How is the power of a capacitive differential pressure transmitter calculated?
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For the capacitive differential pressure transmitters used in the laboratory, the manual states that the supply voltage is 24VDC and the output is 4-20mA. A transformer needs to be purchased to convert it to 24V DC, but it’s not clear whether the power capacity will be sufficient How to calculate the power consumption of a capacitive differential pressure transmitter?
24*0. 020=0. For 48W, even the smallest transformers have a capacity of 3W; it’s totally fine to buy one that is 24 volts and has a capacity of over 3W
It can also be used in conjunction with a secondary meter equipped with power distribution.
The power consumption of the transmitter is very low; a standard 24VDC supply is sufficient
Power is voltage multiplied by current. Current is the output for two-wire differential pressure transmitters; it generally ranges from 22 to 24 mA at most, with a voltage of 24 VDC – you can figure it out by doing the calculations. As mentioned above, this amount is really very small. It’s up to around 0.5W at most.
When operating normally, the power consumption of the transmitter is generally less than 1.5 W; some sources indicate 1.05 W. It depends on the operating conditions on site – higher current levels result in higher power consumption, while lower current levels lead to lower power consumption.
Power hardly needs to be considered; as long as the voltage is sufficient, normal operation is possible. Our factory uses safety barriers to supply power to the transmitters, and the maximum current limit of these safety barriers is generally within 30mA. Think about it: it’s impossible for a 24V DC power supply to provide a current lower than 30mA