Calculate the output voltage of a Wheatstone bridge from its four arms and the excitation, or solve for the unknown resistor at balance. You can also compare quarter, half and full sensor bridges with their mV/V and nonlinearity.
Optional: meter or amplifier across the output.
Output voltage Vo
1.49925 mV
Arms are numbered as in the NI strain-gauge note: R1 top left, R2 bottom left, R3 bottom right, R4 top right. Vo is the right node (R4–R3) minus the left node (R1–R2).
Each arm
| Arm | Resistance | Current | Power | Vo change for +1 % in this arm |
|---|---|---|---|---|
| R1 | 3 kΩ | 1 mA | 3 mW | +14.93 mV |
| R2 | 3 kΩ | 1 mA | 3 mW | -14.93 mV |
| R3 | 3.003 kΩ | 999.5 µA | 3 mW | +14.93 mV |
| R4 | 3 kΩ | 999.5 µA | 3 mW | -14.93 mV |
Treat it as two voltage dividers and subtract: Vo = Vex × [R3 ÷ (R3 + R4) − R2 ÷ (R1 + R2)]. The bridge is balanced (Vo = 0) when R1 × R3 = R2 × R4. With 6 V across four 3 kΩ arms and one arm raised to 3,003 Ω, the output is 1.49925 mV (0.24988 mV/V); for small changes a single active arm gives about Vex ÷ 4 × ΔR ÷ R.
The bridge is two voltage dividers across the same excitation. R1 and R2 form the left divider, R4 and R3 the right one, and the output is the voltage between their midpoints. The arm numbering follows National Instruments’ strain-gauge note, so the formulas can be checked against it.
The output is zero when the opposite-arm products are equal, which gives the unknown arm directly:
Seen from the output, an ideal excitation source is a short, so the bridge looks like R1 ∥ R2 in series with R3 ∥ R4. That is the resistance a meter or amplifier sees:
With all arms equal to R and the active arms changing by ΔR, the four standard sensor arrangements reduce to (TI SBAA532A, equations 4, 7, 11 and 13):
| Arrangement | Active arms | Exact output | Small ΔR |
|---|---|---|---|
| Quarter bridge (one active arm) | R3 = R + ΔR | Vex/2 × ΔR ÷ (2R + ΔR) | Vex × 1/4 × ΔR ÷ R |
| Half bridge, +ΔR and −ΔR on one side | R3 = R + ΔR, R4 = R − ΔR | Vex/2 × ΔR ÷ R | Vex × 1/2 × ΔR ÷ R |
| Half bridge, +ΔR in opposite arms | R1 = R3 = R + ΔR | Vex × ΔR ÷ (2R + ΔR) | Vex × 1/2 × ΔR ÷ R |
| Full bridge (four active arms) | R1 = R3 = R + ΔR, R2 = R4 = R − ΔR | Vex × ΔR ÷ R | Vex × 1 × ΔR ÷ R |
Four 3 kΩ arms at 6 V, with R3 raised by 3 Ω to 3,003 Ω. The left node sits at 3 V and the right node at 3.0015 V, so Vo = 1.49925 mV, or 0.24988 mV/V. The output resistance is 3.001 kΩ and the bridge draws 2 mA from the excitation.
With ratio arms R1 = 1 kΩ and R2 = 10 kΩ, the standard in the R4 position is adjusted until the detector reads zero at 470.5 Ω. Then R3 = R2 × R4 ÷ R1 = 4.705 kΩ. At 5 V excitation a 0.1 % change in the unknown moves the output by 413 µV, which is the signal the null detector must resolve. If the three known arms are ±0.1 % parts, the result is only known to 4.6909 kΩ to 4.7191 kΩ (about ±0.3 %): the reference tolerances add.
A foil gauge with gauge factor 2 changes by ΔR = 350 × 2 × 0.001 = 700 mΩ at 1000 µε. As a quarter bridge at 5 V it gives 2.4975 mV (0.4995 mV/V), matching NI’s figure of about 0.5 mV/V; the bridge’s own nonlinearity is -0.0999 %. Four active gauges in a full bridge give 10 mV (2 mV/V), four times as much and exactly linear.
Output per volt of excitation for each arrangement, with every arm R and each active arm changing by ΔR. The last column is how far the quarter bridge and the opposite-arm half bridge fall below a straight line.
| ΔR / R | Quarter | Half, opposite arms | Half, ±ΔR one side | Full | Nonlinearity |
|---|---|---|---|---|---|
| 0.01 % | 0.024999 mV/V | 0.049998 mV/V | 0.05 mV/V | 0.1 mV/V | -0.005 % |
| 0.1 % | 0.24988 mV/V | 0.49975 mV/V | 0.5 mV/V | 1 mV/V | -0.05 % |
| 0.5 % | 1.2469 mV/V | 2.4938 mV/V | 2.5 mV/V | 5 mV/V | -0.249 % |
| 1 % | 2.4876 mV/V | 4.9751 mV/V | 5 mV/V | 10 mV/V | -0.498 % |
| 2 % | 4.9505 mV/V | 9.901 mV/V | 10 mV/V | 20 mV/V | -0.99 % |
| 5 % | 12.195 mV/V | 24.39 mV/V | 25 mV/V | 50 mV/V | -2.44 % |
| 10 % | 23.81 mV/V | 47.619 mV/V | 50 mV/V | 100 mV/V | -4.76 % |
For an equal-arm bridge the excitation sees a resistance of R and each arm has Vex/2 across it, so every arm dissipates (Vex/2)² ÷ R. 120 Ω, 350 Ω and 1 kΩ are the usual foil strain gauge values.
