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

Andy Wander 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.
>
>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 "near-constant voltage" devices :)
They are diodes, and as such they have a non-linear current transfer
function. ie, the current isn't proportional to the voltage (like a
resistor). Generally, you can think of any diode, including LEDs, as
having three ranges of operation (with respect to voltage) (these aren't
official terms - just easy to grasp):
1) Undervoltage: The voltage is too low, and there's essentially no
current. This is the "off" region - it's not quite off, since there are
still a few microamps or nanoamps flowing. The current won't be
anywhere enough to generate any light from an LED, though.
2) "Normal Voltage": The voltage is pretty close to the "operating"
voltage. Note that manufacturers rate their LEDS as a "xx volts at
If=20mA", in other words, they're telling you that if you put 20 mA
through it, it will be at this voltage. If you decrease the current,
the voltage will decrease slightly, and the LED will dim. If you
increase the current, the voltage will increase slightly. The current
is an exponential function of voltage in this region. (actually, it's
always an exponential, there's just an offset to the exponent - once you
exceed it, the current goes up VERY rapidly.)
3) Overvoltage: Actually, this should be called overcurrent, but oh
well. The LED is dissipating too much power, and will have its life
expectancy reduced, possibly catastrophically (ie, with smoke coming
out). You'd probably have to put an amp through the diode you tested to
get the voltage up another 1/4 volt.

>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.)
>
>

Not really, or at least, it's not significant (the internal
resistance). There is a junction capacitance that can be important in
high frequency circuits, but it won't be an issue here. You can put
current through a transistor as well, but it doesn't mean it's a
resistor inside. The one thing that still does apply, like a resistor,
is that power dissipated by the device is still current * voltage. So,
if you have 2V across an LED, and 20mA, it's dissipating 40mW.

>>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?
>
>

Batteries are usually modeled as a perfect voltage source plus some
internal series resistance - that's what he's referring to. A perfect
battery would be able to supply infinite current with no voltage drop.

>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.
>
>

Nope. Remember - if you plot current vs. voltage for a resistor, you
get a nice straight line. If you plot current vs. voltage for a diode,
you get almost nothing until the breakover point, then you get a steep
line, then you get a nearly vertical line.4

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

Actually, that was a rumor started by a band in San Francisco - Donovan
never intended anyone to think they could get high from that :)

>Andy Wander
>Verrex Corporation
>
>

- Steve

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