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.

Are you sure the output to the buzzer is a steady voltage and not
pulsed? Try putting a small cap. 1uf to start with to see if the
voltage increases.

Homer

"Homer" wrote:

Are you sure the output to the buzzer is a steady voltage and not
pulsed? Try putting a small cap. 1uf to start with to see if the
voltage increases.

It's definitely pulsed, but on my analog meter I see a
pulsing vibration of about 1 sec with 50% duty cycle.
During the 50% on period, it's being pulsed at some
relatively high freq, slow enough to visibly wiggle the
needle, but high enough to limit the swings. The needle
settles in at a wiggling 0.4 volts during the pulsing. At
the end of about 10 secs of this (10 beeps I presume), it
begins to pulse faster, perhaps 2 or 3 Hz for the modulation
for another 5 secs, but still settled in at 0.4 during the
on times.

I dont see how I could put a cap on it, won't it just
discharge during each cycle? Perhaps a diode and cap would
tell me for sure if it was getting more than 0.4 volts. ( I
presume it is higher than that somewhat, since the center is
at 0.4 and it wiggles, but if I was getting anywhere near a
3 volt swing, 0.4 seems like a strange centerpoint. I
suppose it could be a very short on time for the high freq
tone. Maybe I should borrow a scope or stick a diode into a
cap to see how high it charges. Thanks for the suggestion.

I'

"ben norton" nospam.com wrote in message
...
"Homer" wrote:

Are you sure the output to the buzzer is a steady voltage and not
pulsed? Try putting a small cap. 1uf to start with to see if the
voltage increases.

It's definitely pulsed, but on my analog meter I see a
pulsing vibration of about 1 sec with 50% duty cycle.
During the 50% on period, it's being pulsed at some
relatively high freq, slow enough to visibly wiggle the
needle, but high enough to limit the swings. The needle
settles in at a wiggling 0.4 volts during the pulsing. At
the end of about 10 secs of this (10 beeps I presume), it
begins to pulse faster, perhaps 2 or 3 Hz for the modulation
for another 5 secs, but still settled in at 0.4 during the
on times.

I dont see how I could put a cap on it, won't it just
discharge during each cycle? Perhaps a diode and cap would
tell me for sure if it was getting more than 0.4 volts. ( I
presume it is higher than that somewhat, since the center is
at 0.4 and it wiggles, but if I was getting anywhere near a
3 volt swing, 0.4 seems like a strange centerpoint. I
suppose it could be a very short on time for the high freq
tone. Maybe I should borrow a scope or stick a diode into a
cap to see how high it charges. Thanks for the suggestion.

I'

If the drive to the sounder is being driven with a pulsed square wave -
which it almost certainly is - I would suggest that if you got a scope on
it, you would see that it was actually going from 3v to virtually ground.
It's just that the frequency is too high for your meter to respond to. On
its AC ranges, it will be wanting to see low frequency sine waves from the
electricity company, not high frequency square waves from a chip.

You could glue a little AC coupled peak rectifier circuit on the end.
Connect the + lead of a capacitor of say 10uF, to the ' bottom ' end of the
buzzer, and then connect the - lead to the cathode of a small signal diode
ie 1N4148. Connect the anode of the diode to watch ground. Now connect the
anode of another similar diode to the cathode / cap junction, and finally,
connect the + lead of another capacitor of say 47uF to the cathode of the
second diode. The - lead of this cap goes again to watch ground.

Now when the buzzer drive signal appears a DC voltage of the correct
polarity to drive your switching transistor, should appear across the 47uF
cap. Connect this point to your switching transistor, which should be a high
gain type or a FET, via a 1k resistor. Values of the caps are arbitrary '
finger in the wind '. If you can't get enough drive or the drive voltage is
not stable enough, try bigger values.

I'm not guaranteeing that this is going to do the job, but it's only a 4
component circuit, and I've used it many times for similar sorts of
applications.

Arfa

"Arfa Daily" wrote:

If the drive to the sounder is being driven with a pulsed square wave -
which it almost certainly is - I would suggest that if you got a scope on
it, you would see that it was actually going from 3v to virtually ground.

Yep - I did something like this last night. I suppose I
should have known the meter couldn't respond.

You could glue a little AC coupled peak rectifier circuit on the end.
Connect the + lead of a capacitor C1 of say 10uF, to the ' bottom ' end of the
buzzer, and then connect the - lead to the cathode of a small signal diode D1
ie 1N4148. Connect the anode of the diode to watch ground. Now connect the
anode of another similar diode D2 to the cathode / cap junction, and finally,
connect the + lead of another capacitor C2 of say 47uF to the cathode of the
second diode. The - lead of this cap goes again to watch ground.

I drew this circuit out, but I'm not quite sure of the
function of D1 and C1 (see labels above) My simple design
just used the equivalent of D2 and C2, connecting the anode
of D2 to the positive end of the buzzer (B+) and the
negative end of C2 to the negative end of the buzzer (being
switched low). The watch is battery operated and floats
relative to the transistor in the device being switched, so
D2 just charges C2 when the buzzer is being sounded - which
switches the transistor. That's the way the timer was set
up in my first attempt.

Thanks for your comments.

"ben norton" nospam.com wrote in message
...
"Arfa Daily" wrote:

If the drive to the sounder is being driven with a pulsed square wave -
which it almost certainly is - I would suggest that if you got a scope on
it, you would see that it was actually going from 3v to virtually ground.

Yep - I did something like this last night. I suppose I
should have known the meter couldn't respond.

