Fixing a battery charger.

[Tip-C]

Hello,

I have some limited knowledge of electronics, thus im here to ask for help.

I have a normal Ni-Mh battery charger capable of charging AA, AAA and the 9V rectangular batteries.

After doing a few tests with a multi meter, i came to the conclusion that the Transformer inside is broken as no current passes trough it!

My question is does any body know the specifications of a transformer that would go into this type of battery charger???

Input has to be 120V im looking for the output!

Thanks guys!
Dennis

 

[kchriste]

That's a tough one. Can you post a good macro picture of the innards of the beast? It is unlikely we'll be able to tell from just a picture, but it's more than we have to go on right now. You'll probably have to buy a new charger anyway as the transformer is unlikely to be easy to find and purchase.

 

[Willbe]

If you can read the voltage rating of the little fat filter capacitors that follow the transformer's diode bridge, it will give an idea.

 

[Tip-C]

The cap is a 100 microfarad with a voltage of 25V.

Here is some additional info that might help.

There are 5 wires on the output of the transformer...
2 white, 2 red and a black.

the two red go to each extremity of the circuit board where they then each go into a IN400 DC Diode which goes to a 2.7ohm resistor. From this point the circuit splits where on one hand it goes to the AA positive contact pad and on the other it goes yet to another resistor, this time a 12ohm after which is the AAA positive contact pad.

The two white wires connect near each other on the circuit board, they seem to have 2 diodes connecting them with the cap mentioned above...after this im having trouble following the circuit!

 

[kchriste]

Looks like you have 2 different windings. One for charging the 9V battery and the other for AAs and AAAs. With the unit unplugged, what is the resistance of the white windings? The red? The primary (120Vac) winding? Also check to see which pair (Red or white) has continuity with the black wire. Report your findings back here. Also include a clear picture of both sides of the PCB.

 

[Tip-C]

Resistance between white windings: 37ohms

Resistance between red windings: 7ohms

Resistance on primary winding could not be measured as there is no continuity.

All wires have continuity with the black one.

NOTE: I measured using a multi-meter directly on the circuit board and using the solder points of the wire as the contacts for the measurements. Is that correct?

Here are the pics:

Circled are where each wire connects on the back of the board.
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[flat5]

One check before giving up on the transformer is to wiggle it while feeling and looking at the input contacts on the solder side of the board. You want to know if the connections are still good and not "cold".

Just saying...you may have already done this.

 

[transistor495]

Hi Tip,
As you have told that there is no continuity for primary, I suggest you to stick on the primary of the transformer and try to get it's hot ends. Maybe you can get it back if it cuts outside, if its inside I don't think its worth to replace it, better will be buying new one so far in the case of transformer burns.

 

[Tip-C]

I tried what you said but no luck. I guess the damage must be internal. My best bet would be just to replace the transformer!

Then again....what i need to know is the specs of this transformer? More precisely, the output specs!

 

[Andrew Leigh]

Hi,

I see you can charge up to 4 batteries simultaneously? This will determine the output voltage.

Some thoughts, perhaps you can get some clues from the board.

On the one side it gives a voltage of 2.7V which is consitent with charging two cells. Both sides of the board seem identical (input componentry) and are fed by what appears to be identical windings. I think that indeed you have a dual circuit, one charges two AA/AAA and the other side does the same. This is to avoid sensing componentry that will automatically sense how many cells and therefore the voltage that need applying.

Ideally you need 1.4V per cell and I would suggest that the winding (red to white) is as follows;

2.8V (required for charging) + 0.7V (for the drop over the diode) + voltage drop over the 2.7ohm resistor. As this seems to be the one leg of the 9V circuit I thinks they drop the voltage through that resistor?

Can you confirm that the charger works in pairs of batteries?

Check your resistance measurements, I think that the windings are A (Red and White), B (Red and White) and we need to sort out the black wire.

The black winding I believe is to feed the 9V (8,4V) batteries. I think that this winding is also dual having the black as a common. (Red and Black and Red). The voltage of this winding should be 10.5 to 11.9V. Again I believe that each side feed one 9V battery.

By measurement I think you will get the following;
Between the Red and White on both sides, identical resistances.
Between the Black and Red on both sides identical resistances.

Can you check the resistance in both cases of black to white?

PS: I am using logic rather than any of the extremely limited electronics knowledge I may possess.

Lets take it from there.

