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How does an RTD work?

evilsanta

Mar 9, 2014
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One of the questions in my assignment is off how an RTD works with a transmitter and how it interacts with a PLC. Below is what I think happens but if anyone can spot any mistakes in it then please let me know.

An RTD works using resistance and temperature to help control certain things via the aid of a PLC. What this means is the change in temperature detected by the probe equates to an electrical resistance change inside the RTD. By supplying an RTD with a constant current and measuring the resulting voltage drop across the resistor, the RTD’s resistance can be calculated, and the temperature can be measured. An RTD takes a measurement when a small DC current is supplied to the sensor itself, then the current experiences the impedance of the resistor, and a voltage drop is experienced over the resistor. Depending on the nominal resistance of the RTD, different supply currents can be used. An RTD can be connected in a two, three, or four-wire configuration, the two-wire configuration is the simplest and also the most error prone. For more accurate readings its best to use either a 3 wire or 4 wire setup. Then change in resistance is then sent to the attached transmitter where it will undergo a conversion from ohms (resistance) to mA (current now detected). The transmitter works it out by rearranging ohms law to use the equation I=V/R this will then divide the voltage by the resistance to give the current now seen. Once its in the transmitter and converted the range of current is on a scale of 4mA to 20mA. We can set the range between 4mA and 20mA to be different temperatures. We have a start temperature of perhaps 0 degrees at 4mA and a maximum measured temperature of 100 degrees at 20mA, anything in between 4-20mA will be readings between 100-200 degrees on the temperature range. So for 50 degrees to be detected the electrical resistance would be 12mA. We start the range at 4mA as if it started at 0mA then we may not be able to detect a short in the circuit as we may assume its still working, so having it at 4mA means we can tell when its working and not working. The transmitter is connected to a PLC with another resistor connected across it, the current signal is sent from the transmitter to the resistor and converted to voltage again so the that the PLC can then ascertain the reading and select the appropriate response. The reason for the signal being sent as current is because if it was voltage it may have voltage drop over a distance whereas with current the signal is good for up to 1km.


Thanks Alan
 

Harald Kapp

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You have grasped the basic principle correctly.
There are some minor issues I'd like to point out to help you express matters more clearly:

By supplying an RTD with a constant current and measuring the resulting voltage drop across the resistor, the RTD’s resistance can be calculated, and the temperature can be measured. An RTD takes a measurement when a small DC current is supplied to the sensor itself, then the current experiences the impedance of the resistor, and a voltage drop is experienced over the resistor.
That's a tautology. Both sentences express the same thing.


The transmitter works it out by rearranging ohms law to use the equation I=V/R this will then divide the voltage by the resistance to give the current now seen.
Which current are you talking about? The current through the RTD or the current on the transmitter's output?

We start the range at 4mA as if it started at 0mA then we may not be able to detect a short in the circuit as we may assume its still working, so having it at 4mA means we can tell when its working and not working.
Correct. Plus at a minimum of 4mA there is always a certain minimum amount of energy available to power the receiving end from.

The transmitter is connected to a PLC with another resistor connected across it, the current signal is sent from the transmitter to the resistor and converted to voltage again so the that the PLC can then ascertain the reading and select the appropriate response.
Using a resistor is one very common way of measuring the current issued by the transmitter. There are other means, too, e.g. current compensating measurements, although this would not be commonly used in this application. But for the sake of generality you could just refer to the current from the transmitter (by the way, the technical term is "transducer") being measured by the PLC (without stating explicitly how this is done).


The reason for the signal being sent as current is because if it was voltage it may have voltage drop over a distance whereas with current the signal is good for up to 1km.
Right. Plus it makes the setup comparatively insensitive to noise, see here.

Harald
 

evilsanta

Mar 9, 2014
18
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Mar 9, 2014
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You have grasped the basic principle correctly.
There are some minor issues I'd like to point out to help you express matters more clearly:


That's a tautology. Both sentences express the same thing.



Which current are you talking about? The current through the RTD or the current on the transmitter's output?


Correct. Plus at a minimum of 4mA there is always a certain minimum amount of energy available to power the receiving end from.


Using a resistor is one very common way of measuring the current issued by the transmitter. There are other means, too, e.g. current compensating measurements, although this would not be commonly used in this application. But for the sake of generality you could just refer to the current from the transmitter (by the way, the technical term is "transducer") being measured by the PLC (without stating explicitly how this is done).



Right. Plus it makes the setup comparatively insensitive to noise, see here.

Harald

Thanks very much Harald I'll delete one of the duplicate lines in the answer. I was trying to comment on the current being seen by the transmitter and will fix this and the other suggestions you've made I will certainly add in as well. Thanks again for a very comprehensive reply.
 

hevans1944

Hop - AC8NS
Jun 21, 2012
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One of the questions in my assignment is off how an RTD works with a transmitter and how it interacts with a PLC. Below is what I think happens but if anyone can spot any mistakes in it then please let me know. ...

