Hi everyone, I Recently bought some leds and want to know how to convert some voltages with parts please THANKS

LED SPECS
Forward Voltage: 3.2~4.2V
DC Forward Current: Typ. 800mA (Max. Current =1000mA)
Luminous Flux: 120~140 Lumen
http://www.hero-ledstore.com/heroled-3-watt-high-power-cool-white-led-130-lumen-p-8.html?zenid=14c38ade58b7d8933056f2ec5b95def7

1) From 12 volts to 3.2-4.2 volts
2) From 12 volts to 6 volts
3) From 12 volts to 9 volts

 

...want to know how to convert some voltages with parts ...

No, you need to provide ~700ma current to safely light those LEDs. Voltage is irrelevant as long as it's high enough to exceed the minimum. It can be as simple as a single current limiting resistor, although for that current you'll need a large one.

Use Ohm's law (voltage drop = current x resistance, V = I•R) to estimate the resistor you need with any supply voltage. Start with a higher resistance and work down to achieve the desired brightness while staying below the rated current. The power rating of the resistor should be about double the expected power dissipation, which is I^2•R.

 

No, you need to provide ~700ma current to safely light those LEDs. Voltage is irrelevant as long as it's high enough to exceed the minimum. It can be as simple as a single current limiting resistor, although for that current you'll need a large one.

Use Ohm's law (voltage drop = current x resistance, V = I•R) to estimate the resistor you need with any supply voltage. Start with a higher resistance and work down to achieve the desired brightness while staying below the rated current. The power rating of the resistor should be about double the expected power dissipation, which is I^2•R.

Thanks buddy, see the thing is i'm not the best when it comes down to calculating currents... flying a helicopter i can do with my eye's closed. lol. I dont mean to trouble you but if you can just let me know exactly what to use, it'll be very useful. thanks.

 

Thanks buddy, see the thing is i'm not the best when it comes down to calculating currents... flying a helicopter i can do with my eye's closed. lol. I dont mean to trouble you but if you can just let me know exactly what to use, it'll be very useful. thanks.

Suppose your LED drops 3.2v when lit with 800mA. First, this is a high current for an LED, 40X what a "plain" LED takes, are you sure it's rated for this?

If your supply is 12V, that means you need a resistor to drop the other 8.8v. Ohm's law says V = IR , so 8.8 = 0.8 • R. So R = 11 ohms. The power dissipated will be I^2•R or 0.8*0.8*11 = 7W. That's quite a bit, but you can buy 10W or 15W resistors, so it's not impossible. Note however that your LED is using power at I•V or 3.2 * 0.8 = 2.5W, less than what is being burned off in the resistor. That's why other methods of driving your LED, while more complicated, would be far more efficient. A PWM drive would give you more brightness with less power consumption, if that matters.

 

Is this a general aviation application?

The electrical system won't be putting out 12v; more like 13.7v to 14.4v with occasional spikes up to 60v or so. Fixed resistors are just not so great at regulating under those conditions.

Further, if you are powering an LED rated for 3.2v @ 800mA current, you'll be dissipating 3.2v * 0.8A = 2.56 Watts in the LED, and (13.7-3.2)*0.8 = 8.4 Watts to (14.4-3.2)*0.8 = 11.2 Watts in the current limiting resistor; not very efficient - and you'd need resistors rated for ~20 Watts, which would be large, heavy, and somewhat pricey.

Instead, you might consider using a BuckPuck; a pre-made high-powered LED current limiter that you just wire up. These are switching current regulators, so they are quite efficient compared to a simple current limiting resistor or linear regulator.
BuckPucks can be purchased here: http://www.ledsupply.com/buckpuck.php