Voltage Drop On 0.75mm2 Instrumentation Cables

Kim Jeck

I have a VEGABAR 82 pressure transmitter (operating voltage=9.6 to 35Vdc) that connected to the Control Panel with a cable length of est. 150meter. The cable used is a single run of 1x2x0.75mm2 instrument cable. I would like to know if the volt-drop along the cable will affect the operation of the transmitters?

From what i understand from the previous topics over here, the volt drop can be calculated using V=2xIxRxL.
What is the current i should use for the calculations as i can't find any current consumption of the datasheet of the transmitter? Should i use the output current 4-20mA?

David_2

The quick answer - there's no problem using roughly 18g wire for a 2 wire, loop powered transmitter over 150m.

Yes, the current used for the calculation is the loop current, but 4-20ma covers only the measurement range without taking into account the fault indication current which can be as high as 22.8mA (those complying to NAMUR standard use 21mA).

But to do a calculation you also need to know what the load resistance is. The analog input used in the current loop has a shunt/dropping resistor to develop an IR drop. I've seen probably a dozen different resistances used by different AI's, but 250 ohms is widely used in the DCS and PLC world. If HART communications is to be used, then HART requires a minimum of 230 ohms to develop the HART signal.

I didn't look up what the Vega's fault current is but I'll use 23 mA as an example.

A rule-of-thumb chart I have shows 0.75mm/18g wire has a resistance of 23 ohm per 1000m. Your wire run one way is 150m for a total of 300m out and back. 300m is 30% of 1000m, 30% of 23 ohms is about 8 ohms for 300m.

The loop voltage needs to drive 23ma through the load resistance, say 250 ohms and a wire resistance of 8 ohms, a total of 258 ohms.

E = I*R; 0.023A * 258 ohms = 5.9V, the IR drop at the maximum output of 23mA.

If you use a 24V dc power supply then the transmitter will 'see' a little over 18V (or higher), the other 5.9V being dropped across the wiring and load at the max 23mA output). 18V+ is well within the operating range spec of 9.6V to 35Vdc. Good to go.

Jack 1

Hi thanks for the breakdown on formula. I'm new to this so was wondering if the formula V=2xIxRxL is applicable to the above. Im getting different figure and not sure what I am doing wrong?

V = Voltage Drop (Volts)
I = Current (Amps) (0.023)
R = Total Resistance = (258Ohm) (Conductor Resistance ohms/km) 8 + (Load Resistance) 250
L = Distance (Meter) = (150m)

David_2

I don't know. I did my calcs shown above from first principles, not from the V=2*I*R*L formula.

In the V=2*I*R*L formula, if L is resistance (Ohms) of a length of cable (not the length itself), then the multiplier '2' might account for the fact that it takes two conductors to make a circuit therefore the total cable resistance is twice the resistance of the length of one conductor.

But the cable resistance should be added the load resistance (analog input dropping resistor) to obtain a total resistance; the cable resistance is not a multiplier factor, it is added to the load resistance.

So whoever uses V=2*I*R*L should explain what its terms mean; I can't.

My worded formula would be

((cable resistance for length of one conductor * 2) + load resistance) * max current = voltage drop at max current

PhilCorso

Gentlemen...
While the Voltage-Drop equation is correct you have confused unit dimensions:
Vd = 2 x A x Rw x L, where:
o Vd is Voltage across Load, in Volts
o A is Loop-current, in Amperes.
o Rw is Wire-Resistance, in Ohms/km.
o L is Length of cable, in km.
Regards, Phil Corso

David_2

2 * 0.023A * 23 Ohms * 0.15km = 0.16V voltage drop in the copper cable

For purposes of determining whether the DC power supply in a 2-wire loop powered circuit has sufficient voltage to "lift-off" the transmitter (minimum 9V lift-off with no load, 14.2V for a 250 Ohm load and cabling), one has to account for the load resistance in the analog input, 250 Ohms.

For a 24Vdc loop power supply:
((2 * 23 Ohms/km) 0.15 km + 250 Ohms) * 0.023A = 5.9V drop in the loop: 5.7V across the AI resistor, 0.16V across the copper cable, and 18.1 volts drop across the transmitter, which is actively regulating the loop current.