 # Wheatstone Bridge

Mar 10th, 2019
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1. Using a Wheatstone bridge to find matched pairs of resistors
2. ------------------------------------------------------------
3.
4. We want to find a matched pair of 6.8k resistors for use with a phantom power circuit
5.
6. (+5v)---[R_Ref]---vDiv---[R_Test]---(0v)
7. |
8. o
9.
10. o
11. |
12. (+5v)---[10k]-----vRef------[10k]---(0v)
13.
14. R_Ref ~ 6.8k to balance the bridge roughly half way.
15.
16. Various resistors are tested in the R_Test position, and the vDiv-vRef differential voltage is measured in mV.
17.
18. We find two resistors R1 and R2, which give an equal differential voltage down to 0.1mV (the resolution of my meter)
19.
20. The voltage divider has this simple equation, for both R1 and R2:
21.
22. v1 = 5 * R1 / (RRef + R1) - vRef
23. v2 = 5 * R2 / (RRef + R2) - vRef
24.
25. We measured:
26.
27. |v1 - v2| < 0.1mV = 0.0001V
28.
29. Let us expand these equations, and solve for the difference in resistance |R1 - R2| :
30.
31. |(5 * R1 / (RRef + R1) - vRef) - (5 * R2 / (RRef + R2) - vRef)| < 0.0001
32. |(R1 / (RRef + R1)) - (R2 / (RRef + R2))| < 0.0001/5
33. |(R1*(RRef + R2) - R2*(RRef + R1))| / (RRef + R1)(RRef + R2) < 0.00002
34. |R1 - R2| * RRef < 0.00002 * (RRef + R1)(RRef + R2)
35.
36. NB: (RRef + R1)(RRef + R2) ~= 4*RRef^2
37.
38. |R1 - R2| ~< 0.00008 * RRef = 0.00008 * 6800 = 0.544 ohms :D
39.
40. Which is matched within 0.008% (that's 2 orders of magnitude better than matching within 1%)
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