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Cordless Drill Charger.

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sammy004

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Hi guys I was wondering if someone can help me out here. I have a cordless drill and the charger would not charge the battery so I replaced a resister 150 ohm (R1) with a 100 ohm and the 3mm red LED to a 3mm green LED and it started to charge one of the batteries which before it was not charging anything so I figured the burnt out LED was not letting the batteries charge. Then after I put the second battery in which was completely dead the LED indicator blicked and changed from green to a orangey red color and then just went dead so I opened it up again and the big resistor (R2) was not giving me any more reading and when I tested the charger output it was only 4.5 which as before was giving me 15 - 17 volts and the transformer was giving me 17volts. can anyone tell me what might the problem and why it charged one battery with no problems after I replaced the LED and then why did it just die after I put the other battery in? Any help would be greatly appreciated and I added pictures to make it more clear.

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The 150 ohm resistor is a series current limiter and the smaller resistor and LED are in series with each other and in parallel with the 150 ohm resistor. The Resistors keep burning out because you keep trying to charge dead batteries. This charger just isn't high enough quality to survive that kind of load.
 
Put the resistors back the way you found them and quit feeding it really, really dead batteries.
 
what do you mean put them back the way I found them, and that battery is no more then a year old. I know there has to be a way to fix this problem?
 
You changed a resistor from 150 ohms to 100 ohms. You can put that back the way it was designed to be. As for the rest, I've done what I can do. Someone else will answer soon, perhaps with a better method.
 
Batteries less than a year old can go bad.

The quickest way to kill them is to run them to nearly flat and then storing them without a charge. If the battery you're trying to charge is "dead", this means that your battery is in this category.
 
Hi,


Most likely the resistor is failing because it is meant to work with battery packs that charge up quickly to some nominal voltage and then take a while to charge to full capacity. If the pack is bad, that could mean one or more cells act as a short circuit and that means the voltage across the pack is much lower than it should be. That means the resistor sees more of a voltage drop and that means it sees more power and so it gets too hot and burns up.
The only way to fix this is to use a higher power resistor, one that can handle the full voltage of the charger, but a better idea is to replace the battery pack.
You can use a circuit to detect the state of the battery but that's a bit more complicated so i assume you dont want to go through that much trouble.
 
Pardon me saying this, but that is one crappy looking rechargeable battery charger. I guess the last thing on Wagner's mind was to try and provide the customer with a tool that would work beyond warranty as far as the battery goes.

I have seen crappy Chinese chargers with better designs. And that says a lot.

That poor battery. It was hammered from the first charge when it was still new. No sensing circuitry, no stop when fully charged (if it ever got there if used frequently....like daily), no protection whatsoever for the battery if left on an extended charge (forgotten on charge for a Week for example).

Wagner should take a page out of Philips rechargeable shaver book (we all know how small shavers are). And learn how to design a tool that does not abuse rechargeable battery technology. But rather embraces it.
 
Hi,
The only way to fix this is to use a higher power resistor, one that can handle the full voltage of the charger, but a better idea is to replace the battery pack.
Can you tell me what resistor to replace R1 or R2 and can anyone tell me the value of R2?

That poor battery. It was hammered from the first charge when it was still new. No sensing circuitry, no stop when fully charged (if it ever got there if used frequently....like daily), no protection whatsoever for the battery if left on an extended charge (forgotten on charge for a Week for example).

I think thats what might of happed.
 
The problem is one or two cells have shorted and the battery voltage is very low.
Open up the battery and test each cell with a voltmeter. Any cell that has 0v output can be fixed by getting your 12v car battery and quickly zapping it across the faulty cell. This will remove the "short." The battery can now be recharged.
 
Hi Sammy,

If you take a snap shot of the BOTTOM of the board and show that here i can help you more i think, and also the voltage of your battery pack.

Guessing, R1 is for the LED and R2 is to limit charge current. The value of R1 can be as high as 5k or as low as 100 ohms, so it's hard to say there, but for R2 it should be a value that provides for a decent charge rate with your battery pack voltage and charge output voltage.
Lets say the battery pack is 12v and the charger is putting out 18 volts. Divide 12 by 1.2 and multiply by 1.4 and we get 14v, and 18 minus 14 is 4 volts, and 4 volts divided by 150ma
(a typical sub C charge rate) gives us a value of about 27 ohms which is a standard resistor value too. To compute the power assuming we might have a dead batt pack, we take the 18v and square it like this: 18 times 18 which equals 324, then divide that by 27 (the resistor's ohm value is 27) and we get 12 watts, and doubling that for safety we get 24 watts. This means that a 25 watt resistor would do just fine, or maybe you can get away with a 20 watt resistor which could be two common 10 watt resistors in series or parallel.
Two 10 watt resistors in series would have to be about 12 to 15 ohms each, so if you buy two 15 ohm 10 watt resistors and connect them in series it should be ok.
In parallel, you would have to buy two 50 ohm 10 watt resistors and connect them in parallel and that would give you 25 ohms at 20 watts and that should be ok too.
You still have to provide adequate air flow to these resistors however to make sure they dont melt any of the plastic or burn up. That may require mounting them outside the case that the charger is normally enclosed inside of.

