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Main Discussion > An electrical question.

Heli-Driver
Elite Veteran
Location: Arlington, TX

Hopefully someone can help me out here.

I have a spotlight that is a modfied MagLite and I want to run it off my 4.8 navigation lights battery pack. I know the 4.8 volt will blow the bulb as it normally runs at 3v. I want to step down the voltage to the spotlight so it doesn't blow the bulb.

I prefer something simple, light and cheap, can I just put a resistor inline on the wiring to drop the voltage down or do I have to do something else. Just screwing around with it I tried an inline 100ohm 1/2 watt resistor and the bulb didn't even light. I am NOT an electrician and have no idea what to do.

I have a small 270mah/4.8v battery and a larger 600mah/4.8v pack which I could use, I'm leaning towards the 270 as it is much smaller and lighter.

Any ideas?

Thanks,

Raymond

Predator Gasser SE/231 X 2
Century Helicopters

03-12-2002 Over year old.

HOMEPAGE

BladeRunner
Elite Veteran
Location: Ontario Canada. Member of "some sort" s

Question?

I have used Mag-Lite bulbs to make a landing light, I tried a 3.6V cell phone pack I had and it worked great--if a little brighter. Have you tried the 4.8V? It might work only be that much brighter, you may try a volume type switch from a small radio, it would act like a variable resister, just dail it to the brightness you want without blowing the bulb.
BTW, I'm no expert on this stuff--just a hacker, so try what you think will work best for you.

03-12-2002 Over year old.

dchinww
Senior Heliman
Location: Seattle, Washington

Maglite sells flashlights that handle a wide range of cells. You might try looking for a Maglite bulb that's good for four cells (4 cells x 1.5 volts/cell = 6.0 volts). A fully charged four cell nicad pack will top off at close to six volts.

03-12-2002 Over year old.

Heli-Driver
Elite Veteran
Location: Arlington, TX

MagLite

I took my small MagLite, the one that runs on 2 AA batteries (3v) and connected it to my 4.8v pack and blew the bulb.

I don't want a third battery pack in the heli. I have my 2700mh flight pack battery, and a 4.8v battery(270 or 600mah, haven't decided which I want to use) for the nav. lights, I want to use the existing bulb with the 4.8v nav. light system.

Is it the voltage or the amperage I need to worry about the most?

Raymond

Predator Gasser SE/231 X 2
Century Helicopters

03-12-2002 Over year old.

HOMEPAGE

flatspin
Veteran
Location: H.V. Pa.

In this case it's the voltage. The current will be dependent on the resistance of the filament in the bulb.

03-12-2002 Over year old.

slammers
Senior Heliman
Location: Delta, BC

If you have a meter here is what you do.

Connect an amp meter in series with the light using the original power source. Measure the current through the bulb in this case. Now measure the voltage across the bulb when its turned on again with the original power source.

subtract this voltage from the voltage of the new source. If you measured 3v across the bulb then this difference would be

4.8-3=1.8V

So your resistor has to drop 1.8 V. Now voltage = current * resistance so you would need a resistor that is

1.8v / current measured in step 1. If you measured 1/2 amps you would need a 3.6 ohm resistor.

Now the power through that resistor is equal to the square of the current * the resistance so in this example this 3.6 ohm resistor would have to be capable of handling

.5 * .5 * 3.6 = .9 watts

A 3.6ohm 1 watt resistor would suffice (In this example. You need to know your current to calculate it properly)

Shawn

Free Flight Log Software
www.lammers.ca/FlightLog

03-12-2002 Over year old.

HOMEPAGE

flatspin
Veteran
Location: H.V. Pa.

Not so scientific approach to the problem: The reccommended way!
You mentioned with a 100ohm resistor the bulb did not even light up.
Try 50ohm then 25ohm etc.. until it works!

Scientific approach is: Not reccommended in this case (bulb burns out with 4.8 v) and I don't think you can measure the resistance of the filament properly (I think the value changes once the filament heats up)

Knowing or measuring the resistance of the filament gives you the current flowing through the circuit. Knowing the current value and also needing to drop 1.8v across the resistor you simply divide 1.8v by your current (mah) value gives you the value of resistor needed.

03-12-2002 Over year old.

alvinrc
Veteran
Location: Mobile, AL, USA

You could go into the batt pac and run another positive wire from the positive end of third cell from pacs negative wire to the maglite bulb.

Negative batt pac wire will be common to everything.

This will give you the voltage of three cells for mag bulb.

Or tap to just two cells for even less voltage.

The other way with full batt pac voltage and proper resistor for mag bulb voltage prob best way to go once you figure resistor value.

03-12-2002 Over year old.

HOMEPAGE

MikeDD
Senior Heliman
Location: Elkhart, IN

Try adding diodes in series to one leg(get direction right). Each one will drop about .65V so 2-3 should work. Radio Shack should have 1N400x type that will work.

