dual battery basic question

[Dan_68]

Hi all,

This is a basic circuit but i am not getting this. I have just put a second battery in the back of my car it runs through isolator so it doesnt train the main battery.
When the engine is running and the voltage at the front is above 13.2 it starts charging the rear. When i have the rear battery disconnect i am measuring 13.4 volts at the charge wire at the rear when i have the battery plugged in i am only getting 12.6v at the rear. This is the voltage of the second battery but charge wire should be still 13.2 volts so it charges the battery. I am confused if anyone has an idea.
Thanks

 

[(*steve*)]

If the "isolator" is a diode then you would expect a 0.6V difference between the batteries.

 

[Dan_68]

Thanks but i am measuring the voltage after the isolator both times. So the wire goes from the main battery to the isolatpr then to the rear of the car. Its reads 13.4 volts when the second battery is not connected and it only read 12.6 with it connceted.

 

[(*steve*)]

Yeah, sounds like there's a diode in it then.

Is there current flowing? (have you measured it?)

 

[Dan_68]

Havnt measured yhe current. When i measure the cable i am measuring it after the isolator. So if i have just measure thr voltage at the cable with no load attached its 13.4v. Does the .7v drop not happened because the cable isnt earthed anymore? If this is correct how can i charge the battery fully?

 

[(*steve*)]

A 0.7 volt drop is indicative of a diode being used to isolate the batteries.

There is no simple way to avoid it.

One way to get around this is to have a battery switch to allow you to swap batteries over.

 

[Dan_68]

The wire i am measuring is aftrr the diode and i get the higher voltage. If i put a load on after this point it shouldnt change the voltage at that point should it?

 

[(*steve*)]

Yes it will.

You won't see the voltage drop across the diode at very low currents (like the one your meter draws)

 

[Dan_68]

Right thanks for the help then.