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

 

why are you complicating things and confusing yourself?

when we measure voltage, we don't really care about current. of course we don't want our device used in probing (here voltmeter) to alter circuit condition so it has to have large internal impedance (large "enough") but the current flowing through it is really not something we care about. if you use analog input measuring some voltage, nothing changes - we still don't give a rats behind on the current that flow into particular input.

similarly when measuring current, we don't really care about voltage (as long as voltage changes do not affect reading, for example by reaching range limits).

one of popular ranges for analog signal in industrial automation is 4-20mA. in practice, this is accomplished using:
a) standard 24VDC source (already part of any control system),\
b) transmitter/transducer controlling current according to some external variable (temperature, RPMs, whatever) and
c) load (analog input). load is typically 250 Ohm (but it does not have to be).

many analog input cards have precission 250 ohm included and it must be enabled when using current range (0-20mA or 4-20mA). this translates to 0-5V or 1-5V (analog inout cards internally always measure voltage).

back to your question about "extra voltage drop", it is across analog output (transducer or analog output). it adjust automatically to whatever it needs to be based on loop supply voltage (as mentioned, that is usually 24V) and load resistance (within limits).

for example if you use 2.2kOhm as load then 20mA would require supply that is higher than 44V. you can use ampmeter (close to zero ohm) and transmitter is still supposed to maintain same output (4-20mA) and that is where entire voltage drop will be. but if you used load with too large resistance, transducer will saturate at some point (unable to drive current above certain value). in other words for given supply voltage, there is a maximum load resistance (compare that with voltage type analog outputs, they require loads to have resistance greater than some minimum value).

transducer (analog output) checks if the output current is correct or not and adjusts accordingly (this is very fast, PLC scan time is much slower) . in case of 4-20mA, the 4mA are normally used to power transducer allowing 2-wire loop.

if you want to design your own device with 4-20mA output, why not use ready chips designed just for that?

check http://www.ti.com/lit/ds/symlink/xtr117.pdf

 

back to your question about "extra voltage drop", it is across analog output (transducer or analog output). it adjust automatically to whatever it needs to be based on loop supply voltage (as mentioned, that is usually 24V) and load resistance (within limits).

Normally if I had drawn a circuit with a simple current source and a 250ohm resistor, and set the current to 20mA, I would say the voltage across the current source's terminals is 5V.

However, for a PLC analog output, it injects 24V. If I measure the PLC terminals, I measure 24V. But how can this satisfy voltage loop laws if the wiring drops 1V, and my load is a 5ohm resistive element (dropping only 0.1V at 20mA)? Is there an internal voltage drop within the PLC analog output circuitry?

 

I don't know if it helps you at all but the small 4-20mA panel meters I have need at least about 9V to show anything at all on the LCD. You can't drive even 4mA thru them without that voltage. I speculate that it may act like a 5Ω impedance once that voltage hurdle is met, but there is a much larger voltage drop across it than there would be with a 5Ω resistor, say.

Just probe it with your multimeter and you'll see where the voltage goes.

 

Thanks for the reply. This particular device (RTK P607 Trip Amplifer) requires an external 24V supply. So, whilst the PLC also provides a 24V loop powered signal, the device does not draw its power from this.

 

the current source delivers a current at whatever voltage is required, up to it's supply. It will drop the voltage from supply to loop drive requirements. The loop drive voltage will then be dropped across each respective load.

I worked on a Controllogix PID loop most of the day. 4-2ma sensed hydraulic pressures onto the analog input channels, calculated into .PV, The CV was directed to voltage output, which drove a current converter, which drove the hydraulic proportioning valve.

An ammeter is low impedance.

 

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?

@GetDeviceInfo - if what you were saying were true, I'd expect to see varying voltage values across the terminals of the PLC analog output; dependent on line resistances and current output.

 

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?

If you measure 24vdc across the outputs of a current loop, you have an open circuit.

If you meant to say you measure 24vdc as the driver source supply, then yes, the driver drops the voltage.

