4-20mA signal and voltage drop 2

[rockman7892]

If I have a 4-20mA signal going through a 500ohm resistor then I calculate at 20mA I get 10V. This makes sense to me however the source of the 4-20mA signal is a 24V source and since the resistor is the only device in the circuit it would only make sense that the full 24V dropped across this 500ohm resistor.

So my confusion is, when looking at it from a KVL standpoint I see the full 24V dropping across the resistor however when I look at just the current flowing through the resistor at 20ma I only see a max voltage drop of 10V across the resistor. The two terminals of the resistor are then an input to an amplifier card.

In this scenario is the 24V neglegiable and only the 4-20mA is used to cause a voltage drop? Can someone help clear this up for me??

 

[SteveWag]

You last paragraph is pretty much in target. The 24 volts is not negligible though, it is the cause of the current. The device, whatever it is, regulates the current to a value between 4 and 20 ma, depending on the process variable. When 20 ma is flowing, 10 volts is developed across the 500 ohm R and 14 volts is developed across the instrument.
Steve

 

[IRstuff]

Did you forget what you learned about "compliance" of a current source? A current signalling system is usually designed to handle more than one load in series.

TTFN

FAQ731-376

 

[rockman7892]

SteveWag

In my case I am referencing a 4-20mA output. This output is going through a 500ohm resistor which is in parallel with inputs to an amplifier card. I'm assuming the amplifier card is reading the voltage drop across this resistor. Are you saying that in this case 10V is dropped across the resistor and the other 14V is going to the amplifier card however the amplifier card is reading only the 10V? The amplifier card is powered by a 12V DC source seperately.

I know an amplifier card might be an odd case but what about another instrument? Basically what you are saying is that the 10V is dropped across the resistor for signaling purposes, and the other 14V is used to power the insturment? Although the instrument is powered by the 14V its signal reference is coming from the 10V? Does this sound correct?

 

[waross]

1> A 24 volt supply is connected in series with an instrument.
2> The instrument will cause whatever voltage drop across its terminals as is needed to draw the required current.
3> If the measured input is zero the instrument will cause enough voltage drop to draw 4 milliamps.
4> If the measured input is 100% the instrument will cause enough voltage drop to cause 20 milliamps to flow.
5> If you add a 500 ohm resistor, or a 500 ohm device, the voltage across the device will depend on the current flowing.
6> The voltage across your 500 ohm resistor will vary from
2 volts @ 4 ma to 10 volts @ 20 ma.
7> In regards to the value being measured, zero and 100% are defined when the instrument is calibrated and either or both may be elevated or depressed.
8> Your loop will be, A> 24 v supply. B> 500 ohm resistor, voltage drop will depend on current. C> Instrument. current and voltage drop will depend on the measured variable.
9> You may add indicators, controllers, transducers, line resistance, alarm relays, etc. as long as the total voltage demand at 20 ma does not exceed the output of your ppower supply. (Usually 24 volts)

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[SteveWag]

I was assuming a simple series circuit consisting of a loop powered transmitter, a 500 ohm resistor and a 24 volt power supply. Then a high impedance amplifier connected across the 500 ohm resistor. If this is not what your circuit looks like, then I have no answer. You said “and since the resistor is the only device in the circuit”, if so where does the 4-20 ma. signal come from? If you have a 24 volt power supply and a 500 ohm resistor, then, yes, the voltage across the resistor is 24 volts and the current is 48 ma. I thought you had a 4–20 ma. loop because you said you had 20 ma. If you do have a loop powered 2-wire device in the loop along with a 500 ohm resistor and a 24 volt supply, then my answer is correct.
Steve

 

[IRstuff]

I think you're still confused. The 24V supply does NOT appear on the current output UNLESS the resistance exceeds 24V/20mA = 1.2 kohm.

Apparently, you chose to ignore my comment about "compliance," which is how an ideal current source works.

