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PICList Thread
'[OT] Pic Power supply'
1998\02\03@161415 by Charles Laforge

picon face

I am looking to power my device from a wall adapter but would like to
switch over to backup batteries in the event of a power failure or if I
decide to make the device portable for a few days.  I want this thing to
switch over automatically.

I looked at the devices by Maxim but they only seem to switch the
battery for the purpose of powering memory.  Am I missing something?  I
want to still be able to have a functioning device running on batteries.
If anyone knows of a solution for me (preferably 1 chip) I would love to
know about it.  Thanks for your help.


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1998\02\03@162657 by myke predko

Hi Charles,

No can do with the one chip solution...

How about two diodes?

 > +8 V from    Diode
 Wall Wart ------->----+--------------------- }
                       |                      }
                       ^ Diode                }
                       |                      }  To "Typical"
                       _  +                   }  78L05 Regulator
                      ___                     }
                       _   9 Volt             }
                      ___  Battery            }
                       |                      }
  Gnd -----------------+----------------------}

If the power coming from the wall wart is < 8.3 Volts (with the Voltage Drop
over the diode), juice from the battery will be used (and with the diode
between the wall wart and the power supply, you won't back drive the wall
wart if you pull the plug out of the wall, instead of the plug out of the

I've used this circuit in a real time clock I once developed and it worked
quite well, although for an application that you want to have run for a
"couple of days", you better make sure you're putting the PIC to sleep when
it's not in use (and use a 78L05) or don't run it at anything faster than 32

The first diode (at the 9 Volt Battery) cannot be eliminated and the circuit
used to charge the 9 Volt Battery (I don't even think that Ni-Cads could be
used in this case).

Let's see what everybody else has to say about this,

{Quote hidden}

Opus:  There's a 465 pound woman pruning her azelias while wearing a pink
stretch bodysuit.

911 Operator:  So what's the emergency?

Opus:  From a taste perspective, it's a crisis of biblical proportions!

1998\02\03@170030 by John Bellini

What about a small transfer relay like they use to transfer power
from the shore to boats. They are usually in Inverters and are quite
large, but I don't see why you couldn't do the same thing with
a small signal relay.


> {Original Message removed}

1998\02\03@184826 by Morgan Olsson

picon face
{Quote hidden}

Oops! Let«s rewiew the voltages:
1)      A new 9V battery emits slightly over 10V
2)      There is a diode in serier with both Wall side and battery side;
assuming the two diodes voltage drop are equal the breakopoint is when wall
anv battery voltages are *equal*.
3)      There is also higher voltage drop across working diode than the resting

So, to be sure not to draw current from battery, wall side supply shuold
deliver >10,5V.

PS Diodes also protect against backward polarisation. DS

PS2 If battery is rechargeable, connect a resistor in perallel with the
battery«s diode. DS2

Ten years or so ago I used the rechargeable Lithium cell from Panasonic
(first one available). One cell deliver 3V, so I connected it directly to
supply as main supply. The charging technology is constant voltage with
current limiting. Some protection against deep discharge is recommended.

This battery is very nice since it has high energy concentration, is small,
50 years (!) self-discharge, and much less poison than NiCd cells. And much
longer life.

Morgan Olsson, MORGANS REGLERTEKNIK, Sweden, ph: +46 (0)414 70741; fax 70331

1998\02\18@105426 by Larry G. Nelson Sr.

Why not simply use steering diodes. If the dc voltage from the wall wort is
higher than the battery voltage then the diode blocks the current from
flowing out the battery and you run off the power supply. As soon as the
external voltage is less than the battery the current is supplied from the
batteries. No chip needed, simple, reliable, and cheap. All you need are
two diodes that can handle the maximum current.

At 01:12 PM 2/3/98 PST, you wrote:
{Quote hidden}

Larry G. Nelson Sr.

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