12 VDC 500mA Wall Wart Transformer Turnout Control Capacity Question

Hardcoaler May 15, 2018

  1. SP_fan_1951

    SP_fan_1951 TrainBoard Member

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    Sorry - i had a late-night brain fart. The power dissipation is voltage TIMES current (P=IE). 12.3 V * .02 A = 24.6 mW = .0246 W.
     
  2. Hardcoaler

    Hardcoaler TrainBoard Member

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    Thanks SP-fan -- I now see your formula [ 12.3 V * (20/1000) = 0.246 Watts ] which equals 0.02 Amps.

    0.02 Amps = 20 mA, which is the power consumption of the LED alone, right? Doesn't the resistor consume additional current?

    I'm sorry that I'm not picking up on this as quickly as I should.
     
  3. RBrodzinsky

    RBrodzinsky Staff Member TrainBoard Supporter

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    Sort ofr. The equation V=iR defines the voltage/current relationship for the fixed resistor in the system. It sets the current. The power dissipation is split between the resistor and the LED (most in the LED as light, a smidge from the resistor as heat).
     
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  4. SP_fan_1951

    SP_fan_1951 TrainBoard Member

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    The current through the resistor is the same current that flows through the LED. The resistor acts to limit the current. Using your example, the resistor will drop (12.3-1.8=)10.5V. The value of the resistor would be 10.5/.02= 525 Ohms. The resistor will dissipate .02A * 10.5V = .21W. The LED will drop 1.8 V and dissipate .036W, and the total dissipation will be .246W. The total current will be 20 mA, as the current through the resistor and the LED is the same current. If you put 3 LED's in series, the drop across the LED's would be 5.4 Volts. The drop across the resistor would be 12.3-5.4=6.9V. For the same 20 mA current the value of the resistor would be 6.9/.02= 345 Ohms, and the resistor's dissipation would be .138 W. Each LED would dissipate 1.8V * .02 A = .036 W. The total dissipation in the circuit would still be (3*.036) + .138 = .246 Watts. The current drawn from the supply will still be .02 A. All you are doing is changing where in the series circuit the dissipation takes place.
     
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  5. Hardcoaler

    Hardcoaler TrainBoard Member

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    Thanks guys. I'll print out these two posts and spend some quiet time with pad and paper to make it clear in my mind. I think I understand now.

    I fabbed up a test panel yesterday, installed three colored 5mm LEDs and their colored diffusers, then hooked up my 12.3 VDC power supply and played with various resistor combinations to help arrive at a uniform muted light intensity. It worked out pretty well. The yellow and greens glow nicely using a 2200 Ohm resistor, but the red required a 4700 Ohm resistor to properly match the brightness with the others.

    Reading up on the topic, I also learned that various colors are more strongly received by the human eye and red is one. This fact, combined with red LEDs often requiring less voltage might explain why a stronger resistor was needed with my red LED.

    Again, I really appreciate everyone's help with this project!

    2018-08-13 Control Panel LED Brightness Test for Upload.jpg
     
  6. RBrodzinsky

    RBrodzinsky Staff Member TrainBoard Supporter

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    Looking at the photo, I would say the red is emitting significantly less light, even if, to your eye, they are about the same brightness. Why? well, you can basically see the LED's "hot spot", and some concentric rings, which I assume are part of the casing. But, it also isn't worth stressing over; the key is being able to tell that the light is "on" and not to overwhelm.

    Have fun with the project.
     
  7. Hardcoaler

    Hardcoaler TrainBoard Member

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    It's interesting isn't it Rick? I messed around with a number of resistor values with the red LED and noticed the exact same as you with the concentric ring visibility. When I lessened the resistor to brighten the red just a bit, it excessively increased its visual penetration. I guess that's an example of the nature of red light.

    Your summary is perfectly stated and I mentioned the same to my wife, that I'd try to not obsess over it like I do most things.
     
  8. SP_fan_1951

    SP_fan_1951 TrainBoard Member

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    The different colors of LED's are made by using different dopants and base materials, and consequently have different forward voltages. Red has the lowest forward drop, while white (which is actually made from a blue LED with a phosphor coating) has the highest. LED's also are not all created equally when it comes to intensity. A red LED may produce 30 millicandelas at 20 mA, while another may produce 200 mcd, depending on the manufacturer and design. It's also important to remember that while the LED's light output is roughly linear with respect to current, the human eye is logarithmic with respect to intensity.
     

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