On Tuesday 10 June 2003 09:26, Claudio Martin scribbled:
> Hi everybody,
>                     I have a circuit using a 16f874 , with a total powe=
r
> consumption of 500 mA, the circuit have some 7805 voltage regulators in
> parallel that reduce the voltage from 12 V to 5 V. I'd like to add a ba=
ckup
> rechargable battery circuit, that gives me 6 hours of backup when the m=
ain
> power goes off. If somebody knows about some reliable and cheap solutio=
n
> i'd appreciate it. Thanks in advance.


Not sure why this thread went into DC-DC converters, but it sounds to me =
like=20
you already have the circuit working with the 7805, and just need to add=20
battery-backup to this.  Also, if you state the you need a "reliable" and=
=20
"cheap" solution, I'll definitely skip any DC-DC converter mods, as they'=
re=20
relatively expensive (compared to the 7805), and I can't see them being a=
ny=20
more than a trusty old 7805.  That said, let's assume you have this...

           .------.
+12V >-----| 7805 |-----> +5V
           '------'
               |
               |
              GND


Assuming also that you don't need *exactly* 5V, I would assume a few diod=
es in=20
the right places would be able to serve as a battery backup.  You said=20
"rechargeable", but no reference to this circuit actually doing the charg=
ing. =20
First convert the circuit as follows...

           ,------,   D1
+12V >-----| 7805 |---|>|--> +5V output
           '------'
               |
       ,--|>|--'   =20
       |   D2
      GND

D1 keeps the current flowing in one direction, but due to the ~0.7V drop,=
=20
you'd see ~4.3V out.  Instead, using D2 will bring the voltage back up to=
 5V,=20
if they have the same Vf during operation.

Now, add a battery backup in parallel as follows...

           ,------,   D1
+12V >-----| 7805 |---|>|--o---> +5V output
           '------'        |
               |           |
       ,--|>|--'           |
       |   D2              |
      GND                  |
                           |
                    D3     |
Vbat >--------------|>|----'

Assuming Vbat is +4.8V (made up of 4 x 1.2V nicads or similar), and assum=
ing=20
that D3 is a schottky (I'll assume a BAT46 which has about .3V drop IIRC)=
,=20
during normal operation, you'll have +5V at the output, and being higher =
than=20
Vbat, no current will flow back into the battery due to D3.  If the +12V=20
input fails, then there will be 0V from the output of the 7805, and now=20
current will flow from Vbat to the output.  But 4.8V - 0.3V =3D +4.5V at =
the=20
output.

For Vbat, you can parallel/series a bunch of nicads or similar to get the=
 3AH=20
you need, or go with gel-cell as someone else mentioned, but the numbers=20
above will change based on the battery voltage.

For D3, check to make sure the schottky is rated to handle the current/po=
wer=20
you need for the circuit.  Personally, since this is a PIC circuit, I'd t=
ie=20
the 7805 output (at the point before D1) into an A/D or comparator input =
on=20
the PIC to detect power failure, and reduce the operation of the circuit =
to=20
bare minimum, just to preserve battery life.

If you need the circuit to actually charge the batteries as well, you nee=
d to=20
look at the charging characteristics of the actually batteries you use, a=
nd=20
perhaps look at app-specific IC's that handle this function.

On a side note, you mentioned that you have "parallel" 7805's .... ???  I=
f I=20
understand that correctly, then why?  You can get single 7805's that deli=
ver=20
at least the 500mA you require.  And I'm still not sure it's a good idea =
to=20
parallel 7805's.


Cheers,
-Neil.

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