LED's and Resistors

[dcarrell8]

I have purchased some 12v Panel lights

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I'm wondering if anyone else has used these and can help me out.
The info states that no external resistor is needed.  Does that mean one is included within the housing?  If so what is it's rating. 

The following schematic is what I'm trying to accomplish with this LED but I have no Idea what R1 would be.

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I would sure appreciate any help, I'm still very new to electronics.

~Dennis

[M and K Fan]

Yes resistor is in assembly.  I use these both with DC and DCC to indicate power to track.  Works great to see, from across my basement, if power to track is on.
Regards,
Edd

[dcarrell8]

Yes resistor is in assembly.  I use these both with DC and DCC to indicate power to track.  Works great to see, from across my basement, if power to track is on.
Regards,
Edd

Do you know how many ohms the resistor is?  Or how to figure it out by testing?

[peteski]

Well, that lamp assembly is designed to work with 12V.  It will work at 5V (in your circuit), but at reduced brightness.

As for the built in resistor value, why do you want to figure out its value?  Is it to get the total resistance approximately to 10k ohm?  Either way, with a 10k resistor (and 5V), that lamp will not glow very brightly.
If you want to try to figure out the resistor value without taking the lamp apart, you need to hook up an ammeter in series with the lamp.  Then hook this circuit up to a 12V power supply and take a current reading.

The formula to determine the approximate resistor value:  R = ( V - Vf ) / I
Where:
R = resistor value in ohms
I = Lamp's current under its nominal operating voltage in Amps
V = Voltage used to illuminate the lamp
Vf = forward voltage of the amber LED (it will be approximately 2V)

If the lamp's specs provide its operating current, then no measurements are needed at all.  Just use 12V for V, the specified operating current for I, 2V for Vf

If you do not have a 12V DC power supply, you can use a fresh 9V battery and the same equation I provided above, but the result will be less accurate than if using the lamp's nominal operating voltage.

[dcarrell8]

Well, that lamp assembly is designed to work with 12V.  It will work at 5V (in your circuit), but at reduced brightness.

As for the built in resistor value, why do you want to figure out its value?  Is it to get the total resistance approximately to 10k ohm?  Either way, with a 10k resistor (and 5V), that lamp will not glow very brightly.
If you want to try to figure out the resistor value without taking the lamp apart, you need to hook up an ammeter in series with the lamp.  Then hook this circuit up to a 12V power supply and take a current reading.

The formula to determine the approximate resistor value:  R = ( V - Vf ) / I
Where:
R = resistor value in ohms
I = Lamp's current under its nominal operating voltage in Amps
V = Voltage used to illuminate the lamp
Vf = forward voltage of the amber LED (it will be approximately 2V)

If the lamp's specs provide its operating current, then no measurements are needed at all.  Just use 12V for V, the specified operating current for I, 2V for Vf

If you do not have a 12V DC power supply, you can use a fresh 9V battery and the same equation I provided above, but the result will be less accurate than if using the lamp's nominal operating voltage.

Thanks Pete,
I would not be using the 10K resistor in the diagram.  Instead I would use what's already in the bezel of the lamp. (I should have been more clear about that)
Not knowing what the value is (or much about electronics) I'm not sure how to balance the unknown resistor with R2.  This is for my push-button panel

Here's the way this works in LCC.  The node runs the I/O point as an output for 63 milliseconds, and then for a few milliseconds, it converts the output to an input where it polls the input.  So if it detects a "low" or a "high" in this window, it creates the appropriate event.

The I/O point of the LCC node I'm configuring needs to sense the difference in voltage between R1 and R2.  So, the two need to be enough different to assure a good "low" signal when pushing the button.  In the diagram there is a 10K/1K pair, so figuring a 5VDC input, it should have 0.5V when the button is pressed.  Plenty low enough for low.  When the button is not pressed, the 5V level is fed directly to the I/O point via the R1 resistor and LED. 

If my breadboard shows up in the mail today I will run some tests to make sure it will work.  This is all a learning process for me and I'm really enjoying it. 

~Dennis