Electronics/Black & Decker Quattro 4-in-1 Power Tool Set VersaPak VP100 (Silver Coloured) Batteries and VP130 Charger
QUESTION: The battery charger consists of a power adaptor plug connected with twin flex to a charging unit which is designed to simultaneously charge two batteries.
I was unable to charge the batteries so I opened up the charging unit and, with the power adaptor plugged in, tested for voltage with my digital multimeter but got a zero reading. Thinking there might be a break in the flex, I cut it back to within a few inches of the power adaptor but, again, got a zero reading on the multimeter. The adaptor is almost certainly dead (unless there's a break even closer to it in the flex) and I'm treating it as such.
I am exploring the possibility of the replacing of the adaptor with a generic one (as I don't think I can buy an exact replacement from the manufacturer and I'm assuming a generic one will be cheaper and just as good, if not better, anyway) but it doesn't look like I'll be able to get one going by the verbatim specifications on the defunct unit. Also, I'd like to explore the possibility of connecting the charging unit to my current-limiting bench power supply (I have used this to successfully charge the two batteries individually).
Each battery includes the following information on it: 'Type 2 3.6 V [DC]'. Research on the 'Net suggested an mAh of 1200. I thought this was a NiMH (recent research on the 'Net seems to indicate it is a NiCd). Anyway, I charged it as a 1200 mAh NiMH, limiting the current to 1.2 A ('1 C'?), with the voltage set to 3.6 V. After fourteen to fifteen hours charging, both batteries produced good results, brightly lighting the lamp in the supplied Quattro flashlight attachment and slightly less reasonably working the 4-in-1 tool. Multimeter measurements of each battery were: (1st) 2.85 V, 33 A; (2nd) 2.56 V, 24 A. (Most recent measurements: (1st) 1.31 V, 2 A; (2nd) 1.23 V, 2 A.)
The defunct power adaptor has the following information on it: '230V[AC]50Hz 5.0W 2x4.35V[DC]210mA 1.6VA'.
The charging unit is wired according to the graphic I have attached, but I describe it as follows: white-striped lead (electrical positive?) conductor from adaptor going to side of left battery and also to unstriped (electrical positive?) end of a diode, the diode connected on its other side to the central terminal of right battery. The unstriped lead (electrical negative?) conductor from adaptor going to side of right battery and also to unstriped end of another diode, the diode connected on its other side to the central terminal of left battery.
I am not sure how the power adaptor can charge the batteries, given the description, especially re the diode orientation, assuming DC electricity; I imagine it would work with AC electricity (indeed, subsequently, I came upon a message on a Web forum, the author of which stressing that a generic replacement power adaptor would need to ouput 9 V AC--the discussion is here: http://www.howtomendit.com/answers.php?id=235675
). Are the batteries being charged in series or ('isolated') parallel (I heard that series, or, alternatively, individual charging (effectively the same as 'isolated parallel') is best)? What should I set my bench power supply's current limiting and voltage to if I want to connect its (DC-only) output to the input of the charging unit's conductors?
ANSWER: I cannot give you a wise answer to this because the product design engineers are using a charging apparatus that is matching with the battery pack. All batteries are different and charging rates, discharge characteristics are unique to the battery pack. You can often get by with a charger that is close in its specifications but may not be safe. For example charging NiCD batteries a high rate can cause them to explode.
However, here are some general principles. Charging a batter of the NiCD variety requires an input voltage considerably greater than the voltage at the fully charged level. The reason for this is the internal cell resistance which develops across the internal resistance level and leaves less than the charge voltage across the cells internally. And, the current must not be over the specified level for the batteries you are charging. Likewise, their must be a tapered charging current as the battery becomes near its maximum VA level so as not to overcharge the cells.
As a general rule the batteries are more safely charged in parallel or individually if possible. The reason series are sometimes dangerous is because the failure like an internal short of one of the batteries can put all the voltage across one cell - which is dangerously too high.
No diagram was included so I have no ideas about the color coding of the wires.
Hope this helps.
---------- FOLLOW-UP ----------
QUESTION: My apologies: I was so glad to have the text right, I forgot about the graphic (although the wiring and colour coding was described in the question).
Graphic attached; two questions on it, one either side of the schematic.
Very good; thanks for the diagram and photos. Now, I know this model well.
The charger is a bridge rectifier. It works by charging pulse going to the battery on the right in the image you sent. During the next half cycle it charges the battery on the left. You can assume the diodes will have a forward voltage drop of around 0.6 volts.
Therefore, if the secondary of the transformer was 3.6 volts ac rms the peak voltage would be 3.6 x 1.4 = 5 volts peak - which may be about the right level.
This is NOT a dc charger. It is rather, a bridge rectifier using the alternate half cycles of the 50Hz input. So, you cannot charge both batteries at the same time with a dc supply. What you would need to do is charge one battery full then disconnect the power supply and reverse the connection to charge up the other battery. If you had two dc supplies you could do both batteries at the same time.
And, yes you should have a current limiting supply so as not to overcharge the battery and damage it.
A generic supply might be just fine if it meets all the other criteria. The voltage must be at least 5 or 6 volts. The current will limit itself if you put a small resistor in series with the power supply. My opinion is you should not charge at greater than 1 or 2 amps. So,for that you would need a series resistor of around 3 or 4 ohms, 25 W wirewound type. The supply should have a taper which as the voltage on the battery increases to near full charge the supply reduces the charge proportionally such that when fully charged level is reached only a few milliamps of charging current is enough - this will keep the battery from overcharge and over heating.
I think you have all the answers. Cheers.