Re: Simple problem: 0.20 V drop
- From: Joerg <notthisjoergsch@xxxxxxxxxxxxxxxxxxxxx>
- Date: Thu, 18 Jan 2007 17:49:00 GMT
Colin Howarth wrote:
Hi all,
This is a well-known problem (for those using old cameras...):
Mercury oxide cells have a pretty flat 1.35 V open circuit voltage. They have however been banned in many countries. The best replacement would seem to be silver oxide which has a voltage of 1.55 V.
the current drawn (by the exposure meter) is low.
So, I'd like to drop 0.20 V along the way somewhere.
a) Zener diodes: don't go down to 0.2 V
b) Schottky diodes: supposed to have a forward drop of 0.15 to 0.45 V at 1 mA (according to wikipedia) but the drop is current dependent and varies between samples.
c) Germanium diode?
That's how I did it for our old Minolta SRT-100. I had tried a Schottky but that ended up at 320mV. Maybe a larger one would have worked. The OA91 dropped exactly 200mV at the current the CdS circuit is drawing and it is also very small. Sweet. Then I made an adapter to fit a commonly available Silver cell in there.
This has spared me the painful two-point recalibration. And who knows whether those little potmeters would have survived since they are about 25 years old and that camera has been through a lot.
Oh, BTW, if you also have a newer Minolta X-series and the shutter quits it most likely has lost its electrolytic. Don't procrastinate, that nasty thing started leaking in my case.
d) voltage divider with a FET follower? a JFET ?
Is there a standard / best way to do this (with few, smd components)?
There is a surprising amount of space in the SRT-100. Even a larger Ge diode would have fit in.
--
Regards, Joerg
http://www.analogconsultants.com
.
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- From: Colin Howarth
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