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Can 2.5V to light up a typical 3mm LED?

M

Mr. Man-wai Chang

Jan 1, 1970
0
I supposed it should work fine...

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M

Mr. Man-wai Chang

Jan 1, 1970
0
Hmmm, some are more typical than others. If you had Blue in mind it
might be tricky using conventional methods.

Was considering to connect two 3mm LEDs to the 5V of a molex connector
inside a PC chassis. Nothing fancy!

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D

Daniel Pitts

Jan 1, 1970
0
I supposed it should work fine...
Always use a current limiting resistor. If you don't, and you're lucky,
you'll just burn out your LED. If you're unlucky, it'll burn out your
power supply and/or other circuitry in the path.

Diodes work differently than other types of loads. They always drop a
pretty consistent voltage, regardless of current. This means that if
nothing else resists the current, then it basically looks like a short.

That's a simplified model of diodes, but for the most part a very useful
one. As others have mentioned, the voltage that LEDs drop depends on the
chemistry. As does the color.

Usually, when you buy LEDs, the spec sheet tells you the important
characteristics:

1. max forward current, If(max).
2. Typical forward current, If(typ)
3. Typical forward voltage drop. Vf

To figure out how much resistance to use, you want to make sure that the
total current is less than the max forward current, and preferably
closer to the typical forward current.

Let's use a concrete example. From the first google result for "Blue LED
spec sheet" <http://www.lc-led.com/products/500tsb4d.html>

Peak current is necessary to consider if you have capacitor in the
circuit, or some other kind of timing dependent circuitry. Let's ignore
that for now.

Max continuous forward current is 30ma. So we want to keep it below
that. Let's aim for 20ma.

Typical forward voltage drop is 3.6 (already 2.5v is too low).

Take your total supply voltage (5v in your case), and subtract the rated
typical forward voltage drop. 5v - 3.6v = 1.4v

Now we know that the LED drops 3.6v, and the "rest" of the series
circuit needs to drop 1.4v. Using ohm's law V=I*R, we know that we want
R such that 1.4v=20ma*R. 1.4v/20ma = 70ohms. That is the minimum
resistor you need. The next standard size up from 70Ω is 75Ω. You could
also go up to 100Ω and end up with 14ma through the LED, which may be
bright enough.

In professional applications, you'll also need to consider the power
dissipation, but that's another few pages of discussion ;-)

Hopefully this helps,
Daniel.

P.S. I'm not an expert in the field, so I may have forgotten something
or gotten something wrong. Anyone else should feel free to correct me.
 
G

George Herold

Jan 1, 1970
0
Always use a current limiting resistor. If you don't, and you're lucky,

you'll just burn out your LED. If you're unlucky, it'll burn out your

power supply and/or other circuitry in the path.

Diodes work differently than other types of loads. They always drop a
pretty consistent voltage, regardless of current. This means that if
nothing else resists the current, then it basically looks like a short.

Hey, that raises an interesting question. I always figured LED's are like other diodes (I-V curve is an exponential.)
So I've got an amber (590nm) LED... here's the voltage drop vs bias current..

Current Forward Voltage
10nA 1.273V
100nA 1.387V
1uA 1.498V
10uA 1.605V
100uA 1.706V
1mA 1.806V
10mA 1.923V

With a dark room I can just see the light at 1uA. 10uA is no problem.

I've also heard tell you can get a feel for Plancks constant by plotting
forward voltage vs led color (for a bunch of different leds).
e*V = h*freq = h*c/wavlength
But this is not right.... first off, for the above 590 nm LED I get 2.09V
And secondly what current do I pick?

Hey it's gotta be that the LED is just using those electrons at the high energy end of the Boltzmann distribution... Can you see any LED cooling of the junction? At 1mA and ~0.2 V (2.0 - 1.8) that's ~0.2mW

George H.
 
M

Mr. Man-wai Chang

Jan 1, 1970
0
Red, yellow and green LEDs have voltage drops around 2 volts, so two
of them in series, with a current limiting resistor, would work on 5
volts.

It is working! A 3mm green LED in series with a 5mm red LED connected to 5V.
Blue and while LEDs have a voltage drop of 3 volts or more, so you
couldn't run two of them in series from 5 volts, but you could run two
sets of LED and current limiting resistor in series, in parallel from
5 volts.

Thanks!

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M

Mr. Man-wai Chang

Jan 1, 1970
0
https://dl.dropboxusercontent.com/u/53724080/Optos/OsramCurrents.JPG
The greens are visible at 1 uA in office light, barely perceptable to
the close-up, dark-adapted eye at about 1 nA.

