How can I revive this Acer laptop battery?

Thread Starter

rambomhtri

Joined Nov 9, 2015
462
So, I have this Acer ES1-523, and the battery doesn't work. I don't know if it's totally broken or if it simply got over discharged and can't be charged with the laptop IC anymore. The laptop is 3 years old and it has been used just like a regular laptop, nothing crazy, it doesn't get hot at all because components are very weak, it has a 45W charger, that tells you everything (laptops normally have at least 100W chargers). The laptop works just fine as long as it's plugged in to AC, but whenever you remove the AC, instant shut down, as if there where no battery at all. Nevertheless, Microsoft wrongly reports the battery is fully charged, and it detects it, doesn't report any problem. The battery appears to be model 2ICP4/80/104 from Acer I believe,

I disassembled the laptop and took off the battery, and I'm gonna post some pictures and also measurements, wondering if you can help me charge it manually:
1.jpg 2.jpg
3.jpg

Measurements, respecting the sign (polarity):
From orange to red: +1.656V
From orange to yellow: +2.533V
From orange to green or blue: +2.432V
From orange to black: +2.535V
From red to black: +25.0mV

According to my understanding of physics, if from orange to red there is 1.656V, and from orange to black there is 2.535V, from red to black there should be -1.656+2.535= 0.879V. That's my first question.

I don't have a lab power supply, I only have the laptop charger (19V, 2.375A) and AA, AAA batteries. Is there a way I can revive this battery? Like manually charging it somehow?

I know these batteries should start with 0V charging, go up the the nominal value almost immediately, charge at constant Amperage during the first phase while the voltage increases slowly, then about 50% you keep the voltage constant and the amperage drops slowly (second phase).
 
Last edited:

KeithWalker

Joined Jul 10, 2017
2,306
So, I have this Acer ES1-523, and the battery doesn't work. I don't know if it's totally broken or if it simply got over discharged and can't be charged with the laptop IC anymore. The laptop is 3 years old and it has been used just like a regular laptop, nothing crazy, it doesn't get hot at all because components are very weak, it has a 45W charger, that tells you everything (laptops normally have at least 100W chargers). The laptop works just fine as long as it's plugged in to AC, but whenever you remove the AC, instant shut down, as if there where no battery at all. Nevertheless, Microsoft wrongly reports the battery is fully charged, and it detects it, doesn't report any problem. The battery appears to be model 2ICP4/80/104 from Acer I believe,

I disassembled the laptop and took off the battery, and I'm gonna post some pictures and also measurements, wondering if you can help me charge it manually:
View attachment 187251 View attachment 187252

View attachment 187256

Measurements, respecting the sign (polarity):
From orange to red: +1.656V
From orange to yellow: +2.533V
From orange to green or blue: +2.432V
From orange to black: +2.535V
From red to black: +25.0mV

According to my understanding of physics, if from orange to red there is 1.656V, and from orange to black there is 2.535V, from red to black there should be -1.656+2.535= 0.879V. That's my first question.

I don't have a lab power supply, I only have the laptop charger (19V, 2.375A) and AA, AAA batteries. Is there a way I can revive this battery? Like manually charging it somehow?

I know these batteries should start with 0V charging, go up the the nominal value almost immediately, charge at constant Amperage during the first phase while the voltage increases slowly, then about 50% you keep the voltage constant and the amperage drops slowly (second phase).
The batteries are discharged below the level that the charger will accept. If you can get the batteries to read above 3 volts between black and orange,
there is a chance that you can recover them There is a way to do this but be aware that you may damage the electronics built into the battery pack. You may as well try it as you have nothing to loose at the moment. You will need five new AA alkaline batteries and a 39 ohm half watt resistor. Connect them as shown in the attached drawing. Be careful about polarity. I am assuming that black is negative and orange is positive. Double check that and make sure you have + connected to +. Check the voltage on the DMM and when it reads above three volts with the AA batteries disconnected, which may take some time, put the battery back in the laptop. Connect the charger immediately. If you are lucky, the battery will take a charge.
Good luck, Keith
Recover.jpg
 

Thread Starter

rambomhtri

Joined Nov 9, 2015
462
Thanks! Yeah, whatever I can try, I will. These pieces of **** cost about $50!
Never saw a battery that expensive.

Anyway, I posted a picture with the polarity, and you can see it's RED the positive, and BLACK the negative or ground, besides you have 2 red and black cables, so it makes all sense right?

