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?