Li Wang,  I dont know if this provides some further insight, but consider
the following.  When water evaporates the change of state requires a
substantial amount of energy far more that the specific heat of the
material. Inquire into the latent heat of vaporization of water and you
will see a very substantial number.  What could be happening is the
evaporation of the droplet provides a substantial cooling effect. The rate
of cooling will then be dependent upon the humidity in the room as well as
the temperature, especially as the surface area per unit volume of the
droplet will tend to be high. I dont know if this is the only effect only
that it could well be one effect. Hope this helps,

Gary Hillman
Service Support Specialties, Inc.
9 Mars Court
PO Box 365
Montville, NJ 07045
973-263-0640
973-263-8888.



-----Original Message-----
From:   Li Wang [SMTP:liw@andrew.cmu.edu]
Sent:   Sunday, March 14, 2004 11:45 PM
To:     mems-talk@memsnet.org
Subject:        [mems-talk] thermopile

Hello,

I built thermopile (Al/polysilicon pairs) with one junction on a silicon
oxide/nitride membrane (for good thermal isolation) with which I want to
measure the temperature. Usually the output voltage should be zero. I
dropped a room temperature water droplet on the membrane and the output
voltage went down abruptly and went back to zero slowly. But I expected
the output voltage will not change because the droplet has the same
temperature with ambient. Did anyone meet this kind of problem before? Is
it because the temperature fluctuation in the droplet? Is it because the
deformation in membrane will change the output voltage in the bi-material
junction? How can I avoid this systematic error?

Thank you in advance.

Li


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