4-20mA Inputs

[fifty60]

I am trying to understand how 4-20mA analog inputs work. Here is how I see it: 2 terminals from the device sending the signal feeding 2 terminals receiving the signal. The 4-20mA signal is generated from a transistor circuit in the sending device and sent from one of its sending terminal in the receiving terminal in the sending divice, where the current is transformed and interpreted by another transistor/resistor/op-amp circuit.

The next step may seem obvious but I have never read it anywhere to verify it. The current then leaves out the 2nd terminal of the receiving device and is sent back into the 2nd terminal where it is grounded.

A 4-20mA analog output system consists of 2 wires. One wire to send the signal, and one wire to return the signal. The current that leaves the sending device is returned to it, and that is why we need 2 wires.

The reason for my question is that I am trying to find a way to gurantee that I have a low signal (4mA or less) under certain situations, and then during other situations I want the analog output to be proportioned by a PID controller. If I install a relay between the wires of the 4-20mA analog output that will short them together until I need the PID controller. When I need the proportioned control I can open the relay and the 4-20mA will function as normal.

Does this sound reasonable and harmless? If the 4-20mA wires are shorted, all of the current will bypass the receiving device and be sent directly back to the sending device.

 

[Shoe]

Maybe I'm misunderstanding, but why not just install a relay in series with the circuit (rather than parallel) and open the relay when you don't want a signal to the PID, close when you do?

 

[GoldDigger]

I am trying to understand how 4-20mA analog inputs work. Here is how I see it: 2 terminals from the device sending the signal feeding 2 terminals receiving the signal. The 4-20mA signal is generated from a transistor circuit in the sending device and sent from one of its sending terminal in the receiving terminal in the sending divice, where the current is transformed and interpreted by another transistor/resistor/op-amp circuit.

The next step may seem obvious but I have never read it anywhere to verify it. The current then leaves out the 2nd terminal of the receiving device and is sent back into the 2nd terminal where it is grounded.

A 4-20mA analog output system consists of 2 wires. One wire to send the signal, and one wire to return the signal. The current that leaves the sending device is returned to it, and that is why we need 2 wires.

The reason for my question is that I am trying to find a way to gurantee that I have a low signal (4mA or less) under certain situations, and then during other situations I want the analog output to be proportioned by a PID controller. If I install a relay between the wires of the 4-20mA analog output that will short them together until I need the PID controller. When I need the proportioned control I can open the relay and the 4-20mA will function as normal.

Does this sound reasonable and harmless? If the 4-20mA wires are shorted, all of the current will bypass the receiving device and be sent directly back to the sending device.

The only things that you have left out which will have an effect on the operation of the circuit that you need are:

1. The source of the voltage which is driving the current source can either be at the sensor (active sensor) or at the receiver (passive sensor). In the first case you will have three or maybe four wires going to the sensor, namely the current loop + and - and the supply + and -.
The results when you short out the terminals will be somewhat different depending on which setup your control is using.
2. Although one of the two loop wires is usually grounded, this is not a requirement in building a 4-20ma loop circuit. Particularly in the case of a passive sensor the whole circuit may be floating.
3. There are two reasons that the current range from 4 to 20 ma is chosen rather than 0 to 20:
a. So that there will always be current flowing in the circuit to power the sensor in the passive case or the receiver in the active case, and
b. So that an open circuit or other malfunction can be detected by the fact that the current drops below 4ma.

So, if you have an passive sensor circuit, shorting out the loop leads will cause the receiver to sense either a malfunction or a 20ma condition.
For an active sensor circuit, shorting out the leads may damage the sensor and will cause the receiver to sense either a malfunction or a 4ma condition.

A much more predictable and reliable way to get the result you want is to have the relay remove the active or passive sensor from the loop and replace it with an active or passive circuit that always regulates to a 4ma current. You can usually do this with just a single-pole double-throw relay and an appropriate dummy sensor.

 

[GoldDigger]

Maybe I'm misunderstanding, but why not just install a relay in series with the circuit (rather than parallel) and open the relay when you don't want a signal to the PID, close when you do?

The OP's problem is that the PID will be the source of the signal, not the receiver. But it might indeed be simpler to add an "enable" control to the PID which is driven by the switch or relay and causes the PID to generate either a proportional signal or a fixed 4ma output.

 

[steve66]

A signal outside the 4-20ma range is usually considered to be an error, or a fault in the sending device or path.

If you are using a PLC, it would be much better to just send the On/Off signal to a digital input instead of the relay. Then use programming in the PLC to decide when you want to watch the 4-20ma signal, and when you want to ignore it depending on if the digital input is on or off.

 

[fifty60]

Thanks for all of the explanations/suggestions. I understand what I am looking at a lot better now.