Thanks

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M

Mr. Man-wai Chang

Jan 1, 1970
0
Always use a current limiting resistor. If you don't, and you're lucky,
you'll just burn out your LED. If you're unlucky, it'll burn out your
power supply and/or other circuitry in the path.

Do I need that resister if I am to connect *TWO* LEDs in series over the 5V?

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M

Mr. Man-wai Chang

Jan 1, 1970
0

Thanks

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M

Mr. Man-wai Chang

Jan 1, 1970
0
Connect a resistor in series with each one, then connect the two resistor-
LED pairs in parallel. That's the conventional, safe, and easy way to do
it.

What if one LED in the parallel circuit died? Would the remaining LED
receiving more than its share of current, even with a current-limiting
resistor?

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D

Daniel Pitts

Jan 1, 1970
0
What if one LED in the parallel circuit died? Would the remaining LED
receiving more than its share of current, even with a current-limiting
resistor?
No, that's not how parallel circuitry works. In a parallel circuit, the
voltage is equal across the branches. In your case, that's the 5v will
always be across each of the LED+resistor circuits. If you had equal
current going through both branches, and then one of the branches
"died", resulting in an open circuit, your total current would just be
cut in half.

Now, if you had a series circuit, and one LED died, then both LEDs would
appear off, since no current could flow at all.

OTOH, in a series circuit, if either LED failed by shorting, you'd end
up with too much current through the other remaining LED, and likely fry
them both. In a parallel circuit, the current wouldn't likely go
through the other LED.
 
D

Daniel Pitts

Jan 1, 1970
0
Do I need that resister if I am to connect *TWO* LEDs in series over the
5V?
Yes. Unless the voltage drop equals exactly 5v, *and* the voltage source
will never go above 5v, you will have some extra voltage left over. If
you have no resistance (or the very low resistance of the wire), you
will have a run-away current.

It may even appear to work for a while, but you could be damaging your
power-supply.
 
M

Mr. Man-wai Chang

Jan 1, 1970
0
When you connect two or more LED/resistor combinations in parallel,
the current in each parallel branch is independent - adding or
removing branches will not affect the currents in the remaining
branches.

I remember now: 1/R(parallel) = 1/R1 + 1/R2

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M

Mr. Man-wai Chang

Jan 1, 1970
0
Now, if you had a series circuit, and one LED died, then both LEDs would
appear off, since no current could flow at all.
OTOH, in a series circuit, if either LED failed by shorting, you'd end
up with too much current through the other remaining LED, and likely fry
them both.

Would a LED suddenly short itself? Not very likely, I supposed...

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M

Mr. Man-wai Chang

Jan 1, 1970
0
Having said that - I recently aquired a pc case that some muppet had
wired the power LED to a spare drive Molex - it took some dismantling to
discover there was no current limiting resistor. Due to thermal shock,
the LED die bond wire had fractured - as the die thermally expanded and
contracted, the fractured bond wire made or break so the LED flashed
(for a while).

Really? It didn't short itself, did it?

Another question:

LED is a diode.

Can large current make a diode to behave like conductor and hence short
the +5V and GND?

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M

Mr. Man-wai Chang

Jan 1, 1970
0
How long is a piece of string!

How big is the diode - how much current can the PSU supply?

But would the short piece of string melt like a FUSE **BEFORE** the
large current could get through?

If it did melt *FAST* enough, then there was nothing to worry about... . :)

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M

Mr. Man-wai Chang

Jan 1, 1970
0
But would the short piece of string melt like a FUSE **BEFORE** the
large current could get through?

If it did melt *FAST* enough, then there was nothing to worry about... . :)

BTW, are we talking about circuitry or material science? ;)

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M

Mr. Man-wai Chang

Jan 1, 1970
0
Troll detection - and you have been.

Of course not!!!!

**IF** an LED could become a conductor by slight fluctuation in the +5V
supply, a limiting resistor would not help, would it? It would still
short itself even whether you connected it serial or in parallel, with
or without a resistor!

So the question would demand an answer relating 3 objects:

1. diode/LED (properties)
2. conductor
3. fuse

How would you explain the relationship between these words?

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M

Mr. Man-wai Chang

Jan 1, 1970
0
Obfuscation.

Anyway, I agree that a limiting resistor is a safer bet.

That still does NOT explain whether the LED would short itself.... :)

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M

Mr. Man-wai Chang

Jan 1, 1970
0
Even if it did (I've never seen one fail that way) the current limiting
resistor would still limit current in that leg. You'd have to have both
the LED and the resistor in one leg short out -- and resistors just don't
fail short.
You have more to worry about from meteors hitting your noggin.

I acknowledge the safety measure, but is the threat real?

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