Increase Transmission Distance for 4-20mA Pressure Transducer
According to some theories, a pressure transducer using 4-20 mA output signal transmit the current signal on the signal line, so its transmission distance should be unlimited. In theory, the transmission distance of the 4-20 mA current signal can be infinitely only when the impedance of the transmission cable is zero. However, this statement is for specific conditions. When the voltage of the power supply of the pressure transducer is reduced to a certain extent or the resistance of the wire is large to a certain extent, 4-20 mA current transfer signal will create a voltage drop across the line.
This paper discusses the transmission distance of the 4-20 mA signal of the pressure transducer under the premise of ensuring the specified accuracy. The essence is to discuss and determine the transmission distance allowed by the 4-20 mA signal.
The specifications that determine the wire length of the current source pressure transducer are as follows. Load resistance, RL. Resistance of the connecting wire, r. Supply voltage V0 and its fluctuation range ΔV. Maximum output current of the pressure transducer, Imax. The lowest supply voltage when the pressure transducer maintains the maximum operating current, Vmin. We assume that the load resistance is known and set RL = 250Ω.
V0=24V DC, the allowable error is 24V+10%-5%, then the power supply allows fluctuation ΔV=24V*5%=1.2V; Imax=20mA=0.02A.
For various types of pressure transducers, the value of the minimum supply voltage Vmin are different. Since this specification is also related to the characteristics of electronic components. Some pressure transducers have a minimum voltage of 9V (refers to no load). Most pressure transducers are between 12 and 36V.
Now assume that the minimum voltage is 16.28V. That is Vmin=16.28≤24-1.2-0.02(250+r).
The resistance of the connecting wire r=24-1.2-16.28/0.02-250=76Ω.
The connecting wires of the pressure transducer are mostly made of copper wire, and most of the sections are S=1.5 and 0.8mm2. The resistivity of copper wires at 20 ℃ and 75 ℃ can be found on relevant information. It is most suitable to calculate the resistivity at 75 ℃. It is known that when t = 75 ℃, the resistivity of copper is ρ = 0.0217 Ω·mm2 /m.
According to L=Sr/ρ, the maximum length of the copper wire can be calculated.
When using a wire with a nominal cross section of 1.5 mm2, L = 1.5 * 76 / 0.0217 = 5253 m. When using a wire with a nominal cross section of 0.8mm2, L=0.8*76/0.0217=2801m.
Since the wiring of the pressure transducer is two lines, the calculation result should be divided by 2:
When using a wire with a nominal cross section of 1.5 mm2, L1=5253.5／2=2626m.
When using a wire with a nominal cross section of 0.8 mm2, L1=2801.8／2=1400m.
In practical applications, the true length of the wire should be lower than the calculated value because the nominal cross section of the wire is almost always high. If the distance on site exceeds the above length, the wire diameter of the wire can be increased to reduce the wire resistance. It is also possible to appropriately increase the supply voltage of the pressure transducer.
After the above calculations, some users may think that the actual wiring length on site is not very long, but it still cannot be used normally when it is used. What is the reason for this? The reason is possible that when the current transmission signal wire is too long, the necessary shielding measures are not taken. Interference was introduced without following the one point grounding rule.
In the circuit system, the point with the lowest potential should be selected as the common point of the signal. In the pressure transducer with current source output, the negative line potential of the 24V power supply is the lowest, which is the signal common line. Therefore, among the set of signal lines responsible for current transfer, the one with the lower potential is the negative signal line. This line should be connected to the 24V negative line at the output of the pressure transducer. If the wiring is incorrect or missed, it will cause the signal negative line to float and introduce interference.
This paper discusses the transmission distance of the 4-20 mA signal of the pressure transducer under the premise of ensuring the specified accuracy. The essence is to discuss and determine the transmission distance allowed by the 4-20 mA signal.
The specifications that determine the wire length of the current source pressure transducer are as follows. Load resistance, RL. Resistance of the connecting wire, r. Supply voltage V0 and its fluctuation range ΔV. Maximum output current of the pressure transducer, Imax. The lowest supply voltage when the pressure transducer maintains the maximum operating current, Vmin. We assume that the load resistance is known and set RL = 250Ω.
V0=24V DC, the allowable error is 24V+10%-5%, then the power supply allows fluctuation ΔV=24V*5%=1.2V; Imax=20mA=0.02A.
For various types of pressure transducers, the value of the minimum supply voltage Vmin are different. Since this specification is also related to the characteristics of electronic components. Some pressure transducers have a minimum voltage of 9V (refers to no load). Most pressure transducers are between 12 and 36V.
Now assume that the minimum voltage is 16.28V. That is Vmin=16.28≤24-1.2-0.02(250+r).
The resistance of the connecting wire r=24-1.2-16.28/0.02-250=76Ω.
The connecting wires of the pressure transducer are mostly made of copper wire, and most of the sections are S=1.5 and 0.8mm2. The resistivity of copper wires at 20 ℃ and 75 ℃ can be found on relevant information. It is most suitable to calculate the resistivity at 75 ℃. It is known that when t = 75 ℃, the resistivity of copper is ρ = 0.0217 Ω·mm2 /m.
According to L=Sr/ρ, the maximum length of the copper wire can be calculated.
When using a wire with a nominal cross section of 1.5 mm2, L = 1.5 * 76 / 0.0217 = 5253 m. When using a wire with a nominal cross section of 0.8mm2, L=0.8*76/0.0217=2801m.
Since the wiring of the pressure transducer is two lines, the calculation result should be divided by 2:
When using a wire with a nominal cross section of 1.5 mm2, L1=5253.5／2=2626m.
When using a wire with a nominal cross section of 0.8 mm2, L1=2801.8／2=1400m.
In practical applications, the true length of the wire should be lower than the calculated value because the nominal cross section of the wire is almost always high. If the distance on site exceeds the above length, the wire diameter of the wire can be increased to reduce the wire resistance. It is also possible to appropriately increase the supply voltage of the pressure transducer.
After the above calculations, some users may think that the actual wiring length on site is not very long, but it still cannot be used normally when it is used. What is the reason for this? The reason is possible that when the current transmission signal wire is too long, the necessary shielding measures are not taken. Interference was introduced without following the one point grounding rule.
In the circuit system, the point with the lowest potential should be selected as the common point of the signal. In the pressure transducer with current source output, the negative line potential of the 24V power supply is the lowest, which is the signal common line. Therefore, among the set of signal lines responsible for current transfer, the one with the lower potential is the negative signal line. This line should be connected to the 24V negative line at the output of the pressure transducer. If the wiring is incorrect or missed, it will cause the signal negative line to float and introduce interference.