Please let me know how to calculate the voltage drop of 1P X 1.5sqmm Instrumentation cable (BS 5308 Part 2, Type 2) connected between PLC & PT (4-20mA O/P) placed 1500m apart.

I would be glad if some only helps me with the formula, the ref STD (IEE or IEEE) and other tech details for this query.

 

Responding to Sakthi 24-Jul (16:03) query:

Formula for calculating the DC voltage-drop is V = 2xIxRxL, where:

V = Voltage-Drop, in volts.
I = Current, in amperes.
R = Resistance, in ohms/km.
L = Distance, in km.

Contact me if you don’t already have the resistance of the 1.5mmq cable!

Regards, Phil Corso ([email protected])

 

Shakti,
Yes Indeed the formula is correct.

It is basically the same as OHM law like
v=I*R, (V=2*I*L*Z)
& here it is V=2*I*R. can you update why we have 2 in this formula.

I think it would be due to some standard (IEC or something else).

Also can you update me that if I have a HART transmitter working on 24Vdc. If I have more than 2 volt drop in line the voltage at Transmitter terminal would be 22 volt so what effect would there be on the transmitter, whether it would not work or its performance would be degraded.

Regards,
Usman Khan

 

In a cable 100 m long, there are 2 cores connected in series - one for the out run and one for the return.

Circuit length = 2 x cable length.

Bruce

 

i belive that 2 may be used for resistance of 2 cores

 

Responding to Usman's 11-Aug (19:33) comment:

The ‘L’ in the formula represents one-way circuit length. Therefore, the ‘2’ represents two conductors, each being L in length!

Regards, Phil Corso (cepsicon)

 

Usman,

It doesn't look like anyone answered your 2nd question on how voltage drop on the line would impact the transmitter operation. I can answer that for you.

In regards to how the voltage drop of 2V on the cable will impact the Hart transmitter; it is a matter of how the available voltage of the power supply will be divided up by all the voltage drops of the 4-20mA current loop. This is a function of all 4-20mA loop transmitters and does not matter if they are Hart protocol or not. The transmitter is a voltage drop in the loop, the measurement device (with internal or external shunt resistor) is a voltage drop in the loop, and the wire itself is a voltage drop in the loop. The transmitter voltage drop is based on an internal regulator that will power the signal conditioner electronics. If the required voltage drop total for 20mA (transmitter minimum, wire, & instrument) exceeds the available supply voltage; the loop measurement will not reach full span. That will potentially result in the not working or degraded performance.

Now is this a problem for you? I cannot answer for the specific Hart transmitter that you have as to the minimum voltage required to operate, but you can refer to the manufacturers specification for voltage supply range to see what it is. If you use the 4-20ma rtd transmitter specification at http://www.robertoweninc.com/products/hardware/conditioning/4-20ma/resistance/rtd/hardware.html as an example, you will see a specification "SUPPLY RANGE: 7.5 TO 36Vdc". The key number for the minimum operation voltage is 7.5V. If you look at the 4-20mA wiring example at http://www.robertoweninc.com/applic...4-20ma/resistance/thermistor/roi-usb-xma.html and see that; if the supply is 24V, if the wire drop is 2V, then the maximum drop across the instrument can be is Vsupply-Vtransmitter-Vline=24V-7.5V-2V=14.5V. That would mean that the internal/external instrument shunt cannot exceed r=V/I=14.5V/0.020A=725ohms. While I have seen some loop measurement devices with 600ohms & 1000ohm input shunts, it is very unusual; input shunts generally are less than 500ohms (the 20mA USB A-D at http://www.robertoweninc.com/products/hardware/conversion/usb/a-d/current/roi-usb20.html has an input shunt resistance of 250 ohms). Your measurement device would need to have a shunt resistance larger than 725 ohms to cause a problem with the example transmitter I referenced. You will need to work the numbers the way I explained with the range specification for your Hart transmitter to see what your shunt limit would be.

If you have any other questions, you can reach us through http://www.roi-instruments.com and clicking on "Contact Us".

Regards,
Tech

 

But for voltage drop, we need to have voltage available at transmitter end (+ve terminal). This means only one core resistance should be used. The other core is -ve or zero volt line. So practically only one core resistance to be used for calculation.

Pl. correct if I am wrong.

 

You are wrong. The load will see a voltage drop on each wire. The positive wire will not be quite as positive at the far and the the return (grounded) wire will be above zero as the far end. Use "2L" to compute the voltage at the load