What is the relationship between the current signal output by the regulator and the air pressure output by the positioner?
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What is the relationship between the current signal output by the regulator and the air pressure output by the positioner? Does an increase in the current signal output by the regulator result in an increase in the air pressure output by the positioner? If that’s the case, when the regulator loses power, shouldn’t the air pressure signal output by the positioner be 0?
It can generally be used for fault detection, but there is no absolutely proportional relationship. This is how I understand and apply it; I’m not sure if it’s correct.
When the actuator is in the direct mode, there is a proportional relationship; otherwise, it’s the exact opposite!
Determine the forward and reverse actions of the locator, as well as those of the regulator. A comparison will show the difference.
1. If the valve positioner is of the direct-acting type, an increase in the current signal output by the regulator results in a decrease in the air pressure output by the positioner. 2. If the valve positioner is of the reverse-acting type, an increase in the current signal output by the regulator results in an increase in the air pressure output by the positioner
When the regulator loses power, the air pressure isn’t necessarily zero; it depends on the actuator you choose. If the actuator you select remains in an open state, a closed state, or maintains its current state when there is a fault, then the required air pressure will also vary accordingly