| Arm resistance | Current at 2.5 V | Current at 5 V | Current at 10 V | Per arm at 2.5 V | Per arm at 5 V | Per arm at 10 V |
|---|---|---|---|---|---|---|
| 120 Ω | 20.8 mA | 41.7 mA | 83.3 mA | 13 mW | 52.1 mW | 208 mW |
| 350 Ω | 7.14 mA | 14.3 mA | 28.6 mA | 4.46 mW | 17.9 mW | 71.4 mW |
| 1 kΩ | 2.5 mA | 5 mA | 10 mA | 1.56 mW | 6.25 mW | 25 mW |
| 3 kΩ | 833 µA | 1.67 mA | 3.33 mA | 521 µW | 2.08 mW | 8.33 mW |
| 5 kΩ | 500 µA | 1 mA | 2 mA | 313 µW | 1.25 mW | 5 mW |
Measure ratiometrically. The output is proportional to Vex, so if the ADC reference is taken from the same excitation, drift and noise in the excitation cancel (TI SBAA532A, section 3.1). Long leads to a sensor bridge drop part of the excitation; a six-wire connection senses the excitation at the bridge itself.
In a quarter bridge the lead wires are in series with the active arm, so their resistance adds an offset and their temperature coefficient looks like signal. NI recommends a dummy gauge or a half or full bridge for temperature compensation, and 350 Ω or 1 kΩ gauges over 120 Ω to reduce self-heating and lead-wire effects. Keep the arm dissipation low: the table above shows a 120 Ω bridge at 10 V already putting 208 mW into each gauge.
As a resistance-measuring instrument the Wheatstone bridge is limited by its reference arms and the detector: Lessons in Electric Circuits quotes accuracies of ±0.05 % with precise standards. Below about 0.1 Ω the lead and contact resistance in series with the unknown becomes a large part of the reading, which is what the Kelvin double bridge (or a four-wire measurement) fixes.
Two resistive voltage dividers fed from the same excitation, with the output taken between their midpoints. The output is zero when the two divider ratios match, so a small change in one arm shows up as a small voltage around zero.
The Wheatstone bridge equation is Vo = Vex × [R3 / (R3 + R4) − R2 / (R1 + R2)], where R1 and R2 form the left divider and R4 and R3 the right one (R1 and R4 on the excitation side). The output is the difference between the two divider voltages, so it is zero when the two ratios match.
The products of opposite arms are equal: R1 × R3 = R2 × R4, the same as R1 / R2 = R4 / R3. Balance depends only on the resistances, not on the excitation voltage or the detector.
Adjust a known arm until the output reads zero, then solve the balance condition. With the unknown in the R3 position, R3 = R2 × R4 / R1: 10 kΩ × 470.5 Ω / 1 kΩ = 4.705 kΩ.
Use the output formula, which holds for any four arms as long as nothing loads the output: find each divider’s midpoint voltage and subtract. This calculator does that directly. TI’s four 3 kΩ arms at 6 V with one arm raised by 3 Ω give 1.49925 mV.
It is the output per volt of excitation. A 2 mV/V load cell gives 10 mV at full load with 5 V excitation. TI notes that typical bridge load cells are 1 mV/V to 3 mV/V, and NI quotes about 0.5 mV/V for a quarter-bridge strain gauge at 1000 µε (this calculator gives 0.4995 mV/V).
Its output is Vex/2 × ΔR / (2R + ΔR): the changing arm appears in the denominator, so the output falls short of a straight line by about ΔR / 2R. At ΔR/R = 1 % the bridge itself is 0.498 % low, before any sensor nonlinearity. A half bridge with +ΔR and −ΔR on the same side, and a full bridge, are exactly linear.
For small ΔR the output is Vex × ΔR/R times 1/4 for a quarter bridge, 1/2 for a half bridge and 1 for a full bridge. A full bridge is four times as sensitive as a quarter bridge and also cancels temperature effects that change all four arms equally.
The resistance change is tiny. In TI’s example a 3 Ω change on 3 kΩ moves a single divider from 3 V to only 3.0015 V; the bridge subtracts the fixed 3 V and leaves 1.49925 mV that an amplifier can gain up without saturating.
Yes. The output resistance is R1∥R2 + R3∥R4, which equals R for an equal-arm bridge (3.001 kΩ for TI’s 3 kΩ example). A detector with input resistance Rd reads Vo × Rd / (Rd + Rout), so use an instrumentation amplifier with an input resistance far above the bridge resistance.
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