RC Battery Charger Query

[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]

So you think it would be safe to charge at 3 times the mA?

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

[dezza]

Hmm ok well the battery is a JR one, and the manual for the transmitter said 50mA, so I think I'd best stick to that charging rate. I could just go buy a proper charger, but I thought this would be a good little project to try and re-use my existing one!

So using a resistor in the line would bring the mA down wouldn't it?

[DJPhil]

Hmm ok well the battery is a JR one, and the manual for the transmitter said 50mA, so I think I'd best stick to that charging rate.

You would be well advised to do so. This is one of the weaknesses of Ni-Cd batteries. Read through this article on charging basics, and you'll see the problem. Any charging done at faster than C/10 (in your case 60mAh) should be monitored and controlled to avoid overheating, and even in the case of smaller charge rates the cycle is usually timed. Ni-Cd batteries are very fussy about charge characteristics.

I could just go buy a proper charger, but I thought this would be a good little project to try and re-use my existing one!

It's still possible, just might not be worth the effort. You'd have to determine what the details of the charging circuit are. It's possible that the charging is IC controlled by a common chip designed for this purpose, and careful examination of the data sheet may reveal that all you need to do is change a component or two in the circuit. It's hard to say until you know what your dealing with.
Alternately, you could build your own from scratch, reusing whatever parts you can from the old one. Either way, the solution is likely going to involve opening up the charger, and it might be difficult or impossible to do so without destroying the case. They do love to ultrasonically weld those cases together.
You'd learn a lot delving into this, but it's likely to be far easier to just buy a charger.

So using a resistor in the line would bring the mA down wouldn't it?

Possibly, but you'd need to know more about the charger's circuitry to be sure. If there's any regulation in the charging circuit (and there should be) then the circuit may attempt to compensate for the added resistance by adding voltage to maintain a constant current, or it may do something else that's unpredictable. You might be able to test it's behavior with a meter (or two, for simultaneous voltage and current).

Hope that helps.

[Simon]

try it as it is, if it goes over start using resistors, otherwise you could build a current source to do the charging

[Mechatrommer]

if i were you. i'll just change the polarity and charge it and see how the batt heat up. from my very little experience, it will charge the batt ok! the question is just how long the batt will last and can take the treatment. i doubt that your V/A=Ohm analysis and solution is correct, there are other thing to consider as the circuitry may differ. but you may try and check the current using multimeter, if something wrong, then unplug quickly, thats an indication that something inside the circuitry is not working as you've expected. you will learn something from there (try and error? )

[dezza]

Thanks everyone for your info! For some reason I was assuming that it would be a constant power source of 9.6v 150mA, but what DJPhil and others have mentioned about it being complex circuitry and possibly compensating the voltage etc makes sense as the charger has a green LED on it and when the battery is low in charge it shines brightly, and then it gradually fades out to nothing as the battery charges up fully. This to me seems like its measuring the current load etc, so I think I might not try fiddling with it!

I might think it over some more and as shafri said just try it with the polarity reversed and see how it heats up, or if I come across a cheap charger I might just buy one!

Sorry for being such a newb!  

[Mechatrommer]

reverse polarity i mean +ve charger to +ve batt, same as -ve. i used to play around with charger batt combination... from what i found, as long as you give higher +ve voltage to +ve batt, it will charge up, the difference is only how fast and how slow.. there is nothing wrong with being a newb. i'm a newb myself, and always is