Voltage Conversions?

Discussion in 'The Projects Forum' started by parimal27, Mar 4, 2011.

  1. parimal27

    Thread Starter New Member

    Mar 4, 2011
    3
    0
    Last edited: Mar 4, 2011
  2. wayneh

    Expert

    Sep 9, 2010
    12,153
    3,059
    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.
     
    parimal27 likes this.
  3. parimal27

    Thread Starter New Member

    Mar 4, 2011
    3
    0
    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.
     
  4. wayneh

    Expert

    Sep 9, 2010
    12,153
    3,059
    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.
     
  5. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    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
     
Loading...