Wagner Lipnharski wrote:

> Lets especulate about it.
> Suppose you insert a small lamp inside the oven and measure its
filament
> resistance, what can happens?
>=20
> The lamp has vacuum or an dumb gas into the bulb to avoid oxigen to
oxidize
> and burn the filament. Small lamps normally have just vacuum, it is
cheap
> to produce that way.  With vaccum, the temperature will have some
> difficulty to be transported from the oven internal radiation to the
lamp's
> filament.  I mean, the filament will receive just infra-red initially,
> later on the thermal vibration can reach the filament from the lamp's
body
> and glass posts and wires holding the filament. It means, the lamp
> resistence change will take some time to reflect the oven temperature,
time
> not fast enough to allow a good power control.


   I placed it in boiling water - resistance was 155 ohm, pulled=20
it out - resistance dropped to 125 ohm in about 10 seconds.=20
(120 ohm at 25C). At 200-250C this time will be much less,=20
much less then oven's.=20
   It's not a rocket science (Dale used this term recently), so=20
automatic control theory should be implemented at a very=20
basic level.


> Then suppose we build a circuit to measure the actual oven thermal
heating
> elements electric current consume during heating.  A table could be
built
> doing lots of measurements using a thermometer and lots of patience,
what
> can happens?  The resistance of such element will reflect the
temperature
> of the element rubishing like a shy girl, not exactly the oven's
> temperature.  I think that when the oven is at 200=B0C, the element =
red
> probably will be around 1000=B0C, when the thermostat turns it off and
stays
> there, probably seconds before it turns on again its temperature
should
> still much higher than the oven, a guess would say around 300=B0C or
more.
> During this week I will try to measure it with the non-contact
thermometer
> I should receive on Tuesday.


   If you think that heating due to power dissipation of the lamp=20
          P =3D 5V * 20 mA =3D 0.1 W =20
would  spoil measurements take two or more lamps, this=20
will reduce current twice or more and power dissipation=20
will be reduced four times or more on each lamp. You=20
might work at lower voltages as well decreasing power=20
dissipation quadratically. You may get a lamp with the=20
bigger resistance.


> If measuring the actual power consume of such thermal element, I guess
it
> will not help much, since the change in current will not (I guess) be
much
> noticeable from a warm to a hot oven, when you feel 100% power to an
> element, it turns bright red, and this means very high temperature,
for a
> cold or hot oven, I may be wrong.


   Lamp discussed works at very small current not at 100%=20
power. You'll burn AVR with 60V input voltage.=20

   Best regards.
   Mike.





> I remember seen a color table, meaning "color of the temperature". It
was a
> simple color bar printed on a thermal scale (like a ruller), from dark
red,
> going to shine red, redish yellow to bright yellow to bright white in
6 to
> 7 inches. You can guess DARK RED is how your car's brake pads and disk
> goes
> after a strong break at the freeway when you spot a police officer
with a
> radar gun - and WHITE is simply hell's temperature of a star.  By
simply
> using this ruller in front of your eye and matching the color, you can
say
> what temperature the target is, of course, it should be at least dark
red.
> I never found this scale for sale, if someone has any tip, it will be
> welcome.
>=20
> You can relate this issue to a range's spiral heater, the pan
temperature
> could be around 300=B0F, but the spiral will be middle red, and it can
> represent something around 1500=B0F. When the thermostat turns off the
> heater, it probably stays at the pan temperature or a little bit
higher,
> but it happens since there is a direct physical contact between the
parts.
> If there was air in middle, as into the oven, I don't know what the
heater
> temperature could be.  Suppose the minimum it goes is exactly the
oven's
> temperature, then that's ok, but it requires to cools off to oven's
> temperature before taking a valid resitance reading, and that time
could be
> a killing for any temp control process.
>=20
> By other side, any other exposed thermal element, as a simply AWG32
(very
> thin) nickel-chromium wire, or even a disconnected secondary heater
element
> could be used for such measurement.
>=20
> Wait for this week news... :)
>=20
> VV46NER



I wrote earlier:

> >    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=3D1200ohm.
> >    Assuming temperature under 60V is about 2000 C we
> > have (1200-120)/2000 =3D aprox 0.5 ohm per 1grad C. Or this
> > thing should double it's resistance at 250 C.
> >    So needed current I =3D 5V / 250ohm =3D 20 mA.
> >    Power dissipation  P =3D 5V * 20 mA =3D 0.1 W much less then
> > 60V * 50 mA =3D 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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