| ??? 06/14/05 13:07 Read: times |
#94879 - Look at it this way..... Responding to: ???'s previous message |
A wheatstone bridge can be viewed as two resistor dividers - this is evidenced by the equation you provided (and missed an important bracket!). You can rewrite it as this: Vout = (Vin * R2/(R1+R2)) - (Vin * R3/(R3+Rx)) The left hand term is fixed (assuming your Vin and resistors are fixed). The right hand term is the variable term. For a Pt100 lets say all resistors are 100R. Therefore the left term will reduce to 0.5Vin and the right term will reduce to 0.5Vin assuming the Pt100 is exactly 100R (as we assume the other resistors are also). Lets say the Pt100 changes to 101R. What effect does this have? Vout = 0.5Vin - (Vin * (100/(100+101)) = 0.5Vin - (Vin * 0.4975) So for a Vin of 5v we get... Vout = 2.5 - 2.4875 = 0.01244v Whilst I haven't directly answered your question, hopefully I've given you a little insight into how the wheatstone bridge is working. As I mentioned, it is just two resistor dividers - the output is the difference between two ratios (one fixed, the other variable). As for algebraic manipulation - ask your math teacher and/or read about in a math book! You have to learn this sometime! |
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