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

On Tue, 14 Sep 2004 21:09:44 -0400, you wrote:

>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.

That's no big deal, I manage to do that every now and then.

>
>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.
>

They are, the difference is that they are not as "flat" as, say, a
zener diode. The actual voltage does vary a trifle with the current,.
and that needs to be taken into account when you pulse them for
multiplexed displays.

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

Certainly.

>
>>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?

This is slightly tricky. The diode itself has an equivalent forward
resistance, and applying the appropriate voltage across the diode will
result in a current determined by the forward resistance.

I think that in this case, it will draw exactly the rated current at
the rated voltage. The problem comes if you consider it to be a
standard resistor, with a linear voltage/current slope.

>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.

If you set up the power supply to be exactly the forward voltage of
the diode, yes. it's going to drop less and the current will be less,
all according to the curve.

>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.
>

Is ok, I get that.

>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.)

yep, they have an internal equivalent (thevenin) resistance.
Basically, an equivalent battery and a resistor, I think. for forward
bias. This is circuit theory, though, (and only as I remember it).

The resistance is not linear, though, and that's the trick. Very easy
to exceed the max current when driving them with a voltage source that
is not current limited.

>
>>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?

Because the internal resistance is not sufficient to limit the current
to a safe (non-device failure) level.


Perhaps the thevenin equivalent will make sense. Think of a 1.5 volt
battery with a 1 ohm resistor in series... (values not real, but good
for the illustration).

Put exactly 1.5 volts across this, no current flow.
Put exactly 1.520 volts across this, get 20 ma though the battery
(diode equivalent).

put 1.6 volts across this (fresh battery) and get 100 ma through the
battery.

>
>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.

That's it. The main thing is that the effective internal resistance
is lots lower than you think, because of the voltage drop of the ideal
diode, represented by the battery.

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

Nope, just not quite thinking of this in a way that will show you the
problem.

If you neglect the diode (battery) in the example of the part, then
you think the internal resistance is lots higher. It isn't. There's
where the trouble comes in.

>
>Andy Wander
>Verrex Corporation

Harvey

>
>
> -----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.
>
>
>
>
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