Overdriving a MR16 - Voltage drop required, help please

[w01fy]

Hi all,

Bit of a noob here so some help please.

Have an 18v battery I want to drive a MR16 from, over drive to be precise. As I understand 18v is a little to much and will drastically reduce the life span of the bulb. Also understand pushing 15.6v through it would reduce the life span to 200hours, which I will happily live with.

You ask why! well I want to push the lumens up on the light for just as a play. No real reason. Its been a while since I've done any electrics/electronics so starting with something simple that if I blow up it doesnt matter, then going to slowly work up to playing with some LEDs.

Anyway the help I need is working out the resistance to use to drop the voltage from 18v to 15.6v

Anyway

Power source = 18v
Bulb Power = 45w (Rated 12v)

From the above

45w Halogen @ 12v (Presume I calcualte at 12v not 18v)
P = IV
45 = I x 12
45/12 = 3.75A
So it will draw 3.75A


Now I want to drop from 18v to 15.6v
So
V=IR
V/I = R
2.4/3.75 = R
.64R



So .64R Resistor (seems WAY to small)
Power disapaition on the resisitor
P = VI
P = 2.4 * 3.75 = 9w


So a 0.64R resistor rated at above 9w
My maths I think is lacking here


Can someone give a noob a hand?



Cheers


w01fy

 

[UberLumens]

MR16 bulbs can eat 21V direct drive and look amazing.

18 will be fine, just not as bright or white as 21.

Dont worry about bulb life they are cheap $9 for 3 50watters(sometimes on sale at a one dollar for 3) and you can get them everywhere.

I suggest run at 18 until your first bulb goes then decide if you want something weaker.

 

[Illum]

depending on your battery chemistry, capacity, and the load in amps your lamp will draw, expect the voltage to sag by itself somewhat from 18V. Adding a few volts is fine, if any it'll just burn a little hotter

 

[w01fy]

ah cheers chaps! Forget the resisitor then

Was my maths correct though or should I be shot at this point?