7 Apr
2016
7 Apr
'16
6:08 p.m.
"...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:
Err.. unless the voltage of the switcher is
identical to that of the
7805,
then one device will source current, and the
other will sink it.
Like putting two 6V batteries in parallel, where one is fresh and the
other
weak. Current will flow until the potentials are
equalized. But with two
regulated circuits, I don't see how equality can be achieved.
Not saying it's going to smoke-out, but it does seem like a wonky thing
to
do.
On Thu, Apr 7, 2016 at 3:41 PM, wulfman <wulfman at wulfman.com> wrote:
> You should be just fine.
>
> On 4/7/2016 1:38 PM, Bill Sudbrink wrote:
> > If you have a circuit which is normally designed to
> > operate with an unregulated supply, through a regulator...
> > say unregulated +8 through a 7805 to a regulated +5 and
> > you want to test it independent of the +8 supply, if
> > you leave the unregulated rail unattached and put +5
> > switcher straight onto the regulated +5 rail, will you
> > damage the 7805? Clearly the VIN is open, but the ground
> > pin will still be attached. Would this push voltage
> > back through and screw things up?
> >
> > Thanks,
> > Bill S.
> >
> >
> >
>
>
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