TTFN

FAQ731-376

 

[dpc]

http://www.dataforth.com/catalog/pdf/an104.pdf

 

[roydm]

Rockman,
Welcome to the mysterious world of instrumentation.
Your typical instrument transmitter is a 2 wire loop powered device that puts out 4-20 mA. Think of the transmitter as a constant current device, the current depending on the input.
At minimum input the output will be 4 mA
At 50% input the output will be 12 mA (note below)
At 100% input the output will be 20 mA
This current within reason is independent of the supply voltage or load in series.
A typical transmitter requires a minimum of about 11 Volts, the remainder is available to drive the load. E.g. for a load of 250 Ohms you need a supply of at least 16 Volts (11 for the transmitter 5 for the load). The minimum Voltage is given in the transmitter manual.
If you need to supply more loads you need a higher supply voltage, at 24V you will be able to supply 2 x 250 Ohm loads and still have 3 volts to spare for voltage drop in the interconnecting cables.

Note: Not all transmitters have a linear relationship between input and output for example differential pressure flow meters typically have a square root relationship.
Quite often a transmitter will have an offset zero. Sometimes the input to output is inverted

Hope this helps
Roy

 

[rockman7892]

I'm still a bit confused so to make it simplier to understand I'll use a simple example. I have an input card on a PLC that is powered by 24V. On this input card I have a 4-20mA signal coming in from some field device such as a transmitter. If I have a two wire circuit going out to the transmitter but not landed on the transmitter should I measure 24V across the two wires? When I connect the device should I then still read 24V across the transmiter terminals? What if the transmitter has its own 24V supply as well?

How about the same scenario except now the field device is recieving an output from an output card supplied by 24V?

This is still I little foggy to me so I'm hoping to clear it up so I understand.

IRstuff I did take your advice on voltage compliance however I could not find any information on it.

 

[IRstuff]

1. No, the voltage across the resistor is solely a function of the transmitter current, as it should be, if you think about it.
2. No, the voltage across the resistor is solely a function of the transmitter current, being directly proportional to the current being supplied
3. No, the voltage across the resistor is solely a function of the transmitter current, and has little to do with the voltage supplied to the transmitter
4. No, the voltage across the resistor is solely a function
of the transmitter current


TTFN

FAQ731-376

 

[rockman7892]

I think I'm getting mixed up between ideal voltage and current sources. In the attachment the top drawing shows a 24V voltage source in series with a transmitter having a 500ohm resistance. I understand tha the full 24V will be dropped across this transmitter and the current will be driven by the resistance and therfore be I=.048A. Is this correct?

The second figure I used an ideal current source to represent a 4-20mA output signal with the same 500ohm transmitter as the load. This current source is supplied by a 24V supply. I'm assuming that the current will be in the range of 4-20mA and therefore will determine the voltage across the 500ohm resistor. The trouble I am having though is seeing the why the 24V does not appear across this resistor as in the first example? Can someone help me visually understand this?

Thanks for all the help.

 

[xnuke]

The transmitter is acting like a variable resistance in series with your 500 ohm resistor. The resistance of the transmitter varies to only allow a current to flow through the entire series circuit of 4-20 mA. The 24 V is dropped across the series combination of the transmitter's resistance and the 500 ohm resistor. Draw this:

1. An ideal 24 V supply, with the positive end connected to one terminal of a variable resistor (representing your transmitter).

2. The other terminal of the variable resistor connected to one terminal of the fixed 500 ohm resistor.

3. The other terminal of the 500 ohm resistor connected to the negative end of the ideal 24 V supply.

4. Tap two connections off across the 500 ohm resistor as the input to your amplifier.

This is your circuit. Since the current flowing through the 500 ohm resistor will produce a 2-10 V drop (the signal proportional to the 0-100% transmitter range), the remaining voltage of 22-14 V will be dropped across the transmitter's varying resistance. (In the case of the non-ideal voltage source, some of the 22-14 V will be dropped across the source resistance.)

The transmitter is varying its own resistance so that the proper current (proportional to the measured value) flows through the entire loop.

The fixed resistor is used to convert the current signal into a voltage signal, since amplifier or ADC circuits require voltage inputs, not current inputs.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[geekEE]

Your first diagram and calculation is correct.

