If you gauge has a C and a D it's an ammeter. C for charge, D for Discharge. Don't forget there is resistance in the battery cables, both positive and negative sides. As current flows through that resistance it will create voltage drops. The higher the current flow the more voltage drop. The higher the resistance the more voltage drop. Sounds like everything is working fine. Formula is V=IR, Voltage Drop = Current times Resistance. It doesn't take much current across a low resistance to start shaving off volts. For example, charging the battery back at say 2 amps, across a .5 ohm resistance will drop you 1 volt. V=(2)(.5) It's not hard to add up to .5 Ohms by the time you factor in the resistance of the battery cables, the battery lead from the regulator, each wire connection, the internal resistance of the regulator, the internal resistance of the battery...... The reason you are seeing a different voltage reading frame to battery positive post vs reading across the post is that you are only reading the voltage across the batteries internal resistance when reading post to post. Reading frame to post adds in the resistance of the wires, connections, and the internal resistance of the regulator as well. Each component (regulator, cables, connection, battery) adds resistance to the total circuit that the current must flow through. And each component will drop a small amount of voltage across it's resistance.
Last edited by bztguy; 05/27/2015 10:28 PM.
