[time-nuts] Temperature sensors and bridge amps

Florian E. Teply usenet at teply.info
Fri Nov 12 12:29:27 UTC 2010


On Fri, 12 Nov 2010 10:36:15 +0000
"Poul-Henning Kamp" <phk at phk.freebsd.dk> wrote:

> In message <20101112110627.488cbc8f at vz127.worldserver.net>, "Florian
> E. Teply" writes:
> 
> >In a bridge circuit, you don't measure resistance directly, but use
> >the voltage that appears across the bridge. So for a 100 ohms
> >element, you'd usually have ten times the current flowing in that
> >branch compared to a 1kohm element.
> 
> This was one of the things that I wondered about:  How large currents
> are used ?
> 
> Can't be too much because that would lead to self-heating...
> 
Exactly. Usually you want to avoid self-heating. There are two
possibilites: 
a) Use low currents. Still you want to read your voltage precisely
enough to make use fo the superior properties of platinum, so with 1
milliamp of current (roughly 100 uW power in a PT100), the .385 Ohms/°C
make for 385 microvolts of signal for a 1 degree of temperature change.
Yet the power dissipation in the resistor might still be too much,
especially in very sensitive circuits. In a PT1000, 100 microamps will
yield the same voltage swing in your signal but only 10 uW of power 
dissipation in your sensor. Generally speaking, in precision
circuitry one would use as much current as reasonably possible to
maximize the signal. The upper limit to the current is posed by the
power dissipation in the sensor. For one, the system must not be
disturbed by the power introduced by the sensor. And second, the
difference between the platinum resistor internal temperature (that one
sets the resistance) and the temperature to be controlled must be
small enough. On the other hand, you can't get too low with current in
order to not swamp the signal with noise (both thermal noise in the
bridge resistors and noise imposed by the amplifying circuitry).
You could still go lower with the current, but make sure the signal you
want to see, say 20 microvolts for a 0.05 °C temperature change, is
significantly above the noise. So, basically your amplification
circuitry may not introduce more than very few microvolts of noise
referred to the input.

Plan b): Duty-cycling the measurement. That perfectly lends itself to
digital circuitry, but is a tad more complicated in the analog domain.
Basically what you do is power the bridge, measure the voltage and
unpower it again. So if you measure the temperature once every second
for, say, 10 milliseconds, you get only 1 percent of the power
dissipation in the resistor on average. Given the thermal inertia of
typical systems, once every second should be enough. But of course,
you'll have to be careful not to introduce switching noise into your
oscillator. The circuitry will be a lot more complex though.

In both cases, given the slow thermal behaviour of the systems, you can
low-pass filter the signal heavily.

HTH,
Florian



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