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- Using a Wheatstone bridge to find matched pairs of resistors
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- We want to find a matched pair of 6.8k resistors for use with a phantom power circuit
- (+5v)---[R_Ref]---vDiv---[R_Test]---(0v)
- |
- o
- o
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- (+5v)---[10k]-----vRef------[10k]---(0v)
- R_Ref ~ 6.8k to balance the bridge roughly half way.
- Various resistors are tested in the R_Test position, and the vDiv-vRef differential voltage is measured in mV.
- We find two resistors R1 and R2, which give an equal differential voltage down to 0.1mV (the resolution of my meter)
- The voltage divider has this simple equation, for both R1 and R2:
- v1 = 5 * R1 / (RRef + R1) - vRef
- v2 = 5 * R2 / (RRef + R2) - vRef
- We measured:
- |v1 - v2| < 0.1mV = 0.0001V
- Let us expand these equations, and solve for the difference in resistance |R1 - R2| :
- |(5 * R1 / (RRef + R1) - vRef) - (5 * R2 / (RRef + R2) - vRef)| < 0.0001
- |(R1 / (RRef + R1)) - (R2 / (RRef + R2))| < 0.0001/5
- |(R1*(RRef + R2) - R2*(RRef + R1))| / (RRef + R1)(RRef + R2) < 0.00002
- |R1 - R2| * RRef < 0.00002 * (RRef + R1)(RRef + R2)
- NB: (RRef + R1)(RRef + R2) ~= 4*RRef^2
- |R1 - R2| ~< 0.00008 * RRef = 0.00008 * 6800 = 0.544 ohms :D
- Which is matched within 0.008% (that's 2 orders of magnitude better than matching within 1%)
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