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

I hope that this hasn't already been answered but I
see one problem with the 6k resistor. It will of corse
supply the correct current but the risistor will use
about .6 watts of power. most resisters that you get
are 1/8, 1/4, or 1/2 watt. If you use even a 1/2 watt
resistor it may work but get so hot it will melt
plastic and burn you. So you really would want to use
two 3k resistors (1/2 watt) in series and be carefull
because they may still get very hot.

--- Andy Wander <awander@...> 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.
>
> 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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