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What impact does length of the leads have on an LED.

burtamus

What impact does length of the leads have on an LED. I'm working with a live multi-camera system and want to send two LED's to a remote camera operator (Tally Lights). Yellow for "On Preview" and one red for "On Air". I was thinking of using Cat5 cable at varying lengths. Three leads in total (1 ground and the two +). What do I need to consider?

 

Sorry i this is in the wrong location. I'm not sure where it would go.

 

TIA, Burt

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    Hi Burt,

     

    You can use Ohm's Law to calculate how well the Cat5 cable would work along the lines of what Luis has described.

    • Determine the resistance in the cable - Wikipedia gives DC loop resistance of Cat5 as less than 0.188 ohms/meter.
    • Determine the LED current - A typical 5mm through hole LED indicator might have a maximum of 20 mA.
    • Determine the forward voltage of the LED - A typical yellow LED might be ~ 2.1V, a red maybe ~1.5.  The voltage at the LED should be at least this high.
    • Determine how long the cable will be - let's say 100 m.

     

    Assuming only one LED is lit at a time, then V = IR and the voltage drop will be (0.188 ohms/m) * 100 m * 0.02 A = 0.376 V - call it 0.4 V.

     

    Then you would need a voltage source at least equal to the forward voltage plus the drop in the cable = 2.1V + 0.4V = 2.5V for the yellow LED.  In other words you should be able to get it to work with a 3V or 5V source.  You could then put current limiting resistors in series with the LEDs to keep from getting over current.  The current limiting calculation is R = V / I, remembering there is resistance in the cable but a bit of a safety factor is always a good idea.  Or just throw say a 330 ohm resistor on each LED for 5V or less. Or use a constant current source if you want to be fancy.

     

    The assumptions I used are to show how the calculation is made.  If you were to use high power LEDs for example things would change.  All of the assumptions should be checked when ordering or purchasing the parts if you are not sure.

     

    Frank

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  • 0

    Hi Burt,

     

    You can use Ohm's Law to calculate how well the Cat5 cable would work along the lines of what Luis has described.

    • Determine the resistance in the cable - Wikipedia gives DC loop resistance of Cat5 as less than 0.188 ohms/meter.
    • Determine the LED current - A typical 5mm through hole LED indicator might have a maximum of 20 mA.
    • Determine the forward voltage of the LED - A typical yellow LED might be ~ 2.1V, a red maybe ~1.5.  The voltage at the LED should be at least this high.
    • Determine how long the cable will be - let's say 100 m.

     

    Assuming only one LED is lit at a time, then V = IR and the voltage drop will be (0.188 ohms/m) * 100 m * 0.02 A = 0.376 V - call it 0.4 V.

     

    Then you would need a voltage source at least equal to the forward voltage plus the drop in the cable = 2.1V + 0.4V = 2.5V for the yellow LED.  In other words you should be able to get it to work with a 3V or 5V source.  You could then put current limiting resistors in series with the LEDs to keep from getting over current.  The current limiting calculation is R = V / I, remembering there is resistance in the cable but a bit of a safety factor is always a good idea.  Or just throw say a 330 ohm resistor on each LED for 5V or less. Or use a constant current source if you want to be fancy.

     

    The assumptions I used are to show how the calculation is made.  If you were to use high power LEDs for example things would change.  All of the assumptions should be checked when ordering or purchasing the parts if you are not sure.

     

    Frank

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