We commonly use the High density 1746NI16I in our product to measure 4-20 ma analog signals. One of our recent customers has thrown in a set of 1-5 volt signals for us to try to use. Since the input impedence of the card is 250 ohms, wouldn't the fact that the signal is 1-5 volts be transparent to the card? I'm getting a little flack from AB that we can't do this but no reason why... Maybe just to sell more cards?
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Can I measure 1-5 volts with a 1746-NI16I?
- Thread starter finfin
- Start date
No you can't use this for voltage input. The model number is a clue. notice the "I" at the end of the model number. This designates current. A-B also has a 1746-NI16V module (notice the "V" at the end of the model number) which is for voltage inputs. If you want the universal input modules which are field configurable for voltage or current, you are limited to the 8-input modules.
- Thread starter
- #3
The answer is actually "Yes." I was quite aware that the card was a current input card and yes you can drive it with a 1-5 volt input and have great results... as long as the device your driving it with can actually drive the 250 ohm input impedence. That's where the problem lies. The sensors that output a 1-5 voltage signal are usually not capable of driving the 250 ohm input impedence. The one I was looking at was for an input inpedence of no less than 5K ohm.
What we will be doing instead is using a voltage input card (input inpedence of 1M ohm) and applying a 250 ohm (or 500 ohm)precision shunt resistor across the inputs that are being driven by 4-20 ma signals. This way we can still use the high density cards and use either voltage or current.
What we will be doing instead is using a voltage input card (input inpedence of 1M ohm) and applying a 250 ohm (or 500 ohm)precision shunt resistor across the inputs that are being driven by 4-20 ma signals. This way we can still use the high density cards and use either voltage or current.
The answer is not "Yes", as voltage transmitters almost always require a high impedence load.
I have used voltage cards with resistors before, but a better way of doing this is use 2 x 1746-NI8 input modules. The inputs can be individually configured as either voltage or current.
I have used voltage cards with resistors before, but a better way of doing this is use 2 x 1746-NI8 input modules. The inputs can be individually configured as either voltage or current.
If you only have one 1-5V signal, and you dont count on having more in the future, then the easiest solution is to use a signal converter.
I suggest that you use a 0-5V/0-20mA converter. Then you get a signal range 4-20 mA, and you can test for missing signal (<3mA) in your program.
I suggest that you use a 0-5V/0-20mA converter. Then you get a signal range 4-20 mA, and you can test for missing signal (<3mA) in your program.
- Thread starter
- #6
I have received info from Rockwell that indicates what resisters to use for which conversion (simple application of Ohm's Law).
1) A channel of a 1746-NI16V can be configured to function to respond to a current input as follows:
Wanted Select
Current This NI16V
Input Type Added Parallel Resistor
+20 mA* +10 Vdc* 500 Ohm .01%
4-20 mA 1-5 Vdc 250 Ohm .01%
0-1 mA 0-5 Vdc 5000 Ohm .01%
0-20 mA 0-10 Vdc 500 Ohm .01%
***
2) A channel of a 1746-NI16I can be configured to function to respond to a voltage input as follows:
Wanted Select Added
Voltage This SERIES
Input NI16I Resistor Type Type
+10 Vdc* +20 mA* 250 Ohm .01%
1-5 Vdc 4-20 mA No Resistor Required
0-5 Vdc 0-1mA 4750 Ohm .01%
0-10 Vdc 0-20 mA 250 Ohm .01%
1) A channel of a 1746-NI16V can be configured to function to respond to a current input as follows:
Wanted Select
Current This NI16V
Input Type Added Parallel Resistor
+20 mA* +10 Vdc* 500 Ohm .01%
4-20 mA 1-5 Vdc 250 Ohm .01%
0-1 mA 0-5 Vdc 5000 Ohm .01%
0-20 mA 0-10 Vdc 500 Ohm .01%
***
2) A channel of a 1746-NI16I can be configured to function to respond to a voltage input as follows:
Wanted Select Added
Voltage This SERIES
Input NI16I Resistor Type Type
+10 Vdc* +20 mA* 250 Ohm .01%
1-5 Vdc 4-20 mA No Resistor Required
0-5 Vdc 0-1mA 4750 Ohm .01%
0-10 Vdc 0-20 mA 250 Ohm .01%
There is one problem with the suggested approach.
