Dual 12 V battery setup – wiring, grounding and busbars

If you trickle charge the starter battery using voltage from the panels rather than the MPPT controller output you have ample voltage to add a blocking diode and prevent backflow. The current drain is low enough to not significantly affect how the MPPT controller functions.

Long term trickle to a lead acid battery may cause damage. A simple voltage regulator set to a suitable "float" voltage is probably all that's needed to prevent this. This can be cheap simple and compact.

You could join the batteries with a "switch" for charging purposes, with the switch being activated only when the panels were providing energy. The switch could be a transistor but in this case a relay is probably an easy and good solution. Drive relay from panel with a "controller" that senses panel voltage - or use MPPT output or signal to switch relay.

Joining the batteries via a diode prevents back-feed but if the main battery is properly managed by the controller then the lower voltage on the other battery caused by the diode drop can noticeably affect battery lifetime. [I've seen this in practice]. Using a Schottky diode reduces the drop. The diode may try and handle starting currents in some cases so a little thought may be needed.

If you want measure input voltages that may be both negative and positive, using an Arduino ADC where the analog input pins can't read negative voltages, then you might consider the following style of three-resistor divider shown on the left side (the right side we will get towards):

^{simulate this circuit – Schematic created using CircuitLab}

The \$V_i\$ terminal goes to what you want to measure and the \$V_o\$ pin goes to your ADC input (or to a buffer amplifier, if you prefer.)

Your Arduino ADC needs to see an input impedance that is less than \$10\:\textrm{k}\Omega\$, total. With the above divider arrangement, this is \$R_{IN}=R_i+R_p\vert\vert R_g\$. Roughly speaking, since \$R_p\$ or \$R_g\$ will be larger than \$R_p\vert\vert R_g\$, we just need to make \$R_i+R_p\lt 10\:\textrm{k}\Omega\$.

But taking into account your worst case \$V_i=-17\:\textrm{V}\$ and power supply of \$V_{CC}=5\:\textrm{V}\$, it must also be the case that \$R_p \lt \tfrac{5}{17} R_i\$. Another consideration that comes from a separate equation solving for \$R_g\$ suggests that \$R_p \lt \tfrac{2}{9} R_i\$. So \$R_i \lt \tfrac{10\:\textrm{k}\Omega}{1+\tfrac{2}{9}}\$. So that is our more important constraint. \$R_i \le 8.2\:\textrm{k}\Omega\$. That's a standard value of \$R_i =8.2\:\textrm{k}\Omega\$ and I then just picked \$R_p=1.8\:\textrm{k}\Omega\$ as an appropriate standard value, as well.

From this, you can compute \$R_g=V_o\frac{R_i R_p}{\left(V_{CC}-V_o\right) R_i - \left(17+V_o\right) R_p}\$, where you get to pick \$V_o\$ when \$V_i=-17\:\textrm{V}\$. It's not linear, though. Looking at the curve, I think \$R_g=2.7\:\textrm{k}\Omega\$ looks good.

The input impedance is now \$9.3\:\textrm{k}\Omega\$, which meets the needed criteria for the ADC. And at \$V_i=-17\:\textrm{V}\$ you will get \$V_o=670\:\textrm{mV}\$ and at \$V_i=-10\:\textrm{V}\$ you will get \$V_o=1490\:\textrm{mV}\$ and at \$V_i=0\:\textrm{V}\$ you will get \$V_o=2650\:\textrm{mV}\$.

In the above case, I'm assuming you can share grounds, of course.