Replacing external battery

Thread Starter

rambomhtri

Joined Nov 9, 2015
295
Hi, I have this very cheap external battery I got for free in an event, and it's working meh, and I believe it's because the battery is starting to get faulty/wear. Sometimes I've found it doesn't charge itself, may be because it gets to very low voltages. Here are some pictures:




And this seem to be the schematic of the chip:

I want to know if I can replace the original 2000mAh battery and buy a protected battery of 3400mAh like this one:
https://www.aliexpress.com/item/32628168837.html?spm=a2g0o.productlist.0.0.7ffd4dc5yOs2vL&algo_pvid=eb288943-cc15-44b1-92cf-e039c2acb8cc&algo_expid=eb288943-cc15-44b1-92cf-e039c2acb8cc-1&btsid=2000eaa8-dffc-48f5-8f79-bac19a7786a1&ws_ab_test=searchweb0_0,searchweb201602_2,searchweb201603_52

They are a little larger (65mm vs 69mm) but I believe I can make room for it. I guess I can because the IC would cut off the power of the battery when voltages of 3.3-3.4V are reached, which would protect the battery from over-discharging (but I've tested this original blue battery and I read with a multi just 1.5V...). Will it also fully charge a 3400mAh battery without problems?
 

Thread Starter

rambomhtri

Joined Nov 9, 2015
295
That would appear to be an acceptable replacement.
It should, but will take longer to reach full charge.
Sweet news!
I'm worried about the battery capacity not being fully used. By that I mean, as an example, may be the IC knows that it must charge 2000mAh and that's it.
How the IC knows when the battery is full?
By measuring that whatever is plugged is giving 3.7-4.0V?
I believe the battery works like this: there's a "voltage reader" that tells what the battery is giving. If it is giving power and it reaches 3.3-3.4V, the IC stops the discharge and indicates battery dead. If it's getting charged, the voltmeter reads 3.7-4.0V reached, stop charging, battery full. Is it like that?

I don't know much about electronics at this level, but I'd love it if someone explains to me briefly how a battery like this works: what the 4 resistors do, why it has 2 capacitors and 1 coil (or whatever that little tower is).

Thank you!
 

crutschow

Joined Mar 14, 2008
23,291
How the IC knows when the battery is full?
If it's getting charged, the voltmeter reads 3.7-4.0V reached, stop charging, battery full. Is it like that?
Sort of.
It may also look at temperature and the change in voltage as well as the absolute voltage to determine when it's charged.
That is generally independent of the battery capacity, so it should charge your new battery as well as it did the original.
 
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crutschow

Joined Mar 14, 2008
23,291
But just in case there's a problem, I would closely monitor the new battery the first time you charge it to make sure it doesn't overheat.
 

Audioguru

Joined Dec 20, 2007
11,251
www.batteryuniversity.com and battery charger IC manufacturers say that a lithium battery is charged at a constant current with its voltage limited to 4.20V. When its charging current drops to about 1/40th of its capacity then it is fully charged and the charger turns off.

The battery charger circuit first measures the battery voltage and if it is too low (less than about 3V) then it tries charging with a low current and if the voltage does not rise then it turns off and issues an error message. A battery that has a voltage that is too low probably has some Lithium ions converted into solid Lithium that is a dangerous explosion and fire hazard if charged at a normal charging current.
 

Thread Starter

rambomhtri

Joined Nov 9, 2015
295
So, if I understood right, the intelligent part of the IC, which can only be the MP3401A, does this:

When charging, it has an ampere meter that measures the current, knowing that is must read a certain value (but wouldn't you need a memory or something to save that value???). Once the ampere meter reads a very low current (1/40 of the expected current), it determines it must stop charging. I guess this way it doesn't care about the capacity of the battery, which is fine. To avoid damaging overly discharged batteries, the very few moments of the charging process, it measures the battery voltage, and if it's too low, it starts with a low current and increase it over time up to the certain value.

When discharging, it measures the voltage of the 3.7V rated battery. If the voltage drops below 3.3V or something, it stops discharging and says "battery dead".

Is that right? Is that how my IC works, or since it's very cheap it lacks some components and therefor my IC works worse, skipping some security measures?
 

Thread Starter

rambomhtri

Joined Nov 9, 2015
295
Hi, I just wanted to point out that I just received the batteries today, and replaced my external battery with one of the ones I bought (Panasonic whatever, protected and all). It's working nice, so far!!!

So I'm very happy. These external batteries are actually quite expensive, a 5000mAh battery costs $10 or more, and I just "created" mine for just $2. I had to work around some problems, like the new battery with protection IC is bigger and larger, and I had to modify the case, ended up soldering the battery to the IC... but the resulting result results nice.

Anyway... one quick question... If I wanted to have a bigger capacity external battery, like more than 10000mAh, could I simply join the positives of 4 batteries, then the negatives, solder that to the IC, and get a total of 4*3400mAh = 13600mAh?
In other words, connect in parallel 4 batteries. Would that work?

Also, I just thought... what if you put 4 batteries in series?
I know you get a 4*3.8V = 15.2V battery, but what about the energy stored?
Would you still have 13600mAh?

Say the batteries can give a maximum of 1A. If that's correct, the power of the parallel would be 3.8V*4A = 15.2W
The power of the series would be 15.2V *1A = wow, the same, interesting...

Oh, I'm a little upset with the performance of the circuit. My smartphone is 3100mAh, and this battery I put is 3400mAh. I've been charging my smartphone for 3h or so, and it got up to 71% before the battery died. I've used it a little bit while charging, but I expected the battery to at least fully charge once my phone. So, I guess the IC is to blame, right? There's a lot of power lost because of the chinese basic, bad IC, right?
That, or my Philips battery is crap, but I don't believe that.
 
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