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# Re: Need to confirm resistor selection

ehsjr
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Posts: n/a

 08-27-2011, 05:04 AM
SklettTheNewb wrote:
> I don't like to start a post with a disclaimer, but in this case I
> must; I am a programmer, not a EE or hardware guy. However I can
> handle some very simple tasks and do enjoy the rare chance to work
> with electronics.
>
> So, with that said I find myself in a situation where I am taking over
> on a circuit design for a project we are working on. Logic problems I
> can handle, but it's some of the more rudimentary electronics issues
> that I stumble on. At this point I need to specify a resistor network
> for use with a SSR (http://www.clare.com/home/pdfs.nsf/www/
> CPC1014N.pdf/\$file/CPC1014N.pdf) that draws 2mA. The SSR requires
> 1.2v for it's input.
>
> The resistor networks are needed to bring the 5v source down to the
> required ~1.2v range for the SSR input. An engineer that was (no
> longer is) involved on this project specified a 1k ohm resistor.
> Today I was faced with the question of which power rating was needed
> for this use. I did some research and found the formula:
> P = V2 / ohm
> or
> P = 25 / 1,000 = 0.025mW

^^
Not mW, Watts. P = .025W - but that's not the best way to figure
the R value. See below.

>
> Is my understanding of the calculation correct? Did I provide enough
> information?
>
> Steve

A low wattage 1K resistor will be great for you.

The SSR uses an LED with a Vf of 1.2 and an If of 2 mA.
That If figure - 2 mA - is the maximum current needed to
guarantee that the SSR will be energized. You could run
the LED current a lot higher than that. What you want to
do is ensure that the LED gets _at least_ 2 mA and limit
the current to something a lot higher.

With a 5V supply, and ~1.2 volts dropped across the LED, you
want to drop ~3.8V across the resistor. At 3.8 mA a 1K resistor
will drop 3.8 volts. The LED inside the SSR will happily draw
the 3.8 mA - in fact, it will draw as much current as it can get.
The 1K resistor limits what it can get to 3.8 mA. The LED could
draw a helluva lot more current without damage to itself if the
resistor was a lower value, but there is no need to have it do
that, as the 1K will give it well over the 2 mA maximum it needs
for guaranteed correct operation.

The power dissipation for the resistor is computed with P = I^2*R,
or .014 watts in this case, so use .03 watts or higher for the
resistor wattage.

Ed

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