On Thu, 22 Oct 2020 08:32:25 -0700 (PDT), Tim R
wrote:

Two thoughts, from a mechanical engineer.

1. Cheap solution: write DOORBELL WORKS! on a 3x5 card and tape it to the door.

Worth a try ... except the batteries do go flat and so the bell
doesn't work .. ;-(

I was going to include an LED in the bell push that lights or flashes
when they press the button (discharging cap) but again, that's
assuming they look for such (and most deliveries are in the day and
this would be South facing).

2. Also cheap solution: if you're like 90% of people you have a smoke alarm that doesn't work because you've changed the battery. Take out the little piezo speaker and put it on the other side of the door.

That was the sort of idea of my plan. ;-)

Wire it to the buzzer. Include the 9 volt battery as needed. (not sure exactly how your doorbell is powered.

6V in series with the solenoid and bell push.

I apologize in advance for violating the Terms of Use of this forum, which include to provide the most complicated DIY electronic design project possible for any problem that might have a simpler solution.

;-)

The thing is, I'm not sure a simple solution works in this situation.
What I was hoping for is a 'subtle' (electronically) solution that
does what I need.

So, under normal circumstances there will be 6V DC across the open
bell push.

I have a cct in parallel with that that charges a cap via a diode and
resistor I have a small 'remote' PSU that wouldn't be affected by the
bell push shorting it out (because of the diode).

If something could latch and discharge the cap though a buzzer and
release once it has, everything else the doorbell will 1) work as
designed and 2) the 'repeater' should work without much interference
to the existing solution, albeit other than a slightly increased load
on the battery.

If I could find a buzzer that works from 1.2V, I could put a couple of
fwd biased diodes across it and went back to alkaline cells, that
should work and be as simple as it can get?

Cheers T i m