There is a resistor in the input channel on the PLC card for converting the current signal to a voltage signal for the AD converter. This resistor is probably 150 ohms or something like that.
What you have then is a voltage splitter, as any resistance will split the voltage between the resistances.
That means that the wire resistance from the sensor to the input will generate an error. Maybe not so much, but an error indeed.
A dedicated analog voltage input will have a high impedance and will not have such a problem.
Unless you are really on a budget, why waste more time on this resistor story. Get a signal converter, or get a voltage input card.
There is a resistor in the input channel on the PLC card for converting the current signal to a voltage signal for the AD converter. This resistor is probably 150 ohms or something like that.
What you have then is a voltage splitter, as any resistance will split the voltage between the resistances.
That means that the wire resistance from the sensor to the input will generate an error. Maybe not so much, but an error indeed.
A dedicated analog voltage input will have a high impedance and will not have such a problem.
Unless you are really on a budget, why waste more time on this resistor story. Get a signal converter, or get a voltage input card.
To JesperMP
If you are going to reply, you really should read the postings.
finfin has already said that the internal resistance of the current input card is 250 ohms, and the list he posted already considers that if you work it out.
He also said they are using the voltage card,
and who isn't on a budget. Signal converters are generally quite expensive, though I do agree it is the 'ONLY' option for a voltage signal to a current input module.
The problem with the current input option is still voltage signals normally require high impedance load, however, the 0-1mA option solves that, but with the tradeoff of a very low resolution on the AD converter, making this a bad option.
I don't see why use SLC anyway. Compact logix system is cheaper and produces a better result although you need 4 analog input modules to get 16 inputs. The input modules have both current and voltage on them.
If you are going to reply, you really should read the postings.
finfin has already said that the internal resistance of the current input card is 250 ohms, and the list he posted already considers that if you work it out.
He also said they are using the voltage card,
and who isn't on a budget. Signal converters are generally quite expensive, though I do agree it is the 'ONLY' option for a voltage signal to a current input module.
The problem with the current input option is still voltage signals normally require high impedance load, however, the 0-1mA option solves that, but with the tradeoff of a very low resolution on the AD converter, making this a bad option.
I don't see why use SLC anyway. Compact logix system is cheaper and produces a better result although you need 4 analog input modules to get 16 inputs. The input modules have both current and voltage on them.
killemall,
sorry I didn't notice finfins second posting. So my post is a bit off the mark - but not that much !
A converter will be approx 15% the cost of an analog voltage input card.
Finfins method of using a shunt resistor for his 4-20mA inputs appears to me to be a cludge, even if he dont need high precision.
There are readymade boards with shunt resistors available, but price and size is more than comparable with that of a single converter.
sorry I didn't notice finfins second posting. So my post is a bit off the mark - but not that much !
A converter will be approx 15% the cost of an analog voltage input card.
Finfins method of using a shunt resistor for his 4-20mA inputs appears to me to be a cludge, even if he dont need high precision.
There are readymade boards with shunt resistors available, but price and size is more than comparable with that of a single converter.
- Thread starter
- #10
Interesting that you should mention ControlLogix. We have gone through that donkey ride about three times here. It always turns out to be more expensive, not to mention that it would cause us to have to move up to a larger enclosure.(we commonly have 34 analog inputs) There is also the battery problem. Many of our systems are not installed quickly enough to keep the program from being lost. I know there is a larger battery for it but at quite a hefty cost as I remember.I think it moves the battery life up from a few weeks to a couple of hundered days. (Very poorly engineered.)
There is also a problem with being able to leave cards out of one of our systems for customers that don't want all of the options. With the SLC we have programmed auto detection of missing cards and deal with it on the fly. This way we have a one size fits all program. The control logix would force us to re-map all of our subsequent outputs each time we leave out a particular card.
I understand that they are working on a "empty slot" filler card for the ControlLogix as well as a 8 input analog card.
There is also a problem with being able to leave cards out of one of our systems for customers that don't want all of the options. With the SLC we have programmed auto detection of missing cards and deal with it on the fly. This way we have a one size fits all program. The control logix would force us to re-map all of our subsequent outputs each time we leave out a particular card.
I understand that they are working on a "empty slot" filler card for the ControlLogix as well as a 8 input analog card.
jonesy29847
Electrical
The resistor solution works, I used it many times when I designed control systems for functional test systems.
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