Lets say that I needed to sense temperatures around 0C to 50C without introducing any magnetics (or at least no DC field), any Ideas?
All thermistors appear to be magnetic so I can't use the thermistor fed with AC.
Lets say that I needed to sense temperatures around 0C to 50C without introducing any magnetics (or at least no DC field), any Ideas?
All thermistors appear to be magnetic so I can't use the thermistor fed with AC.
A conventional alcohol thermometer + optical readout via the fiberoptic line?
Vladimir Vassilevsky DSP and Mixed Signal Design Consultant
April 1 came early?
How about a back-to-back dual diode with AC drive?
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal ElectroOptical Innovations
55 Orchard Rd Briarcliff Manor NY 10510 845-480-2058 hobbs at electrooptical dot net
Platinum or copper RTD, very low level DC or AC excitation
Fiberoptic thermometer
Infrared
Diode, at pulsed nanoamp currents
Varicaps have huge dC/dT
FR4, about 900 PPM/K capacitance TC
IC sensor, LM45, LM71
What's a room temperature SQUID? The "S" stands for "superconducting."
John
Thermistors are resistive.
Why do you think they are magnetic ?
Also, why does it matter ?
don
Two problems:
1) just try to find a diode that really is nonmagnetic. I bought some "nonmagnetic diodes" intended for MRI use. They are magnetic. They read about 25nT at 2 inches.2) The rectified current in the diodes flow in a little circle working like a little dipole.
I hadn't thought of that. It is an interesting idea because it also removes the copper wires completely.
Probably because they stick to a magnet. Lots of parts do.
Some parts are repelled by a magnet, which can be just as bad a problem.
John
Do you know of a maker of nonmagnetic ones?
Vladimir beat you here.
This will be a bit costly because it would need calibration. Won't it?
This is why I bought the "nonmagnetic diodes" that turned out to be magnetic.
Voltronics as an "extensive line of nonmagnetic diodes" which turns out to be exactly one type of back to back Schottky. I pointed the issue out to them. They have neither introduced another type of diode nor fixed the wording the last time I looked. I still haven't ruled them out because it wasn't their diode that turned out to be magnetic.
On a sunny day (Sun, 13 Sep 2009 07:46:05 -0700 (PDT)) it happened MooseFET wrote in :
Would it be incorrect to think that making an opposite current next to it, would cancel that magnetic field? Like a reverse wire loop in series (sort of folded back) directly above it?
I know they are magnetic. I have measured several types They are a strange sort of metal oxide semiconductor thing. Among the oxides are magnetic ones
It will cause a large error in the measured magnetic field.
No, but all it takes is platinum and copper, so it should be available somewhere. Or just all-copper, a small coil of fine wire with maybe a
4-wire connection.Cryo-con sells an RTD with platinum leads, no nickel. I think Omega may, too. Try Lakeshore Cryo, too.
Copper is faintly paramagnetic, so messes up fields at the 1e-9 sort of levels. The NMR guys plate their copper antennas with a secret sauce to null that out.
I meant the kind with a temperature-sensitive fluorescent end. They input short-wave light and bounce back a temperature-dependent spectrum. Not cheap.
Maybe they have lots of them.
I pointed the
Sounds like a thermocouple is the best idea so far.
John
How about a mirror mounted on a bimetalic strip?
What sort of metals are used on typical bimetalic? Can you get one that is sufficiently non-magnetic? Or make one out of something you like?
-- These are my opinions, not necessarily my employer\'s. I hate spam.
Kovar leads, for the glass bead type.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal ElectroOptical Innovations 55 Orchard Rd Briarcliff Manor NY 10510 845-480-2058 hobbs at electrooptical dot net http://electrooptical.net
Would you tell us what your real application is? (There's no such thing as quantum interference at macroscopic scales at room temperature, nor are there superconductors there, so all that's left of the 'squid' is 'device'. Or maybe it's for oceanography?
That would make it a bit easier to understand what your actual requirements are. (5 nA going around a 1 mm diameter loop is a pretty small magnetic moment.)
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal ElectroOptical Innovations 55 Orchard Rd Briarcliff Manor NY 10510 845-480-2058 hobbs at electrooptical dot net http://electrooptical.net
Ok it is just a "device". The active area is about 3mm in size. I want to know the temperature of the active part or the assembly without creating a DC field greater than about 0.3nT. Because of the physics involved frequencies above several KHz, are not seen by the sensor.
I only really need to know the temperature to an accuracy of about
0.5C. I can wait a large fraction of a second for the measurement so very low currents are ok.
One of the two or perhaps both metals will be magnetic. The only copper it seems I can trust is made thanks to the audio nuts.
Cheap stuff and other cheap stuff. They are optimized for low price.
Can you get
You can make a 'bimetal' strip out of two kinds of plastic.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal ElectroOptical Innovations 55 Orchard Rd Briarcliff Manor NY 10510 845-480-2058 hobbs at electrooptical dot net http://electrooptical.net
Cute idea. The all optical nature of it appeals to me in a different application.
If I can use a fiber to measure the deflection, it would be an all non conductive connection.
I already have a method for measuring the field with no electricals. It involves three light pipes. Adding a couple more wouldn't be a bad thing to do. The active can be inside a sealed glass with a vacuum.
Over what range of temperature(s) do you need it to work?
Some of the LCD thermal paints are non-magnetic and accurate to 0.5C with a relatively slow response time. The other way would be to put the sensor into a temperature controlled flow of dry nitrogen, argon or helium. The latter gives the best thermal transfer.
Or are you saying the part dissipates real amounts of power?
Regards, Martin Brown
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