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Help with illuminated switch.
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Please be positive and constructive with your questions and comments.

Help with illuminated switch.

by john on Thu Nov 25, 2010 3:02 pm

I'm not sure if I need a resistor for the LED in an illuminated SPST switch; here is the data sheet
http://www.lumex.com/specs/CLS-RC11A125191B.pdf
The data sheet indicates 12V for the LED but no resistor or current rating.
I hooked it up to 12v supply for a minute or two, first with no resister, and then with a 1K, and 2.8K and the LED worked in all three cases.
Any suggestions?
Aint nothing like the smell of flux in the morning

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Re: Help with illuminated switch.

by stinkbutt on Thu Nov 25, 2010 3:53 pm

Sounds like its already got a resistor in it. But put a multi up to it. You'll be able to determine the impedance of the circuit, and if it's not zero in at least one direction then there's already a resistor in there.
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Re: Help with illuminated switch.

by john on Fri Nov 26, 2010 11:00 am

Thanks for the help. I though as well and it's nice to get second opinion.

Before I posted my question here I tried using an meter to measure the impedance between the switch LED terminals. In one direction I get open loop indication and in the other direction I measure 7.419M. Then I compared the switch LED measurements with a couple LEDs I had kicking around in my parts cabinet. Those measured 7.404M - 7.622M. After the investigation I was still unsure.

There is always (non)destructive testing methods. I was ready to hook up 12V on the switch and see if/how long it took to die -that's one way to find out :lol:

Thanks again.
Aint nothing like the smell of flux in the morning

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Re: Help with illuminated switch.

by stinkbutt on Fri Nov 26, 2010 2:16 pm

Whats happening is that your multi's sourcing a tiny amount of current through the led and reading teh voltage drop. Because the voltage drop is the same no matter what, it reads as a very high impedance. Which totally defeats the purpose of trying to read the impedance, I suppose.

You'd need to measure the apparent impedance of an equivalent LED (same voltage drop) and then subtract out the baseline impedance to definitively say if there's a resistor in series there. Of course, at this point we're all pretty sure there is.
Red M&M, Blue M&M: They all wind up the same color

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Re: Help with illuminated switch.

by zener on Fri Nov 26, 2010 3:51 pm

The schematic on the data sheet indicates there is no resistor. The only evidence supporting the resistor theory is that it allegedly did not blow up after being connected to 12V.

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Re: Help with illuminated switch.

by john on Fri Nov 26, 2010 4:02 pm

Zener you make a very good point.

I don't mind adding a resistor, but then I will not have 12V across the switch terminals since voltage will drop across the resistor. :roll:
I'm curious if I can conceivably under-power an LED.
Aint nothing like the smell of flux in the morning

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Re: Help with illuminated switch.

by ImaginaryAxis on Fri Nov 26, 2010 4:03 pm

I am curious. What is your 12V source?
An LED does not require a resistor in series to work. An LED is a current controlled device, so what really is required is the forward current, I(f), and forward voltage, V(f).
Granted, depending on the nature of your driving source, say a power op-amp, then you may need a current limiting resistor. But you can certainly drive an LED with a uC using PWM with no series resistors. Now if you want to say the rds(on) of a CMOS output is the resistor... :wink:
I looked at the datasheet and it is very poor. It really is a mechanical diagram rather than a datasheet IMO because is has no characteristic curves or specification tables.

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Re: Help with illuminated switch.

by john on Fri Nov 26, 2010 4:22 pm

Source is an op-amp.
http://www.datasheetcatalog.org/datasheet/panasonic/AN1358S.pdf

I tried the switch LED with 2K resistor and it was sufficiently bright.
Am I playing it safe just using the resistor; or could less than 12V damage the LED?
Aint nothing like the smell of flux in the morning

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Re: Help with illuminated switch.

by zener on Fri Nov 26, 2010 5:14 pm

I am guessing the op amp limited the current to a safe level. If you don't give an LED enough current it just gets dim. A more definitive test to determine if there is a series resisitor in there or not is:

1) Put a resistor in series with the LED connections on the switch, about 390 ohms.

2) Apply 12V and confirm that it isn't loading down.

3) Measure the voltage across the LED connections of the switch.

4) If it is 3 to 4 volts then there is no resistor inside.

5) If it is around 8 volts then there is a resistor inside.

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Re: Help with illuminated switch.

by zener on Fri Nov 26, 2010 5:17 pm

John wrote:could less than 12V damage the LED?

NO. Too much current will.

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Re: Help with illuminated switch.

by john on Fri Nov 26, 2010 5:47 pm

1) I put 384Ohm in series with the switch LED
2) Supply is 11.86V at source nodes, no load, just meter. 11.86V across source nodes with Resistor and LED in circuit = no drop in supply voltage - is this what "load down" means, if there was a drop in Vs when circuit was added?
3) Voltage across LED connections is 9.64V

Conclusion: poor data sheet and probably an internal resistor.

Thanks Zener, stinkbutt and L.A.W.
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Re: Help with illuminated switch.

by zener on Fri Nov 26, 2010 5:58 pm

Yes. I am guessing there is a 1K in there, based on your test data.

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Re: Help with illuminated switch.

by ImaginaryAxis on Fri Nov 26, 2010 6:24 pm

The op-amp is limited to sourcing 40mA of current when in open loop rather than closed loop. Or at least that is the way I read the testing condition which is nothing more than specmanship.
The op-amps maximum output swing with a 2k ohm load is Vcc-1.5V on a 5V supply meaning that it can only swing up to 3.5V while sourcing 1.78mA. This indicates a bipolar output stage instead of CMOS.
What is happening is the op-amp is providing a given current limited only by the emitter resistances of the output stage and the static resistance of the LED. If you are driving the LED with the anode at the output of the amp and cathode to ground, then you should measure the typical V(f) of the LED or something higher if there really is an internal resistor. This op-amp will be able to give you the current required but the output voltage will droop lower than the input voltage.
It works, but not the ideal circuit; i.e., op-amp.

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Please be positive and constructive with your questions and comments.