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. The0.4 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.