10W led outdoor spotlight to work on 12V

I know you can get them in a 12V version, but I allready have them in 220V so would rather mod them than to buy new one's. Is there a way to make them to work off 12V?
Thank you

What voltage does the LED need? Does the LED use a driver?

Do you have LEDs that run from a 220v AC source and want to run them off a 12V DC source? You might need a car transformer or something.

I don't have answers to your questions. All I think is that I don't believe the led runs at 220v, somewhere the voltage might be stepped down.

Here are some fotos to show the light, not mine but basically simmular.

Just an update, there is a driver, obviously
Specs: 180-265V input.
Output: 6-12V max 900mah

If I would like to run the light of a lead acid. What type of transistor should I use to limmit the current? I've read that directly coupling the led to the battery will overpower it, severly shortening its life.

See if you can remove the current driver. If you can, then a 12v driver with similar output characteristics could work for you. There are some MR16 LED drivers on eBay and they are meant for a 12v source. Not sure about durability for some of those drivers though.

If I would rather want to use a resistor, what ohm should it be?

What's the forward voltage of the LED at 900 mA?

Calculating the resistor needed is pretty easy. The Equation is V = I * R, "V" being the voltage dropped, "I" bring the current in amps, and "R" being the resistance needed in ohms.

To find the resistance needed, just rearrange the terms and get R = V / I

Since you know I already (900mA, or 0.9A), the only other part you need to know is the voltage drop. That would be 12v - Vf, Vf being the forward voltage of the LED @ 900 mA.

A 10 watt rated resistor should be more then adequate to handle the heat. It is an inefficient way to loose voltage though.

Thanks for trying to help, but I'm really not clued up on these things and this is my first ever attemp at something like this.
To answer some of your questions. Yes the driver is easy to remove. The ''forward voltage'' Sorry I don't know what that is.
I have not measured the voltage drop or anything as yet. I did not know there will be much of a voltage drop anyway as I plan to use a 11.1V lipo battery pack. They can deliver lots of current so I assume the voltage drop will be minimal?

LEDs run a certain voltage, for example a white LED is usually 3-3.7v, the actual voltage will depend on the drive current and the LED itself.

If a LED is given too much voltage, it will try to draw a lot more current then it is supposed to take, so that could burn out the LED.

For some of those larger LEDs, they have a few LEDs in series, think LEDs chained together one after another. This will cause the voltage the LEDs need to change. For example, if two LEDs are wired in series, then the voltage the whole chain needs is now double that of one LED.

If you have a power supply or some batteries, you could try different voltages until the LEDs light up. For example, start with 1 Li-ion battery and hook it up to just the LED. If it lights, then the forward voltage of the LED is around that of the battery voltage. If that doesn't work, try something with a slightly higher voltage. Keep stepping up the voltage until the LED lights, and that should be close to the forward voltage of the LED. A adjustable power supply would be best, but a chain of AA batteries or something could be a decent substantiated. Just add one more AA battery each time the LED doesn't light.

Someone please step in if that is a bad idea.

Once you get the battery voltage needed, you can plug the voltage of the battery into the equation I gave above, and figure out the resistor you need.

The resistor doesn't have to be exactly the value you calculate, lower resistance = more current = more light, higher is the opposite.

I think I've put this in the wrong section as this is not a flashlight. Maybe that is why others are not also helping, even with your question to them
I will start a new thread in the correct section and ask the question differently.
Thanks for your willingness to help.