In my test I put the orange probe as a "reference" because it's the highest voltage among all pins, not because it's the positive. It's also the one that gives me solid results. Read what happens when you compare black-orange-red triangle of voltages. I can't find an explanation.

Where can I find a 39 ohm half watt resistor? Why do you give me the power of a resistor, if the only spec of a resistor is its ohms?

About the test:
5x1.5V = 7.5V aprox (when they are fresh new, it's 5x1.65V = 8.25V)
The internal resistance of the battery between black and red is 8.15MOhm. So, those 7.5V would drop mostly at the battery. Why we need a resistor?

Is this correct?
 
Last edited:

KeithWalker

Joined Jul 10, 2017
2,306
Thanks! Yeah, whatever I can try, I will. These pieces of **** cost about $50!
Never saw a battery that expensive.

Anyway, I posted a picture with the polarity, and you can see it's RED the positive, and BLACK the negative or ground, besides you have 2 red and black cables, so it makes all sense right?

In my test I put the orange probe as a "reference" because it's the highest voltage among all pins, not because it's the positive. It's also the one that gives me solid results. Read what happens when you compare black-orange-red triangle of voltages. I can't find an explanation.

Where can I find a 39 ohm half watt resistor? Why do you give me the power of a resistor, if the only spec of a resistor is its ohms?
The orange wire has the highest positive voltage with respect to black so this is a direct connection to the battery positive terminal, which is what you need to access.
You need a 39 Ohm resistor that can handle 100 mA without overheating. With a four volt drop across it, it will be dissipating 0.4 Watts. I don't know where you can get one from. Depends on where you live. You could make it up from a number of higher value resistors in parallel. For example, four tenth watt 150 Ohm resistors in parallel will give you 37.5 Ohms at 0.4 Watts. That would be close enough to do the job.
 

Thread Starter

rambomhtri

Joined Nov 9, 2015
462
But in the picture I see really clear red is the positive, it even has 2 cables. I though the orange would be a thermistor, clock or something like that. I don't understand why it's higher than the red, seriously no idea.

How do you know how many V will drop at the resistor?
 
Last edited:

Thread Starter

rambomhtri

Joined Nov 9, 2015
462
Is the battery bulging?
If so then it is terminally defunct. Do not attempt to recover it, buy a new one.
Not at all, it's dead flat, flatter than earth actually, hahahaha. The battery really, looks perfect, and as I told you guys, this laptop has very weak components, there's almost no heat.
 

Thread Starter

rambomhtri

Joined Nov 9, 2015
462
A resistor is a resistance, and that's measured in ohms. The wattage is determined by where you connect that resistor to, not by the resistor itself, that's what I wanted to mean. I guess when you talk about the watts of a resistor, is the maximum wattage is designed to support, right?
 

Thread Starter

rambomhtri

Joined Nov 9, 2015
462
I suggest you read the basics about resistors to not repeat such a question.
OK, this is going kind of off topic, but, instead of posting that, can you lead me to "the basics" of resistors?
Or simply explain to me what the power means in a resistor spec sheet?

I guessed it was a safe wattage for a given resistor (before it burns or behaves differently as designed), but you seem to discard that by telling me to learn the basics. I think I know the basics, but one never knows...

But hey, this post is about reviving my battery, keep that in mind, hahaha.
 

KeithWalker

Joined Jul 10, 2017
2,306
Thanks! Yeah, whatever I can try, I will. These pieces of **** cost about $50!
Never saw a battery that expensive.

Anyway, I posted a picture with the polarity, and you can see it's RED the positive, and BLACK the negative or ground, besides you have 2 red and black cables, so it makes all sense right?

In my test I put the orange probe as a "reference" because it's the highest voltage among all pins, not because it's the positive. It's also the one that gives me solid results. Read what happens when you compare black-orange-red triangle of voltages. I can't find an explanation.

Where can I find a 39 ohm half watt resistor? Why do you give me the power of a resistor, if the only spec of a resistor is its ohms?

About the test:
5x1.5V = 7.5V aprox (when they are fresh new, it's 5x1.65V = 8.25V)
The internal resistance of the battery between black and red is 8.15MOhm. So, those 7.5V would drop mostly at the battery. Why we need a resistor?

Is this correct?
You need the resistor to limit the current to about 100 mA when you first connect to the battery.
 

KeithWalker

Joined Jul 10, 2017
2,306
But in the picture I see really clear red is the positive, it even has 2 cables. I though the orange would be a thermostat, clock or something like that. I don't understand why it's higher than the red, seriously no idea.

