First, thank you all who replied to me.

The comparators are feeded at the voltage provided as 12V, so if it is
10V, they are feeded at 10V, so I think it could be a voltage drop
bellow 12V.
You suggested that I put the battery first, after the transformer and
the rectifier bridge, if I understood well ? Shouldn't it be some
charging supervisory circuit there ?
It's not very clear to me how the schematic looks if I admit that I
don't need exacty 12V, but it can drop to 10V.
13.8V unrectified may power some sensors ? Isn't this dangerous for them
?

Lucian
-----Original Message-----
From: pic microcontroller discussion list
[mailto:PICLIST@MITVMA.MIT.EDU] On Behalf Of Byron A Jeff
Sent: 23 martie 2004 21:26
To: PICLIST@MITVMA.MIT.EDU
Subject: Re: [EE]: 12V and 5V Power Supply

On Tue, Mar 23, 2004 at 12:57:02PM +0200, Lucian wrote:
> Ok, the source I need is for a security system I'm designing. if it
> works well, it could be produced in high volumes.
> The system is powered from the wall outlet to a 13.8V step-down
> transformer, which feeds the regulators. This would be simple if I
> didn't needed a battery backup. From the sensors point of view, the
12V
> could be between 9V and 16V, but I have also some comparators which
work
> at 12V and if the voltage drops bellow, the result isn't accurate
> anymore.

OK. This helps.

Here's a key question: Why do the comparators work at 12V? As you know
comparators compare voltages. Using resistor dividers, possibly built
into
an opamp divider, it should be possible to take the original 12V
requirement
and cut it to something much less, such as 4V. Once you've done that,
then
you can ditch the 12V requirement, still keeping the unregulated  higher
voltage supply to power the circuit. You then have:

1) 13.8V input supply feeding...
2) A 12V gel cell which serves as the input to...
3) A 5V switcher which then using dividers drives....
4) the comparators.

12V is then rendered unnecessary.

> When the power is off, the battery must backup both the 12V and the
5V.

Again using the comparators at a lesser voltage obviates the need for a
regulated 12V supply.

> I thought of using 2 regulators, and when the power fails, a relay to
> connect the battery to the 12V directly and to the 5V regulator.

Bad idea. It'll glitch when the power fails. The only way for this to
work
effectively is for the battery to be the primary power source while
actively
being charged from the line. I have a sunrise/sunset light controller
that
is battery backed, and when developing I tried both ways. I never got
the
relay method to work without the board resetting. But with the battery
feeding
the regulator, it "switches" smoothly, because it doesn't switch.

> Is this suitable ? I would also need a charging circuit for a
lead-acid
> battery, can you indicate me a good circuit ?

The UC3906 I alluded to earlier. However if you have a PIC onboard,
there's
not necessarily a need for it. Also if your input voltage is only 13.8V
you can only float charge anyway as you need a minimum of 14.4V to
bulk/absorption charge.

BAJ

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