Voltage drop in a 4-20mA loop

[PhaseShift]

Lets say I have a 4-20mA loop where my PLC is looking at an input from a field transmitter. My PLC is supplying the 24V for the loop.

So as far as I can tell the 24V dropped across this loop will be dropped in 3places, the wire, the transmitter and the input. Assuming 20mA current flow then the drop in the wire will be small lets just say 2V and the drop at the input (usually 250 ohm resistor) will be 5V. Does this mean that the remaining 17V is dropped across the transmitter? If lets say there was only 4mA flowing, does this mean that 21V would drop across transmitter? Does the transmitter care how much voltage is dropped across it?

What about if my PLC was supplying an output to a device in the field? Now the output transmitter is located in my PLC with the field device most likely having the 250 ohm resistor? So for the same case above, with 20mA flowing, will the 17V drop across the output transmitter in my PLC card? So if I measure between the two analog wires in the field I would see less than 24V (maybe 5V or so) if this current is flowing? Obviously without current flowing I'd seel all 24V between wires in field?

 

[iwire]

I thought the entire reason for using a mA signal was to eliminate errors caused by voltage drop?

 

[jumper]

I thought the entire reason for using a mA signal was to eliminate errors caused by voltage drop?

Me too, this is series wired?:-?

 

[eric9822]

A transmitter will regulate the amount of current to maintain the output in proportion to what the transmitter is measuring. There is a maximum loop resistance that must be taken into account and this is usually specified by the device the transmitter is connected to. Voltage drop in most instances is not something you need to be concerned about in a 4-20mA loop. For a PLC output the card regulates the current and the maximum loop resistance is specified by the PLC manufacturer. The resistance of the controlling device varies and may not necessarily be 250 ohms. A 250 ohm resistor is used to convert 4-20 ma to a 1-5 volt input for devices that require it.

 

[james_mcquade]

The plc is designed to operate using 4-20 ma. The voltage can vary from as little as 8 volts
(picked a number) to as much as 30 volts (AB slc 500) if memory serves correctly.

The point is that this is a current loop and the voltage can and will vary depending on the
transducer.

regards,
james

 

[StephenSDH]

Lets say I have a 4-20mA loop where my PLC is looking at an input from a field transmitter. My PLC is supplying the 24V for the loop.

So as far as I can tell the 24V dropped across this loop will be dropped in 3places, the wire, the transmitter and the input. Assuming 20mA current flow then the drop in the wire will be small lets just say 2V and the drop at the input (usually 250 ohm resistor) will be 5V. Does this mean that the remaining 17V is dropped across the transmitter? If lets say there was only 4mA flowing, does this mean that 21V would drop across transmitter? Does the transmitter care how much voltage is dropped across it?

Yes, the majority of the voltage drop is over the transmitter and can vary. The transmitter adjusts it's impedance to maintain a current flow reg. The voltage at the transmitter is not critical as long as it is in the operating range.

What about if my PLC was supplying an output to a device in the field? Now the output transmitter is located in my PLC with the field device most likely having the 250 ohm resistor? So for the same case above, with 20mA flowing, will the 17V drop across the output transmitter in my PLC card? So if I measure between the two analog wires in the field I would see less than 24V (maybe 5V or so) if this current is flowing?

In this case you might supply the plc card with 24vdc, but the voltage you would record at the +/- terminals of both the PLC and the Field Device would be approximately the same (V = Current * 250 ohm). If the circuits resistance is too large the signal can be clamped because you exceed the available voltage . (ex .020 A * 2000 ohms = 40V so this exceeds 24vdc and the signal would likely be clamped around 24V/2000ohm = .012A)

Obviously without current flowing I'd seel all 24V between wires in field?

In your first scenario yes, in your second scenario 0v.

 

[PhaseShift]

In this case you might supply the plc card with 24vdc, but the voltage you would record at the +/- terminals of both the PLC and the Field Device would be approximately the same (V = Current * 250 ohm). If the circuits resistance is too large the signal can be clamped because you exceed the available voltage . (ex .020 A * 2000 ohms = 40V so this exceeds 24vdc and the signal would likely be clamped around 24V/2000ohm = .012A)

So you are saying that the voltage output from the PLC might not be 24V even though you are supplying it with 24V? Is this due to the fact that the transmitter in the output will adjust its impedance accordingly to maintain loop current as you described earlier and therefore drop different voltages across this ouput? This would then read as a different voltage across the output of the PLC and field device?

