4-20 mA problem with impedance or generic problem?

[micapp]

Hello. I try to use an Unitronics V200-unit (on a 570) 4-20mA output to operate a Samson 4-20mA-operated positioner. To start this positioner I must have 5 V across input (-impedance). When I use a Fluke 4-20mA simulator these 5V is generated at 3.8mA. When I connect the V200 to the positioner I must send between 8 and 9 mA to aquire the same 5V. This is a big problem!! Can anybody tell me how to solve this?

[Simon]

It sounds like maybe the Fluke simulator is introducing additional resistance into the circuit, hence the lower current required to generate the 5V. This also depends on where you are measuring the 5V.

It seems odd that a 4...20mA device requires a specific voltage across its terminals to operate. Nornmally the voltage must be free to change, based on the current and series resistance.

What is the impedance of the 4...20mA input on the Samson positioner?

[Dominic]

I recently had a customer who had an issue with a 4-20mA transducer input to an expansion I/O module. The problem was that the transducer's output was out of scale. For example, when it should have been sending 10mA it was sending 13mA, when it should have been sending 20mA is was sending 25mA. The problem was resolved by moving the input module's ground wire to the machine earth ground point where all the other ground wires were. So proper grounding is very important. I suggest making sure the grouning is done properly on all the components and controls.

[TT_ZX]

I have the same issue with the analogue outputs not being able to drive a Spirax Sarco SP400 positioner. These positioners require 7V at 4mA. The positioner would not power up until the output is close to 12mA. The impedance of the positioner is inversely proportional to the current. As the current to the positioner increase, the voltage actually reduces. You can test this yourself by putting a voltmeter across the positioner and drive it through the range with the fluke signal generator.

I have been using Seneca K109UI or K109S isolating transmitters between the PLC and the positioner to overcome the problem. Expensive, but I don't know of better way. You could range the positioner to work from 9-20mA. This is a bit of a hack that I used with the SP400's until I got the isolating transmitters.

[Simon]

That's an interesting characteristic of those positioners, something I will remember for future occasions.

If that is the case, then it seems that the isolating transmitter is the best option.

It would be interesting to understand why the Unitronics output can't drive that type of input, whilst the signal generator and the isolating transmitter can.

[Simon]

Are those positioners loop-powered, or do they have an auxiliary power supply?

[TT_ZX]

My issue is with loop powered positioners. I wouldn't expect problems with self powered positioners. It seems as though a minimum amount of power is required from the loop for the internal circuitry. Therefore at lower currents a higher voltage is required. Ie 7V x 4mA = 28mW. The impedance in my case is 1750 ohms at 4 mA (7V / 4mA = 1750). Way above the maximum of 500 ohm. Yet at 20 mA the impedance was something like 350 ohms.

I think signal generators have a maximum voltage which is available at any current. 10V is all that is required to supply 20mA to a 500 ohm load. 10V could supply 4 mA in to a 2500 Ohm load. I don't know why the Unitronics analog output modules can't supply 7V at 4 mA.

[Simon]

Generally when selecting analogue outputs, it is necessary to be aware of any loop-power requirements. I have dealt with other brands of I/O systems, and the default offering is usually for auxiliary powered devices. If loop power is required then a specific analogue output module must be used. This is at least the case in industrial automation, it may be different in other industries where loop power may be more the norm.

As far as I know, Unitronics devices do not provide analogue loop power on their outputs.