Diode between buss
 

Please bear with me, did not build my plane and not an electrical genius. My EMS is reading my battery voltage at 12.5, however bat the battery and main buss it's 13.5. chased it down to a 1 volt drop when it goes through a square diode between the main and essential buss? There are 2 wires from the main buss to 2 prongs on this diode and a single wire going to the essential buss leaving 1 prong empty, does this sound right?

Welcome to VAF!
 

Peter, welcome aboard the good ship VAF.

Not familiar with the setup you described, but I believe it is from the Aeroelectric Connection book------you can check it out here. http://www.aeroelectric.com/

Take a look here, bottom of the page. http://www.aeroelectric.com/articles...Components.htm

Yes, that sounds right. There are 4 diodes inside of the square case. The assembly is called a BRIDGE DIODE. The way that it is wired in your plane uses 2 of the 4 diodes. A 1 volt drop is normal.

The one volt drop does not hurt anything, except a little wasted energy that heats the diode. The avionics will run just fine on 13 volts instead of 14. Turning on the E-Bus switch (if there is one) will bypass the diode.

Why would a diode cause a voltage drop? Thought resistors did that? And I set my low voltage alarm at 12.3v and it comes on when using a lot of the electronics. And in actuality it's 13.3v. Is it necessary to use that bridge diode?

The function of the diode is to allow the alternator to charge the aux battery, but to block reverse flow from the aux battery to the main buss. If you need aux power presumably the main battery or buss has a short, and you do not want your aux battery shorted too!
Diodes are basically the junction of two dissimilar materials. At the junction there exists a potential 'hill' due to atomic forces, typically about 0.6 volts for doped silicon diodes. You get no forward current unless you have that 0.6 volts or more. In your 'bridge' set up you have two diodes in series, so you loose 1.2 volts. There are better ways to isolate your system so you only loose 0.6 volts. There are also 'Schottkey' diodes with a forward drop of only 0.3 volts or so, that are popular for this useage.

This is what I used to connect the main and aux buss in my plane.

Schottkey diodes used to be rather expensive, but they seem to have dropped in price well. Be cautious with their current rating.......

I am using dual alternators and batteries, so the placement of the diode in my plane is not designed to deal with the aux battery charging, but it is to allow either battery to feed power to the aux buss--------my EFIS and radios are directly off the aux buss, all other electron eaters are on the main buss. You could also call my aux buss the essential buss.

It has functioned well, and I would do it again if building another plane with similar electrical needs.

Anytime current flows through a conductor, there is a voltage drop. That is why the wire size must be increased when current must travel a long distance. The longer the distance and the smaller the wire, the greater the voltage drop.
What kind of alternator and voltage regulator do you have? Ideal voltage output at cruise should be closer to 14.2, not 13.3.
This SCHOTTKY DIODE will drop 0.625 volts when conducting 20 amps according to the DATA SHEET graph. Power diodes should be mounted to aluminum heat-sink with heat conductive paste or film. Heat is the enemy of electronics.

It's hard to take EE101 via the forum...suggest you get Bob Nickoll's book and do some reading...he does a great job of explaining basic electronics used in aircraft without getting too deep in theory and math.

Also Bob came up with the essential bus scheme used by many RVs and explains the rational for the design and several others in the book.

Order the book, "The AeroElectric Connection" from B and C Specialities.

Bob Nuckolls even makes his book available for FREE download.