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4-20ml,0-10V and similar sensors

Kal_B

Member
Hello everyone,

I would like to know how analog sensors work. Let's say a pressure sensors, some have 4-20ml and others have 0-10VDC output signals and of course there are other variety but let's stick with those two.

Regardless of how the output is produced we have both current and voltage at the same time. That's how electricity work.
So if I have 0-10 voltage does that mean that the current is constant and only the voltage changes?
And by the same logic 4-20ml output will have a constant voltage output?
How?

Other questions that follow an answer to the above are:
How do I measure both types of those outputs, voltage and current?
How does a PLC analog input of 0-10VDC measure that coming signal?
Why am I able to convert a current signal to voltage signal by placing a resistor across the PLC input? How does that work?

Too many questions I know.

Thanks
 

ronsimpson

Well-Known Member
Most Helpful Member
So if I have 0-10 voltage does that mean that the current is constant and only the voltage changes?
If you have a very old analog meter it might consume 100uA to measure while a new meter might only need 1uA. Either way the meter will measure correct. The 0-10V output will hold the voltage constant. (I don't know what the max current is but lets say 10mA so from 0 to 10mA the voltage will be right)
And by the same logic 4-20ml output will have a constant voltage output?
This is done so you can run long wires and not have noise. Notice it does not go to zero. It is hard to measure near zero with out noise getting into the measurements. In this mode the current is held constant, or I should say the information is in the current not the voltage. If you put a mA meter on the terminals it will measure in the 4-20mA range depending on what the sensor is doing. If your PLC can only measure voltage put a resistor on the end of the wire. (ohms law) 4-20mA & 1 ohm resistor = 4-20mV, 10 ohms=40-200mV, 100 ohms=0.4 to 2 volts, or 1k ohms=4 to 20V (if you use a 500 ohm resistor you should get 2 to 10V which matches the PLC's input range.
 

picbits

Well-Known Member
4-20mA is still susceptible to noise but has some advantages over 0-10v.

If your loop is broken, it can be detected at the other end and a fault raised. You can also use other current on the line to signal a remote device to recalibrate or go into test mode (i.e. 3mA = recalibrate and 22mA = test mode).

Lastly, the biggest advantage with a current loop is that no matter how long your cable is, you have the same current in any part of that cable (Kirchoffs law). If your cable is a mile long and you feed it with a constant 4mA at one end, you'll measure 4mA at the other end. It you terminate the other end with a 250R resistor you'll now have a 1-5V output or as above, with a 500R resistor you'll get a 2-10V output.
 

picbits

Well-Known Member
* oh and a 4-20mA sensor can be line powered over two wires. We've got loads of pressure sensors that we've just installed at work which only require two wires to work. The current loop powers them and they provide a 4-20mA load based on their reading.
 

SPDCHK

Member
What picbits said...
0..10V usually used very short distances. Example comms between racks etc.
But in all my life with instrumentation I've never used 0..10V. Always 4..20mA.

Don't get confused thinking a transmitter will source 0..10V and that is proportional to 4..20mA. No. It's either the one or the other, never both together (combined). Maybe it's like comparing distances in miles or kilometer. You cannot use both units at the same time (well, maybe you can), but as not to confuse the Russians, you stick to one unit.

As far as the PLC I/O card is concerned, you have to select the signal input. (0..20mA, 4..20mA, 0..5V, 0..10V etc). With 4..20mA you also have the 2 or 4 wire option. 2 Wire usually means that the Analog input card also supplies the (industry standard) 24V. With 4 wire, usually the instrument is powered externally.
Examples of 2-wire. PT100 temperature transmitter, DP-Cell.
Examples of 4-wire. Magflow flow transmitter, Densitometer, Conveyor belt weigh bridge etc.
 

Reloadron

Well-Known Member
Most Helpful Member
I would like to know how analog sensors work. Let's say a pressure sensors, some have 4-20ml and others have 0-10VDC output signals and of course there are other variety but let's stick with those two.
I believe, for the most part if we start with a pressure sensor in the heart of the sensor we will find a Wheatstone Bridge which when provided an excitation voltage will provide a voltage out proportional to an applied force. One or more legs of that bridge will have a Strain Gauge which makes it work. It starts with a voltage proportional to an applied force. So the bridge is the heart of the transducer.

What we do with the voltage out is pretty much up to us. I can amplify the voltage and scale it to measured pressure or I can convert the voltage to a current which is proportional to the applied pressure. A Google of "voltage to current circuit" will bring up a few dozen examples of such circuits.

