Analog Input/Output on 4 to 20mA - Where Does the Extra Voltage Go?

[mriksman]

Hi,

I was pondering the workings of 4-20mA loops for both inputs (transmitters to a PLC), and outputs (PLC driving an I/P or even a Trip Amplifier).

A loop powered instrument say has a PLC driving a 24V supply to the instrument. Lets say there is 1V drop on the cabling, and the instrument needs a minimum of 12V to run. It then regulates the current to provide the signal back to the PLC. The PLC say has a 250ohm resistor to convert this current to a voltage for measurement; there goes another 5V (20mA * 250ohm). So where does the extra 6V disappear? It just dissipates in the transmitter? Where? Would the transmitter, if I measured across the terminals, show a voltage drop of 18V?

Now I have an analog output. For my pondering it is simply transmitting a 4-20mA to a Trip Amplifier, which on the specifications says has a 5ohm input impedance (the device is a RTK P607). I have measured the voltage with the Trip Amplifier disconnected, so the wires are simply connected to the PLC. I am reading 22.5V. I presume the losses are due to the wiring. OK, so the PLC has a 24V output signal, and can drive a 4-20mA signal. Well, at 20mA, where is the voltage dropped? If connected to the Trip Amplifier, wouldn't there be a drop of 0.1V? (20mA * 5ohm)? Where is the rest of the voltage dropped?

I'd like someone to tell me if I had my voltmeter - where I would find each and how much voltage drops at what location.

Oh, and a side question. The 'senior' guy said to me, 'you can measure the current on the analog output wires using an ammeter because it is high impedance'. Ummm, isn't the ammeter a low impedance device? Essentially a dead short? I didn't check, but I can only surmise the current whizzed through the ammeter, and none went on to the Trip Amplifier at that point. If the Trip Amplifier had actually have been in operation with a low trip point, I can only guess it would have tripped the pump!!

Thanks for any information.

Mike

 

[jayanth.devarayanadurga]

Post your circuit.

 

[mriksman]

Errmm. Nothing crazy, just a transmitter to PLC for the analog input example, and a PLC to resistive shunt for a Trip Amplifier.

An example circuit for the AI is here; https://www.allaboutcircuits.com/vol_1/chpt_9/3.html

This is a generic theory question, not anything in particular to a circuit I may or may not have.

 

[chuckey]

As I see it, you have the wire resistance (two wire but summed as one), a current monitoring resistor (at your receiver) and the variable resistance of the transducer. So select your monitoring resistor, so it and the wire resistance at 20 mA will still leave more then 12V for the transducer. So Vout at transducer zero input = 4 mA X mon resistor, Vout max = 20 mA X mon resistor.
Frank

 

[mriksman]

You have described the Analog Input circuit; a transmitter to the PLC. Let's say the wires drop 1V total, and the PLC has a 250ohm shunt. At 20mA then, we have dropped 5V at the PLC, and 1V on the line. The PLC provides 24V loop power. HOW does the transmitter dissipate this extra voltage?

And what about a PLC providing the 4-20mA signal and 24V. Let's say it simply output a 20mA signal to a resistive element of 5ohm. 0.1V would be dropped there (at 20mA), and 1V in the lines. So where is the other 22.9V dropped/dissipated?

 

[chuckey]

Across the terminals of the transducer.
Think about a 1.5V battery, it will deliver 1.5V from a current of 0mA to 100 mA (or more). What you have is a CURRENT battery that will deliver 20 mA across a wide range of voltages, 0 -> 12 V.
Frank

 

[mriksman]

Thanks. What every one is saying makes sense; but try and clarify/focus on this point;

I measure 24V across the terminals of a PLC analog output. Line resistance drops 1V, and my Trip Amplifier uses a 5ohm impedance; so at 20mA it drops 0.1V. Where is the other 22.9V dropped? Why can't I find anywhere on the loop to measure a 22.9V drop? Is it internal inside the PLC circuitry?

 

[kripacharya]

A 4ma-20ma loop is driven by a controlled current source, not a fixed voltage source.
A current source would ideally have the capability to pump 4-20ma even into an infinite (i.e. very large) resistance, but this would require an infinite (i.e. very large) voltage. Obviously this is not practical. A practical system would use a voltage rail of some highesh value - in your case a 24v supply.

With such a supply, the largest loop which can be properly driven would be ~24v/ 20mA = 1200 ohm.
For anthing less than this, the current-control feature would automatically limit the current, and this would be visible on the output as a reduced voltage. The 'extra' voltage would be dropped inside the current-source circuitry. So - yes - its dropped internal to the PLC circuitry.

I suspect you measured the voltage (if at all you actually did) with an open loop, where the CCS is trying to drive the output to 4-20mA, but cannot, and hence hits its ceiling voltage.

see also Chuckey #6 above

 

[mriksman]

Thanks. I checked a few voltage measurements today, and have a bit better understanding. For the analog input, the PLC has a return/common, and signal. Something like my attached drawing.

Where would the 24V be in this case if I am measuring the 0.9V drop across return and signal? 0.9V is the 3.6mA my transmitter is sending multiplied the shunt of 250 ohms of the PLC.

 

[kripacharya]

Why this fixation that there has to be 24v someplace ?
Lets clarify this point first and foremost.

thanks

 

[mriksman]

Well... if the instrument requires a minumum of 12V. There has to be a voltage drop of 12V on the instrument?

And let's say there was an external power supply, and you connect the 24V supply between CH1 (signal line), and the transmitter to supply the power, then there is 24V there.

... Hence my confusion...

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This further helps my confusion; http://iamechatronics.com/notes/general-engineering/267-troubleshooting-current-loops

About half way down;

The only voltage measurement that directly and accurately correlates to loop current is the voltage directly across the 250 ohm precision resistor. A loop current of 4 mA will yield a voltage drop of 1 volt, 12 mA will drop 3 volts, 20 mA will drop 5 volts, etc.
A voltage measurement across the transmitter terminals will show us the difference in voltage between the 26 volt power supply and the voltage dropped across the 250 ohm resistor. In other words, the transmitter’s terminal voltage is simply what is left over from the source voltage of 26 volts after subtracting the resistor’s voltage drop. This makes the transmitter terminal voltage inversely proportional to loop current: the transmitter sees approximately 25 volts at 4 mA loop current (0% signal) and approximately 21 volts at 20 mA loop current (100% signal).

 

[kripacharya]

this is all very simple ohms law/ kirchoff/ et alia. Does it really take a 12 msg thread to work out ? I think not.
You need to sit and read the posts, and figure this one out for yourself now dude.

LoL !!

 

[mriksman]

Great post.

If it is so simple, how about you post up a simple schematic showing how a PLC can provide 24V, at the same time I cannot measure 24V anywhere (only the 0.9V, 3.6mA across the signal and return line). I get that a current source regulates voltage, but that doesn't explain how if an instrument needs 24V, then I shouldn't be able to measure 24V somewhere around the loop. Use the simple laws, and show at what points will I see what voltage drops on a simple diagram.

Picture is worth a thousand words eh?

Cheers.

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Something like this;