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09

Transmitter to an analogue input

A two-wire transmitter is wired to a PLC analogue input as a current loop. The program converts the 4–20 mA signal to a measured value and detects a broken wire.

Transmitter B1 measures the level in the tank over a range of 0–4 m. Loop: +24 V to the + terminal of the transmitter, from its − terminal to PLC analogue input AI1, and from there to 0 V.

How it works

1Tank empty

The level is at the start of the measuring range, 0 m. The transmitter lets 4 mA flow in the loop.

2Half full

12 mA is the middle of the range, which is 2 m.

3Full

The end of the range, 4 m, is 20 mA.

4Broken wire

No current flows in the loop. 0 mA is below the 3.6 mA fault limit, so the program knows that the measurement is faulty.

A transmitter gives the measured value as a current of 4–20 mA. A two-wire transmitter takes its supply from the same loop it signals on, so it needs only two wires.

Wiring

+24 V to the + terminal of the transmitter.
From the − terminal of the transmitter to the PLC analogue input.
From the analogue input to 0 V.

The current is the same at every point of the loop, so even a long cable does not change the signal, as long as the supply voltage is high enough. Use twisted, screened cable for analogue signals.

Scaling

value = start + (I − 4 mA) / 16 mA × (end − start)

Limit values can also be worked out in milliamperes: on a 0–4 m range, 3 m is 4 + 16 × 3 / 4 = 16 mA.

Fault limit

According to the NAMUR NE 43 recommendation, a current of 3.6 mA or less, or 21 mA or more, means a fault. The program checks the fault limit before it uses the value. Otherwise a broken wire looks like an empty tank or a low pressure.

Common mistakes

The limit is calculated without the 4 mA start of the range. 3 m is not 15 mA but 16 mA.
The transmitter is wired to a voltage input. For example, the analogue inputs of the LOGO! base module measure only 0–10 V, so a 4–20 mA transmitter needs a current input.

Sources