The Cyber-Spy.Com Usenet Archive Feeds Directly
From The Open And Publicly Available Newsgroup
This Group And Thousands Of Others Are Available
On Most IS NNTP News Servers On Port 119.
Cyber-Spy.Com Is NOT Responsible For Any Topic,
Opinions Or Content Posted To This Or Any Other
Newsgroup. This Web Archive Of The Newsgroup And
Posts Are For Informational Purposes Only.
From: C.D.H.Williams@exeter.ac.uk (Charles DH Williams)
Subject: Re: Quickie Thermal Probe
Date: Fri, 13 Dec 2002 17:34:39 +0000
Organization: speaking for myself...
References: <3DF8FDBE.34F2@Spam.Bots> <3DF93D82.3F6FDCB2@earthlink.net>
NNTP-Posting-Date: 13 Dec 2002 17:37:44 GMT
X-Newsreader: MT-NewsWatcher 2.4.4
In article <3DF93D82.3F6FDCB2@earthlink.net>, Robert Baer
> I really like that reference.
> However, for better balance and maybe sensitivity, shouldn't R0 be the
> same sensor as R(T) only shielded or otherwise protected?
The short answer is "no".
I think you may be mis-seeing (if there's such a word) the circuit. It is
not a bridge followed by an amplifier, the bridge is inside the feedback
loop. Anyway, the longer answer is as follows:
At switch-on the bridge is biased only by offset errors (and noise) but
R(T) < R0 so the the output of the op-amp increases due to what is
essentially positive feedback. If R(T) were constant the op-amp output
would hit the supply rail and the circuit would not be interesting.
However, as the voltage across the bridge increases, so does the
dissipation in R(T) so its temperature rizes and so does its resistance
until R(T) == R0. The circuit reaches equilibrium with a DC bias V0 at the
output and the non-linear sensor at a temperature T0 above ambient.
Now the clever bit:
In the low-frequency limit, anything that changes the net heat loss from
the sensor by deltaP will cause the sensor temperature to change, which
causes the op-amp output to change by deltaV in such away as to restore
the constant sensor temperature equilibrium point. Since you know the
resisance of the sensor is R(T0) = R0 the total dispation of the sensor is
P_tot = P0 + deltaP = Vout^2 / ( 4 * R0 ) = (V0 + deltaV )^2 / (4 * R0 )
therefore to first order
deltaP ~= V0 * deltaV / (2 * R0 ) and deltaV = deltaP * R0 / (2 * V0 )
Things like the heat capacity of the detector and the precise functional
form of R(T) don't matter in the DC limit. Howver, they do play a role
when one analyses the frequency reponse and noise; as one would guess, the
best sensitivity comes from sensors with small heat capacity and large
(1/R)*dR/dT figure of merit.
In the case of the draft detector detaP comes mainly from the fact that
the air is moving past the hot sensor and thereby force cooling it. The
temperature of the air is a relatively small correction. There's an
intesting trade-off between increasing deltaP by increasing the
temperature of the sensor (i.e. a larger V0) and increasing the
responsivity of the circtuit which requires a smaller value of V0.
There are numerous variants of this cirucit which involve assymteric
bridges and puting impedances in the feedback branch.
Anyway, I'd argue that this detector qualitatively unlike the OP's scheme
which I think is intended to measure the temperature gradient in the air.
Go Back To The Cyber-Spy.Com
Usenet Web Archive Index Of
The sci.electronics.design Newsgroup