LED blacklight voltage Q

[Downunderwonder]

The wizzard rounds things off a bit.

If I put a 130 Ohm resistor in with 2 of 3.1V LED @9V I make it about 22mA like so:

9-(2x3.1)= 2.8V left after LEDs turned on

2.8/130 = 0.022A

Mysterious things these LEDs. Waiting on seller to confirm spec.

 

[UncleFluffy]

The wizzard rounds things off a bit.

"a bit" is fine with this stuff. 10% either way won't hurt anything.

 

[Bob Lee (QSC)]

To drive LEDs, use a current source (relatively high source Z, voltage can vary but current is fairly stable), not a voltage source (relatively low source Z, voltage is stable but current can vary widely).

The resistor in series with the LEDs helps approximate a current source.

 

[Stumbo]

It seems like either I'm missing something or you guys are simpletons.

[OT]

Lighten up, Francis.

[/OT]

I suggest a wee bit more humbleness in your replies.

Calling fellow TB's names does not lend itself to winning and influencing people(and getting assistance)

Rewrite: Maybe I'm a simpleton or just missing something here.

 

[Downunderwonder]

Rewrite: Maybe I'm a simpleton or just missing something here.

You're missing the words "either" and its mate "or"

Very happy to have been led from the wilderness that is LED in an Ohms Law mindset.

So I heard back from the seller.

1. 23mA should be fine, as long as there's no spikes.
2. My resistor values are calculated towards the safe and reliable side, always accounting for the 0.7V voltage drop of silicon diodes, and just using Ohm's law (V=IR)
3. The UV LEDs are quite sensitive to high Vs and current spikes, hence the resistor to counter both.

He originally suggested three LED to one 130 ohm resistor. Is he looking at the same graph that was posted above where I'm supposed to substitute 3.1V as the corner voltage?

 

[Downunderwonder]

ha, ha, I geddit now, "rewrite", yeah you got me good.

Hey if someone just say's "you're wrong" but doesn't say why then I tend to fall back on what I do know, which was wrong in the case of LED not being straight ohmic like a simpleton would assume.

 

[UncleFluffy]

He originally suggested three LED to one 130 ohm resistor. Is he looking at the same graph that was posted above where I'm supposed to substitute 3.1V as the corner voltage?

Let's run the numbers...

is that 3 LEDs in parallel? if so ...

Voltage across resistor: 9V - 3.1V = 5.9V
Current through resistor: 5.9/130 = 45mA

interesting ... possible ... need more info

 

[Downunderwonder]

It was three in series.

 

[Downunderwonder]

From my OP

Thanks for buying my items.

Just want to let you know that you definitly need resistors for your application. Each LED only drops the voltage by 0.7V, so the 'lightest' LED in the chain of 3 will still be getting 6.9V, not good

Anyway I will include 3 130 ohm resistors on the house, ideally if you can drop the voltage or obtain some resistors close to the value of 200 ohm (and still do the R-LED-LED-LED) that'd help out the circuit's reliability big time. Oh and these go open circuit from memory once blown.

 

[UncleFluffy]

Ah, I only saw the "3.1V" part.

No LED has a Vf as low as 0.7V. To be blunt, he's talking out of the wrong hole, and confusing you (and us) in the process.

 

[UncleFluffy]

Best thing is to ignore his numbers. Grab a 1k resistor and a 9V battery, put them in series with the LED. Measure the voltage across the LED. That's the number you care about.

 

[Downunderwonder]

Thanks, yeah I got another nonsense email. I think he gets them from his cousin in China.

The datasheet shows a bunch of specs in the chart which differ from the graphs. All very micky mouse.

I just hope they aren't purple when they show up! 413Nm ain't UV.

 

[Stumbo]

You're missing the words "either" and its mate "or"

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