Wheatstone Bridge with Resistor Tolerances

Thread Starter

TimoG

Joined Nov 2, 2017
17
Question: to determine the minimum value of R4 given that each of the resistors has a tolerance of 5% and Vout = 0.
upload_2017-11-9_16-34-0.png
To get the minimum value for V4, I assume that R4 is in series with R1 and that the potentiometer wiper is all the way at the bottom.
When Vout = 0: (R1+R4)/R2 = R3/R4
Rearranging to solve for R4:
R4 = (R2R3/R4) - R1
To achieve the highest value for R4, R2 and R3 should be maximum values(105 ohms) and R4 and R1 minimum values(95 ohms). In addition R4 should be 0.95R4 in the equaion.
Solving gives that R4 = 22.16 Ohms.

Would this in fact be the minimum total value of the resistor R4 to accommodate for the tolerances? Is my thinking correct or would there be a better solution that would be more convincing?
 

Attachments

WBahn

Joined Mar 31, 2012
32,823
You need to look at the worst case situations in both directions. First, remove R4 and ask under what conditions of resistors at the edge of their tolerance will Vout be the greatest. Then, if you could add a single resistor in series with either R1 or R2, what value would it need to be to get Vout back to 0 V? Then do the same, but asking about the conditions that will make Vout the least (as in most negative) and again determine what single resistor added to either R1 or R2 will allow you to null the output (make it 0 V). How can you accomplish both of those things with a single potentiometer?
 
Top