Electronic power switching?

[gez]

Does anyone know of a simple way to 'switch off' a battery (electronically not mechanically) when a 9V power plug is inserted into its jack (battery to mains switching).  

The only solutions I've come up with so far are pretty convoluted.  Anyone?

[Mike Burgundy]

I believe RG had something on that in his advanced power protection with mosfets article, somewhere at GEO.

[gez]

I believe RG had something on that in his advanced power protection with mosfets article, somewhere at GEO.

I just skimmed through it and couldn't find anything (I'll take a closer look tomorrow).  I just want to kill the battery by electronic switching when a mains power plug is inserted into its jack.  

The only way I've come up with so far is convoluted, it uses a couple of logic gates and a couple of switches.  I was just wondering if there was some chip out there or some simpler method of doing this?

[Nasse]

I must check my archives, I have a faint memory about reading something relevant to this few days ago, it is somewhere in my table. It was automatic power off circuit for battery powered equipment, and used BS250 fet, but the switch was in neg terminal... so maybe not so useful...

[smoguzbenjamin]

You need a DC jack with a normally closed switch on the pin. There's a third lug which is connected to the pin when there's nothing in there, when you plug in a DC jack this third lug is disconnected

[gez]

You need a DC jack with a normally closed switch on the pin.

No I don't, I need a way of doing this ELECTRONICALLY not mechanically.  I don't want to use a break-contact (I'm fed up of mechanical failure...I'd much prefer electronic failure!)  

Nasse, that sounds interesting.  Anything to give me some ideas will be much appreciated.

[Gilles C]

But it can give you an idea on how to do it.


http://www.maxim-ic.com/appnotes.cfm/appnote_number/1136/ln/en

[Nasse]

I did quick clance trough few books and did not find any ready solutions. But there must be simple solutions with easy to finfd components, thank for Gilles we have info what problems there might be.

I´ll check if I can find few circuits like simple digital clock or burglar alarm battery backup schemos.

But of course it must be p mosfet in a positive line, like BS250. It can pass about 100 mA

[gez]

But it can give you an idea on how to do it.


http://www.maxim-ic.com/appnotes.cfm/appnote_number/1136/ln/en

Thanks!  I suppose this could be simplified by using a p-channel mosfet - source to batteries + connection, gate to the junction of the 100k resistor and diode, drain to circuit's  +connection, then do away with the chip.  

I'll do some experimenting.  Thanks again!

[puretube]

how about electro-mechanical: relay ?

[smoguzbenjamin]

gez what do you do to your dc jacks for them to fail?

[gez]

On second thoughts, the drain to batteries + would be better so that there’s protection against accidental reversal of the battery.  Plus it would avoid the channel conducting if the battery is flat (edit: ignore this, I was forgetting it's a p-channel NOT a n-channel).

This set up should work ok without the chip;  switching noise wouldn’t be a problem as it’s unlikely anyone is going to plug in a 9V adaptor mid solo!  I’ll mess around with this in the next week and report back.

Puretube, I'm looking into relays - haven't yet decided which way to go.

Ben, with the jacks I use the body is sometimes damaged if it's over-tightened causing shorts and a lot of wasted time!  I can get some nice heavy duty jacks but they don't have break contacts.  Please, no links to other suppliers for 'better' jacks.  I prefer to order stuff in the UK as it's cheap and fast.  I also like the idea of doing everything with switching circuits, even bypass. (heretic!!!)

[smoguzbenjamin]

Ah. That makes sense. Anyway I don't know any better suppliers So don't worry. I buy everything locally.

[gez]

It works!  I connected the gate to the junction of the 100k and D1 via a 1M resistor, changed the 100k to 10k, did away with the chip and wired up a forward biased diode in series between the source and the + connection of the load.

Maybe I’m being paranoid, but if the mains supply is spot on 9V and the battery is slightly higher (possible if a fresh battery is used), it’s just enough to bring the substrate into conduction, hence the additional diode.

The battery is killed dead when the mains power supply is connected and there’s hardly any voltage drop across the FET’s channel.  There’s also no problem if either  power source is accidentally reversed - full protection is provided!

All in all, a result!  Thank you Gilles - give that man a round of applause everyone!