Wall wart through a bridge rectifier?
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I was asking on usenet about combining wall wart power with a 9V battery,
using diodes to protect the battery and detect when the wart was supplying
power (can't use the jack's auto-battery ground disconnect feature.
I figured I could run the wall wart + through a diode (1N4148) to the 7805.
The battery + would go through another diode to the 7805 as well. When the
wart is plugged in, the battery's diode would be reverse biased and shut off
the battery.
I could then run a trace from between the wall wart and its diode to the MCU
and detect when power is coming from the wall wart (for power saving
features when not present). Still have to figure out how to protect the MCU
from the wart voltages.
Anyway, a suggestion came back to use a bridge rectifier for the wart
instead of a single diode. The idea being that then it wouldn't matter what
polarity the jack is or even if it was AC - everything would work.
Anyone know of any gotchas with this or is it a good idea?
using diodes to protect the battery and detect when the wart was supplying
power (can't use the jack's auto-battery ground disconnect feature.
I figured I could run the wall wart + through a diode (1N4148) to the 7805.
The battery + would go through another diode to the 7805 as well. When the
wart is plugged in, the battery's diode would be reverse biased and shut off
the battery.
I could then run a trace from between the wall wart and its diode to the MCU
and detect when power is coming from the wall wart (for power saving
features when not present). Still have to figure out how to protect the MCU
from the wart voltages.
Anyway, a suggestion came back to use a bridge rectifier for the wart
instead of a single diode. The idea being that then it wouldn't matter what
polarity the jack is or even if it was AC - everything would work.
Anyone know of any gotchas with this or is it a good idea?
Comments
dev. kit uses on its input. It seems to work really well.
- Robert
Original Message
From: "pm" <pmeloy@s...>
To: <basicstamps@yahoogroups.com>
Sent: Wednesday, July 16, 2003 9:48 AM
Subject: [noparse][[/noparse]basicstamps] Wall wart through a bridge rectifier?
> I was asking on usenet about combining wall wart power with a 9V battery,
> using diodes to protect the battery and detect when the wart was supplying
> power (can't use the jack's auto-battery ground disconnect feature.
>
> I figured I could run the wall wart + through a diode (1N4148) to the
7805.
> The battery + would go through another diode to the 7805 as well. When the
> wart is plugged in, the battery's diode would be reverse biased and shut
off
> the battery.
>
> I could then run a trace from between the wall wart and its diode to the
MCU
> and detect when power is coming from the wall wart (for power saving
> features when not present). Still have to figure out how to protect the
MCU
> from the wart voltages.
>
> Anyway, a suggestion came back to use a bridge rectifier for the wart
> instead of a single diode. The idea being that then it wouldn't matter
what
> polarity the jack is or even if it was AC - everything would work.
>
> Anyone know of any gotchas with this or is it a good idea?
>
>
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(120 HZ). The 'dual diode' approach would prevent the 9V
from providing power ONLY when the wall-wart output was
above 9.4 volts. This works fine for a DC wall-wart. The
ripple from an AC wall-wart would let the 9V conduct many
times a second.
--- In basicstamps@yahoogroups.com, pm <pmeloy@s...> wrote:
> I was asking on usenet about combining wall wart power with a 9V
battery,
> using diodes to protect the battery and detect when the wart was
supplying
> power (can't use the jack's auto-battery ground disconnect feature.
>
> I figured I could run the wall wart + through a diode (1N4148) to
the 7805.
> The battery + would go through another diode to the 7805 as well.
When the
> wart is plugged in, the battery's diode would be reverse biased and
shut off
> the battery.
>
> I could then run a trace from between the wall wart and its diode
to the MCU
> and detect when power is coming from the wall wart (for power saving
> features when not present). Still have to figure out how to protect
the MCU
> from the wart voltages.
>
> Anyway, a suggestion came back to use a bridge rectifier for the
wart
> instead of a single diode. The idea being that then it wouldn't
matter what
> polarity the jack is or even if it was AC - everything would work.
>
> Anyone know of any gotchas with this or is it a good idea?
>I figured I could run the wall wart + through a diode (1N4148) to the 7805.
Use a bigger diode - any of the 1n400x family is good. The 1n4148 is only
good for about 75 mA and will not withstand large surge currents (such as
when the capacitors charge up).
We generally use 1n4005 but any of that family will work.
>The battery + would go through another diode to the 7805 as well. When the
>wart is plugged in, the battery's diode would be reverse biased and shut off
>the battery.
>
>I could then run a trace from between the wall wart and its diode to the MCU
>and detect when power is coming from the wall wart (for power saving
>features when not present). Still have to figure out how to protect the MCU
>from the wart voltages.
Use a resistive voltage divider. The resistor chain should be low enough
value (total resistance) to ensure that the input voltage collapses all the
way to ground. Figure on 10K total.
The actual resistors in the divider depend upon the wall wart voltage which
you have not specified. You want the voltage to the stamp pin to be well
above the 1.4V threshold but below the 5V supply. For example, with a 12V
nominal input voltage, 7k5 series resistor and 2k4 shunt resistor will work
well (about 4:1 divider ratio). Input reads a logic 0 below about 6V and
will withstand 20V before foward biasing the pin ESD protect diodes on the
stamp.
>Anyway, a suggestion came back to use a bridge rectifier for the wart
>instead of a single diode. The idea being that then it wouldn't matter what
>polarity the jack is or even if it was AC - everything would work.
Good idea. Only 'gotcha' is the extra diode drop which is not usually a
problem. You will also need 2 more diodes for your wall wart detect
circuit (feeding the resistive divider) - 1n4148 is OK here since the
currents are low. Or the bridge could feed the 7805 via another diode
(wall wart sample point is junction of bridge and output diode) - but now
your total drop is 3 diode drops (~2.1V).
dwayne
--
Dwayne Reid <dwayner@p...>
Trinity Electronics Systems Ltd Edmonton, AB, CANADA
(780) 489-3199 voice (780) 487-6397 fax
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volts per diode. 1N5819 is a 1A Schotkey diode and a 1N5822 is a 3A
Schotkey diode. There leakage current is about 500 uA so keep that in
mind.
Jason
"but now your total drop is 3 diode drops (~2.1V)."