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Answer Posted / carl_ellis
Answer 1 is correct. The ISA SP50 standard established
4-20mA worldwide as the analog control signal decades ago.
But what the answer 1 and the others miss is that using 1 or
2 mA at the low end would not provide sufficient power to
power 2 wire loop powered instruments.
2 wire loop powered instruments use about 3.6mA at 24Vdc to
power the electronics which makes the measurement and
controls the output circuitry. That's about 85mW, not a lot
of power.
1mA or 2mA of current would provide only a fraction of 85mW
available at 4mA for powering the instrument itself, and
reduce the functionality of 2 wire loop powered instruments.
Carl Ellis
Measure First
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Q1: Consider part of a control loop, which excludes the transmitter, consisting of a process, a controller and a control valve which may be represented by two dead times of 0.5 min each and three exponential lags of 0.8 min., 1.0 min. and 1.5 min. respectively. We wish to express this system as an overall first order plus dead time (FOPLD) model ie gain, time constant and process dead time. (We will see later that this is often done, to simplify controller tuning). For this exercise, gain is considered to be 1.0. (A) If the transmitter is a flow transmitter whose behaviour can be described by a dead time of 0.2 min. and an exponential lag of 0.5 min. in terms of the overall dead time and overall first order lag how can the system behaviour be approximated ? Overall dead time = Overall time constant = (B) If the transmitter is a temperature transmitter with a temperature sensor in a protecting well whose behaviour can be described by a dead time of 0.7 min. and an exponential lag of 15 min. how can the overall system behaviour be approximated now? Overall dead time = Overall time constant =
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