4-20ma signals

shoelesscraig

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Join Date
Apr 2009
Location
LA
Posts
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Someone help me out here...
I've dealt with a lot of 4-20ma signals that come from a transmitter (for instance a pressure transducer) back to a PLC analog input card. The transducer requires 24vdc to operate. So, when I measure voltage across my 2 wires I get 24 volts. When I hook up in series, I get 4-20ma. No problem.

But, recently I've been dealing with variable frequency drives (VFD'S) and they have been fed a 4-20ma output from the PLC (analog output card) to the drive to control the speed at which a motor runs. I get my 4-20 signal when I measure current in series, but when I measure the voltage across the 2 wires, I get something very low like 1 to 3 volts depending on the VFD. I have found this senario on multiple VFD's in my plant. All of them are working just fine, I was just playing around teaching myself more about them.

I understand how a 0-10v signal would be interpreted by the VFD, as it would see a potential of somewhere between 0-10 volts between the 2 wires. But how it the 4-20ma signal working??

Can someone please explain how this works. And feel free to talk to me like I'm a 3 year old. I won't be offended and I'll understand it a lot better!!
 
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If you disconnect the wires entering your device and measure the resistance of that devices input you will find normally a 250 ohm or 500 ohm input.

Now if you have 4 mA flowing, or 20 mA flowing, and now apply ohms law you will derive the voltage needed to drive the current into the resistive load.

Example 0.004 x 250 = 1 volt
0.020 x 250 = 5 volts

The results if the resistive load is 500 ohms will be 2 to 10 volts

The device reacts to this voltage that has been generated across this load resistor and controls the device in your case the VFD.

When you do an open circuit voltage test you will find you have normally 24 volts available that is because the max voltage available is 24 volts, and because of the open circuit it still has not reached a minimum current flow.
5 volts or 10 volts would have been enough to fully drive the load.

In you case of 1 volt there would have been 4 mA flowing into a 250 ohm load
at 3 volts there would have been 12 mA flowing into a 250 ohm load
 
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If I might add a comment to explanation given by Gil47.

Often within the specifications listed for a device with a 4-20 ma output, you will find a table listing the required minimum supply voltage based on the input resistence connected to the transmitter. The higher the total resistance, the higher the supply voltage. This then allows the transmitter to calculate and use the voltage needed to provide the proper 4-20 ma output. Just as Gil47 explained.

All this also explains how you can wire more than 1 reciever in series, and still get the properly proportioned signal.
1,2,3,4 devices, doesn't matter as long as it falls within the maximum resistance allowed for a given supply vpltage.

It's the type of thing that is often referred to as P.F.M. technology! :p
 
Ok, so let me see if I'm getting this...

The 4-20ma input terminals of the VFD have, in your example, a fixed resistance of 250 ohms. Therefore the PLC output will raise the voltage to increase current flow to 20ma and decrease it to lower the current to 4ma?
 
All this also explains how you can wire more than 1 reciever in series, and still get the properly proportioned signal.
1,2,3,4 devices, doesn't matter as long as it falls within the maximum resistance allowed for a given supply voltage.

It's the type of thing that is often referred to as P.F.M. technology! :p

Ok, so you're talking about a group of receivers that are in series and therefore all receiving the same current signal, correct?
 
Ok, so let me see if I'm getting this...The 4-20ma input terminals of the VFD have, in your example, a fixed resistance of 250 ohms. Therefore the PLC output will raise the voltage to increase current flow to 20ma and decrease it to lower the current to 4ma?
Well, I'd say that the PLC's AO changes its resistance, not that it raises its voltage, but that depends on what one measures and where in the circuit.

The external loop represents a 'fixed resistance', consisting of wire resistance and AI input resistance

The reason I shy away from saying 'the voltage is varied' is that a DC power supply provides a fixed voltage to the AO circuitry (power supply can be internal for 'active' outputs or external for 'loop powered output').

The AO circuitry changes its 'resistance', as though it were a transistor, to source a given current for whatever loop resistance exists.

Since the DC voltage to loop is 'fixed', and the AI resistance is 'fixed', and the wiring resistance is 'fixed', the AO's circuitry has to change its resistance in order for current in the loop to vary.

The voltage supply needs to be sufficient to drive the current through the total resistance, as in any DC circuit.

Ok, so you're talking about a group of receivers that are in series and therefore all receiving the same current signal, correct?
correct
 
Ok, so let me see if I'm getting this...

The 4-20ma input terminals of the VFD have, in your example, a fixed resistance of 250 ohms. Therefore the PLC output will raise the voltage to increase current flow to 20ma and decrease it to lower the current to 4ma?

The voltage source at the PLC is probably fixed at 24V, but
includes a variable resistance in series with this voltage
source as part of the current loop. The complete loop is made
up of the voltage source, the variable resistance in the PLC,
the wiring, and the fixed resistor of (for example 250 ohms)
in your VFD. While the PLC does not know the resistance of the
wiring, or the value of the resistor in the VFD, it can monitor
the loop current. This is because in a current loop, the
current is ALWAYS the same at all points in the loop. The PLC
can adjust the variable resistor to achieve the desired loop
current. Example calculations below ...


Ok, so you're talking about a group of receivers that are in series and therefore all receiving the same current signal, correct?

Right, because it is a current loop, the current flowing across
each receiver and through the transmitter are all the same. The
limit to the number of receivers is a function of the
resistance of each receiver, and the voltage (and resistance)
at the transmitter. With a source voltage of 24 V , the total
resistance must be less than 1200 Ohms (see below for how this
calculated). For example, with receivers with 250 ohms
resistors, we could have upto 4 receivers, but not 5.

