AC Adaptor in Parallel With Battery?

You could have done that in less time than it took to write that post. So what do you think the answer is?

"Paul"

Not only will the output circuit of the adaptor draw current from the 12 volt battery - but it may in fact place a dead short across it !!!

This switching PSU is stated to have over-voltage protection that shuts down the output ( in case of internal failure) - so very likely it has a SCR " crow-bar "

circuit. This will trigger at some voltage a little above the nominal 12 volt level and hard short the output PLUS any connected battery supply.

Use a series diode after the adaptor, for god's sake.

...... Phil

daptor we will use:

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the impedance of the AC adaptor (when it's

e adaptor looks anything like the following:

at the current draw going into the adaptor

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ust solder in a series diode in-line with the

Don't ask advice from a numbskull...

=EF=BF=BD!!!

How "in fact" do you know this?

wn

very likely it has a SCR

INCORRECT. This is just a wall AC adaptor. It's just a linear regulator.

And the crow-bar would limit the current FROM the unit, not limit current INTO the adaptor.

e the nominal 12

What the hell are you talking about??

The over-voltage circuit is separate from the over-current protection, and a short-circuit protection will do anything BUT short out the output!

Why would you have a SHORT-circuit protection SHORT out the output???!!?

I probably will, but certainly NOT from anything you have typed!

What's ONLY EITHER mean? I thought I knew but then you mentioned a diode, which wouldn't be necessary if only either but not both would be simultaneously hooked up.

How about a power input jack that switches one off (out of circuit) when the other is plugged in? We don't know enough about the device, usage, and power consumption. Depending on these another option might be a relay flips when AC-DC adapter is present.

As Phil not so gently put it, that is a switching PSU. May or may not hav ea crowbard, some of this small size are fairly crude and yet still probably sufficient, unless you were running an analog amp.

You'd have to pop 'er open and reverse engineer a bit to know this, so while some would turn their nose up at a series diode because of the fraction of a volt power loss (from an efficiency standpoint), it is a cheap, quick, reasonable solution.

It might be a bit more elaborate, different than that, but you shouldn't leave the adaptor and battery in parallel.

Another option that hasn't been covered would be if the adapter were behind the battery as a charger with a peak voltage under the maximum tolerable by the powered device, but without an expressed need for this, the series diode would be the more reasonable solution, or the relay.

I suppose it depends on what you consider a hassle, a relay isn't all that complex but there's something to be said for altering only the cheap little AC-DC supply internally with a diode.

But even if only the battery is hooked up, the AC adaptor may still have a not-so-high impedance looking into it. In which case, a reverse-biased diode will certainly help things.

Yes, but we'd like to avoid this if possible.

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dance of the AC adaptor (when it's

Yeah, i'm also gonna use an external power supply, and measure the current going into the adaptor when it's unplugged, and see what the current is.

That should tell me if I'll need the series diode or not.

looks anything like the following:

rrent draw going into the adaptor

r in a series diode in-line with the

Yeah, but a relay would draw current, which i can't do if i want the batteries to last. And even just a simple manual switch (SPDT) would be a bit bulky, which is what i'm trying to avoid.

Thanks for your input.

High impedance or not, it's generally a bad design to have unnecessary parts putting drain on the battery, even moreso when a cheap, easy diode can switch the situation around making the far more abundant AC-DC supply suffer a minor loss instead of the battery.

I don't know this battery you're using, but personally I would make the first attempt at this current-limited with inline resistor or rheostat. Probably unnecessary, but if one has a resistor lying around...

You're looking at the opposite relay function I mean, DPDT relay which switches circuit to AC-DC supply when energized by the AC-DC supply, or closes battery circuit when unpowered by AC-DC supply.

How small does this really need to be? Given a 1A AC-DC supply we already know the absolute max your project could consume, a 1A switch only need be big enough that the human flipping it can manipulate the lever to use it.

IMO, which is the better solution depends more on the user and scenario. Maybe there's no point in spending the time, if the cheap easy diode will suffice, you're practically done.

This discussion has degenerated into a farce.

If, you, Paul, knew enough about electricity to design something that MIGHT work you would know enough to not ask this dumb question about connecting the battey and an outside supply.

A simple ordinary external supply connector as used on many, many devices has a mechanical switch to switch from the internal (battery) supply to the outside supply whenever it is connected or a properly placed diode would allow the battery to do its job whenever there is no external input.

John G.

Or respond to questions from one.... Your thick, armor-plated skull is resistant to not only the physical beatings you must endure every day in real life, but to all forms of sense people here are trying to tell you. You are an ignorant f****it. You are of the dangerous class of people who are convinced they know everything and therefore don't need to learn anything. The only good thing here is that you are playing with mains voltage, one can only hope that you die a quick, crispy death.

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Gee, thanks for repeating what i already typed, moron!

You are misunderstanding me. If the AC adaptor is not hooked up, the impedance looking into it may not be very high, in which case a reverse biased series diode would help out.

That would be a good idea, to current limit with a series resistor, if i expected a complete short, but i don't think it's THAT low an impedance.

Oh, ok, then you meant normally closed when the battery is hooked up, and open when plugged into the wall, when i don't care about energizing a relay from the wall.

I'd still rather not do this, but point taken.

Yeah, i hope to not need even the diode.

Again, thanks for your input kind sir.

Paul

...... Phil

I wouldn't do that, it is much safer, and ultimately cheaper, to use an autonomous electronic mux, where the line power overrides battery power. There's not enough adapter voltage to charge the battery, but including a boost charger in it would not be inappropriate. This assumes you work with electronics: View in a fixed-width font such as Courier.

. . P . AV ====+=======D S==========================. . (+) | G || .(adapter) | __ | || . +--| \\ | || . | | o--+ =====(+) . +--|__/ | || . | | || . | | P P || . BV =====================D S=+=S D=========' . (+) | | G | G .(battery) | | | | | . | | +---|---' . | | __ | | . | '--| \\ | | . [100K] | o-' | . | .--|__/ | . | | | . | '------------' . | . GND=====+============================================GND . . . 1N914 . AV--|>|--+--------. . | | . | | . BV--|>|--+ |VDD . 1N914 | --------- . |0.1U | | tie unused . === | 4011 | inputs to GND . | --------- . | | . '---------------+ . | . -+- . GND .

You speak for the group, you sorry-assed, worthless, useless, sniping troll, ignoramus? Talk about delusional...

You don't know anything about electronics, and you're too stupid to learn, so why do you post to SED?

Hi Fred,

I'm trying to wrap my head around that topology, but I need a nudge to figure out something.

When the 'A' voltage is removed, the top NAND goes high (1), and sends this to the top leg of the bottom NAND.

What I can't see right now, is how the bottom leg of the bottom NAND gets a high (1) signal, so that it will then go to a low (0) output and turn on the two FETs to conduct the 'B' voltage towards the desired output.

I know I must be missing something really obvious here?

Thanks for any insight,

Jon

It gets that from the forward biased substrate diode, D-S, of the left P-channel. Turns out it's a good way to protect against a reverse battery connection...

He likes to master-bait with Jim. >:->

Cheers! Rich

The battery has built-in diodes to prevent a re-charging current.

Thanks, Fred. Looks like I need to learn a bit more about FETs.

Thanks,

Jon