HCBBS Forum (English)
Submit Chemical Projects / Find Solutions
Amplify Your Requirements on a Broader Chemical Platform *Engineering · Technology · Equipment · Solutions*
Submit Request

Seeking help with PLC program issues (analog-to-digital conversion)

2017-11-08View Original

Thread Content

Please refer to the image above. The program shown in the image is one for collecting current data. The specific questions are: 1. It’s not clear why the input signal (LD14) is divided by 32000 2. What does Business (LD18) multiplied by Range (LD2) mean? (Are these two steps meant to convert the quantities into engineering values?) 3. The most confusing part is: what does it mean to divide the next value by 4 using the range (LD2)? 4. What does the final step, which involves subtracting the quotient from step 3 (LD26) from the product of step 2 (LD22), mean? I hope engineers familiar with PLC programs can help answer my questions; thank you all in advance!
Reply #22017-11-08
The signal processed by the PLC is converted into an internal value; 32000 represents the maximum value of this internal value, which corresponds to the range. Doesn’t the actual measurement value from the instrument also need to be converted into 4-20mA data? 32000 corresponds to 20mA, so the range is from 20mA to 32000
Reply #32017-11-08
I’m clear about this; it mainly comes down to the third point – why do we need to divide the range by 4? Thank you for clarifying it
Reply #42017-11-08
In the S7200, 4MA corresponds to 6400, and 20MA corresponds to 32000. In a typical configuration, it is necessary to set the upper and lower limits of the range, which means two LD register variables are required. You only have one here; perhaps he assumes that the lower limit of the range is always 0. Is this current a communication point or an AIW hard point?
Reply #52017-11-08
What is the type of signal? Write the program in mathematical form and take a look.
Reply #62017-11-09
Is it the multiplier of the current transformer? I don’t think it is
Reply #72017-11-09
The current transducer measures the current in the secondary coil; without applying a scaling factor, it is not possible to obtain the current in the primary coil. It’s unclear why subtraction is used in the fourth step. If you can tell me what type of device’s current is being measured
Reply #82017-11-09
The pump of dust removal equipment, ordinary motor current
Reply #92017-11-12
No one else wants to share? Thank you all

Submit a Project

**Looking for Chemical Technology, Equipment & Solutions?** No Registration Required Broader Platform Exposure | Global Chemical Service Provider Connections

Submit Request — Free Consultation

Disclaimer

This is an automated machine translation of the original thread. Some technical terms may have inaccuracies; the original text shall prevail. Click "View Original" at the top right to access the source page, which supports IP-based automatic real-time language translation. Please watch out for contact details and sales inducements to prevent fraud. All content and translations are for reference only, representing solely the poster's personal views. For enquiries, email service@hcbbs.com.