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Martaine2005

May 12, 2015
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Hi guys, I am a little discombobulated.
I don’t have the data sheet for these.
I have 5mm RGB common anode and common cathode LEDs. I am only using RED and GREEN, BLUE leg is snipped off.
I want GREEN on constantly dim and RED to over power the GREEN by being brighter when switched on.
I’m supplying 14V and current limiting RED with 1K (11mA). This is bright enough and I’m assuming Forward voltage is 2 to 3 volts.
Now, the for the GREEN. I have gone through my resistor values to get a nice dim glow. Calculations seem pointless now, 1mA was still too bright. I am now satisfied with 3MΩ. But wow, 0.00003μA?.
Could this be a mistake?
It’s the same with common anode and common cathode.

Martin
 

Audioguru

Sep 24, 2016
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You made an arithmatic error.
If the green LED is a very bright modern 3.4V one then it has the same chemistry as a white and blue LED. If the green LED is a dimmer old one then it is about 2.2V.
Why 14V? In a car?
(14V - 3.4V)/3M= 3.4uA, not 3.4pA. Maybe your 3M resistor is actually only 30k ohms?
 

Harald Kapp

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I second Audioguru's assumption about the resistor value. 3.4 µA is way too little for an LED to glow visibly. Very good LEDs require around 1 mA minimum. Why not measure the actual current and/or resistance instead of assuming?
I’m assuming Forward voltage is 2 to 3 volts.
Why assume? Measure it.
 

Martaine2005

May 12, 2015
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0.00003μA
Oh dear me.
That should have been 0.000003, I omitted a 0.

I have done some measuring this morning and double checked resistors. Forward voltages are 1.8V red, 2.1V green. (Blue 2.9V).
The result for a 2MΩ are:
(14-2.1)/2M=0.000005μA.
The GREEN LED is visible in daylight and a perfect glow in the dark.

What am I doing wrong?. I never would have believed a LED would be visible at such a low current.

And yes, it’s for a door open and closed indicator in my sons car.

Martin
 

kpatz

Feb 24, 2014
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Assuming a 2.1Vf on the green LED and a 14V supply, that leaves you with 11.9V across the resistor. 11.9/2M = 0.00000595 Amps, or 5.95 μA (not 0.0000595μA). Remember in Ohm's Law, the units are Amps, Volts, and Ohms, and you have to convert units accordingly.

Modern LEDs are efficient, they can glow (dimly) at these currents.

Those insanely bright blue LEDs you see everywhere are running on just a few mA of current. Gone are the days you need to give a small indicator LED 20mA.
 
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Harald Kapp

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I agree to a few mA, but a few µA? Well, Martin measured so it seems to work.
How dim is dim? Still visible in daylight? Or from an angle other than 180 °?
Personally I'd use a higher current, especially in a car. What with changing lighting, temperature etc.? But of course if your satisfied you're entirely entitled to use as small a current as you like.
 

Martaine2005

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Yes, visible in daylight on my breadboard and more so inside the car on a black dashboard. At night, it’s a perfect glow to be seen but not dazzle.
They are diffused so visibly good all round.
An example of dim/glow is a TV standby LED or smoke detector LED. Although these obviously vary from manufacture etc, they are quite visible indicators.
And yes, 0.00000595A not μA. My mistake.

Martin
 

Tha fios agaibh

Aug 11, 2014
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Interesting.
Doing the math the other way would suggest a 3.4Vf although at current that low, i would expect Vf below 2v given the low current.
I believe the 2.1Vf is based at its current rating of 10ma or so, not a miniscule 5.3uA.

Try measuring the voltage drop across the led.
 

Martaine2005

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Hi John,
I’ll do more measurements tonight.
It’s my math I doubt. I get light headed when moving the decimal point!.

I will try less resistance too, but 820K was too bright in the car at night. Very strange.

Martin
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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The LED could be visible in daylight due to UV exciting the phosphor (assuming it's a LED with phosphor)
 

Martaine2005

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It’s clearly off without a power source and once connected, it’s clearly on.
Later I will try my bench supplies as I’ve been using my RS powered breadboard.
Although I was keeping a close eye on voltage, a second test won’t hurt.

Martin
 

(*steve*)

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Well, your eye is most sensitive to green, and depending on the wavelength of the red LED, you're eye is quite insensitive. Also, your perception of brightness is not linear, so this may have some bearing on it.

I've connected a LED and resistor across a supercap and it was visible for several days, although only in darkness after a day or so. The current through that LED was extremely low, although I can't remember if I measured it (or tried to) when it was very dim.
 

Martaine2005

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Well, I have measured several scenarios and came to a happy compromise.
I’ve added an extra power source to the GREEN LEDs via diodes and via the car lights auto dimming circuit. We now have brighter LEDs in daylight which dim when the car lights are turned on. And all are dimmer with the engine off (12.7V).
Daylight (14 - 2.1)/820=14.51μA. Actual measurement 12.68μA.
Dark 2MΩ and 5.95μA. Actual measurement 5.71μA.
I am still very surprised at the brightness at such low current.

Martin
 

Tha fios agaibh

Aug 11, 2014
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I think you mean 14.5mA.

6uA surprises me too. It seems hardly enough current to get it to turn on.

Also, If you look at a Vf vs current chart, the voltage drop is far below 2.1v @ 6uA. How is it that the math works out?
 

Audioguru

Sep 24, 2016
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The math is wrong again. (14V - 2.1V)/820 ohms= 14.5mA, or 14.5 thousand micro-amps.
Maybe he used 820k ohms for 14.5uA?
 
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