dvc 6010 positioner

[shkoko2000]

I measure the voltage on dvc 6010 and found it 10 V but input supply voltage is 24V why did it happen? How the voltage drop from 24 to 10?

 

[Calistodwt]

Remember this is a current loop. Your output device to the positioner (PLC / DCS) adjusts itself to the required value of current between 4-20mA and also supplies the required voltage to the positioner which in your case should be a minimum of 10.5 volts. The balance of available voltage (13.5 volts) is being dropped across the source (PLC / DCS output card) internal resistance and any other item in the loop (example zener barrier if any).

 

[shkoko2000]

Can you explain that more clearly

 

[danw]

The positioner is a load. It does work which requires energy. It uses the energy supplied by the analog output current loop.

The current in a current loop is driven by a power supply capable of some level of voltage output. A load in an electrical circuit is a resistance. Some of the power supply's voltage is dropped across the positioner, because the position is a load.

 

[shkoko2000]

why the transmitter is not load as i mesured the voltage on it and it 24v as input

 

[Calistodwt]

The transmitter is a load but the current through it is set by your process measurement via the transmitter electronics . The voltage at the transmitter will also depend upon what else is in series with it in the loop such as zener barrier (has resistance) and PLC / DCS input card (has resistance). If all three of these items are in the loop then each will have a voltage drop. The total voltage in the loop would then be 24Volts = input card volt drop + zener barrier volt drop + volt drop in signal wires + (remaining voltage = transmitter voltage).
I think that you need to provide details of exactly what items are in each loop to proceed further if the above does not help you.

 

[shkoko2000]

I don't understand why point no. 3 at notes in the attached file happened plz help me

 

[Calistodwt]

What point 3 is saying is that you need a minimum open circuit voltage from the current source (output card) of 10.5 to 11 volts (according to the positioner specifications) before you connect the positioner. Once you connect the positioner this voltage will fall to some value between 9 and 10.5 volts. The voltage difference after connecting the positioner will be dropped across the current source and any other items which are in the loop. If this 10.5 / 11 volts is not available to the positioner then you will not be able to drive 20mA through the positioner.

 

[danw]

Do not put 24Vdc from a power supply directly into the (+) and (-) terminals on a positioner because it will damage a positioner.

A positioner is a load driven by a 4-20mA signal; it is not the loop's current regulator like a 2 wire transmitter, the analog output is the loop's current regulator.