what happens if you provide 78xx less input voltage than xx?

[fatecasino]

I am building battery powered guitar amp and I am using a 7812 voltage regulator (in the case I power it up by the plug)
When the battery voltage goes less than 12V what actually happens? Does the regulator simply works as a bypass? or like switch?

[TejfolvonDanone]

My knowledge suggest that it simply won't give enough voltage. But i'm not really sure so we have to wait for an "expert".

[Mattnezz]

No "expert", but the voltageregulator seems to track the battery voltage minus a dropout voltage of about 2V.
So if your battery is at 12V, a 7812 will probably output something like 10 volts.

Maybe bypass the regulator when using the amp with battery?

[fatecasino]

if the battery goes lower? let's say 11V, will the output of the regulator be 9V? 11V? other?

[MrStab]

i use a resistor and zener diode for basic over-voltage protection on charge pump circuits, where voltage regulation itself isn't the goal. voltages under the zener's rating are only affected by the series resistor, and there's no big drop. if you expect to be using a higher input voltage for longer than an "oh, s***!" moment, though, maybe a regulator is better.

maybe you could do something clever using the switched battery lug on a DC barrel jack? like Batt+ could be sent AFTER the 7812, but Adaptor+ could go BEFORE, and cut the connection to After when a plug is inserted? you might need a diode to prevent Batt+ flowing back into the output pin of the 7812. or maybe use transistors somehow.

just an idea.

[R.G.]

Mattnezz is right. The output follows the input down, minus a minimum dropout voltage. The DOV is about 2V for 78xx devices, may be quite small for LDO devices. The minimum dropout voltage is the voltage the innards of the regulator need to work.

If you use a regulator, use diodes to protect the regulator against a suddenly shorted INPUT voltage, which will kill the regulator chip.

[antonis]

Because of R.G.'s mood between voluble and terse is unpredictable... 

What tells you is that if there is a significant voltage difference between OUT and IN (reverse polarity) the regulator is prone to damage because of the internal discharge of OUT capacitor..
(same case as if the OUT is connected to a power source with voltage greater than the unregulated..)

An easy way to overcome this case is to connect a diode facing from OUT to IN and bypass any possible voltage difference..
(exactly the same case as for n-p-n Vbe protection for Emitter's voltage greater than Base's one..)

This trick will actually not work for voltage difference lower than diode's Fvd, but the regulator (or the BJT) doesn't mind a lot for such a small reverse voltage difference..

[MrStab]

An easy way to overcome this case is to connect a diode facing from OUT to IN

it's been more than a few times i've read about diode protection being mentioned as an afterthought, eg. in debugging threads after a failure. same with MOSFETs. just something that surprises me.

[antonis]

it's been more than a few times i've read about diode protection being mentioned as an afterthought, eg. in debugging threads after a failure. same with MOSFETs. just something that surprises me.

It depends on the "point of view"..

An unexperienced person should only take in mind the normal (ideal) conditions for which is called to design something..
(mainly because of the difficulty to design  something that is not a "rule of thumb" or a "piece of cake" for himself..)

[MrStab]

i guess i can appreciate the importance of getting a raw concept into someone's head, so long as the external components/caveats are emphasised when a beginner tries to build the thing.