How to Program Analog inputs 4 to 20mA with TIA portal S7-1200 PLC || Lesson# 10
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- Опубліковано 19 жов 2024
- #analog #scaling
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Analog 4 to 20 mA Scaling in PLC
siemens s7 1200 analog input scaling
Analog Signal Scaling Concept
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S7 1200 Analog input example
tia portal analog input 4 to 20mA
tia portal analog input scaling
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UP DOWN COUNTER Lesson#9 || TIA PORTAL PLC programming for beginners
• UP DOWN COUNTER Lesson...
Excellent vedio it's fc105 for step 7 and norm x and scale x for tia...
Excellent vedio..
Excellent video. Do you have a video showing how to program in TIA a Level Transmitter (Profibus) on TIA? For example: Deltapilot M FMB50. Thanks in advance!!
Very good.
Wonderful information many thanks
I think you forgot to compensate for the fact that the sensor is calibrated at 4-20mA output but the analog input is 0-10 volts NOT 2-10 volts which is the voltage equivalent of 4-20. To be accurate you have to lower the minimum value of the x-scale parameter to correct this
Kindly make video for telegram communication between plc and drive in detail to explain telegram concept
Thank you so Much Sir !
In the configuration there is voltage not current.. You must change it or use another terminal input analog
you missed to type max correctly in video i think its 27648 instead of 62748 which appears in video titles
nice
Please share all the video with simatic manager programming.
WINCC SIMATIC MANAGER: ua-cam.com/play/PLkpA_8uYEhjAfSzpT-uGrPEGMiSw2_1dv.html
Check here
I am not able to understand int to real, and why ' Max ' value is 27648 and which is correct temp showing MD4 or MD10
You need to read a documentation on input module. It's a ADC code for a current 20mA.
Just wow.
500 ohm resistor ..But what is the power rating of the resistor??? .5 or 1 W?
5 watt
@@EASYPLCTRAINING .5 or 5 watt??
5 watt .... dear
@@EASYPLCTRAINING Can you please clear it ?
Since Power=current * current * Resistance =.02*.02*500= .2watt.. \
required is .2 watt ..If we use 5Watt in place of .2Watt will it make eratic values??
please guide me
@@ansalkm Don't mix things AKM, the power rating of the resistor (W) is equivalent to the amount of current which can go trough it. The sensor only outputs 4...20mA, if we do the max power that the resistor will dissipate, P = I^2*R = (20x10^-3)^2*500 = 0,2 W. This is the minimum power rating needed in order to not burn the resistor. The resistor is 5 W, which means it's 25X more what we need in order to not have magic smoke. Was I clear?