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

Use this as a guide to caculate the droping resistor for a LED. The color
is important.

Red LEDs drop about 1.7 volts when conducting so subtract 1.7 volts from the
supply voltage and apply Ohm's law to the resulting voltage to arrive at the
correct value of the resistor.

Green LEDs drop about 2.5 volts when conducting.

If your supply voltage is more than 10 times the voltage drop across the led
you can just use the supply voltage ans ignor the LED voltage drop.

Blue and amber LEDs have different voltage drops but I don't remember them
right now

Most 5 mm LEDs can withstand 40 - 50 ma

Hugh

----- Original Message -----
From: "Andy Wander" <awander@...>
To: "'AbbyKatt '" <cnc@...>; <CAD_CAM_EDM_DRO@yahoogroups.com>
Sent: Tuesday, September 14, 2004 5:28 PM
Subject: 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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sister site to the CCED group, as many of the same members are there, for OT
subjects, that are not allowed on the CCED list.
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DON'T POST IF YOU CAN NOT ACCEPT THIS.....NO EXCEPTIONS........
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