Transformer current and a NiMH battery pack

You built a power supply by using a power supply? Hmmm.

Power supplies, unless designed with current limiting to protect themselves, will put out enough current to damage themselves. The question you want to ask is what current can be taken from the supply without degrading its reliability to a level worse than you would like or need.

It is a rare power supply using linear regulation that can deliver (for very long) a DC current equal to the RMS current rating of the transformer. A typical ratio is half as much DC current as RMS tranformer rating, due to the cresting factor in the transformer current waveform.

I doubt that will help. Your charge times are likely going to be longer than the thermal time constants of the components that would be over stressed.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

I meant to say transformer.

Well I guess you havent got too much to lose by connecting up to the battery with the volts set low and then increasing until the charge rate or the max output of the supply is reached (current). At that point you should be checking the heat generated on the heat sink of the supply and of the battery. This would be a check to see if it would be possible to do what you want. If successful then you will have to consider how to make it work for practical purposes, ie when the battery is discharged it would be easy to overload the supply if you set the volts too high. Probably some form of current limiting could be used and constructed easily if it's not in your supply anyway.

-- Regards ..... Rheilly Phoull

On 17 Feb 2005 09:51:34 -0800, computer snipped-for-privacy@hotmail.com wrote:

Well, you can't set it at 6V or else it will never charge because a 5 cell NiMH has 1.45V/cell under charge. Therefore your pack voltage under charge will be

5 x 1.45 = 7.25VDC You don't say what the capacity of your pack is but IIRC from my days in R/C racing they used 1.4AHr packs so that is what I'll base all the following calculations on. If you wanted a slow charge you could charge at C/10 rate which would be 140mA. If you set your power supply to 12VDC and then hooked a current limit resistor between the power supply and the battery you would calculate as follows R = (12-7.25)/.14 which equals 33.9 ohms. Figuring your power dissipated in the current limit resistor P = .14x.14x33.9 which equals .664 Watts so for a C/10 charge you would be fine with a 33 ohm, 1 watt resistor. To summarize set the power supply to 12VDC and put a 33 ohm, 1 watt between the power supply and the battery pack. It should take about 15 hours to get your pack charged completely at this rate but at C/10 you will not cause any damage if it goes longer. If you want a faster charge C/5 (.28 amps) then half the resistor value to 16 ohms and double the power rating to 2 watts. This should have you charged in about 7 to 8 hours. Not knowing anything about the internals of your power it's hard to say whether either of these current rates would cause damage to your supply but I doubt that .28 amps out would cause any grief unless you have some kind of low current fuses or really low power devices in your internal power supply circuity. Really it's not possible to know without more info. If you want a really fast charge, on the order of 2 and 1/2 hours you can put a 6.8 ohm, >= 3.5W between the 12VDC power supply and the battery and charge the pack at C/2. If you do this, I would get the battery off charge pretty soon after the 2.5 hours. I would also watch my power supply carefully to make sure it can take it.

-wlb