I know this is simple and stupid
I have a small light bulb that drops 4.5 volts and uses 1.5 watts (as specified on the package)
I have a 12V power supply
What size resistor should I use?
Ive blown up 2 bulbs already because I thought 50 ohm was needed. Obviously not.
What is the calculation I need for this?
Thanks,
Is this an LED or an incandescent bulb? While you can use a resistor with the latter, I don't think I've seen this done much.
Are you sure it isn't a 4.5 volt bulb (if incandescent)? Have you tried using a much lower voltage and ramping it up (once again, if incandescent).
Or - alternatively, have you tried a really large resistor (or maybe a potentiometer acting as a rheostat), and adjusted down the resistance until the bulb glowed as bright as needed - then read the resistance used?
My ohm's law is fairly rusty from underusage; though I have a feeling not enough information is being supplied...
:-?
Like cr0sh said, you don't usually operate incandescent lamps this way, although it ought to be possible. Unlike LEDs, incandescent lamps will limit their own current if provided the proper voltage. but...
4.5V and 1.5W means .333 Amps (P = I*V or I = P/V)
Ohms law says V = I * R, and you want to drop (12-4.5 V) across the resistor.
12 - 4.5 = 0.333 * R
7.5/.3333 = R
22.5 = R
So your 50ohm resistor ought to have been fine (if a bit dim.) It should be at least 3Watts, though. (P = I2R) (which makes sense - you have about twice the voltage across the resistor as you do across the lamp, so it will be dissipating twice the power.)
Since it WASN'T fine, then either part of your data is wrong, or you didn't connect things properly, or your resistor shorted out under a too-high load. Or something else...
It's a small little incandescent you would use for a flashlight or maybe a night light.
I calculated the value same as westfm so feel better. I tried 1 Kohm and it was still to much power, so probably don't have the correct info on the light (maybe it was 1.5V at 4.5watts :o ).
I'll pick up a potentiometer today and give that a try; interested to see what value is needed.
Thanks guys,
I tried 1 Kohm and it was still to much power
You're wiring these in series, right? PWR-->resistor-->bulb-->GND ?
I'll pick up a potentiometer today and give that a try
Beware that most potentiometers are very low wattage devices. If your bulb is indeed 1.5 Watts, then you're likely to burn out the pot.
Can you say any more about the bulb? Like where you bought it, especially if it was an on-line seller that you can link to? And can you show us a diagram of the wiring you're using with the bulb and resistor?
Well, it is early and I'm barely into my first pot of coffee, but give the below a snicker test for figuring out what is needed, assuming the bulb is incandescent.
(volts)x(amps)= watts, so amps = (1.5 watts)/(4.5 volts) = .333 amps.
volts = (amps)(resistance), so the resistance of your bulb is (4.5 volts)/(.333 xmps) = 13.5 ohms.
Noting that your bulb requires .333 amps of current for it to illuminate properly.
12 volts = (.333 amps)x(13.5 ohms + X ohms)
(12 volts)/(.333 amps) = 13.5 ohms + X ohms
36 = 13.5 ohms + X ohms
X ohms = 36 - 13.5 ohms = 22.5 ohms
You need to add a 22.5 ohm resistor in series with the bulb to have .333 amps flowing at a supply voltage of 12 volts. Note that the resistor will need to be rated for the .333 amps at 12 volts,
(12 volts)x(.333 amps) = 3.9 watts, so you will need a 22.5 ohm resistor rated for at least 4 watts.
That seems an unnaturally large voltage to burn off with just a resistor...
That seems an unnaturally large voltage to burn off with just a resistor...
Well, yes. The "proper solution" is "buy some 12V 1.5W bulbs instead of 4.5V 1.5W bulbs."
The brightness overall should be about the same (based on 1.5W), but All the power will go into the bulb rather than 2/3 into the resistor...
Alternately, put 3 bulbs in series...
The easiest way to handle this would probably to get a variable regulator, readily available for < $3.00 or so at any radio shack: