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

On Wed, 15 Sep 2004 12:45:02 -0700 (PDT), you wrote:

> 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.

Half as hot as they would before. You can also take two 12K 1/2 watt
resistors and parallel them.

Harvey

>
>--- 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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>http://groups.yahoo.com/group/jobshophomeshop I consider this to be a 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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