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Means of dropping watch battery voltage by .2 Volts

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I am a watch collector & repairer. I also have a fair, though somewhat
dated, electronics background. I have several early '60's vintage
Bulova Accutron wristwatches. These watches operate using a very basic
oscillator circuit which energizes a tuning fork via a pair of coils.
The tuning fork then drives the mechanical movement of the timepiece. A
basic description can be found here:
http://www.timezone.com/library/horologium/horologium631672882451976629

These are very cool watches to the collector. They are stylistically
very much of their era, and because of the tuning forks, they hum,
rather than tick. The problem is, they were originally designed to run
on a 1.35 volt mercury oxide cell (343) which is no longer available.
There is a 1.55v silver cell which will fit, but it can cause some
Accutrons to run very fast, and possibly even damage them. At least one
supplier sells modified cells with a component to drop the cell's
voltage to 1.35 volts. I think this is accomplished with a surface
mount type diode. Here's a thread on these cells in a watch discussion,
with pictures of the component added
http://bdwf.net/forum/showthread.php?t=75397 I would like to simply
insert an appropriate diode (or other component) into the circuit of the
watches as a one time modification, after which I can use the much
cheaper/readily available silver cells. But, as I said, my electronic
component knowledge is late 70's vintage, so I'm not sure what component
to add. I know in general silicon diodes have a forward voltage drop on
..7V, germaniums, .3V, so I'm not sure how to produce .2V. A zener? A
Schottky? Anyone have any suggestions on what would lower the output
of the 1.55 volt cells to 1.35 volts, over the life of the cell? It
would have to be fairly small, about 1/2 the size of a 1/4 watt
resistor, or less.

TIA

Dan
 
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nospam

D said:
to add. I know in general silicon diodes have a forward voltage drop on
.7V, germaniums, .3V, so I'm not sure how to produce .2V. A zener? A
Schottky?

Diode voltage is very dependant on diode current at low levels so it would
help to have an estimate of battery current, perhaps calculated from
typical battery life and capacity.

A small Schottky will drop around 0.12v at 10uA.

I would get one, stick it in and see what happens. If you have a decent
meter you could measure the drop across the diode.
 
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