8 Apr
2016
8 Apr
'16
8:54 a.m.
https://www.fairchildsemi.com/datasheets/LM/LM7805.pdf
http://www.ti.com/lit/ds/symlink/lm78l05.pdf ( page 8 gives you the
internals for a low power version )
Nothing in the data sheet saying you cant apply voltage to the output.
As per the low power version circuit diagram i cant see how you can do
any damage to it
I was not able to find an internal diagram for the higher power version but
i am sure its close to the same circuit just larger pass transistors.
I took a new regulator and measured the output pin to ground with my
trusty ohmmeter
on the diode setting and had no reading. If you think about it the
output on the regulator is a pass transistor
and the output to ground will be in effect a diode that will not allow
current to pass from output to ground.
My last comment still stands. i doubt you will cause any ill effects to
your regulator.
I personally have done this to some old arcade boards with no ill effects.
On 4/7/2016 6:08 PM, drlegendre . wrote:
"...if you leave the unregulated rail
_unattached_ and put +5
switcher straight onto the regulated +5 rail..."
My error, I read that as "attached".
In any event, just lift both the 7805 IN and OUT pins, and then supply
known-solid +5DC between the OUT and GND pads on the board.
No, you can't feed the IN pin with +5V, for as others have mentioned,
the 7805 has a minimum dropout of 2V or so.
On Thu, Apr 7, 2016 at 4:03 PM, William Donzelli <wdonzelli at gmail.com>
wrote:
> Per his description, the 7805's input will be open. It will not try to
> source any current, as it will have none to give.
>
> I suppose there might be a little leakage.
>
> --
> Will
>
> On Thu, Apr 7, 2016 at 4:58 PM, drlegendre . <drlegendre at gmail.com> wrote:
>
--
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