Newbie Question Help - Power Supply Switch

I am developing a circuit and want to do something that I think is relatively simple. I am very much a novice at designing my own circuits. I have taken pieces from other circuits and adapted them to my own use. On my current project, I want to have an automatic power supply switch.

Meaning, under normal conditions I would run off of a 9V battery throuh a power regulator to +5V to drive my circuit. However, in cases where my battery diesor I want to preserve it, I want to be able to run off of a +5v Power adapter.

What do I do circuit wise to provide an automatic switch that senses when the adapter is plugged in and does not draw power from the battery.

I have doen several searches, and probably do not have the right terminology for this to provide an accurate result.

Many thanks appreciated!!

A simple 3 terminal 2.5mm DC jack will do what you are looking for. These are normally set up so that when no plug is inserted, the Negative connection would be provided from your battery. When you plug your adaptor into the jack, it opens the Negative connection between your battery and circuit, thereby isolating the battery, and connects the Negative pole from your adaptor to your circuit.

Just do a Google search for 'DC jack' and I think you will find what you are looking for.

HTH -- Mike

Thanks!!

But one follow up question to that solution. Will it actually work in the scenario I am employing. Meaning that I will have a +5V adapter and use a 9V battery through a 7805 power regulator. Meaning is it safe to put the 5V current from the adapter through the 7805 power regulator circuit, or should that be a separate path?

Well, since your 7805 requires about 2 VDC of overhead in order to provide any regulation, I would suggest that you use a 7.5 to 9 VDC adaptor. If you need a wiring diagram or anything, send me a private message and I'll email something to you.

Mike

I gave a little more thought to your question last night and, if you want to use your 5VDC adaptor, there is an easy way still using a DC jack. Excuse the scribbled drawing but here's the way to do it. The DC jack has a breaker switch built in between pins 1 & 2 on the drawing. With no plug inserted into the jack, pins 1 & 2 are shorted. When a plug is inserted, it opens the connection between these pins and the barrel of the plug (usually the negative pole) will be in contact with one of these two pins. Be sure and check which pin is in contact with the barrel when you insert the plug into the jack since not all jacks are constructed the same - depends on type and manufacturer. Just make sure you get one with an internal switch.

If you want to use a 9VDC adaptor, just move the positive connection from the DC jack from the 5V output of the 7805 on your circuit to the same point that the positive pole of the 9V battery connected.

In effect what you are doing is connecting one negative lead or the other, depending on which power source you are using, to ground. With the adaptor plugged in, it removes the 9V battery completely from the circuit.

So, hope that helps you with your project.

Mike

WOW!! Thanks a bunch!

I'll be making some posts in the future about my project this is for, which is a MIDI Note trigger pedal. I'll use it to trigger samples based on my laptop via MIDI.