Power issues with two-wire instruments

The power of the transmitter is only this much.

4~20mA is the signal current. Is the current consumed by the instrument itself also added to this 4-20mA signal? I think it should be applied on top of 4-20mA. Asking experts too?

This post was last edited by qugd on 2015-12-17 at 10:01. The actual power of a two-wire transmitter is not calculated using the simple formula of current multiplied by voltage, as the maximum supply voltage for two-wire instruments can reach 32VDC, while the minimum voltage can be less than 10VDC; moreover, the current can exceed 22mA. The power of a two-wire transmitter remains essentially constant, while its current output can be regarded as the amplified output current of a transistor; however, the total power of the transistor stays within a certain range and is basically constant. 24VDC power supply refers to the open-circuit voltage, that is, the voltage when the resistance between the two terminals is infinite; it is also known as the rated voltage. When a two-wire transmitter is connected to the circuit, it can be regarded as a variable resistor, and the voltage drop across this resistor is not 24VDC; you can use a meter to verify this if you doubt it. For two-wire transmitters, the power is generally between 0.2 and 1 W, as there may be other resistive loads in the transmitter’s circuit. Such as sampling resistors, indicating instruments, etc. The diagram below shows the schematic of a common pressure transmitter. When powered by direct current, the instrument contains various functional components responsible for power conversion, modulation and demodulation of internal measurement signals, amplification, and display. http://www.sokyowh.com/uploads/allimg/140314/8-140314154GKP.jpg

It can’t be calculated that way; what you have here is only the power required for the transmission output, as well as the power needed for data acquisition by the sensor and for the DSP processing itself. Although it won’t be very high