How do you know how many V will drop at the resistor?
You stated that the maximum voltage you measured was 2.535 volts between orange and black. That has to be the battery terminals. Five AA cells in series give approximately 7.5 volts when supplying 100 mA. The difference between the two is about 4 volts.
 

Thread Starter

rambomhtri

Joined Nov 9, 2015
462
You stated that the maximum voltage you measured was 2.535 volts between orange and black. That has to be the battery terminals. Five AA cells in series give approximately 7.5 volts when supplying 100 mA. The difference between the two is about 4 volts.
Yes, orange is the highest, but it's weird, isn't it?
Don't you think that the orange is a thermistor or something?
In all batteries, the positive and negative are opposed, so, here, black and red, and also they happen to have an extra cable. It all indicates that they are the + and -, besides the sign in the battery...


I mean, the red cables are right above the + sign of the battery, indicating the positive.

I thought the orange is giving the highest because the unnatural state of the battery.

By the way, I've just found a B500K pot:
https://www.digchip.com/datasheets/parts/datasheet/682/RV16AF-10-15R1-B500K.php

Is that able to be set to 39 ohm and handle the watts?

I happen to have as well a bunch of these 331 Ohm resistors:

http://www.yampe.com/product/detail...roduct_id=2064&product_type_id=38&brand_id=34

If I connect 8 of those in parallel, I'd get 1/R = 8/331, R = 41.25 Ohm
Each resistor is rated 0.25W max and 331 ohm with 5% tolerance (315-345 ohm), so I guess it would work, right?

Give me green light and I will do it, except I still am not sure about the orange being the positive, it doesn't make any sense.
 
Last edited:

djsfantasi

Joined Apr 11, 2010
8,256
Yes, orange is the highest, but it's weird, isn't it?
Don't you think that the orange is a thermostat or something?
In all batteries, the positive and negative are opposed, so, here, black and red, and also they happen to have an extra cable. It all indicates that they are the + and -, besides the sign in the battery...


I mean, the red cables are right above the + sign of the battery, indicating the positive.

I thought the orange is giving the highest because the unnatural state of the battery.

By the way, I've just found a B500K pot:
https://www.digchip.com/datasheets/parts/datasheet/682/RV16AF-10-15R1-B500K.php

Is that able to be set to 39 ohm and handle the watts?

I happen to have as well a bunch of these 331 Ohm resistors:

http://www.yampe.com/product/detail...roduct_id=2064&product_type_id=38&brand_id=34

If I connect 8 of those in parallel, I'd get 1/R = 8/331, R = 41.25 Ohm
Each resistor is rated 0.25W max and 331 ohm with 5% tolerance (315-345 ohm), so I guess it would work, right?

Give me green light and I will do it, except I still am not sure about the orange being the positive, it doesn't make any sense.
In a power supply, there may not be just one positive supply. Multiple voltages can be supplied, each with a positive polarity (and some maybe having a negative polarity).

So the idea that orange must be something like a thermostat is false. Or the idea that orange having a different voltage is “weird” also is false.

Laptop battery packs have built in electronics. Consider that orange might be the raw battery power and red is somehow processed.
 

Thread Starter

rambomhtri

Joined Nov 9, 2015
462
In a power supply, there may not be just one positive supply. Multiple voltages can be supplied, each with a positive polarity (and some maybe having a negative polarity).

So the idea that orange must be something like a thermistor (I meant, not thermostat xd) is false. Or the idea that orange having a different voltage is “weird” also is false.

Laptop battery packs have built in electronics. Consider that orange might be the raw battery power and red is somehow processed.
I said thermistor just cause, but I thought it could be clock, data or switch...

So, should I proceed?
Where did you get the info that 5 batteries would be required?

I read that it should deliver 7.7V, from the first picture, but in that picture, it says down below: charging current 3A/ 8.8V. What that means?
 

Ylli

Joined Nov 13, 2015
1,057
Put a resistive load across the voltmeter leads* and re-measure as you did in post #1. Drawing some current may give a better indication of which wire goes directly to the battery +.

* one of those 330 ohm would be fine.
 

Thread Starter

rambomhtri

Joined Nov 9, 2015
462
Put a resistive load across the voltmeter leads* and re-measure as you did in post #1. Drawing some current may give a better indication of which wire goes directly to the battery +.

* one of those 330 ohm would be fine.
Wow, with a 331 ohm resistor in parallel, I get exactly 0V between every single pin. It's solid 0.00V between red-orange and orange-black. Every thing gets 0.00V. Whenever I remove the resistor in parallel, I get the values in the first post.
 
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