On ALL/BOTH you will not see the voltage disparity because the batteries are in parallel. You will only see a minor cable drop variance at higher currents..
Alternator B+ @ 15.4V - House Positive Terminal @ 14.5V =
0.9V drop @ 65A & this is just too much especially when you realize you've not even tested the negative side of the circuit and that at 80A or 90A this is going to be even worse.
Undersized wire, bad terminations and a diode isolator in the path = BAD...
If you take it off
BOTH this 0.9V will now be stacked onto the start battery because you are sensing the house bank and it's path has a LOT of drop under high current. The start battery has almost no voltage drop, because it is accepting
minimal current, most of that 0.9V will be seen on the start battery..
Remember it takes CURRENT (house bank) to create voltage drop. The higher the current the more voltage you are dropping. The lower the CURRENT (start battery) the less voltage drop.
This is one reason diode idolators have not been used by
competent installers in over 20 years......
Rx
Larger alt B+ & B- wires (You want minimum voltage drop eg: less than 2%)
Feed alt B+ & B- direct to house bank & place a fuse within 7"of positive post
Fuse is sized @ 150% of alternator rating and not above wires max ampacity
Sense voltage at house bank
Place an ACR between house and start banks
Problem solved...
Again, study this drawing below paying close attention to the voltage drops on each section of wire and you will see how easy it's to over charge a start battery with a diode isolator, especially when it is installed at the wrong end of the alt B+ cable and you're sensing the house bank.....
On of the major mistakes installers often made with diode idolators is installing them close to the alt not close to the batteries. Course if you are running cable large enough for the alt output, all the way to the bank, then the best option is to feed the alt direct to house, or a house charge bus, and then use an ACR to charge start...