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Need to confirm resistor selection

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

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$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

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

Thanks for reading, Steve

Reply to
SklettTheNewb
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25/1000 =3D 25 mW
Reply to
Globemaker

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

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

Reply to
ehsjr

Where did the 25 for V2 come from?

The important values are: LED current to operate: 2 mA (you want to supply somewhat more than that) Absolute Maximum control current: 50 mA (you want to stay well below that)

The resistor has to drop [supply voltage] - [LED "on" voltage], or 5 -

1.2 = 3.8 volts.

I'd plan for about 10 mA, which would be R = E/I = 3.8V/.010mA = 380 ohms (390 ohms is a standard value)

The suggested 1000 ohms would give 3.8 mA, which is still OK.

The power dissipated in the 1000 ohm resistor is about .014 watts, so the lowest-power resistor available would be much more than adequate.

--
Peter Bennett, VE7CEI  
peterbb (at) telus.net
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca
Reply to
Peter Bennett

I
k

Ah yes, typo. Good catch.

Reply to
SklettTheNewb

more

well

standard value)

The 25v came from 5^2 - I had used "v2" because my superscript didn't work, I forgot that I could use the "^" character instead.

Thanks for providing the additional formula and explanation on how to achieve the required current. This is all new to me, I love learning this stuff!

Reply to
SklettTheNewb

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