Analog

[damaged 0]

I have an analog card that Im going to implement with 2 6 wire scales. Will I need to supply the 10 vdc or will the card supply the current? I know this is probably a dumb question but I have never dealt with an analog input card. The number that shows up in the IO table is SIOU/C200 AscII Unit (A)(1)(0)(out: 5, in: 5) Help and info would greatly appreciated

[Sleepy Wombat 27]

do u mean 6 wire weigh sensor ? Yes you provide an external voltage to those sensors. The Analogue input card will read the signal, not output the signal. Secondly, i would use a weigh scale to 4-20mA convertor. I would not wire the sensor directly into the PLC analogue. Also, get you self a copy of the manual for the particular analogue card...you hav'nt specified the model in your post. They are avaiable for free from Omron sites.

[BobB 58]

If using strain gauges, use a strain gauge to 4-20ma converter, preferably isolated. It will supply the power to the strain guage. Edited 30 Aug 2005 by BobB

[Guest Guest_damaged]

It is a CJ1W-MAD42...So you are saying to use something to isolate the load cell (intertechnology model 1250) sensor from the analog card right? May I ask Why. Also I have never used a strain guage what exactly does this do? The load cell diagram is this ______+Input and +Sense (2 wires) -Output ____ /\_____ +Output \/______ -Input and -Sense (2 wires) I would take a 10v supply to +Input and -Input and then what? Sorry about all the questions but like I said I have never used an analog in. This is just the beginning. I still have to know how to take this input and with Cx-Prog turn it into a usable input for Floating point math (which I already have written with the exception of the conversion of the input). I am laughing now because I am the only one that can at my place that can even halfway understand how ladder logic works and yet i feel so dumb. I love a challenge.

[Guest Guest_damaged]

Sorry that diagram got messed up. Here is the link http://www.intertechnology.com/Tedea_Huntl...h/pdfs/1250.pdf

[BobB 58]

See my last post

[Jay Anthony 118]

The rated output of the load cell is 2mv DC. That's .002 volts! The lowest input voltage for a CJ1W-MAD42 is 0-5 VDC. That's 0 to 5000mv! The load cell will never even make a blip at 12 bit resolution. Like BobB and Sleepy Wombat have suggested, get a 4-20ma amplifier which will raise the signal to the level that the analog module can read and it will also supply the necessary 6 wire connections that deliver the excitation voltage.

[Nibroc 4]

Hi, Not sure where I found this, but why not use one of these... Load Cell input for CJ1 CJ1_Loadcell_Unit.pdf

[Nibroc 4]

... There is also one for the CS1 series... Nibroc CS1_Loadcell_Unit.pdf

[damaged 0]

Thanks Guys Seems like the Omron rep would have taken all this into consideration since he had all the information That you guys have. Everything makes sense now thanks for your help. Let me see what I can do. One last thing anyone have an idea on the conversion that I mentioned in my earlier post?

[Sleepy Wombat 27]

Depends on how you plan to connect to the load cell.... 1. If via 4-20mA convertor then you know what the minimum weight value will be which will correspond to the 4mA and you will also know the value taht corresponds to the max value = 20mA. This in turn will correspond to a value in the PLC of 0 for 4mA and 4000 for 20mA. That is to say your alanlog value for weight has be broken down to a digital coresponding value. Ie in simpliest terms, lets say the weigh cell is calibrated for 0 to 4kg. Through the convertor this would correspond to 4mA for 0kg and 20mA for 4kg. This in turn translates to a values of 0 and 4000 in the PLC. Therefore you could then say that you had a +/- 1g accuracy. 2. If connected to the special card as Nibroc pointed out then your internal PLC decimal value would be 1 to 10000.

[Nibroc 4]

... The CJ1W-AD0x1-V1 analog input units can be set to a resolution of 0 to 8000 if required. You will need to look carefully at the quality of the weight transmitter used to ensure this has equivalent or better resolution and/or accuracy. I note that in the real world, this level of resolution is difficult to acheive in the industrial environment due to noise, calibration, linearity etc. (1/8000th of 16mA is only 2 micro amps!! - there is generally more noise/flutter than this in typical industrial analog signals!!) For applications requiring accurate weighing, I usually use an independant electronic scale with a serial communications link to the PLC. The PLC can read/write set points, weights etc from the electronic scales to suit the application. One advantage of this approach is that ongoing calibration checks are outside the responsibilty of the PLC integrator!! The PLC simply reads the scales!! Nibroc