-------    [OT]: => [EE]:   -------

   I'm looking at a small incandescent  lamp 60V-50mA,
lying on my desk. (Even don't ask me what device I took
it from :-) When cold it shows 120ohm, when under 60V,
I guess, it's current should be 50mA, so resistance
60/0.05=1200ohm.
   Assuming temperature under 60V is about 2000 C we
have (1200-120)/2000 = aprox 0.5 ohm per 1grad C. Or this
thing should double it's resistance at 250 C.
   So needed current I = 5V / 250ohm = 20 mA.
   Power dissipation  P = 5V * 20 mA = 0.1 W much less then
60V * 50 mA = 3W.

   So voltage across the lamp should change from ~2.5V
when cold to 5V when oven is heated to 250 C.

   Why not use this as "Temperature Sensor" ?
20 mA current source is not a big problem, I think.

   Mike.
-------
Correct me if I'm wrong, please.

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