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I have 10 x 1.2V Sanyo NiCd batteries. They are connected in series. I would like to charge them. The thing is, I don't have a charger. I looked up chargers and they are all quite expensive(I live in India). I would like to know what voltage AC to DC adapter should I get to charge them.

What are the risks involved in doing so?

Could you also suggest me some cheap ways of getting the batteries charged?

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Something occurred to me. If you can separate the batteries from the battery pack you can charge them in groups. Lower voltage sources like 5 and 12V are much cheaper to find than larger ones. –  carveone Jul 30 at 11:52
    
they are all soldered, i cant remove them –  harvey_slash Jul 30 at 11:59
    
That's unfortunate. 5V adapters are very cheap and easy to find because of USB and a 12V car battery is too low a voltage. You'll have to dig up a larger source I'm afraid. –  carveone Jul 30 at 12:01
    
C is the mA rate corresponding to the cell mAh capacity so eg for a 1200 mAh cell, C = 1200 mA. If they are NiCd you cam safely trickle charge them at C/10 - notionally for ever but preferably for about 1 day overall. 24 hrs x c/10 ENSURES all cells are well charged and helps balance the pack. Even about 18 hours at C/10 is enough from fully discharged. Overcharging at > C/10 will damage cells. || NB NiCd and NimH differ - Modern high capacity NimH MUST NOT be trickle charged past full capacity. (eg an AA NimH of 1800 mAh or more is best never trickle charged. ) –  Russell McMahon Jul 30 at 13:20

1 Answer 1

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The fully charged voltage on 10 NiCd batteries is about 1.4V * 10 = 14V. So the first thing you will need is a voltage source higher than that. To keep it safe you really need 1.5 * 10 = 15V minimum.

NiCd batteries are charged with a constant current. You don't say what the rating on your batteries are, but let us assume they are 500mAhr. Trickle charging them is C/10 or 50mA. That's nice and safe and will fully charge your batteries in 15hrs (you add 50% to the time because charging is not 100% efficient)

schematic

simulate this circuit – Schematic created using CircuitLab

To keep things cheap you can use a resistor to supply this current and calculate the resistor value as:

$$R = \frac{V_{source} - V_{batpack}}{I}$$

Let's say you find an 18V source. Then 18 - 14 = 4V. And you want to charge at 50mA so your resistor value is 80 ohms. 100 is the closest you will get which will charge at: 4/100 = 40mA.

Edit:

I should add that the current will vary as the batteries charge. So a discharged battery at 1.1V will charge at (18 - 11) / 100 = 70mA. This current will fall to 40mA as the battery comes up to full charge.

With a C/10 charge rate, 14 hours is safe even if they are partially charged to begin with. You are "trickle charging" at a "charging rate" and the batteries are designed to full cope with this rate. Any faster than this you will have to verify they are charged - the battery pack voltage will be above the nominal 1.25V * 10 = 12.5V when you disconnect the charger and check with a meter. At 1.45V/cell (or anything above 14V for your pack), they are fully charged. Expensive chargers do all this for you.

You can trickle charge at a "top up" rate by adjusting to C/20. So for 500mA batteries, trickle charge at 500/20 = 25mA. You can do this indefinately without damage. Although to charge at this rate from discharged will take, well, days!

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Thanks to Ricardo for Mathjaxing the equation. Must read the help more ;-) –  carveone Jul 30 at 11:33
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I'll edit my post in a minute and add some more details on charging rates. –  carveone Jul 30 at 11:36
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Maybe read this. –  Rev1.0 Jul 30 at 11:37
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Yes. I've added a circuit. It always turns out huge but you get the idea. –  carveone Jul 30 at 11:58
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That assumes identical cells. In real life the cells have slightly different capcity and you would have in the 10 Volt pack for example 5 cells at ~1.2V, 4 cells at ~1.1 Volt and one (the weakest cell) taking damage at -0.4 Volt. Thats why you don't want those packs to discharge too depply. –  Turbo J Jul 30 at 13:09

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