I was always told that the field device in this case of an ouput would always have the 24V across it that was supplied in the loop, but it looks like here it is obvious that the voltage across the field device will be a function of resistance and output current only.

In your first scenario yes, in your second scenario 0v.

Why would it be 0V in the second scenario?

 

[StephenSDH]

So you are saying that the voltage output from the PLC might not be 24V even though you are supplying it with 24V?

It won't be. If it applied the full 24v then you would have 96ma (24v/250ohms)

the voltage across the field device will be a function of resistance and output current only.

The voltage is only the result of current * impedance.

Why would it be 0V in the second scenario?

Zero current * 250 ohm = 0 Volts

 

[PhaseShift]

It won't be. If it applied the full 24v then you would have 96ma (24v/250ohms)



The voltage is only the result of current * impedance.


Zero current * 250 ohm = 0 Volts

Is the 250 ohms you are referring to in this situation at the field device?

So then at the PLC output, it will adjust the voltage to supply the required amount of current to the field device?

So depending on output current we will read whatever voltage is being outputted by PLC which will be less than 24V and be adjusted by PLC to drive appropriate current?

 

[steve66]

Is the 250 ohms you are referring to in this situation at the field device?

So then at the PLC output, it will adjust the voltage to supply the required amount of current to the field device?

So depending on output current we will read whatever voltage is being outputted by PLC which will be less than 24V and be adjusted by PLC to drive appropriate current?

Its not the PLC output that adjusts to maintain the right current flow, its the sensor. Don't think of the sensor as a fixed resistor. It is an active network that takes power from the supply, and uses that power to run circuits that drive a transistor. That transistor controls the current flowing through the circuit.

You can think of the transistor as a variable resistor. Its calculated resistance would be controlled by two separate variables. Variable #1 is whatever the sensor is measuring. Variable #2 is the circuit parameters including the supply voltage, and the external circuit resistance.

Lets try an example that shows what happens when we change variable #2 (the supply voltage and resistance.)

Assume a PLC input that supplys 10 volts, and has 250 ohms of resistance. Assume a sensor that measures 0-100 PSI, with 0 PSI being 4 ma, and 100 PSI being 20 ma. Also assume it has a linear output so that at 50 PSI the output is 12 ma. (Thats halfway between 4 ma and 20 ma.)

Draw the 250 ohm input resistance in series with the 10 volt supply. Then draw two wires going over to a resistor to represent the sensor resistance.

Now assume the sensor is measuring 50 PSI. It will adjust its resistance so 12 ma flows in the circuit. Using ohms law, 10V /.012 = 833 ohms. So the total circuit will have 833 ohms of resistance. The equivalent resistance of our sensor will be 833-250 = 583 ohms.

Again, this isn't a true resistor, its actually a transistor adjusting itself so that 12 ma flows in the circuit. But we can calculate an equivalent resistance for this transistor, which is what we just did. This "equivalent resistance" helps us understand what is going on.

 

[steve66]

Continuing the example above, we found that with 10 volts and a 250 ohm PLC input, and 50 PSI on the sensor, the circuit had 12 ma flowing, and the equivalent resistance of the sensor was 583 ohms.

Now lets assume the PLC input changes to 24 volts, and 300 ohms.

The sensor still sees 50 PSI. So it adjusts itself to keep 12 ma flowing. Now we have 24V/ .012 = 2000 ohms of total circuit resistance. Subtracting the 300 ohms of the PLC input, we see that the sensor has adjusted its resistance to be 2000-300 = 1700 ohms. (Again, its really a transistor, but we can calculate an eqivalent resistance.)

Now, our calculation has assumed a perfect sensor (no change in output current for a change in input voltage or resistance), but most real sensors are pretty close to ideal. But I think you get the idea.

