# Wheatstone Bridge Calculator

## This calculator helps determine the resistance of a variable resistor within a Wheatstone bridge circuit or the differential voltage created across the bridge terminals.

### You must input everything except for either Rx or Vout

This calculator can be used to either design a Wheatstone bridge or to interpret the output of a sensor within a Wheatstone bridge. As shown in Figure 1, the Wheatstone bridge is a simple circuit consisting of two sets of series resistors connected in parallel.

**Figure 1. **A Wheatstone bridge circuit

**Figure 1.**A Wheatstone bridge circuit

The Wheatstone bridge configuration allows for high accuracy measurements of unknown resistance.

### Balancing a Wheatstone Bridge Circuit

This type of circuit is balanced when:

$$ V_{out} = 0 $$

To achieve this balance, the resistance ratios in the circuit must be balanced:

$$ \frac{R_1}{R_3} = \frac{R_2}{R_x} $$

Even a small change in resistance (R_{x}) causes the bridge to become unbalanced. These changes result in a non-zero voltage at V_{out} with the polarity indicating if R_{x} is smaller or larger than the balanced value. Utilizing a differential amplifier across the V_{out} terminals to extract the differential signal while rejecting common-mode noise makes precise detection of very small changes possible.

This calculator with either solve for V_{out} in the figure above given the R_{x} resistance or solve for the R_{x} resistance given V_{out}.

### Wheatstone Bridge Equation—Using KVL and KCL

To solve for output voltage, V_{out}, given the R_{x} resistance, we use Kirchoff's Voltage Law (KVL) and Kirchoff's Current Law (KCL) to arrive at the following equation:

$$ V_{out} = V_{in} \centerdot (\frac{R_x}{R_3+R_x} - \frac{R_2}{R_1+R_2}) $$

Where:

- V
_{out}= differential voltage across the Wheatstone bridge, in volts - V
_{in}= input voltage to the Wheatstone bridge, in volts - R
_{n}= resistor values, expressed in ohms

Similarly, to solve for R_{x} given the output voltage V_{out}, we'll use KVL and KCL to arrive at the following equation:

$$ R_{x} = \frac{R_2 R_3 + R_3 \centerdot(R_1+R_2) \centerdot \frac{V_{out}}{V_{in}}}{R_1 - (R_1+R_2) \centerdot \frac{V_{out}}{V_{in}}} $$

1 CommentThe equations used do not match the resistor designators in the circuit causing an error in the calculations.