When using a current loop there are a few considerations. Compliance voltage is one consideration. Just for an example if I have a 1,000 PSIG pressure transducer which outputs 4-20 mA I can place a 500 ohm resistor in series with the output and get 2-10 volts or a 250 ohm resistor and get 1 to 5 volts. both being popular analog inputs but that does not mean I can place a 10000 ohm resistor in there and get 40 volts to 200 volts. Most sensors with a 4-20 mA scaled output will specify their working voltage and their maximum R shunt resistance.

Be it a PLC with an analog input or any other data acquisition device getting the voltage or current input is then a matter of writing the code.

Ron
 

Kal_B

Member
I believe, for the most part if we start with a pressure sensor in the heart of the sensor we will find a Wheatstone Bridge which when provided an excitation voltage will provide a voltage out proportional to an applied force. One or more legs of that bridge will have a Strain Gauge which makes it work. It starts with a voltage proportional to an applied force. So the bridge is the heart of the transducer.

What we do with the voltage out is pretty much up to us. I can amplify the voltage and scale it to measured pressure or I can convert the voltage to a current which is proportional to the applied pressure. A Google of "voltage to current circuit" will bring up a few dozen examples of such circuits.

When using a current loop there are a few considerations. Compliance voltage is one consideration. Just for an example if I have a 1,000 PSIG pressure transducer which outputs 4-20 mA I can place a 500 ohm resistor in series with the output and get 2-10 volts or a 250 ohm resistor and get 1 to 5 volts. both being popular analog inputs but that does not mean I can place a 10000 ohm resistor in there and get 40 volts to 200 volts. Most sensors with a 4-20 mA scaled output will specify their working voltage and their maximum R shunt resistance.

Be it a PLC with an analog input or any other data acquisition device getting the voltage or current input is then a matter of writing the code.

Ron
Thanks for replies. Much appreciated.

Ron, your entire reply is very interesting and I will read the links you posted and likely come back with more questions and the part of your reply I changed to bold is exactly what is confusing me. In the attached snapshot from SICK transducer it shows Maximum ohmic load R(a), is that the shunt you mentioned?

Let's say I wire that transducer (the 4-20ml version with two wire connection) with 24VDC then the maximum resistor value I can use is 700 Ohm?
What does the A subscript mean?

Thanks
 

Attachments

Reloadron

Well-Known Member
Most Helpful Member
Yes, when they say A maximum ohmic load RA Signal output and RA in Ohm 4 … 20 mA, 2-wire RA ≤ (L+ – 10 V) / 0.02 what they are getting at is what I mentioned. I have seen it expressed several ways and maximum ohmic load is one of them. There will always be a slight caveat or bug to this. Since the 4-20 mA sensor is loop powered (works off the applied power) it will have its own voltage drop. This is the 10 volts we see above. Just as an example if I power the transducer with 24 Volts it will look like this. 24V - 10V = 14V / .020 = 700 exactly as you calculated. There will be approximately a 10 volt drop across the transducers internal circuitry. :)

As to the subscript I assume the NEC Class 2?
What is a Class 2 power supply?
"In general, a Class 2 circuit (operating at 24V with a power supply durably marked "Class 2" and not exceeding 100VA) is the type most commonly used. The NEC defines a Class 2 circuit as that portion of the wiring system between the load side of a Class 2 power source and the connected equipment".

Ron
 

KeepItSimpleStupid

Well-Known Member
Most Helpful Member
With 4-20 mA then sensor is USUALLY loop powered. The device constantly uses 4mA at whatever that voltage might be to power itself.

I did some lab stuff, where you needed isolated voltage inputs. They were used to measure the voltage and current of a power supply,
One measured the voltage across an internal resistor. The other, the voltage divided output voltage (0-40VC) to (0-5). You needed isolatio because the reference to the setpoint is the positive output.

In another case, current output across a resistor made a nice 0-5V input. Current output would not generate a ground loop.
Devices had voltage outputs which had a common ground which could be read pseudo-differentially. Both used together prevent ground loops without the expense of isolation.

One instrument had the unusual capability of moving it's 0V reference.

I messed up too when converting a home made manual readout/setpoint controller for computer operation.

The readouts were powered by +5 and could read differentially. 6 instruments were powered by a single +-15V supply.
The computer could read the output pseudo-differentially.

There were issues providing the setpoint signal without interaction.
 

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