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Charging a valve regulated lead acid battery? Answered

A while ago (a few months maybe?) I bought a cheap robot vacuum cleaner from a car boot sale. I decided to take it apart as it was pretty useless as a vacuum cleaner. When I took it apart I kept the battery and the charging plug (wall wart) together. Now I want to use the battery but I am not sure how to connect it to the charger. (I tested the battery and it still works.)

The battery is a 6V 2.8Ah valve regulated sealed lead acid battery. On the front of the battery it says it needs a constant voltage charge of 7.25-7.45V and less that 0.84A. 

The plug that came with it is a 7.5V 600mA wall wart. 

I have a vague recollection of how it was connected up. I remember that there was a diode in the mix. The positive terminal was connected to a switch and then the main control board. The negative terminal connects to the main control board and there is also a small wire that was connected to something else. I would like to know how to connect this up in such a way that is safe for charging the battery?


Here are the pictures as requested. The first is the battery and wall wart together. The second is the text on the battery, the third is the adapter and the fourth is the connectors that came with the battery. The thin white wire I mentioned form the negative terminal can be seen.

I would also like to know whether or not the battery needs to be under a load when charging. This is because I would like to use the charging plug to turn off the circuit when it is charging.

Some more information I remembered, there was an indicator LED that was not connected to the main board.


The Red & Black stripe wire colors are standard for Red=Positive, Black=Negative.

The connector thing is a small fuse holder.

As a rule batteries are not under load when being charged.

The LED may have been an indicator for if the fuse was blown or just a power ON.

The Diode across the motor makes even more sense now that I see the charger is made for a German "Only for domestic use" ...

My main concern here is that the wall wart doesn't look like a dedicated charger. I am scared that if I connect it up the battery will explode. Is there a correct way of charging or should I just connect the positives together and the negatives together?

If you charge the battery at a low enough current, it won't "explode". The usual rule of thumb would be to charge at the 12hour rate.

See what current the fuse is ( fuse holders are pushed and untwisted a quarter turn to open and release the fuse)

When you charge the battery measure the voltage to verify it is under 7.46 volts..

A lead acid battery gives its state of charge by its no load voltage always waiting 5 minutes after the charge is stopped. You can Google that.

charger(+) -----|>|---------Battery +

charger(-)-------------------Battery -

the ---|>|---- represents the diode, which is used to prevent a problems if the charger is hooked up in reverse of its intended configuration.

I would normally agree about the diode placement But the required constant voltage charge of 7.25-7.45V does not allow for the forward diode drop which would be more then o.25v.

BTW this sounds very much like an extra points school problem....

1) It sounds little like a homework question to my shrewd eyes. I could be wrong, but meh if so. I'm usually peg on and have no trepifdations about my answering this one..

2) He said he 'remembered there was a diode in the mix'. I suspect that diode was a reverse voltage protective device. I also suspect that if the forward voltage drop was is a problem, they would have specified a diode with a lower forward drop, like Schottky, instead of grabbing a TV repairman's limited solution of using a general purpose 1n400x series high forward drop diode for a situation that calls for low forward drop.

I think we can reckon the wallwart output is only 7.5 V on full load.


2 years ago

Pictures are useful but it sounds like the small wire could have been a current sensor or thermal battery control. The wall wart will probably drop into the 7.4v when under the control board load. The diode would not make sense except to go across the DC vacuum motor to suppress the turn off inductive kick.