I want to know is a simple diode is enough to protect non rechargeable batteries from getting charged while used as a memory-saving backup.

The use case is: I have a Car stereo head converted to home use by means of simply using a 12v power adapter. I have a simple switch connected to the cable that normally goes to the acc switch.

It works fine, but here we have frequent power outages so I want to add a backup battery to preserve memory values thus preventing my settings (EQ, radio stations, current track) from getting deleted ever so often.

I want to use 8 AA batteries to hold

I have battery holders lying around and diodes salvaged from old equipment, and part of my purpose is to recycle this stuff.

The following is the intended circuit.

Will the diode (1) achieve this purpose?

Diode 2 is supposed to prevent the battery pack from feeding the ACC line, thus preventing the unit from fully turning on from battery power.

enter image description here

Edit: Corrected battery symbol polarity.

Update: The proposed scheme worked fine. The head unit draws about 0.07 A in standby mode and not much above 1A when turned on and at volume 20 out of 40 using one 8 ohm speaker. The diode used in the +12v Constant line (1N5406 3A 600V) got a bit warm to the touch during operation with 2 8Ohm speakers. Other than that, the contraption works as expected: The head unit keeps its settings and the voltage from the power supply does not reach the battery pack. Also, the battery does not feed the ACC line.


1 Answer 1


Yes, it could work for you. Be advised that the +12v Constant input will draw current from whatever voltage source is higher.

I think it is quite likely that you really don't need +12V at the Constant input. I would see if your settings are remembered if you apply, say, only +9V at that input. In that case you could use a lower voltage battery pack and that would ensure that power would always be drawn from the 12V supply when it was connected and turned on.

I would also measure the amount of current that the +12V Constant input draws. It's likely to be very small. The diodes will reduce the voltage reaching the +12V Constant input, but if the current draw is small the voltage drop will also be very small.

(btw, the battery symbol in your schematic is the wrong way around.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.