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dual battery basic question

Dan_68

May 1, 2011
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May 1, 2011
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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*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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25,510
If the "isolator" is a diode then you would expect a 0.6V difference between the batteries.
 

Dan_68

May 1, 2011
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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*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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Yeah, sounds like there's a diode in it then.

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

Dan_68

May 1, 2011
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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*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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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

May 1, 2011
36
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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*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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Yes it will.

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