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I am designing a Wheatstone Bridge, and would like to calculate the transfer function of the bridge input to the bridge output.

The Wheatstone Bridge is designed as follows: Wheatstone Bridge

Three of the resistors are identical, with the fourth being a potentiometer.

In this case the bridge input is 5V, but I'm not sure how to relate this to the bridge output.

UPDATE: There are two voltage dividers, where:

\$V_- = 5V(potentiometer/ (potentiometer + R1))\$

\$V_+ = 5V(R1/ (R1+ R1)) = 5/2\$

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  • \$\begingroup\$ These are two voltage dividers. V- = 5V (POT/(R1+POT)). The second is left as an exercise. \$\endgroup\$
    – Mario
    Sep 21, 2016 at 18:54
  • \$\begingroup\$ @Mario so the second would be: \$V_+ = 5V(R1/ (R1+ R1)) = 5/R1\$ \$\endgroup\$
    – tibsar
    Sep 21, 2016 at 19:02
  • \$\begingroup\$ R1/(R1+R1) =1/2 \$\endgroup\$
    – Mario
    Sep 21, 2016 at 19:03
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    \$\begingroup\$ Now we could define Vout as Vout = V+ - V- = ... and that's it. \$\endgroup\$
    – Mario
    Sep 21, 2016 at 19:05
  • \$\begingroup\$ @Mario of course! My mistake. So then \$V_o = 5V(potentiometer/ (potentiometer + R1)) - (5/2)\$ \$\endgroup\$
    – tibsar
    Sep 21, 2016 at 19:05

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Tansfer functions are written Vout/Vin, so in your case V-/Vin = pot/(pot+R1) V+/Vin = R1/(R1+R1) = 1/2

Usually the outputs of a Wheatstone Bridge are put into an instrumentation or differential amplifier. Then the transfer function will change to something like the following:

tf_inst_amp = (V+/Vin - V-/Vin) * gain_of_diff_amp = (V+-V-)*gain_of_diff_amp/Vin

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  • \$\begingroup\$ Transfer functions are typically Vo/Vin. \$\endgroup\$
    – Daniel
    Sep 21, 2016 at 19:57
  • \$\begingroup\$ whoops nice catch \$\endgroup\$
    – klamb
    Sep 21, 2016 at 19:58

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