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Battery backup for dc voltage power supplies...

Hi,

I've got this garage door alarm that makes a hideously loud 5 beep noise whenever it powers up. It runs on a 9v power dc adapter (brick) that plugs into the wall.

What I would like to do is make it so it could tolerate a mains power loss and prevent it shutting down, then back on, and beeping its startup beeps.

I realize i could put the brick on a cheap UPS, but fitting a UPS at its location would be a pain.

My question is, could a 9v nimh rechargeable battery with some diodes be used to provide backup power?

The circuit would have two operating modes:

ac on, battery charging (at a low rate such as 10ma), ac providing power ac off, battery discharghing, battery providing power.

What is the easiest way this could be done?

Thanks,

Alan

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Reply to
Default User
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Yes, two diodes and a bleed charge resistor off the brick will work just fine.

Reply to
TT_Man

Hi,

Thanks, I found this:

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What is the purpose of the diode between the battery + and - ???

Also, is there a formula for the charge rate of the battery? Obviously the difference between the battery voltage and supply voltage is one factor and the bleed charge resistor, is there another factor such as the internal battery resistance? Is there a formula for this?

Thanks,

Alan

Reply to
Default User

Blows the fuse if you connect the battery backwards.

No. As you point out, the charge rate will vary with the battery voltage.

I would use an LM317 to set the current to a few mA more than the device draws, as long as that draw is below C/10 for the NiMh. Run the output through a diode to the NiMh, and power the device from the NiMh. That way, the NiMh will always be charging while power is on, but at a very low rate so it won't be damaged.

Ed

Reply to
ehsjr

That diode is for reverse battery connection protection- it blows the fuse! Forget battery internal resistance. Just use V supply minus Vbatt/ I Charge to give the bleed res. value.

Reply to
TT_Man

"Default User" skrev i en meddelelse news:485923e9$0$20568$ snipped-for-privacy@news.astraweb.com...

....

Glue in the "beeper"!

Reply to
Frithiof Andreas Jensen

Hi,

I guess I should rephrase the question, why connect the + and - terminals of the battery together at all via the diode/wire? Why would we want to allow current to go in reverse to the battery? Even without this connection wouldn't the current charge the battery by going into the positive and then back around through the negative (let's not talk about edison and the real direction of electrons!)? I don't see the reason for the diode and connection of the two battery poles...

Thanks,

Alan

Reply to
Default User

It's been answered twice, but perhaps not in enough detail for you.

The diode will allow current flow ONLY if you connect the battery _backwards_. That current will blow the fuse, thereby disconnecting the battery. That prevents the equipment and the power supply from being exposed to reverse polarity.

Ed

Reply to
ehsjr

Hi Ed,

You are right, I wasn't getting that its sole purpose was for reverse polarity protection.

So, am I right in thinking that if I did not desire reverse battery polarity protection, I could eliminate it?

Thank you,

Alan

Reply to
Default User

Yes, you could eliminate it with no deleterious effect, except that you would have no reverse polarity protection. I would recommend that you keep the diode in there.

Ed

Reply to
ehsjr

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