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I want to connect two motors on an On-Off-On switch. On position 1, I want motor 1 to me active, on position 2, I want motor 1 and 2 active.

Here is the diagram I was thinking of implementing

enter image description here

My second question, if this work, what kind of Diode should I be buying? on the circuit lab I've just used the basic diodes.

My motors are 12V 25W water pumps

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2  
You can replace D2 with a wire. Sure they DC motors? You need a diode that can handle 3A continuous, but to be safe during starting of the pump I'd personally rate it at least at 6A continuous. With this application, the cheapest diode (that can handle > 12V reverse and mentioned current) will probably already do the job perfectly well. – jippie Jul 19 '12 at 13:04
Thanks, for the info. I wasn't sure about the D2. For the motors, yes I'm sure they are DC motors. I've updated my sketch. – allo_man Jul 19 '12 at 13:14
1  
@jippie - You should make your comment into an answer - it's fine. You are welcome to copy any material from my answer if wanted (eg diode part number etc). If you copy enough to cover my points I will delete my answer. – Russell McMahon Jul 19 '12 at 13:22
@RussellMcMahon - interesting how you get extra upvotes for noting that I should have turned my comment into an answer, while my original comment isn't upvoted at all ;o) – jippie Jul 19 '12 at 13:59
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@jippie - I never vote on comments - make it an answer and you can have my +1 :-). My answer has slightly more in it than your comment BUT if you answer and add diode information I'll delete my answer, as I said. You are welcome to the answer votes BUT you have to answer :-). – Russell McMahon Jul 19 '12 at 14:22

1 Answer

up vote 3 down vote accepted

Jippie's answer (put as a comment) is largely correct.

Current = Power/Volts = 25/12 ~= 2A.
Substantially more current flows at startup and the diode must handle this.

Diode D2 is not needed to achieve your functionality (just replace it with a wire, as he says). Using D2 may be useful to provide both motors with the same voltage (as both then have 12V - 1 x diode_drop but you will probably not notice the voltage drop across the motor when the pump is pumping.

Many diodes will meet your need. Using Digikey's parametric selection guide I specified a diode of at least 5A rating and 40V reverse rating (safety margins never hurt) and asked for the cheapest silicon diode in a leaded package. Surprisingly it was [this 600V 5A BY500] (http://www.vishay.com/docs/88544/by500.pdf)

This is far more capable than needed but cheapest is cheapest :-). About 53 cents/1 at Digikey. (I tried specifying a >= 2.5A and >=20V part and the above was STILL cheapest :-) ). You can actually get a Schottky diode for less, and these are in fact superior technically BUT are more easily killed when playing so specifying a standard Silicon diode is safer here.

Be sure to be sure that the switch will handle the 12V inductive load of the motor.

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Thanks, my switch is rated 10A 250V. I think it should be enough. I'm planing on adding a fuse in the battery connector. I was planning on putting 10A fuse. – allo_man Jul 19 '12 at 13:54
10A 250 V will be an AC rating. DC is harder on the switch as there is no reverse polarity to extinguish arcing SO current and voltage ratings for DC are lower. It will probably be OK BUT be sure to look for DC specs and not to assume that AC specs are enough. – Russell McMahon Jul 19 '12 at 14:24

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