I want to control a device to turn it on once a day. It's
got a complicated microprocessor startup, so I can't just
turn on power - I have to close a switch for a few seconds.
I've already put a transistor across the switch, and turning
it on turns on the device, so that part is working. For a
test, I put the output of a simple battery operated kitchen
timer (previously used to run the piezo buzzer) onto the
base of the transistor, and this system works fine. The
output provides up to 1.5 volts (single AA cell) and
switches my transistor on.

The problem is that the timer only allows 20 hours of
advance setting and has to be reset every day. So now I
wanted to use an old digital watch with a daily timer
setting. I thought it would be easier - the watch uses a 3
volt hearing aid battery. The buzzer part was dead, but I
measured the input to the buzzer, and it does have a signal
at the alarm time.

The problem - The dead watch buzzer is sonnected to the
battery B+ and the circuit drives the other end towards
ground. However, the other end just drops down 0.4 volts
from the +3 volt battery B+, i.e. to only 2.6 volts. The
0.4 volt differential won't drive the base-emitter of the
switching transistor.

My simple circuit design skills don't see an easy way to
drive the transistor with this. Ideally, the solution would
let me stay battery operated, although I could easily use a
couple of large D-cells instead of the tiny battery. I've
considered a simple comparator or a transistor amplifier to
get enough voltage to drive the base of the switch, but I'm
looking for cheap and easy. I admit I'm confused as to why
the voltage across the buzzer is so small when it has 3
votls available - could it be the defective buzzer is
causing this? It looks hard to get the buzzer out of the
watch, so I'm just measuring across it. Alternatively,
perhaps the watch drive circuit is bad and the buzzer is
good.

Ideas or suggestions appreciated. Thanks.