L7809 - can it get damaged if...?

[ribolovec2]

Hello, I'm designing a device which is supposed to run on 9V DC supply. My idea is to make it run off a 9V battery, and have a dc input jack for an external 12V 'wall wart' supply. A L7809 regulator will lower that 12V to 9V. At first, I was going to put the regulator directly on the main board with the rest of the device's components (and the regulator's output should be directly connected to the 'supply' rail of the device; a switch on the DC jack should switch the battery on and off) but then i realized that when the device was operating off the battery, then supply voltage (9V) would be present at the regulator's input. My question is: Will that damage the regulator, and should I think of another method of switching?

[sofaspud]

It is assumed you meant voltage present at the regulator's output.
With the battery connected using a switched DC jack, here's two options to modify the 7809 circuit for a wallwart supply. The first maintains the regulated 9V; the second uses a single Schottky diode, though a 1N4001 would probably work as well (with a bit lower Vout).

[ribolovec2]

Yes, I meant 'voltage present at the output', sorry. And thank you for the ideas. As far as i've understood it, the arrangement with the 2 diodes keeps the regulated voltage (in my case 9V), since the diode on the common pin lifts the output *forward voltage of the diode* volts above the original value, but then the other diode causes a drop of exactly the same voltage, which effectively cancels out the lift by the first diode? And the other arrangement with the one diode produces an output which is *forward voltage of the diode* volts below the regulator's nominal output? Have i got it right?

[sofaspud]

Yes, that is exactly correct. My understanding though is that since you're also running from a 9V battery, your circuit doesn't require a specific tightly regulated supply voltage. So I offered some choices you could pick from with convenience in mind.

[kikikaka]

Just be careful: if both the battery and the 'wall wart' are connected, the battery will be 'charged', and can be damaged....

To make it safer I'd suggest to insert a schottky diode in the (+) line of the battery.

[ribolovec2]

Well, the idea is that the switch on the DC supply jack turns the battery off when the wall wart is plugged in.

[vixr]

yeah, those DC jacks usually allow the battery ground to open when the plug is inserted...

[KatieandDad]

As a picture paints a thousand words.

Here is the data sheet for a typical power socket.

Pin 1 is the centre pin of the Power Socket and it is normally connected to BOTH the +ve of the battery and the +ve of the circuit.

Pin 2 is the SWITCHED CONTACT. When there is no plug inserted pins 2 and 3 are connected together.

If the battery is connected to the switch contact, it will be connected to pin 3 when there is no plug inserted.

When the plug is inserted the battery (pin 2) is disconnected and the power jack body is connected to the -ve of the circuit.

Thus the battery is disconnected when the power plug is plugged into the socket.

[kikikaka]

Originally Posted by ribolovec2

Well, the idea is that the switch on the DC supply jack turns the battery off when the wall wart is plugged in.

I did not read your initial post carefully...
Usually the simple solution is the better. Like in this case