Cheers
Andrew

 

[Tip-C]

Yes the charger works in pair...you have to charge either two batteries on the left or on the right. Sorry, i knew this before, i guess i forgot to mention it. When you put it 2 batteries it closes the circuit and lights up the LED.
Im just not sure if this applies for the 9V batteries as i never use those. and to be honest, if it is as you say that the black winding is for charging the 9V, to me it doesnt really matter if it works or not.

And you were right as to the red and white windings. They have nearly identical resistances.

On one side, there is a resistance of 23ohms and on the other there is 21.7ohms.

but when you say "2.8V (required for charging) + 0.7V (for the drop over the diode) + voltage drop over the 2.7ohm resistor. "
How can i calculate the voltage drop over the 2.7ohm resistor to get the correct voltage of the winding?

As for the black one...you were saying that i should get the same resistances with black and the 2 reds and black and the 2 whites.
This is the case!

Black/RED = 3.7ohms on both sides
Black/White = 19.6ohms on

But as i said before, i don't need to charge the 9V. I need this for charging AA mainly. So if its too much of a headache to find the voltage of the black winding, dont bother!

 

[Andrew Leigh]

Hi Tip-C,

I wish the boys who really understand this stuff would chip in now. As I said my go at it was logic more than anything else. Anyways it proves that the system is identical on either side so this makes life much easier.

I am lost as to how to get to the correct voltage mathamatically. What I would have done is to take a ~ 5V transformer and connected the output to the red and white wires. I would have then checked what the voltage was at the points where the battery was and then calculated how much voltage I needed to drop / increase in order to get to the 2.8V required. Doing it this way keep all the components in circuit and the 5V should be low enough as not to damage anything for the short duration you need to measure the voltage.

Not very elegant but would work for me.

Hopefully one of the clever boys can give you a more definitive answer.

Cheers
Andrew

PS: What is the Ah rating of your cells?

 

[Tip-C]

can anybody else point me into the right direction on this one?

Thanks

 

[kchriste]

What is the model number of the charger? What we need to figure out is the specified charge rate of the charger for the AA or AAA cells. This should be in the little manual that came with the charger. From this we can deduce the secondary winding voltages. I suspect you have two center tapped windings where the black wire is the center tap for both the white and red pairs. The red pair charges the AA and AAAs. The white pair charges the 9Vs while also powering the charge controller IC. So you'll need both center tapped windings to make it work even though you only want to charge AA and AAAs.

 

[Tip-C]

Here is some information provided on the back of the charger:

Class 2 Battery Charger

MODEL: CHM4AA
INPUT: 120V AC 60Hz 4.6W
OUTPUT: AA- 2 x (2.8V DC 150mA)
AAA-2 x (2.8V DC 16mA)
9V cell-2 x (9.0V DC 16mA)

Charging Time: 15 hours by timer control.

 

[transistor495]

Bravo! This specs sound like a common one. I think you can simple replace it with any transformers used for this kind of chargers. But I'm afraid of the part availability.

 

[kchriste]

It sounds like you need a 6 Vac 400ma center tapped transformer for the AA/AAA side. This would give you 3 Vac between the black wire and either red one. This would result in 4.2Vpk - 0.7V (diode) - 2.4V (battery) = 1.1Vpk halfwave across the 2.7Ω resistor. This should give a nominal 137ma RMS to the AA battery. These calculations ignore the winding resistance/loading effects of the transformer. They also assume that the AC input to the transformer is at a nominal level. A 6.3Vac center tapped transformer would give you apx 167ma and would be the better choice due to losses from winding resistance etc.
I can't estimate the voltage for the 9V/IC supply because the circuit configuration is unknown to me.
This charger doesn't appear to be worth saving to me. It is a slow charger. The worst part is that it simply uses a timer to turn off charging so it will under charge batteries with a capacity higher than apx 1600mAh. It will over charge lower capacity batteries.

 

[Tip-C]

WOW...Great Explanation.

Truly appreciated!

I will try to get the part on Friday and I will let you know how it goes!

Thanks again.

 

[kchriste]

Don't forget: you must also power the circuitry on the 9V charger side because the white windings also power the timer chip which turns on/off the charging current for the AAs! Just adding the 6.3 Vac 400ma center tapped transformer won't be enough to make it work.

 

[transistor495]

Don't forget: you must also power the circuitry on the 9V charger side because the white windings also power the timer chip which turns on/off the charging current for the AAs! Just adding the 6.3 Vac 400ma center tapped transformer won't be enough to make it work.

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Great!