Beginner question on reed relay voltage

[ynicorns]

Hi,
I'm working on a small circuit to change a pedal mod from being a normally open footswitch jack to a normally closed footswitch jack with a reed relay.  I'm going to use this one: http://www.mouser.com/Search/ProductDetail.aspx?R=W171DIP-17virtualkey52830000virtualkey528-171-17 and am trying to find out how exact the 5V needs to be.  The data sheet says the voltage range is 5-25V, and I've got a 9V power supply running the pedal that I could take power directly from or I can build in a voltage regulator if needed to get to 5V.  Anyone have any advice on this?  Thanks!

[PRR]

> The data sheet says the voltage range is 5-25V

No, they *make* them 5V to 25V. All on one data-sheet, but different parts.

What you want is the line: "Operating Range % of Nominal -- 80% to 110%"

So the 5V Part must be given 4V to 5.5V.

> build in a voltage regulator

Bah. A relay is not fussy.

Figure its resistance, add a series resistor to waste-off the excess voltage.

How?

Page 3 says that part takes 5V and has 500 ohms coil resistance. This is easy. You need to lose another 4V, use a 400 ohm series resistor (390 is nearest common value). You should compute the resistor's dissipation; in this case you could even use a 1/8W resistor (but 1/4W is cheap and more rugged).

[ynicorns]

Thanks for the explanation, and for the direction on how to get the voltage right!  I'm trying to wrap my head around the math and falling embarrassingly short, if you have time I would love to know which formula and where the values go to come to this answer.  Wish I had paid more attention in high school these days.

[GibsonGM]

1)  You need to know the current the *item you are dropping voltage to* requires.  Your relay will operate at 5V/500 ohms = .01A (10mA).   Just Ohm's Law, using the RATED voltage and known resistance to get the operating current.   Don't confuse the voltage with your source V, use the V you are dropping to.

2)  The formula for calculating a dropping resistance is:    Vs-Vd/IL    or, Source voltage - the drop you WANT /  load current in Amperes.

So, to GET 5V at your relay from a 9V source, we must DROP 4V.   9V-4V=5V, which we desire.  The key thing here is the 4V.   The Drop.     

Answer:
Your dropping resistor should be:   4V/.01A = 400 ohms, as PRR stated.   Handy little thing to know!   

This will take care of your relay issue no problem.   If it were something more critical, a better way of dropping & regulating would be required, but this is a very non-critical application so we can use this simple, easy and fast way to do this.

[ynicorns]

Thanks a ton!!  I really appreciate the help on this PRR and GibsonGM!  Baby steps to understanding simple circuits better.