Michiana R/C Choppers

03-12-2002 Over year old.

HOMEPAGE

mtj4300
Heliman
Location: Arlington, TX

hey raymond its miah

I think slammers is right. i go to school for computer and electronics engineering and thats what id do! its just ohms law V=IR or change the formula to V/R=I voltage devided by resistance equals current. so the first one would be voltage is eqaul to current times resistance. i figured out you would need a 3.3 ohm math mathmaticly and the color bands on it are orange, orange, gold, gold so check in some old electronics for it but i would be easier to go but it for $.25 but if the resister is off buy 10% it will have a silver band instead of gold

03-14-2002 Over year old.

Virtual1
Senior Heliman
Location: Waterloo, Iowa - USA

I'd go with Mike's idea and use diodes. They produce a reliable voltage drop, regardless of load so no need to haul out formulas and try to calculate resistances. 0.6 volts are dropped by each 1N4001. Besides, the resistance of the bulb isn't constant because it changes as battery voltage changes, so those formulas are less than ideal.

Diodes and resistors will both produce heat though, you will lose some of the pack's power needlessly. The most efficient solution would be to use a voltage regulator chip. I don't think rat shack would stock a 3v regulator though, but you can always look. Get one that can handle the current your light will draw. (1 amp should be plenty)

Oh, and have you considered just removing a cell from your battery pack? Sometimes the simple solutions are the hardest to see. A three cell battery pack would kick out 3.6v, add a 1N4001 in series with the light and you'd drop it right down to 3.0 volts with a minimum of power lost to heat, plus it'd be one cell lighter. The lowered battery supply voltage should also lower the current draw and lengthen battery life.

OK last suggestion I just thought of... if you have a few spare bucks or a stocked parts box, use an ESC! They're used to regulate power to DC motors in electric helis, they hook right to the throttle channel. You could use a small one like for a Hornet, and just run it right to a spare channel on your receiver, and you could turn on or off the light by remote, or set it at any brightness you wanted if you have a spare dial. And that has a regulator so it would be very efficient and could use as big of a pack as you'd care to hook up. Just be careful, you can dial in enough juice to blow the light since you're setting the voltage.

03-14-2002 Over year old.

HOMEPAGE

donlynn
Key Veteran
Location: New Zealand

Do it with diodes reliable and simple-KISS
probably three will be ideal -0.7 v each one therefore 3 times 0.7 =2.8
perfect.allow a little breeze ,don't wrap them but still insulate.If you want I can post a address remote switch you can operate from a spare ch to turn light on/off from tx

good thing to remember with halogens which may be what is in your torch
for each10% overvoltage results in a 50% yes 50% reduction in life.
another option for nav lights is to use superbright LED'S I use 8 in a glider flying wing 3mm in the coreflute tips (canterbury sailplanes Jazz xtreme)and can quite comfortably fly it 200+m away from me at night no sweat and I run two led's in series at 15 mA so therefore total current is 15 times 4 =60 mA total say a 3w tungsten or halogen P=V*I therefore I=P/V
so four bulbs at 3w = 12w =12/3=4A=4000mA
4000mA as against 60mA-quite significant

03-14-2002 Over year old.

slammers
Senior Heliman
Location: Delta, BC

Voltage Regulator ( a little off topic )

Virtual1,

FYI - a voltage regulator would also have to disipate the same heat (or power) as the diodes. With any linear voltage regulator you still draw the same current from the battery as you do through the final load. Therefore you still have extra power that must be disipated in the form of heat.

The benifits of the regulator is a constant supply to the load while the battery voltage is dropping.

The only efficient way of doing something like this is to use a dc to dc converter. These are typically about 80% efficient and will have the least losses in terms of wasted power. Of course they are very complex to build.

Shawn

Free Flight Log Software
www.lammers.ca/FlightLog

03-14-2002 Over year old.

HOMEPAGE

Maxx
Key Veteran
Location: Shreveport Louisana

Raptordriver, Dchinww gave the easist answer, go the Wal-Mart and buy a Maglite replacement bulb for their 4 cell flashlight( 4.8v), it will fit the same blub holder as your two cell light has and will run off your 4.8 pack... "walk into the light..." Chris

03-14-2002 Over year old.

donlynn
Key Veteran
Location: New Zealand

Get that Maxx a beer now!

03-14-2002 Over year old.

Heli-Driver
Elite Veteran
Location: Arlington, TX

Thanks guys...!

Next time I'm at the "Mart" I will get some 4 cell replacements, think they'll handle a hot pack.

I did try a 10 ohm resistor inline and it light the MagLite bulb and didn't blow it. So I think I'm safe there but I certainly will get the 4 cell spares in the future, makes sense.

thanks guys for all the input, I'm getting closer to finishing this thing and this was one of the bumps I had to address.

Why is it 86 degrees here and the wind is blowing out of the south at 30 mph..... that's soooo cruel.