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:
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\$
Best Answer
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