# Odd LED question

Discussion in 'LEDs and Optoelectronics' started by donkey, Jul 28, 2012.

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Feb 26, 2011
2. ### donkey

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Feb 26, 2011
I really hate bumping threads, but can someone help me with the above please?

3. ### duke37

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Jan 9, 2011
I presume that the microamp figure is the leakage current taken in the reverse direction. You should limit the voltage to a few volts in this direction.
The led emits light when current is passed in the forward direction.

Last edited: Jul 29, 2012
4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Normally you have a resistor to limit forward current and simply do not apply a negative voltage.

Having said that, I'm not exactly sure what that thing is.

5. ### Mitchekj

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Jan 24, 2010
Duke and Steve are correct. The datasheet is just saying that at 5V reverse polarity you'll get about 10uA of leakage current. It's just a metric of what to expect, not that you should do it. If you don't have a chance of applying reverse polarity, then there's no need to worry about it.

The forward voltage (Vf) and forward current (If) is what you'd base your resistance value on. The Vf is 2.0 to 2.4 volts at 20mA, this all depends on the individual construction of the LED and is usually never consistent across LEDs produced, hence the min/max tolerance. Worst case (best we have is typical, according to the datasheet) would be 2.0V, so with a 9V supply:

9V - 2V = 7V.
7V / 20mA = 350 Ohms.

6. ### CocaCola

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Apr 7, 2012
Since it's a common cathode LED array, in this case it has to be taken into consideration... If you are for example using a micro and setting a pin low to turn an LED off, you 'might' get a reverse flow when another/other LED(s) is/are turned on in that array...

7. ### Mitchekj

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Jan 24, 2010
Good point! I must admit I didn't pay much attention to the package.
Edit: But what are the chances that you'd end up with a potential difference of something close to 5V? (A serious question, not too familiar with micros.)

Last edited: Jul 29, 2012
8. ### Harald KappModeratorModerator

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Nov 17, 2011
How's that supposed to happen?
If the µC's pin is low it is at ~0 V, the same potential as the common cathode, regardless of other LEDs being turned on or off (I'm neglecting a few mV voltage drop due to the resistance of the common wire),
If you out a current limiting resistor in series with each LED and connect the resistor to a µC's pin, then there is nothing to worry about (least not reverse current).

Harald

9. ### CocaCola

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Apr 7, 2012
I said it should be taken into consideration, yes when you add all the correct "ifs" together (as you have listed) it won't be a concern, but that doesn't necessarily negate it in all instances or designs where the "ifs" might not be held true...

10. ### donkey

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Feb 26, 2011
its a bargraph of sorts. but rather than being vertical or horizontal its circular. the issue is it cost \$12 and I don't want it to blows up so if hyperthetically i was to hook it up to my arduino (only cos I can't find anything else right now) what size resistor should i look at, and should it go on 5v or 3.3?
and to whoever moved this thread thank you so much

11. ### Harald KappModeratorModerator

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Nov 17, 2011
The voltage is not critical.
Connect the common lead(s) to ground.
Connect each LED's anode to a series resistor.
Connect the other end of the resistor to the µC's (arduino's) output pins.

The resistor is calculated as follows:
R=(V(µC)-V(LED))/I(LED)
Where:
V(µC) = µC's High output voltage (either 3.3 V or 5 V, depending on the circuit)
V(LED) = LED operating voltage (around 1.6 V for a red LED - I can't read the DS because I don't want to download the necessary font)
I(LED) = LED operating current (see DS). You will have to take into account not only the LED's datasheet but also the driving capability of the µC's pins.

Harald

12. ### donkey

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Feb 26, 2011
nvm blonde moment

13. ### donkey

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Feb 26, 2011

hope that helps. I assume its classed as the bright red orange as the company I bought it of said red and there is no colour that matches besides this one

14. ### donkey

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Feb 26, 2011
I see the common cathode is proving an issue, would this affect me if I was to use a 350ohm resistor on each and then light up more than 1 LED?
also is it better to go with 330r or 390r? these I have

Last edited: Jul 29, 2012
15. ### Harald KappModeratorModerator

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Nov 17, 2011
If you want to drive the LED with 20mA (as is nominal in the datasheet) your µC will most probably not be able to drive the LEDs. I suggest you insert a simple driver circuit (common emitter PNP with suitable base resistor) between µC and LED. In this setup a LOW on the µC's pin will turn on the PNP and thus the LED.

With a forward voltage of 2 V, Vcc=5V and ILED=20mA this gives R=150 Ohm. If you use 330 Ohm you will have ILED~10mA which should be no problem at all. Using 390 Ohm will just dim the LED a bit more.

Could you explain why the common cathode should be an issue? I can't see the problem there. This is fairly straightforward and has been used for years (cf. common cathode 7-segment display).

Harald

16. ### donkey

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Feb 26, 2011
I have no issue with it, just read coca colas response. and then I found out the most interesting part which is the schematic is wrong. the circle is broken up into 4 sections. then each section has 4 LEDS.
so in total the are 4 grounds (1 for each section) and 16 LEDS/anodes