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I've got a battery pack for one of my peripherals that recently died, so i took it apart to find half the (nicad) cells were dead (I tested each cell's voltage both with and without a load - the dead ones dropped off to zero with load).

So what i wish to do is replace them with nimh's. Apparently the easiest way to charge nimh's is at 10% of C (C being 2500mAh) for 15 hours at 1.5V per cell (i will have 10 cells in series).

Now the question: I have a 15v power adaptor (correct voltage) but it is 400mA. I want to 'throttle' it to 250mA somehow. Is there some way i can do this with a diode or resistor or transistor or something?

Cheers

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closed as off topic by Bob, Journeyman Geek, Sathya Jun 7 '12 at 12:19

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2 Answers 2

up vote 2 down vote accepted

There are IC chips like the LM317 that vary the voltage to keep the current constant. But you can't do it with out building at least a small circuit. An alternative would be to put a potentiometer is series with the battery, but you would have to periodically adjust it to keep the current constant.

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Is that similar to this? zen22142.zen.co.uk/Circuits/Power/add-on.html –  Chris Aug 28 '09 at 0:15
    
Found a good example of that circuit: sv1bsx.50webs.com/charger/charger.html –  Chris Aug 28 '09 at 0:48
    
Chris, The second one's a good example. I might replace the fixed resistor with a pot or better yet a pot in series with a fixed resistor to provide an absolute limit. With the pot you can adjust the current. –  Jim C Aug 28 '09 at 11:55

Simplest method to limit current is with a series resistor. Be careful with the power rating of the resistor, though.

My understanding of battery charging is that the best method is to use a control circuit. Here are some examples of battery charging circuits that would perform better than a series resistor:

Linear Tech LTC4060 NiMH/NiCd Battery Charging Circuit

Maxim MAX712 NiCd/NiMH Battery Charging Circuit

National Semiconductor LM317 (as mentioned in another answer)

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If i use a series resistor, would the resistor soak up part of the voltage, leaving less voltage for the batteries? –  Chris Aug 27 '09 at 23:18
    
Yep. There would be a measurable voltage drop across the resistor. –  hanleyp Aug 28 '09 at 2:57
    
Are we talking a lot of the voltage (eg 3 volts) being used by the resistor? Or only a little bit (eg < 1volt) ? –  Chris Aug 28 '09 at 6:11
    
Oh, yeah. To answer your question, V=IR, so the voltage drop across a 60-ohm resistor at 250mA would be ~15V. :-) A resistor limits current, but is not best for a battery charging application. –  hanleyp Aug 29 '09 at 14:08
    
I don't recommend using only a series resistor for a constant current supply for use when changing batteries. The cost of a correct circuit is too cheap to accept unnecessary risk of damaging the batteries (which can have dangerous failure modes from over-charging or over-voltage while charging). –  mctylr May 14 '10 at 21:12

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