On 22/10/2020 21:37, T i m wrote:
Hi all,

I appreciate this is a design rather than repair question but I
appreciate there are some very skilled people here. ;-)

Scenario. I have a wired bellpush on the front door (frame) that runs
to a battery 'Ding-dong' type solenoid chime further back in the
house, the idea being that 'we' can hear it anywhere in the house (we
can) and also in the back garden if we are lucky (we also sometimes
can).

The 'problem' is that you can't generally hear it outside the DG front
door and so couriers often ring the bell *and* repeatedly bang on the
glass door or flap the letterbox flap, sometimes winding the dog up
thinking it's an aggressor. ;-)

So, I was just wondering ... if I had some sort of small
electro-mechanical buzzer (possibly to also give physical feedback),
just inside the front door (on the inside door frame, directly behind
the bell push for best feedback and easy wiring) that you could hear
from the outside, that might give anyone on the outside some level of
positive feedback that the bell had worked (hopefully anyway)?

So, I am limited to 6V DC via the 4 x C cells (currently Ni-Mh's so
only 4.8V) and don't want to add any real parasitic drain, especially
if I go back to alkaline cells to get the full 6V.

So I was thinking (but am no electronics design engineer) of some form
of diode fed - series resistance cap that could be charged by the 6V
supply seen at the back of the bell push that could then be discharged
via a low voltage buzzer when the button is pressed (even if only once
and for ~1s) ... and would then reset ready for another press say in
30 seconds time (given the original doorbell will continue to work as
normal etc).

Things I believe are relevant ... you would need a cap big enough to
provide as good a quality level of power long enough to give a
significant 'buzz' to be heard.

The charging resistance would need to be low enough to not cause the
bell to hang on the 'Ding', once the button was released (the chime
solenoid just returns under spring power when no current flows but
might be 'held' by a much lower current (so no proper 'dong').

I didn't know if the buzzer trigger could / should be done by a self
latching circuit of some sort or if one should use a zener to manage
the voltage allowing a better release action?

Do you even need any trigger/latching circuit? Often doorbells only make
a sound when the button is pressed, so the visitor should not be
bothered by that, and it would not require such a trigger circuit.

Why not do pretty much exactly what you already suggested:

Get a big capacitor (perhaps a couple of 1 Farad 5.5V supercapacitors in
series, they are quite small and should withstand 6V even without
equalising resistors), charge the capacitor from the existing doorbell
wiring via a 1N4004...1N4007 diode, and put a piezo or mechanical buzzer
connected in parallel with the diode in the direction with the positive
of the buzzer connected to the cathode of the diode. During charging,
the buzzer will experience 0.7V of reverse bias which probably won't
bother it, and when the button is pressed, the diode is reverse biased
and the buzzer receives whatever voltage the capacitor was charged to.
For simplicity, don't bother with any extra series resistor to limit the
charging current, as that would expose the buzzer to more reverse bias,
and it is probably unnecessary after the first charging cycle. The
capacitor won't have time to discharge much when the button is pressed,
which means it won't take long to recharge either. When the circuit is
initially installed, it might take a while to charge the capacitor,
perhaps a minute or so.

So you need capacitor, diode, buzzer.

If the existing bell has a lot of inductance, when the button is
released there might be a large voltage kick across the switch. The
diode ought to catch this so I would not expect it to fry the new piezo
buzzer, but it might alter the sound of the existing doorbell when it
releases.