On Thu, 29 Nov 2012 21:51:12 +1100, Franc Zabkar
put finger to keyboard and composed:

power on

Vin ~111 V

Vout to 0 V ~200 mV

This is consistent with an open circuited series resistor. The zener
is not shorted, as per your previous measurement.

That said, I'm surprised that Vin is so high. I can't imagine that the
zener voltage would be more than 10V, so this would mean that the
series resistor would need to drop a lot of voltage. It seems too
small for that kind of wattage. Something doesn't look right to me,
but I can't see the circuit clearly enough (too many components
obscuring the PCB). I don't think I can help you without a circuit
diagram.

I'm wondering if there may be a series capacitor whose function it is
to losslessly reduce the supply voltage ...

C2
bridge 1uF
rectifier 250VAC
______ | | Vin
AC1 o--| |----| |----o-- R --+----+---o Vout
| | | | | |
||| | ZD1 C
| | | |
AC2 o--| |-----------o-------+----+---o
|----| 0V


As for the zener voltage, perhaps we can guess at its value by
examining the battery charging circuit.


D1 100 ohm
Vout o---||---- R4 ----+-o
|
3.6V NiMH
110mAH battery
|
0V o------------------+-o


If we assume that the charging current is 11mA (mAH rating / 10), and
if we allow for a charging voltage of 4.5V, say, then Vout would be
....

4.5 + (100 x 0.011) + 0.6 = 6.2V


- Franc Zabkar
--
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