01-21-2020 05:32 AM
Hi LabView Board,
Please refer to the image shared on : https://forums.ni.com/t5/Multifunction-DAQ/Inductive-proximity-sensor-reading/m-p/2616659#M79464
I am reposting as a fresh question.
I was considering a similar implementation for my circuit using the proximity sensor. However, given with my basics of the voltage divider and considering the following:
Vi as the input voltage
R1, R2 as divider resistances
V1, V2 respective voltage drops
then, V1= Vi * (R1/(R1+R2)) and V2 = Vi * (R2/(R1+R2)).
the circuit configuration marked as the solution has R1 & R2 indicated as 1000 Ohm and 1500 Ohm. The output across R2 is given input to 0-10V range DAQ.
Now calculating V2 as per given values and above formula:
V2 = Vi * (R2/(R1+R2)) = 24 * (1500/(1000+1500)) = 24 * 1500 / 2500 = 14.4 V
definitely 14.4 V shouldn't be the input to the DAQ!
Higher resistance has a higher voltage drop across it and vice versa.
then why DAQ is connected to higher resistance R2?
Kindly clarify.
Solved! Go to Solution.
01-21-2020 05:16 AM
hi,
I was considering similar implementation for my circuit using proximity sensor. However given with my basics of voltage divider and considering the following:
Vi as input voltage
R1, R2 as divider resistances
V1, V2 respective voltage drops
then, V1= Vi * (R1/(R1+R2)) and V2 = Vi * (R2/(R1+R2)).
the circuit configuration marked as solution has R1 & R2 indicated as 1000 Ohm and 1500 Ohm. The output across R2 is given input to 0-10V range DAQ.
Now calculating V2 as per given values and above formula:
V2 = Vi * (R2/(R1+R2)) = 24 * (1500/(1000+1500)) = 24 * 1500 / 2500 = 14.4 V
definitely 14.4 V shouldn't be the input to the DAQ! Kindly clarify
01-21-2020 05:42 AM
01-21-2020 05:49 AM
I had a headache wondering why why why? thanks to you.
01-21-2020 05:53 AM