12V UPS, How to ?

Question

I want to build a circuit to switch between two different 12V DC sources. One from Mains power and the other from a battery (Just manually). I tried a 12V relay(SPDT).Coil connected to the 12V battery.See schem.diagr. down below Not active relay gives 12V to the equipment. Manually connect 12V battery to the relay coil gives 12V(battery) to the equipment. The change gives a 5ms break of 12V to eq. but can handle this with a big capacitor. But when release of the battery connection the voltage to the coil drops too slow, and the break of power to the eq. therefor is to loong(100ms) (the release (drop out) of the relayconnectors is at far lower voltage then 12V)

Have to speed up the release, but how ? Two relays ? Or switch the relay via a transistor ?

Info or links , is appriciated.

/Ake

. Manually connect `- . `------|---_ | ) Relay coil 12V | | )| --- | _)| Battery - | | | |---o __--o- Out to eq. ----o .-----------. | | | | | -----| Mains | | | |12V '-----------' DC

Genome · Accepted Answer

That sounds like a problem with the relay. Have a look at the specifications for the make and break times in the data sheet. If you haven't got one then look for a relay with a data sheet that gives that information and buy one of those.

DNA

Tim Wescott · Answer

Also look at how you're driving the relay. The way one finds most often is to do circuit A, with a diode across the relay coils to suppress the inductive spike when you turn off the transistor. This is good, but it holds the current in the coil for a long time (remember that the current change is driven by voltage). Circuit B speeds up the relay turn-off by allowing the relay voltage to go higher (as high as you want, depending on the selection of the resistance R). This has the up side that it uses a nice low-tech resistor. Circuit C also speeds up the relay turn off, but it uses a zener. It has the advantage that it doesn't suck extra current the way circuit B does, but you have to select the diode carefully.

+V +V +V --- --- --- | | | .---o \\ .----o \\ | \\ | '-._ o o | '-._ o o '-._ o o - )| .-. )| )| ^ )| Relay | | R )| Relay )| Relay | _)| | | _)| _)| '---o-' '-' .-' .-o------. | | | | | | '----o | z | | | A |/ |/ |/ | -| -| -| | |> |> |> === | | | GND | | | === === === GND GND GND

(A) (B) (C) (created by AACircuit v1.28.6 beta 04/19/05

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Genome · Answer

This is why I specifically did not mention the things you have mentioned.... meaningful though it might be.

I was assuming that someone else would read it and blither on about clamp diodes and reset voltages and other such stuff like what I almost did.

However, being clever I spotted that this bloke/girl is wiggling a wire on the coil of the relay...... so, at the moment, this is not they problem they are experiencing.....

DNA

default · Answer

Is there some reason a steering diode won't work? With your circuit and assuming some basic stuff like the battery is really 12.5 volts for a lead acid type and a regulated 12 mains supply . . . just add a diode between the mains supply and the relay's movable contact.

This should keep power to your application until the relay has time to drop out.

And 100 ms seems awfully long. Make sure it is a DC relay - AC ones use a shading pole to slow the response to keep it from buzzing. And you might try tweaking the spring a tad.

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Jamie · Answer

use 2 diodes, rated to hold the current you expect to use for at least twice the time it take's for the unit to switch.. have each diode by pass the contacts of each source to the center contact of the relay.. this will prevent back flow of current and only drop .6 volts on the average, if you use the cap as you have been, this will most likely smooth it nicely.. so, for each diode, the anode side comes from each source. the cathodes of each diode are coupled together to the CT of the switch. i don't know what your constant load is, but i think something like 6A100 diodes (6 amp 100 volt) SI diodes should do it. since the load will be minimized for a short time.

Ban · Answer

It is because when you interrupt the switch, the relais gets fed back from the input caps of the eq. until they are exhausted. switch only the coil on and leave always 12V on the relais-contact.

default · Answer

Damn - that never occurred to me. He's got some big assed cap sitting out there keeping the relay powered - now I can believe 100 ms drop out.

If the cap is necessary to keep uninterrupted power to the load it will still take a diode or two to keep it isolated.

I don't understand why the relay - why not just a switch and a diode or two?

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Homer J Simpson · Answer

No. Use a diode from each supply to the load. No relay or capacitor. Turn off the supply you don't want.

default · Answer

That was my first thought too.The op is conversant enough to send a schematic in ASCII so I figuredhe must have some reason for wanting to use a relay.----== Posted via  - Unlimited-Unrestricted-Secure Usenet News==---- The #1 Newsgroup Service in the World! 120,000+Newsgroups----= East and West-Coast Server Farms - Total Privacy via Encryption =----

Tim Wescott · Answer

-- snip --"Just manually".  Oh, yea.-- Tim WescottWescott Design ServicesPosting from Google?  See "Applied Control Theory for Embedded Systems" came out in April.See details at

Homer J Simpson · Answer

I believe in KISS. This method meets the supplied specification.

12V UPS, How to ?

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