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I will add a small auxilary battery (lead acid 12V 2.4Ah) on my car for me head unit to be able to crank the engine and not turning off the system. Now, if I crank the engine while the unit is on, it will turn off due to the voltage drop of the car battery.

The small battery will be connected behind the dashboard using the cable that would normally go to the stereo. I will add a relay between the two batteries that gets signal from the alternator so the second battery won't be connected with the main battery while the engine is off.

I found an old circuit that was using Ni-Cd batteries, two zener diodes and a resistor. I added the relay on that circuit and I came up with this

enter image description here

I think that D5 is there to ensure that no current will flow from the small to the large battery. R1 limits the charge current and D6 allows current to flow from the small battery to the car stereo. Which battery will power the car stereo while the engine is running? Is this safe?

My main question is about the resistor. How could I calculate the resistor value? Now my battery reads 12.8V and if connect it to a power supply at 13.7V it draws about 100mA. If I connect a 10Ω resistor the current drops to 50mA. What will happen if the battery voltage drops to 10-11V? How much current will it need and what resistor would I choose to be safe?

What type of zener diodes should I use? Are these ok? My battery is this one.

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  • Isn't this basically the same as your first question?
    – CharlieRB
    Mar 12, 2018 at 12:43
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    The head unit shuts off during cranking because most ignition switches cut accessory power while cranking, not because of the voltage drop. It may be easier to go to a key-on source instead of an accessory source.
    – vini_i
    Mar 12, 2018 at 12:45
  • I made many tests and the problem is not only the accessory line that turns off the unit. It's also the voltage drop, so I need the second battery. Eventually the second battery will power also the accessory line of the stereo using a short delay off circuit. But my concerns are about the charging current and how to limit it. That's why I made another question. I need more info about the current limiting resistor and how to calculate its value.
    – John
    Mar 12, 2018 at 13:27

1 Answer 1

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You should use a split charge relay, as this will provide the power to charge both batteries and separate the batteries during starting which is what you want, leaving the stereo with power when starting.

Here is a link to one and there are many many ... No affiliation just one picked from the search result:

Split charge relay link

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  • Do I need this 12V/100A relay for my tiny 2.4Ah battery? The output of the relay will power only the head unit and charge the small battery. My battery needs just a few hundreds mA to be charged - "Initial Charging Current less than 0.69A"
    – John
    Mar 12, 2018 at 19:22
  • You can find others - I gave that one as an example ...
    – Solar Mike
    Mar 12, 2018 at 19:25
  • If you look on my schematic I connected a relay with the alternator signal so my auxilary battery will be connected with the main only when the engine is running. I did that to avoid using a voltage sensitive relay. I don't want to change anything on the car's main battery wiring. I think I could use a simple SPST 12v/10A relay for this job. Am I wrong?
    – John
    Mar 12, 2018 at 19:59
  • Well, split charge relays are designed to connect to charge the aux when the sense voltage is above approx 13.6V, which avoids all the fancy diodes that you are including, and diodes cause an inherent voltage drop. Anyway, I offered a solution which I have used in the past for similar and different situations - up to you what you choose to do.
    – Solar Mike
    Mar 12, 2018 at 20:06

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