Or I could have 2 receivers with 250 ohms resistance and one
of 500 ohms in the same loop.


Here is the "below stuff" :)


Hi, I'm new to PLCs (touched one for the first time yesterday),
but I do understand 4-to-20 mA signaling.

There is a fixed relationship between Volts, Amps, and Ohms:

V = I * R

In a typical 4-to-20 mA signaling system, there must be a
voltage source and at least 1 transmitter and 1 receiver.

Lets look at your VFD scenario.

A common situation is a fixed 24 V source combined with a
series resistor (which is part of the receiver) and a series
resistor (or equivalent) that is variable which is the
transmitter. All connected in a loop. Remember, the current
through all parts will be the same.

In your VFD example, the voltage source and the variable
resistor are in the PLC, and the fixed resistor is in the VFD.
There is also an unknown resistance in the loop which is the
wire.

Let's try some values: Resistor at receiver (VFD) end is 250
ohms and since we know we want 4 to 20 mA, we can also say
something about the voltage across this resistor:

At 4 mA, the voltage across a 250 Ohm resistor
= I * R = .004 * 250 = 1V

At 20 mA, the voltage across a 250 Ohm resistor
= I * R = .020 * 250 = 5V

Now this resistor at the receiver (VFD) can't "Control" the
voltage across it, this voltage ( 1 to 5 V) is a function of
the current running through it.

At the transmitter end, there is the variable resistor that is
controlled by PLC.

Lets say it wants to transmit 4 mA (actually, what it does is
pass 4 mA). To do this, it must adjust the resistance it
presents to the loop, so that the current is 4 mA. Since we
just calculated that at the receiver, the voltage drop is 1V,
then the voltage drop for the rest of the loop must be 23V.
If you have short, low resistance wiring, then the voltage
drop is mostly inside the PLC, but a long higher resistance
wiring may represent a non-trivial voltage drop, for example
2000 ft of 0.01 ohm/foot wire (2000 * 2 * .01 = 40 Ohms)

the formula V=I*R can be rewritten: R=V/I

So the variable resistance is =(24-1)/.004 = 5750 Ohms

At a loop current of 20 mA (5 V across the 250 Ohm resistor
at the transmitter end) the remaining voltage is
=(24 - 5)/.020 = 950 Ohms

So the transmitter changes its apparent series resistance
between 950 and 5750 ohms to control the loop current between
4 and 20 mA.


Now.... Let's see if I answered your questions:

Ok, so let me see if I'm getting this...

The 4-20ma input terminals of the VFD have, in your example, a fixed resistance of 250 ohms. Therefore the PLC output will raise the voltage to increase current flow to 20ma and decrease it to lower the current to 4ma?


Not quite. The combination of the fixed PLC voltage and the
variable resistance presents an apparent variable voltage, but
it does this by monitoring loop current, so the actual voltage
is also dependent on wiring resistance and the VFD resistance.

Rewriting the second part of your question:
Therefore the PLC output will decrease the variable resitor
to increase current flow to 20ma and increase the variable
resistor to lower the current to 4ma.

Ok, so you're talking about a group of receivers that are in series and therefore all receiving the same current signal, correct?

That is correct. In a current loop all devices receive the same
current.

The water pump and plumbing analogy goes like this:

A pump plus flow regulator puts out 1 gallon per minute (the
voltage source plus the variable resistor). The fluid then
flows through multiple different pipes of varing diameter
(receivers, with differing resistances). While the fluid
velocity through each pipe segment (voltage) depends on the
diameter of that segment (resistance), the volume of fluid
per fixed time interval is the same (current) through all
segments.


How's that for a first post?
 
How's that for a first post?

Excellent! Welcome to the forum. If you have any questions about those newfangled PLCs, ask away. There are many people here who will bend over backwards to help member like you.

I do want to mention one caveat with regards to your explaination...

That is correct. In a current loop all devices receive the same
current.

For an academic discussion, this is all that needs to be said; but before you go connecting 4-20mA receivers in a string willy-nilly, you need to do some research.

Some 4-20mA receivers (VFDs, panelmeters, PLC inputs, etc.) are internally wired such that the input is not floating. For example, some receivers (I believe I have some Hitachi drives here) have the negative side of the 4-20mA input tied to ground. Now, this is not a problem until you want to send the same speed signal to two of these drives at once.

Referring to the diagram attached to this post, the left string operates as you described a 4-20mA signal chain. The right string is made up of non-floating receivers and operates differently. The dashed lines represent the internal connections to ground that some receivers have. Note how the second receiver effectively has both signal lines tied to ground. This means that only the first receiver will see the 4-20mA signal, while the second will see 0mA.

This is just a real-world gotcha similar to the load limits of a 4-20mA loop that I thought would be worth mentioning.

Looking forward to you second post!

Brian

4-20mA.PNG
 
Thanks for the responses guys. As far as the ohms law stuff goes, I appreciate the response, but I consider myself an accomplished electrician. I could write a book on electrical theory. I just haven't dealt with VFD 4-20ma signals very much recently, and wanted to make sure I understood them good. I figured this was the quickest place to find out! Thanks.
 
Just to complicate things, some manufacturers use "active" inputs. Instead of a fixed resistor such as 250 ohms, they use a solid state input that you can't measure with an ohmmeter since its resistance changes depending on current.
 
Just to complicate things, some manufacturers use "active" inputs. Instead of a fixed resistor such as 250 ohms, they use a solid state input that you can't measure with an ohmmeter since its resistance changes depending on current.
Who might these manufacturers be? I haven't run into this; don't quite understand how the input resistance could vary for a current signal circuit; and would like to check it out.

Dan
 

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