Wally,

Continuity also confirms this, in the even you don't have any voltage to use to check. You'll only get continuity if the current from the meter is flowing in the right direction.

Thanks. Brad.

Now you made me do math and OHM's law. You are right, those 5 watt ones will not last.

No biggie, the new ones are only $8.. cheapest thing on the boat!

Been doing electronics on/off for some 35yrs now and still have to look it up to make sure i've got the right values in the correct locations when doing the calculations!

LOL, I think I have forgotten more than I know. Got my degree back in the mid 80's, sooooooooo a hint to my age!

Funny story, I was in high school auto shop when electronic distributors came out. I said I need to take some electric classes these cars may get more of this stuff. Ha, who would have known.

Im about 10yrs behind ya. Got mine mid 90's then went to work for Motorola (back when they were still decent) in their R&D dept for cell site infrastructure. I got out of there around 2000 when they started doing global layoffs. Now i'm a "Lead Engineer" (in title only LOL) for a company that makes mag-drive pumps. The only electronics i get to play with now a days is fixing my kids broken camera equipment (just changed out an NPN transistor the other day) or playing on my own stuff around the house Still fun to start up the ole brain cells every so often and get a refresher

The "zener" diodes are different than "regular" or rectifier diodes. Zener diodes allow current to free flow in one direction with a small voltage drop (~0.7V) like a regular diode. However zeners will flow in the reverse direction when the voltage reaches the rated value (12V in this case). In this case, the current would flow pretty much unimpeded in both directions.

Wattage rating depends on how much resistance is in the relay coil and the voltage used.

For example, if the relay coil measures 250 ohms, take 12V and divide it by the resistance in ohms to find how much current will flow. In this example 12V divided by 250 equals 0.048 amps. Multiply the 0.048 amps by 12V to find the wattage. 0.576 watts in this example.

Another example would be if the relay coil resistance measures 25 ohms, 12V divided by 25 equals 0.48 amps. 0.48 amps times 12V equals 5.76 watts.

There's another reason diodes may be needed in the proposed circuit besides preventing power from switches backfeeding into the blue-tooth gizmo. When a coil is energized, it produces a magnetic field that closes the relay. When power is removed from the relay coil, the magnetic field collapses back into the coil. This collapse creates a very brief voltage spike that can be quite high (>100V). If too high, it can damage the blue-tooth gizmo if it uses transistors as opposed to mini relays in it's output section.

I started with Moto in their semiconductor R&D lab (APRDL) in the mid 90's. Moto SPS later became Freescale and then NXP. Bailed from NXP abt 7 years ago. So sad to watch Moto be dismantled. It's now just a shell of what it used to be.

So it wont work? I`ll just send the box back to CP , this is too much fuc king around

A 12V zener diode won't work. Need a rectifier diode like Mbam posted.

Edit: technically speaking a zener diode rated well above 12V reverse breakdown would work if it was big enough to handle the current. I say well above 12V cause with alternator working, voltage at trim pumps would be around 14V to 15V.