How to make battery backup for circuit?

[Axe_Murderer]

I want to include a battery backup for my alarm clock project. However I can't find any circuits for this, nor can I figure out quite how to do it. It sounds simple... until I try to figure it out.

I intend to use a standard 9v alkaline battery to run the chips and power the alarm circuit (if needed) until mains power returns.

Thanks.

 

[bmcculla]

You should be able to just use a diode to keep power from flowing from the mains into the battery. When the mains supply goes away the diode conducts powering the circuit.

Brent

 

[Noggin]

might be a bit more difficult than that though, he's got a 9v battery, and I'm guessing that his circuit may be done with 5v logic. But of course if he's using a >9v input with some kind of regulator then yeah, the 9v + diode would work.

And past that, if there is an enable line for the display, if he uses another diode from his main input line he can pull a signal off of that to use to turn off the display if main power fails. Leaving the clock and alarm on, and display off. A little button could be used to jumper 9v to the signal to turn the display back on too to see what time it is...

 

[Axe_Murderer]

My clock circuit is running 9v.

I think you guys are over-simplifying it. With just a diode, what's to keep the battery from being drained when not needed? Adding a 9v battery with a diode on it would be just like putting two 9v power sources in parallel. They would both supply power equally. No?

 

[Sebi]

The two-diodes gate work fine, if the mains PSU out 150...200mv higher as batt. When asolutely equal, apply a scottky diode seriel with mains PSU, and a normal Si diode seriel with batt. But my question: what type the display? LED or LCD? If LED, You need disable it when the cct run from batt, as Noggin suggested.

 

[Noggin]

My clock circuit is running 9v.

I think you guys are over-simplifying it. With just a diode, what's to keep the battery from being drained when not needed? Adding a 9v battery with a diode on it would be just like putting two 9v power sources in parallel. They would both supply power equally. No?

Consider the drop on a diode, I'm not familiar with the behavior of all types of diodes, just the normal PN junction. It'll have a drop of .7 volts or so, so ensure that your clock can run off of 8.3 volts. Once you confirm that, you can connect up the 9v w/ diode to the circuit. The battery will only supply current if the main source drops to 8.3v or below. If you follow the drawing I posted, there is a diode on the main supply. It is there to allow for the display enable line. If the battery and the supply are both the same voltage, then yes, with both diodes in place the battery will source some current.

If that is the case though, leave the diode off of the main supply, then only when it fails will the battery supply current.

 

[Noggin]

Or you can do as sebi says and use a schotkey diode with the main for less of a voltage drop

 

[nettron1000]

Using a diode is the most commonly used simple method ive seen but a relay-capacitor combination would work as well. This scheme is used in some emergency lighting pacs and exit lights in public buildings. The relay is normally open when there is regular mains power but with a power failure the relay contacts close and power is supplied from the batteries ( to DC bulbs only of course).

 

[killjoy]

is this circuit fine for back up supply?
**broken link removed**

 

[ericgibbs]

hi,
Trouble is, its 6 years too late.

 

[Mikebits]

As a side note, the OP is doing fine now in prison, after his serial axe murdering spree. Prozac is doing wonders for his demeanor.

 

[killjoy]

would this circuit work?

 

[ericgibbs]

would this circuit work?

hi,

It should work OK, if you allow for a diode forward volatge drop 0.7V and that the 7805 requires at least 2V higher
on the input [7V] in order to give a 5V output.

So thats about 5V+2.7V = 7.7V minimum from the battery as it discharges.

 

[Mikebits]

would this circuit work?

I guess I did not get the secret decoder ring memo. I don't see anything posted other than your text.

 

[killjoy]

the 12v supply current to the regulator, so if you cut off the 12v supply then the battery will supply?or they both supply current to the regulator? can you explain it to me?

 

[Mikebits]

I guess I did not get the secret decoder ring memo. I don't see anything posted other than your text.

Oh, okay, I did not get the read the earlier post memo...

 

[ericgibbs]

the 12v supply current to the regulator, so if you cut off the 12v supply then the battery will supply?or they both supply current to the regulator? can you explain it to me?

While the 12V is connected, the diode from the battery is reversed biassed, so no current flows from the battery.

When the 12V is off, the reverse bias is removed and the battery diode conducts.

Do you follow OK.?

 

[killjoy]

I guess I did not get the secret decoder ring memo. I don't see anything posted other than your text.

ive inserted an image. why cant see it?

 

[ericgibbs]

ive inserted an image. why cant see it?

Dont know. I saw your first image.

Use the 'manage attachments' button further down the screen when you post.

Try that.

 

[killjoy]

While the 12V is connected, the diode from the battery is reversed biassed, so no current flows from the battery.

When the 12V is off, the reverse bias is removed and the battery diode conducts.

Do you follow OK.?

now thats clear to me. ^_^
who cause the diode to reversed biassed?