4-20 mA maximum loop distance

4-20 mA maximum loop distance

Really long, you can do more than 10km. However, at this distance, other problems may start to creep in (ground loop if not isolated, EMI, so on).

The distance depends on your voltage source and your actual instrumentation voltage span, and cable diameter of course.

Maximum voltage drop is when the instrument consumes 20mA. However, you want it to work up to 24mA, so that your PLC system can detect ‘over-range’ (maybe due to short circuit, or actual over-range on the instrument side).

Let say, for our example:

  • Loop powered, with 24VDC at source,
  • Our particular instrument only works down to 10 Volts,
  • No HART interface.

Due to the above, you only have 14VDC leeway for transmission loss, or 580 ohms. That equates to 8-ish km for 20AWG cable (66 ohm/km, accounting both cores).

I assume that the cable between the two points is already installed or you’d be asking about how to do it wirelessly.

Is the current signal from a 2 wire loop powered transmitter? Most have a chart like this that shows what voltage is needed to drive a certain loop resistance.

A web source states that the resistance of 18 gauge wire is 0.00751 ohms per foot.

4km = 4,000m
4,000m * 3.281 ft/m = 13,124 ft
13,124 ft * 0.00751 ohms/ft. = 98.6 ohms
98.6 ohms * 2 = ~198 ohms

If the analog input resistance is a typical 250 ohms, then the loop resistance is 250 ohms + 198 ohns = 448 ohms

According to the table above (for that particular transmitter) current in a loop resistance of 450 ohms can be driven by a standard 24Vdc power supply.

Another source gave 21.8 ohms/Km, which over 8 Km (round trip) would give a copper wire resistance of only 175 ohms, so you can shop for answers. I’d pick the conservative value, but that’s me.

Substitute values for your gauge copper wire, your analog input resistance, and check against your transmitter’s loop power requirements.

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