You could glue a little AC coupled peak rectifier circuit on the end.
Connect the + lead of a capacitor C1 of say 10uF, to the ' bottom ' end
of the
buzzer, and then connect the - lead to the cathode of a small signal diode
D1
ie 1N4148. Connect the anode of the diode to watch ground. Now connect the
anode of another similar diode D2 to the cathode / cap junction, and
finally,
connect the + lead of another capacitor C2 of say 47uF to the cathode of
the
second diode. The - lead of this cap goes again to watch ground.

I drew this circuit out, but I'm not quite sure of the
function of D1 and C1 (see labels above) My simple design
just used the equivalent of D2 and C2, connecting the anode
of D2 to the positive end of the buzzer (B+) and the
negative end of C2 to the negative end of the buzzer (being
switched low). The watch is battery operated and floats
relative to the transistor in the device being switched, so
D2 just charges C2 when the buzzer is being sounded - which
switches the transistor. That's the way the timer was set
up in my first attempt.

Thanks for your comments.

It's a slightly unconventional way of doing it, but if it works for your
application, then that's fine. My way of doing it kept all the grounds
together - the watch and the external circuit, but ensured that the two
circuits were isolated from each other, and the correct polarity of drive
signal was always produced at the highest level, irrespective of the duty
cycle of the drive waveform. C1 AC couples the high frequency drive, which
is then floating with respect to ground. D1 then behaves as a basic shunt
rectifier, reclamping the voltage pulses about 0.6v above ground. D2 and C2
then form a charge pump to largely remove the low frequency pulsing of the
high frequency drive ie the reason that the buzzer goes BEEP BEEP BEEP.

Buzz '-' ------| |-------||----------^^^^-------- Drive to transistor
C1 | D2 |
-- --
D1 ^ -- C2
| |
------------------------------------------------- Gnd

Don't know if that will come out ok. Gives you the idea if it does.

Arfa

If you go to any of the big hardware stores, they sell a fair number of wall
switch timers. Some of these can be very sophisticated. They are very well
engineered, and are very reliable.

--

JANA
_____


"ben norton" nospam.com wrote in message
news 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.

"Arfa Daily" wrote:

It's a slightly unconventional way of doing it, but if it works for your
application, then that's fine.

I really appreciate the help.

My way of doing it kept all the grounds
together - the watch and the external circuit, but ensured that the two
circuits were isolated from each other, and the correct polarity of drive
signal was always produced at the highest level, irrespective of the duty
cycle of the drive waveform. C1 AC couples the high frequency drive, which
is then floating with respect to ground. D1 then behaves as a basic shunt
rectifier, reclamping the voltage pulses about 0.6v above ground. D2 and C2
then form a charge pump to largely remove the low frequency pulsing of the
high frequency drive ie the reason that the buzzer goes BEEP BEEP BEEP.

Buzz '-' ------| |-------||----------^^^^-------- Drive to transistor
C1 | D2 |
-- --
D1 ^ -- C2
| |
------------------------------------------------ Gnd

Don't know if that will come out ok. Gives you the idea if it does.

Thanks for taking the time to draw it out. I drew it based
on your last message, and it appears I drew it correctly.
Let's see if I understand. "Buzz" is normally high, but
switches to and from watch ground (or near it) during
buzzing. When Buzz is high (off), D1 and D2 will be not
conducting and C1 will be charged. C2 will be discharged.

As Buzz goes low, D1 clamps at 0.6 volts and C1 discharges.
As Buzz then goes high D2 charges C2 through C1 and D1 is
reverse biased. If I get that right, then I see the
advantage of your design, in keeping the common ground (that
did worry me). Thanks again.

"ben norton" nospam.com wrote in message
...
"Arfa Daily" wrote:

It's a slightly unconventional way of doing it, but if it works for your
application, then that's fine.

I really appreciate the help.

My way of doing it kept all the grounds
together - the watch and the external circuit, but ensured that the two
circuits were isolated from each other, and the correct polarity of drive
signal was always produced at the highest level, irrespective of the duty
cycle of the drive waveform. C1 AC couples the high frequency drive, which
is then floating with respect to ground. D1 then behaves as a basic shunt
rectifier, reclamping the voltage pulses about 0.6v above ground. D2 and
C2
then form a charge pump to largely remove the low frequency pulsing of the
high frequency drive ie the reason that the buzzer goes BEEP BEEP BEEP.

Buzz '-' ------| |-------||----------^^^^-------- Drive to
transistor
C1 | D2 |
-- --
D1 ^ -- C2
| |
------------------------------------------------ Gnd

Don't know if that will come out ok. Gives you the idea if it does.

Thanks for taking the time to draw it out. I drew it based
on your last message, and it appears I drew it correctly.
Let's see if I understand. "Buzz" is normally high, but
switches to and from watch ground (or near it) during
buzzing. When Buzz is high (off), D1 and D2 will be not
conducting and C1 will be charged. C2 will be discharged.

As Buzz goes low, D1 clamps at 0.6 volts and C1 discharges.
As Buzz then goes high D2 charges C2 through C1 and D1 is
reverse biased. If I get that right, then I see the
advantage of your design, in keeping the common ground (that
did worry me). Thanks again.

Yep, that's about it, I think. The first part of the circuit just isolates
the watch, with D1 just re-establishing the reference point - in this case,
nearly ground. If you want to get nearer to ground than 0.6v, use a Schottky
diode, or a germanium one. The second part is just a charge pump. Provided
that the following stage doesn't draw too much current off the capacitor,
you should get a fairly constant voltage across C2, with the gaps between
beeps not causing much of a drop. The higher the gain of switch transistor
you use, the higher you can make the value of the output R, and the less
will be the loading on C2. You can make the value of C2 as big as you need,
within reason, but it may take a couple of beeps to ' pump ' the charge on
C2 up to the level you need, if the value gets too big.

Arfa