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 . <[email protected]> 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 <[email protected]> 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. >> > >> > >> > >> >> >> -- >> The contents of this e-mail and any attachments are intended solely for >> the use of the named >> addressee(s) and may contain confidential and/or privileged information. >> Any unauthorized use, >> copying, disclosure, or distribution of the contents of this e-mail is >> strictly prohibited by >> the sender and may be unlawful. If you are not the intended recipient, >> please notify the sender >> immediately and delete this e-mail. >> >>