If you dont know the voltage of your battery pack then tell us the number of cells inside the pack, being sure to count every single one (some may be hiding in the neck of the pack).
 
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Hey sammy004

Thank's for your comment.

Philips manage to package a reliable rechargeable battery circuit in a small hand held device. The battery on my shaver is around three years old now. And still works. Reliably.

I have shaved my ugly mug tonight with my old Philips shaver on batteries. And I am good to go tomorrow. Customers and all.

I know I am digressing. Sorry mate.....but there is no point in a Manufacturer trying make things better by using Rechargeable batteries when the charger was made and designed in hell.

No matter what you try and do to try and rectify the charger, it will always be a POS.

It was designed that way.

.
 
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Thank you so much guys for helping me out on this I really appreciated it. The battery is a 14.4 volts and here are a few pictures of the board, battery and transformer.
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If you guys need any more just let me know and thank you again.
Sammy
 
Hey, good job with the pictures! It's nice to not squint at fuzzy pics.

Wow, that is one dead-simple battery charger!

If these resistors keep burning out, just use higher wattage ones. It's obvious that they are dying because they are trying to charge REALLY dead batteries, not because of some fault.
 
Thank you I like to use the macro option at close-up. Well I replaced the LED and the battery that was half dead went to full so it charged that with no problems, but when I tried to charge the completely dead battery the R2 now give me a reading that is open (I think you call it?) or infinity and the LED just died. So I will try to replace them with a little higher values, so R1 was 150 I'll change that to 200? I also don't know the value of R2 but I will try to use the idea MrAl told me sounds kinda complicated I hope he can help me with that math part :)
 
If you keep increasing the value of resistors until this charger will survive TRYING to chage a completely dead battery, you will get the power down low enough that the heat triggered shut off device on the good batteries will not open and you will ruin them by overcharging them to death.

just a suggestion
 
A completely dead battery will get a "floating charge" very quickly. There is obvioulsy something wrong with one or more of the cells.
 
Yeah I will not increase it that much where it is suppose to be 150ohm I will just change it to a 200ohm I will not go over that. As for the dead battery I will open it up and see if I can find the dead cell and try to zap it. If someone can just tell me the value of R2 that would be greatly appreciated as I can not figure out what that is.
 
Hi again Sammy,


Oh yeah, very nice pics! It's nice to be able to see the entire circuit now as that helps a lot in trying to figure out what is going wrong.

After looking at the pic's I sill have to agree with the other poster's here in that the dead or near dead pack is causing the problem.
However, we can still adapt the power resistor to take the full charger voltage if you like.

The power resistor limits the current to the pack, and the smaller resistor limits current to the LED who's brightness varies a bit with the charge current level.

The pack voltage is 14.4v as you indicated, and the wall wart is 19v that can handle 500ma, but we'll start with a design for 250ma which is a nice comfortable charge level.

We start by taking 19 and subtracting 14.4, and that gives us 4.6 volts. That's the voltage that the resistor has to drop when the cells are just starting to charge.
The resistor value will be 4.6/0.250 which comes out to 18.4, so we'll round up to 20 ohms for now. Now the power is 4.6/20 which is only around 1 watt, but that is when the pack is working perfectly which we are going to assume isnt always the case. With a pack voltage of zero volts, the power can be as high as 19*19/20, which is close to 18 watts.
That means we would really have to use a power resistor able to handle at least 30 watts, and that means at least three common 10 watt resistors in series or three in parallel. Lets say we use four in series, that would mean four 5 ohm resistors, each rated for 10 watts, which would give us 40 watts which would definitely handle the required power. The charge current near the end of charge will be lower however.

Now to calculate the other resistor. Since the voltage can be anywhere between 19 volts and 4.6 volts, we really have to use a 1k ohm 1/2 watt resistor. The LED may be a little dim near the end of charge, so you may want to use a high brightness LED rather than a regular LED, and it should be Red.

Note also that you will have to mount the power resistors where they can get free air flow too. This might mean mounting them outside of the charger case. For testing, you can wire them up and let them hang from something and see how hot they get while charging a dead pack.

I do have to caution you however in that if the pack voltage is that low then you should really fix it before charging. That's the best way to do this.
 
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