On Thu, 10 Jul 2003 23:54:59 -0400, "KILOWATT"
put finger to keyboard and composed:

Hi Sam. Nice to have some knowledgeable people here!
But assuming you have the zeners backwards, that circuit
is actually a constant current source[...]
Ok Sam i've double-checked and the diodes are not shown
backwards on the schematic,they're really oriented this way.
I've tested two of them with about 30mA of current flowing through
them to be shure they where indeed 6.2V zeners (C6V2 is shown
on their casing) and...they are. In fact i got a reading of 6.4V with
the 30mA current.

The circuit is that of a current source whose value is given by

I = (2 * Vf - Veb) / 47

where Vf is the forward voltage drop of the zener diode, probably
around 0.7V at 5mA. Assuming Veb is about 0.65V, then I = 16mA.

I'm not sure why zeners are used as they are. I know that two 6V2
zeners have a lower combined temperature coefficient than a single 12V
diode, at least in the reverse direction. But I'm not sure whether
this is relevant in this case.

Anyway, here are two datasheets:
http://www.dectel.ru/analogues/stabil/pdf/BZX79_2.pdf
http://www.fairchildsemi.com/ds/BZ/BZX79C22.pdf


- Franc Zabkar
--
Please remove one 's' from my address when replying by email.