In your second diagram, you first say it's an "ideal" current source. If this were the case, then it would not make sense to say that it's supplied by a 24V supply because it would be able to drive 20mA through any resistance. In the real world, there are no ideal current sources. The 24V supply determines the maximum resistance that you could put in your chain and still maintain a 4-20mA loop. Consider your current source to be 24V supply in series with a potentiometer and a guy that stands there to turn the pot and maintain the current between 4mA and 20mA. This is possible with a 500-ohm load. This would be possible with any load between 0 ohms and 1200 ohms. However, if the load were something like 2k, there's no way you can get 20mA through it with a 24V supply. If you look up "compliance" as IRstuff mentioned, you will find a more info on this.

Glenn

 

[roydm]

Ah, but the transmitter isn't a 500 Ohms resistor it changes it's resistance to match the system voltage and the transmitter input for example as you show with zero input it will be 24/0.004 Ohms and at full scale it will be 24/0.02 Ohms.
The 24 volts will appear across the transmitter if there is nothing else in the loop as you show however usually the signal is going into a PLC, DCS or controller which has a resistor (typical 250 Ohms) in series. Draw one on your top sketch just below the power supply then you will see how the transmitter by changing the current is able to change the voltage across the 250 Ohm resistor.
I think you may be getting confused as to where the signal is controlled from. It's the transmitter that sets the current in the loop the power supply just supplies the source of power
In your second figure, you don't use an output to supply a transmitter, the output will go to a reciever e.g. control valve for (simple) example if you send the control valve 4 mA(0%)it will be closed, 12 mA(50%) it will be half open, 20 mA(100%) and it will be fully open.

Note: Not all PLC and DCS inputs are 250 Ohms, 50 Ohms is quite common also (0.2 - 1 Volt) but the transmitter doesn't care it just puts out a mA signal based on the input if the loop resistanc or supply voltage changes it just corrects itself. Thats the beauty of a current loop.
Send me your address, i will send you a sketch
roy_matson@yahoo.ca
Roy

 

[rockman7892]

Xnuke

I understand your circuit example, but what if I didn't have the 500ohm input to an amplifier and only had a transmitter in series with the current source as shown in the second figure?

The 500ohm resistor in the diagram was supposed to represent the transmitter however I guess I should have shown this as a varying resistance.

 

[xnuke]

There is no ideal current source in the circuit. The combination of the supply voltage and the transmitter is the "current source," if you must think about it that way. Draw a box around both components in the sketch I had you make, and label it "current source." The 500 ohm resistor is the only load, as in your second diagram.

The 500 ohm resistor does not represent a transmitter. As I stated earlier, it simply is a current-to-voltage converter.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[geekEE]

Roy, I think we have different interpretations based off of a question that is not clearly posed. In earlier posts, he said that the 500 ohms was a resistor across the input to an amplifier that was reading the voltage across it. I was still assuming this to be the case. However, the fact that he has started calling this 500 ohms the "transmitter" has led to your answer.

Rockman, do you have an actual circuit that you can draw rather than these theoretical ones? It will probably lead to better answers.

Glenn

 

[itsmoked]

If you just have a supply tied to a transmitter the entire supply will be dropped across the transmitter. But it will still be controlled to between 4-20mA.

If you supplied 24V to a 10 ohm resistor you have 24V across it. Your circuit would have 2.4A running thru the circuit.

If you supplied 24V to a 10,000 ohm resistor, you have 24V across it your circuit would have 2.4mA running thru it.

The transmitter doesn't care, and doesn't know, it just changes its internal resistance to be whatever resistance is needed to maintain 4-20mA.

In our moron circuit of just a transmitter and 24V supply the transmitter will have to maintain its resistance to values between 6,000 ohms (4mA) and 1,200 ohms (20mA).

When a 500 ohm resistor is added to the circuit the transmitter will have to alter its controlling resistance to between 5,500 ohms (4mA) and 700 ohms (20mA).


Keith Cress
kcress -

http://www.flaminsystems.com

 

[roydm]

Rockman,
I have posted a few sketches