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hamza		 What is the percentage non-linearity error with a potentiometer? View All


16 years ago - 1 week left to answer. - 1 response - Report Abuse
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shai
 If I understand your question, you have a 550 Ohm pot with cursor at center, and the lower half of it loaded with a 10 Ohm resistor:
Let's call the top half of the pot R1 = half of 550 or 275.
Calculate the parallel of 275 and 10, call this R2.
Now calculate the voltage Vout across the 10 Ohms using the voltage divider equation: 10/(R1+R2)=Rout/R2.
With no 10 ohms load the Vout would be 5Volts. With the load it would be as you calculated it. The error is the ratio of the two voltages Vout/5 times 100, expressed as a percentage.

16 years ago

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