Electrical – Determining the Transfer Function of a Wheatstone Bridge

circuit analysistransfer functionwheatstone-bridge

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\$

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