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RC Battery Charger Query

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dezza:
Hi everyone, I'm new to the forum but have been watching the EEVBlog for awhile now! I'm a newb when it comes to electronics as I work in the computer industry (Network Administrator) but I have an interest in electronics and that's why I watch the blog I guess!

Anyway here's my query. I'm interested in radio control gear and I have been using a 'Futaba Skysport' controller with a charger that has an output of 9.6V 150mA.

Now I've recently purchased off a friend his old 'JR Max 66' controller but didn't get any charger with it. I tried plugging in the Futaba charger but it didn't work. So I looked up the manual for the JR and it says that the polarity of the plug is opposite to other makes of controllers, but I also noticed that the rating for the JR charger is supposed to be 9.6V 50mA. I have looked at the battery's in the controllers and although they are different shapes, they are both Ni-Cd 9.6V 600mAh.

So my newb question is, if I made a lead up for my Futaba charger that changed the polarity of the plug, would it still be safe for me to use this charger even tho its rated for 3 times the amount of amps?

Any help/suggestions would be very much appreciated!

Simon:
whithout knowing how the charger works a deffinite answer cannot be given but essentially something will only draw as many amps or mamps as it requires, I can have a 100 A supply but if I only need 1 A from it it will only get 1 A (till something goes wrong) voltage has to be exact and in you case it is, you will probably be ok but don't quote me

Kiriakos-GR:
The charging current are always less than the batteries max  mAh ,  1/10 its the standard for " slow charge "

So yes it would work ..

dezza:
Thanks for the replys!

So you think it would be safe to charge at 3 times the mA?
Otherwise is there a way I could put a resistor or something in the line to reduce the current output to 50mA?

My very basic electronics knowledge leads me to this, forgive me if I'm wrong as I'm not sure if its correct?

V / A = Ohms
9.6v / 150mA = 64ohms
9.6v / 50mA = 192 ohms

So therefore I need a resistor or something that will give me 128ohms of resistance and then put that in the crossover lead that I will make? If this is the case should I put it in the positive or negative side of the lead? And also what colour bands would I need on the resistor?

Thanks again!

Kiriakos-GR:

--- Quote from: dezza on May 24, 2010, 10:22:36 am ---So you think it would be safe to charge at 3 times the mA?

--- End quote ---

In such cases we do not think , we read the specs of the battery it self ,
manufacturer  web site --> technical documentation .

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