Steve

 

[PhaseShift]

Continuing the example above, we found that with 10 volts and a 250 ohm PLC input, and 50 PSI on the sensor, the circuit had 12 ma flowing, and the equivalent resistance of the sensor was 583 ohms.

Now lets assume the PLC input changes to 24 volts, and 300 ohms.

The sensor still sees 50 PSI. So it adjusts itself to keep 12 ma flowing. Now we have 24V/ .012 = 2000 ohms of total circuit resistance. Subtracting the 300 ohms of the PLC input, we see that the sensor has adjusted its resistance to be 2000-300 = 1700 ohms. (Again, its really a transistor, but we can calculate an eqivalent resistance.)

Now, our calculation has assumed a perfect sensor (no change in output current for a change in input voltage or resistance), but most real sensors are pretty close to ideal. But I think you get the idea.

Steve

Steve your example was excellent.

Will the voltage across the field transmitter also change as it varys its impedance? I ask this because we mentioned before that since the whole 24V must drop across the loop then depending on the mA signal the voltage across the transmitter will change?

Can you give a similar example as above for how an analog ouput from a PLC output works?

 

[steve66]

Steve your example was excellent.

Will the voltage across the field transmitter also change as it varys its impedance? I ask this because we mentioned before that since the whole 24V must drop across the loop then depending on the mA signal the voltage across the transmitter will change?

Yes the voltage will change, but not necessarily by alot. It's really just a matter of applying ohms law and basic circuit principles to the circuit. (basic principles, like the one you mentioned -the entire input voltage has to be dropped across the 3 things you mentioned). The thing to remember is that the sensor will do what it can (within its design and specifications) to keep the right current flowing in the circuit.

Can you give a similar example as above for how an analog ouput from a PLC output works?

Its very similar to the above example, and I'll bet you can work it out. Just consider the output to be a source voltage, and a "variable resistance" that again is probably a transistior. Then add a load resistor. Again, the transistor will vary its resistance in an attempt to maintain some set output current. As the load resistor increases in value, the source resistance will decrease.

And thanks for saying it was a good example. I'm glad it helped.


Steve

 

[stevenj76]

4/20 sucks.

2-10V rules!

 

[PhaseShift]

What if there was a loose connection in the 4-20 loop that caused an increased impedance? Would this effect the input signal at all.

Since you are saying the field device always adjusts itself to keep the correct mA current in the loop, then wouldn't it recognize this increased impedance and adjust its impedance to still supply the correct mA current. Wouldn't the signal then still be the same?

 

[steve66]

What if there was a loose connection in the 4-20 loop that caused an increased impedance? Would this effect the input signal at all.

Since you are saying the field device always adjusts itself to keep the correct mA current in the loop, then wouldn't it recognize this increased impedance and adjust its impedance to still supply the correct mA current. Wouldn't the signal then still be the same?

Yes, it will compensate to a certain extent. But there are limits. If you have a 24V supply, and 10K of resistance, there is no way the field device can push 20 ma.

I'm also sure there are specifications on accuracy too - the accuracy of a device is usually specified at a certain loop impedence. As that impedence changes, you may loose accuracy.

 

[Cmdr_Suds]

4/20 sucks.

2-10V rules!

Try running a 2-10v signal 5 miles down a phone line. you will have a lot of noise and quite a bit of voltage drop(good by accuracy!). with a 4-20mA signal, your distance is really only limited by the excitation voltage on the loop (you know, that external power supply you have hooked up). If your sensor is simple (i.e. pressure or temperature as opposed to a mag flow meter) you only need your two wires. Even in a 4-20mA loop, the way the current is measured is by running it through a shunting resistor and measuring the voltage across it. That is why if you have a PLC with a voltage input, you can place a resistor across its terminals and now you have a current type input! Actually, 2-10v is what you get when you run your 4-20mA through a 500 ohm resistor.

Think of current as water flowing in a pipe with the current being the flow (gal/min) and the voltage as the pressure required to push the water through the pipe. The amount of water that goes in one end must come out the other end, no matter how long the pipe. Whereas the pressure at the supply end will be much greater then the pressure at the outlet end. But in the end, they(E,I & R) all have to be there.