Look at it this way, whatever drives the current loop, acts as a voltage regulator across a fixed shunt, which in turn translates into a current regulator. This internal shunt is in series to the current loop. The driver drops whatever supply voltage is required to maintain signal current through the shunt, and subsequently the loop.

@GetDeviceInfo - if what you were saying were true, I'd expect to see varying voltage values across the terminals of the PLC analog output; dependent on line resistances and current output.

Given a fixed signal level, yes, you will see fluctuating voltage in relationship to line resistance.

 

Ahhhh - that's exactly what I found in my tests today! An open circuit, or reverse polarity (I assume my Trip Amp has a protection diode), and I see ~22.8V. When all was connected correctly, I saw a voltage that would roughly be line voltage losses and through the Trip Amp impedance.

Now, back to the Analog Input scenario. My PLC has a common signal return for all instruments, and can supply 24V. When I measure the voltage across the return and signal line, it equals the voltage drop across a 250ohm shunt (so for 3.6mA, I read 0.9V). My question is, can anyone supply a diagram of what the internal circuitry would look like? I can't fathom how the PLC can provide the voltage to the transmitter, and I am only reading 0.9V across the terminals.

 

Ahhhh - that's exactly what I found in my tests today! An open circuit, or reverse polarity (I assume my Trip Amp has a protection diode), and I see ~22.8V. When all was connected correctly, I saw a voltage that would roughly be line voltage losses and through the Trip Amp impedance.

Now, back to the Analog Input scenario. My PLC has a common signal return for all instruments, and can supply 24V. When I measure the voltage across the return and signal line, it equals the voltage drop across a 250ohm shunt (so for 3.6mA, I read 0.9V). My question is, can anyone supply a diagram of what the internal circuitry would look like? I can't fathom how the PLC can provide the voltage to the transmitter, and I am only reading 0.9V across the terminals.

your transmitter will be regulating the voltage across its shunt, in proportion to whatever it's sensing. This then becomes the current loop to your input terminals. Look at your transmitter and will identify what specific output it has.

 

Thanks for trying to work through this with me. Perhaps you could sketch up a schematic describing how the PLC can provide 24V to the instrument, and how I can measure 1V across the return and signal lines. I have attached the start of the diagram, but I don't understand how I am measuring the 1V at the terminals when the PLC has a 24V supply on the signal line;

 

... when the PLC has a 24V supply on the signal line;

It won't have 24vdc on the signal line. What would be the point of that? The 24vdc would be a power source for the transmitter to operate. The transmitters operation is to output a current.

sorry, don't have time for drawing, but you could list the components involved with links to their datasheets, along with how they are connected.

 

here is a typical circuit for loop powered 4-20mA sensor.

the 24V power supply is fixed.
when sensor operates normally it's output is in range 4-20mA.

it is common that PLC analog inputs have a switch to connect resistor. when resistor is connected (Sw. closed) input is used to measure current. when Sw. is open, input is used to measure voltage. this is only hardware setup, additional parameters need to be set in the PLC configuration to specify voltage or current, range, filter setting (50/60Hz) etc. in some cases resistor has to be applied externally (again, only in case we are measuring current). some inputs are meant to measure only current and cannot be configured to measure voltage (resistor is permanently on, there is no switch). standard value for resistor is 250 Ohm and it used to be 1%, it is probably tighter tolerance these days. however - there is nothing to say that resistor value must be 250 ohm. you can use your multimeter (as a miliamp meter) or make own current sensor with different internal resistance. 250 Ohm is common for one reason - it allows same internal range to be used for both V and I measurements.

Standard voltage ranges are:
a) -10...+10V
b) 0..10V
c) 0..5V
d) 1..5V

Standard current ranges are:
e) 0-20mA
f) 4-20mA

using 250 Ohm resistor, internal hardware of the analog card uses same range for 'c' and 'e' (0..5V) and for 'd' and 'f' (1..5V).

When you measure 1V across analog input, loop current is
I=V/R=1V / 250 Ohm = 0.004A = 4mA
and remaining 24V-1V=23V is across sensor.

When you measure 5V across analog input, loop current is
I=V/R = 5V/ 250 Ohm = 0.020A = 20mA
and remaining 24V-5V=19V is across sensor.