about 12 A, which is basically right; the formula didn't account for parasitic resistance.


While testing, I noticed the contacts were getting unusually hot, so I reinforced them as well.
I connected the ones that were already joined with a jumper. Where the contacts connect to the board, I added leads from some diodes, pushing them through the existing holes. Of course, the toy won't draw this much current, but it does have startup currents and there are losses, so why waste energy?


1. While I was experimenting, my neighbor opened up an old battery and took out its innards—not very carefully, but that's how it was.
2. Since the batteries are in a holder, they stick out a little beyond the case, but that doesn't really matter; the cover still closed properly.
3. Solder the wires from the board to the output terminals. The contacts inside are lead, but they solder just fine.
4. Just out of curiosity, I checked how much room there was inside. Six 21700 cells fit perfectly, with plenty of space left for a protection board.
So, you could easily build a battery with a capacity of around 120 Wh in this space, especially if you use 21700 5600 mAh cells.
This kind of battery would easily leave even the familiar 7 Ah, 12 V batteries in the dust. Those have up to 85 Wh in ideal conditions, but closer to 60 Wh in reality. In practice, a lithium battery made from 6 V, 4 Ah cells can provide the energy of two 12 V, 7 Ah batteries. That's exactly why I wrote earlier that lead-acid batteries are gradually dying out.


I didn't glue the case together yet; I just wrapped it with tape, charged it, checked that the voltage was normal, and handed it over for testing.


The next day, my neighbor brought the battery back and said it had lasted for two hours of riding. After that, the toy started cutting out occasionally when he pressed the pedal—in other words, the protection was kicking in. He also asked me to measure how much capacity had been used during those two hours.
I connected it to a battery tester and discharged it. It turned out that about 50% of the capacity had been used. So the toy draws only about 6–7 W in normal operation, or less than 1 A.

He also said the toy runs noticeably better, and the runtime is fine too.

While discharging it, I checked the voltage at which the over-discharge protection tripped. It was about 2.9 V per cell, but since the battery wasn't discharged, I suspected the overcurrent protection had kicked in instead.



After the test, I charged the battery. I noticed that toward the end, the charging current dropped to 80 mA and stayed there. Apparently the balancer had started working, which is fine.


Just to be safe, I removed one shunt, leaving 5 mΩ plus the transistors. That should have brought the maximum current to about 17 A, though I didn't measure it.
So why was the protection tripping? It's simple: if you go back to the description of the transistors, you'll see that their resistance depends on the gate voltage. As the batteries discharge, the overcurrent protection threshold drops too. Shunts don't have this problem, of course, but as I mentioned, the original circuit causes additional losses. The protection should now trip somewhere around 12–17 A.
I could have added a capacitor to the protection circuit to help with current spikes, but in the event of a short circuit it could burn out the board, so I left it as is.

So, I changed the shunt value and also put together a charger for it. The charger is the simplest kind. It’s rated for 2 A, but in reality it’s probably closer to 1 A. It’s cheap, of course, but experience has shown that these chargers are quite reliable - link.

The only issue is that the output voltage is higher than it should be. I only thought of it afterward: I could have simply added a Schottky diode in series with the output instead of opening up the charger. But I’ll see; I might still modify it.
The connector for the car and the polarity are the same. In principle, even if you connect the old charger, nothing bad will happen; the battery will just be undercharged because its output is 7.8 V instead of the new charger’s 8.4 V.


Conclusions.
As for the battery conversion itself, it has nothing but advantages: the car goes faster and runs longer, the battery is lighter, it charges quickly, and it should last longer too. What’s more, if you use standard 21700 cells that can be spot-welded or soldered, you can make a battery with a really high capacity.
As for the protection board, I’d say it’s decent in principle, but why they put an 80 A shunt on it is still a mystery to me. I bought this board back in 2018, and they’re still selling it with the same resistors. Anyway, when the one I ordered arrives, I’ll check what chip it uses and what current trips the protection.

In short, if your child uses a ride-on car with a similar power setup, I recommend replacing the lead-acid battery with lithium batteries. The lead-acid battery costs about the same, but performs worse and doesn’t last as long.

That’s all from me. I hope you found it useful.