Thanks Alan
Alan, it appears that you may not have distinguished an RTD from a 4 - 20 mA RTD Transmitter. An RTD is typically made from plantinum wire with a resistance of 100 Ω at 0° C. Other values are possible, depending on wire diameter and length, but 100 Ω is typical. The resistance varies in a very precise manner described by an empirical second- or third-degree polynomial equation whose coefficients are well known for platinum. Other metals, with different coefficients, are also used, but rarely.

The resistance of a platinum RTD that is 100.00 Ω at 0° C increases to 138.50 Ω at 100° C. To measure this change in resistance, the sensor is typically connected with four wires to a current source and an analog differential amplifier. The current source, typically 1 or 2 mA, excites the RTD using two of the four wires. The voltage drop across the RTD is sensed on the other two wires, which carry very little current because the input impedance of the analog differential amplifier is very high, usually several megohms. For a 100 Ω sensor excited with 1 mA constant current, the voltage drop will be 100 mV at 0° C. Greater excitation current would produce a larger voltage drop at any given temperature, but would also cause increased self-heating in the RTD. Lower excitation currents are desirable, but there is a practical lower limit that depends on the noise and offset characteristics of the analog differential amplifier.

The integration of the RTD with the current source and analog differential amplifier can occur locally (near the RTD) or remotely (at the data acquisition system or PLC). If local integration is desired, some means must be provided to power the analog differential amplifier and transmit the temperature signal to the PLC. The easiest way to do this is to make the transmitter, consisting of the RTD current source and the analog differential amplifier, part of a 4 to 20 mA current loop. The compliance voltage of the current loop is provided by an external DC power supply, typically in the range of 12 to 48 volts. It does not have to be precisely regulated, but it should have low ripple voltage. Transmitter devices with 4 to 20 mA output are commercially available. They generally include span (gain) and offset (zero) potentiometers for calibration of the 4 mA and 20 mA end-points to a specific temperature range.

Remote integration is a more expensive option because either a 3-wire or 4-wire connection to the RTD is necessary to compensate for the effects of connecting wire resistance. It also requires that the PLC have an RTD data acquisition module, which is much more complex and expensive than a simple analog-to-digital input module. All that is needed for a 4 to 20 mA current-loop "receiver" interface is a fixed-value resistor to convert the loop current to a measurable voltage.

Hop
 
Last edited:

evilsanta

Mar 9, 2014
18
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Mar 9, 2014
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Alan, it appears that you may not have distinguished an RTD from a 4 - 20 mA RTD Transmitter. An RTD is typically made from plantinum wire with a resistance of 100 Ω at 0° C. Other values are possible, depending on wire diameter and length, but 100 Ω is typical. The resistance varies in a very precise manner described by an empirical second- or third-degree polynomial equation whose coefficients are well known for platinum. Other metals, with different coefficients, are also used, but rarely.

The resistance of a platinum RTD that is 100.00 Ω at 0° C increases to 138.50 Ω at 100° C. To measure this change in resistance, the sensor is typically connected with four wires to a current source and an analog differential amplifier. The current source, typically 1 or 2 mA, excites the RTD using two of the four wires. The voltage drop across the RTD is sensed on the other two wires, which carry very little current because the input impedance of the analog differential amplifier is very high, usually several megohms. For a 100 Ω sensor excited with 1 mA constant current, the voltage drop will be 100 mV at 0° C. Greater excitation current would produce a larger voltage drop at any given temperature, but would also cause increased self-heating in the RTD. Lower excitation currents are desirable, but there is a practical lower limit that depends on the noise and offset characteristics of the analog differential amplifier.

The integration of the RTD with the current source and analog differential amplifier can occur locally (near the RTD) or remotely (at the data acquisition system or PLC). If local integration is desired, some means must be provided to power the analog differential amplifier and transmit the temperature signal to the PLC. The easiest way to do this is to make the transmitter, consisting of the RTD current source and the analog differential amplifier, part of a 4 to 20 mA current loop. The compliance voltage of the current loop is provided by an external DC power supply, typically in the range of 12 to 48 volts. It does not have to be precisely regulated, but it should have low ripple voltage. Transmitter devices with 4 to 20 mA output are commercially available. They generally include span (gain) and offset (zero) potentiometers for calibration of the 4 mA and 20 mA end-points to a specific temperature range.

Remote integration is a more expensive option because either a 3-wire or 4-wire connection to the RTD is necessary to compensate for the effects of connecting wire resistance. It also requires that the PLC have an RTD data acquisition module, which is much more complex and expensive than a simple analog-to-digital input module. All that is needed for a 4 to 20 mA current-loop "receiver" interface is a fixed-value resistor to convert the loop current to a measurable voltage.

Hop

Thanks very much Hop that was amazing and very easy to follow, took a couple of reads to sink in but I think I have got it now.

All your help has been very much appreciated, Alan
 
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