youdoo

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I recently bought a Hobbico starter pack with 12 volt 1200 MAH battery. I need to charge at 120 MAH, (10% rate). I have a transformer that puts out 250 MAH at 12 volts. What size resistor do I need to put in the line to drop current to 120 MAH?

Combat Addict

Being a computer geek as I am, I am embarresed not to know. However, IMO, I would just charge it. Instead of takeing 10 hours for a full charge(120mAh x X = 1200mAh, X = 10 hours), it would take 4.8 hours (250mAh x X = 1200mAh, X = 4.8 hours).

higheronnitro

Voltage=Current x Resistance

R=V/I 12 / 0.12 or 100 Ohm

Power rating

P = V x I 12 x 0.12 equals 1.44 Watts

In this case though I would say that the increase in charging current will be ok so long as you are not going to leave it on overnight.
BR Steve

Combat Addict

If you get a resistor, make sure that it will not get too hot. Use a power resistor so you don't have a core meltdown .

joetsunami

youdoo,
There's not enough info there to answer your question. Here's why: You could try to charge your pack with a car battery, which can put out hundreds of amps. This car battery would not charge your 12 volt battery very much at all. Ya see, the most important thing when determining the current level is the VOLTAGE of the charger.
The voltage of your car battery as a charger is too low. If it hovers at 13.8, it is not really high enough to charge another 12 volt battery very well.

Second most important is internal resistance. While you read the nameplate of the charger and it stated 12 volts, you can be sure that it is nowhere NEAR 12, but more like 14.5 or more. These ratings on transformers are a little deceptive. What 12 volts/ 250 ma REALLY means is that if you load this transformer at 250 ma, it will put out AT LEAST 12 volts. In other words, it is giving a power rating of sorts. I'd bet you could charge a 14 volt battery at 50ma easily, or a 10 volt battery at 300ma; up to some point, where the transformer over-heats or a diode smokes due to over-current.

I could go on, but here's how to accomplish it:
Hook battery to 'charger' through a resistor, say about 10 ohms, to determine current. Use ohms law stated in another post. Change resistor value, larger for less current, till you deem the value to be correct. I'm fairly certain that charge current should be measured on a fully charged pack.

This will give you a general idea, because this is a deceptively complicated business. The more current you demand from the transformer, the lower it's voltage will be and you'll find that the current does not go up as much as you anticipated it would. Likewise for the battery. The more current you feed it, the higher it's voltage will be, and the current will go up less than you expected. This is because the battery and the transformer both have an internal resistance which messes with your calculations. You can figure these values out experimentally, but it is much easier to swap resistor values till it's right! I do this, and I've been a tech since 1982. (I would, however, use math to figure a starting value, and after changing the resistor a few times, use math again to find the best/closest fit)

As far as the power rating of the resistor goes, don't sweat it too much, at 150ma, you'd have to drop about 2 volts before you were taxing a 1/4 watt resistor. If you get to that point, just double the required value, and place two resistors in parallel.

Good Luck,
email me if you need further help.

Joe Myers