RTDs, 1000 ohm vs 100 ohm

Hi All. I have looked and looked, and I can't seem to find a good reference on this. What are the pros and cons of a 100ohm vs 1000ohm
platinum RTD?
Appreciate the info, DanW
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Differencies between PT1000 and PT100
A PT1000 is however a temperature-dependent resistance. It has 1000 ohms with zero C, with 1C 1003.85 ohms with 10C 1038.5 ohms and so on. The temperature coefficient amounts to depending upon platinum material +3.85 Ohm/C with the PT1000 and +0.385 Ohm/C with the PT100. A PT1000 has therefore a larger slope and makes possible thereby a higher resolution. The sensor has small currents, a self-heating of sensors small is minimized. With a PT100 it is easy to develop thermometers for a large temperature range.
Reference: http://www.amplifier.cd/Technische_Berichte/PT100/Temperature_measurement.htm
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At one time finding measuring instruments would have been far easier for PT100 elements although I suspect that might have got better. Because of their wider use PT100 elements are likely to be cheaper
Steve
Steve YATES
MTL Instruments Luton, UK www.mtlgroup.com www.mtlblog.com
Dan Wright wrote:

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Dan Wright wrote:

Omega Engineering makes 1t least 100-ohm units. Why not ask them? http://www.omega.com /
Offhand, I'd imagine that the 1000-ohm units have higher output for the same excitation but are less robust. Electronics to read 100-ohm devices are probably more common. I use four-wire devices whenever I have a choice.
Jerry
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Thanks for the reply, all. I am assuming (I know, ass-u-me) that the most common way to make a measurement with the device is to provide a low constant current source and measure the Vdrop across. I guess a trade off of the higher resolution would be a higher I2R loss?
DanW
Dan Wright wrote:

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Dan Wright wrote:

It's not the power loss, but the self heating. In one application, I pulsed the excitation, turning it off except when a reading was needed.
The simplest way to read the drop across only the RTD supplies the excitation current through one pair of wires and reads the voltage with another (Kelvin connection). It is possible to use three wires and separately measure the drop in one excitation lead and the combined drop in the RTD and the other. Subtraction gives the desired result. I can't think of a good reason for choosing a three-wire device, but there may be one.
Jerry
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Dan Wright wrote:

Apropos self heating and your original question:
Assume, just to have definite numbers, a 100-ohm device and 10 mA exciting current. The heating power is 100 uA^2 * 100 ohms = 10 mW. We get the same sensitivity with a 1000-ohm RTD and 1 mA. The power is now 1 uA^2 * 1000 ohms = 1 mW. The high-resistance RTDs are less robust, but suffer less from self heating. With good thermal bonding to a large object of high thermal conductivity, self heating rarely is an issue.
Jerry
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