I had to solve the same task on one of my projects last year. I think, the range for the sensor is 4-20 mA.
You wil get for 4 mA - 8000 counts and for 20 mA - 32000 counts. So, the working range for 4-20 mA will be 24000 counts. To get the voltage you have to write next PLC formula:
-from the counts number from PLC substract 8000 counts and divide everything with 24000.
The simplest way would be to buy a signal converter from Phoenix or various other suppliers. For this signal range you probably need one of the configurable models.
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Michael Griffin
London, Ont. Canada
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What about the impedance of the input, and more important, what about the output impedance of the sensor? Neither was specified in the original message. This method may not work without the right conditions.
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Michael Griffin
London, Ont. Canada
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... Hmm, I think I misinterpreted this and got it the wrong way 'round. Assuming the signal is 0-1 volt and the available input senses 0-20 mA, Ohm's Law still dictates a 50 ohm resistor, but it's highly unlikely that the 0-1 V output can source that much current. Amplification is then then needed, whether done using a fabricated op amp circuit (for a few dollars) or a commercial signal transmitter (for 100 or more dollars). If you inspect the 0-20 mA input circuit there's a good chance you'll find a load resistor in there, and in fact the circuitry is sensing a voltage after all. Whether you can reasonably "hack" the circuit to skip the current conversion is another question.
If you really mean what you say the simple way to convert 0 to 1 volt to 0 to 20 ma is to insert a 50 ohm resistor in series with the 20 ma input. Of course I have no way of knowing if your 1 volt source is capable of driving 20ma since you didn't state what your driving source was. Sorry I flunked mind reading 101.
Not sure what the "counts" business is, and you have a sensor that is sending a 0-1 Vdc signal, therefore a resistor is going to do nothing for you here. If you had a current output, obviously you could convert it to a voltage signal. Anyway, the best way to convert your signal is to purchase an Action Pak AP4000 Signal Conditioner (http://www.actionio.com). You will need to make sure the unit that you specify has the resolution/accuracy that you need. You could design and build a circuit, but it is safer and less time consuming to just purchase this off the shelf. Good Luck.
That's what I thought, but didn't get to try it first hand. A co-worker told me it didn't work. We decided to ditch the 0-1 vdc sensors; we found 4-20mA sensors that were cheaper anyway.
That probably wouldn't work even if the output could source the current. It would, if the current input were a virtual ground, but since the input more than likely has a resistance of 250 ohms and they are in series, you would get 0-3.3 Ma. I suspect what is needed is to invert the 0-1 V and refer it to a positive reference voltage, then use it to drive a current source referred to that same voltage to achieve a
current source that sources current in the proper direction. I'd have to draw it out to check my thinking for sure. But, that's just a first pass at a general class solution. A dual op amp, a few resistors and a precision reference for the compliance voltage you wish.
In the meantime, I like checking the card to see if it, indeed is a 0-5 V input with a 250 resistor across it. One would still need a non-inverting amplifier with a gain of 5, but it would
be easier than the general class converter. A single supply Instrumentation Amp with a fixed gain of 10 and a 250 ohm resistor in series with the output would work also in that case and would
probably be the simplest solution. The gain of 10 would give 0-10 V and the 250 ohm resistor would form a voltage divider with the existing 250 input resistor to give the 0-5 V, ground referred
signal needed.
Problem with this method is that standard 20mA loop inputs will have too high a resistance, typically 500ohm, with no other resistance in the circuit you will need 10v to produce 20mA.
You need some kind of amplifier circuit located near the sensor to prevent noise influencing the circuit. The sensor is probably designed to work with a matching amplifier (check manufacturer spec).
How do you measure the current with the 50 ohm resistor. I though most 4-20ma inputs converted the current signal into a voltage signal across a (normally) 250 ohm reisitor, thus it is really a voltage reading. I would have though that there would have been an input for voltage measurement as well on the input card.
Like I said, it was easier (and it turns out cheaper) to ditch the 0-1 vdc sensors and go to loop-powered 4-20 mA. I think the PLC input had to much impedence for the sensors to drive a signal through (he used a 50 ohm resitor in series, but the input is 250 ohms).
