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

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.

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.
This means R=1/.01=100 ohms. If it looks good at this point, the design for
2V across the LED.

Whichever way you decide to go, 1V or 2V, measure the current drawn. I am
assuming it will be 10mA for the equations below. Otherwise, just plug in
what you measure for the current.

Then, you want Vd(design voltage-my own term) across the LED, so the voltage
across the resistor will be the 60V "minus" Vd.

So for 1V, it is 59V across the resistor, and 10mA thru the resistor, so
R=59/.01=5.9K.

For 2V, it is 58V across the resistor, and 10mA thru the resistor, so
R=58/.01=5.8K.

You can see that it really doesn't matter, and you can just use the "simple"
equation as you first stated it. Just check on what current you really want,
and use that.

I don't know on the other question.

Sure took me a long time just to say, "Yes", didn't it?

Andy

-----Original Message-----
From: AbbyKatt
To: CAD_CAM_EDM_DRO@yahoogroups.com
Sent: 9/14/2004 6:16 PM
Subject: [CAD_CAM_EDM_DRO] On LED's and Gecko Tunings

Two quick, and perhaps silly questions:

Will a 6K resistor be enough with my 50-60V power supply, to light an
LED and not fry it?
I'd like to have a neat light telling me if my power-supply is on, and
the old external HD case I'm using already has one built in... So..

R=V/I
R=(60/0.01)=6K?

Goodmath? Wrongmath?

.. And.. Which is better, tuning gecko-drives with the motor attached,
or not attached to the leadscrew (ie. Bench or field?) I mean, attached
it might break something if it starts violently oscillating, but then,
wouldn't it be better if you adjusted it when it was actually moving the
table? Or is a combo of both best? Your experience, please!

OkayThanks!
~Abby




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