RE: [CAD_CAM_EDM_DRO] On LED's and Gecko Tunings

Harvey:

As usual, there I go shooting my mouth off and saying stuff that's not
REALLY what I meant-at least not all of it.

What surprised me in some tests I recently did with some LEDs was just how
much latitude in series-resistor value(and hence current) there was, while
still maintaining a nearly constant voltage drop across the LED.

Recentl, with a 24VDC supply, and a rather nondescript green LED, I got the
following results(where R is the series resistor value in Kohms, I is the
current in mA, and V is the voltage measured across the LED):

R I V
1.0 21 2.31
1.2 18 2.27
1.4 15 2.23
1.6 13 2.20
1.8 12 2.17
2.0 11 2.15

Thought it was interesting-that's why I said I thought they were "constant
voltage" devices-makes sense, too, if you think of a forward-biased diode,
like you mentioned.

A few questions on what you wrote, if don't mind:

>Not exactly, they will draw whatever current their internal resistance
>will allow them to draw. That's why you need the dropping resistor,
>to limit current flow.

By this, do you mean that if I apply a voltage of exactly the LEDs rated
voltage across it, it will draw more current than it is rated for? It seems
to me that if I add an external series resistor to limit the current(which
will also drop the voltage to the LED) then I will no longer get it's rated
voltage across the LED. Of course, getting a power supply to give me exactly
3.4V or whatever would not be easy, but this is more so I can understand,
rather than something I would really try to build and use.

And do LEDs really have an "internal resistance", or does it just look like
they do? (I guess if I put a current through it and see a voltage drop, I
have no idea whether I'm measuring a resistor or an LED, so I guess they do
have an internal resistance.)

>Ok, LED directly across battery, not good. Some do it, but they
>depend on the internal resistance of the battery to limit the current.

Note that this test was recommended only to see if 1.5V will light the LED
at a good level-not as a recommended way to wire it up in a circuit-but
again, why do we need the battery to limit the current if there is an
internal resistance?

I am starting to think that it is not as simple as an "internal resistance",
but is more of an "internal voltage drop", based on my tests above which
gave nearly constant voltage for a halving of current-so the LED isn't
acting as a resistance, at least not according to Ohm's Law.

Or have I just been smokin' too many banana peels?

Andy Wander
Verrex Corporation


-----Original Message-----
From: Harvey White [mailto:madyn@...]
Sent: Tuesday, September 14, 2004 7:11 PM
To: CAD_CAM_EDM_DRO@yahoogroups.com
Subject: Re: [CAD_CAM_EDM_DRO] On LED's and Gecko Tunings

On Tue, 14 Sep 2004 18:28:08 -0400, you wrote:

>On the resistor question, yes that should give you about 10mA. I usually g
o
>for about 20mA thru an LED, but if you have the specs on yours, you should
>use that number.
>
>The math is not really "correct", because you are ignoring the voltage that
>will appear across the LED, but that will be comparatively small(on the
>order of a volt or two), so it really makes no practical difference.
>
>It seems to me, though, that LEDs are really "voltage" devices, menaning
>that they want to see their rated voltage, and will then draw their rated
>current.

Not exactly, they will draw whatever current their internal resistance
will allow them to draw. That's why you need the dropping resistor,
to limit current flow. Below the rated voltage, some small current
will flow, so you can get a very dim light. Think "forward biased
diode"


>
>What I'm getting at is you may want to approach this a different way:
>
>Hook a 1.5V battery to your LED directly-most modern LEDs(I think) are
rated
>at between 1.5 and 2V. If it looks ok at 1.5V, you may want to design for
>1.5v, if it is too dim, try 2 1.5V batteries in series for 3V, and hook the
>LED up to this 3V with a series resistor, to drop 1V at 10mA.

Ok, LED directly across battery, not good. Some do it, but they
depend on the internal resistance of the battery to limit the current.