Using diode on battery line

[kiteman]

My Carvin came with active controls so my guitar have a battery compartment. I rewired the guitar with Blackouts and now Carvin M22 passives.

Now I want to put the Blackouts back in and I'm thinking about the battery compartment and the possibility of putting in the battery backward. Too easy to do that so I remembered reading about putting a diode in line on the positive side.

So do they cause the voltage drop? Will two in parallel reduce the voltage drop? What diode for the 9V battery?

I know something about electronics but not enough. :help

 

[jim p]

Use a Schottky diode

If you add a Schottky diode in series with the positive lead to the battery it will drop approx 0.45 volts with a few milliamps of current draw. The preamp for the active pickup is probably a FET so it will only draw a few milliamps of current. As a ball park idea of the current draw a 9 volt battery is good for approx 650mA/hours so if you can recall often you needed to change the battery it will give you an idea of the current draw. In the circuit the banded side of the diode the cathode will connect to the preamp while the anode will connect to the positive terminal of the battery. Having diodes in parallel will not reduce the voltage drop so you just need one diode. The Bat-43 and BAT-42 look like good choices for what you want.

 

[kiteman]

Thanks mate. Looks like I'm going for it.