On Sat, 30 Sep 2006 19:45:31 GMT, Al wrote:

In article ,
John Fields wrote:

On Sat, 30 Sep 2006 14:12:59 GMT, Al wrote:

In article , jasen
wrote:

On 2006-09-29, Al wrote:

If your LED is specified at 10mA, then get two diodes whose forward
voltage is 0.75V at 10ma. Put them in series with your LED. If the
supply voltage goes up and the current goes up, the forward voltage of
the diodes will go up also thereby dropping more voltage across them.

yeah, if the current doubles the voltage may increase maybe 10%
how's that going to help?

Thus the dynamic resistance of the dropping diodes serves as a voltage
regulator. Really good diodes with a low dynamic resistance won't do.

it seems to me that regular resistors would work better,

If your current doubled for some reason, you have other big problems.

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I'm not trying to be rude, but the problem would rear its ugly head
if your suggestion was followed. Take a look at what it takes to
double the current through pretty much _any_ diode once it's on the
far side of the Vf knee and you should be able to see that your
suggestion, if followed, would be inviting disaster. The LED(s)
need to be fed from a constant current source. Period. End of
story.

A series resistor is not a constant current source.

---
No but it's a hell of a lot closer than forward biased diodes.
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And, this method is not for manufacturing millions of items. It's for
solving a special home problem. Like I said, I do it and it works for
me. No, I don't bake the circuit nor do I freeze it. And as I mentioned
above, diodes with soft knees should be used.

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Ridiculous. Take a diode and straighten out its knee and what have
you got? a resistor. The point is that the softest knee you can
find won't help you regulate (or even limit) the current into the
LEDs if the source voltage varies.



--
John Fields
Professional Circuit Designer