Library / Circuits, electronics and noise / ELN-03
Circuits, electronics and noise
Unloaded Wheatstone bridge output voltage
Difference between the two midpoint voltages of a bridge made of two resistor dividers across the same supply, read by an ideal high-impedance instrument.
- Ref
- ELN-03
- Rev
- v1
- Field
- Circuits, electronics and noise
- Result
- Va-Vb [V]
A bridge is two dividers across the same supply. The difference between their midpoints is Vs[R₂/(R₁ + R₂) − R₄/(R₃ + R₄)]. It is zero when R₂/R₁ = R₄/R₃, even if the resistor values on the two sides are different. A positive result means point a is at a higher voltage than point b. Use an instrument with a high input resistance.
Difference between the two midpoint voltages of a bridge made of two resistor dividers across the same supply, read by an ideal high-impedance instrument. It is zero when R₂/R₁ equals R₄/R₃.
Calculator
- Left midpoint voltage Vₐ5 V
- Right midpoint voltage Vᵦ6 V
- Balance condition, R₂R₃ − R₁R₄ (zero when balanced)−1,000,000 Ω²
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The two midpoints sit at 5 V and 6 V, so the bridge reads −1 V.
- If R₂/R₁ = R₄/R₃, such as 1 kΩ, 2 kΩ and 3 kΩ, 6 kΩ, the output is exactly zero.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (V) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Vs Supply voltage across the bridge | 0.1 to 100 V | −10 | −0.01 | 9.99 | +10% | +10 % |
| R3 Upper right resistor | 100 to 100,000 Ω | −4.677 | 4.709 | 9.386 | +10% | −23.1 % |
| R4 Lower right resistor | 100 to 100,000 Ω | −4.804 | 4.524 | 9.328 | +10% | +22.6 % |
| R1 Upper left resistor | 100 to 100,000 Ω | −5.901 | 3.091 | 8.992 | +10% | +23.8 % |
| R2 Lower left resistor | 100 to 100,000 Ω | −5.091 | 3.901 | 8.992 | +10% | −23.8 % |
Sweep of Lower right resistor
| Lower right resistor (Ω) | Bridge output voltage (V) |
|---|---|
| 100 | 4.524 |
| 237.1 | 3.94 |
| 562.3 | 2.805 |
| 1,334 | 0.9997 |
| 3,162 | −1.126 |
| 7,499 | −2.895 |
| 17,780 | −3.989 |
| 42,170 | −4.547 |
| 100,000 | −4.804 |
Sweep one variable
Variables and how each one contributes
Bridge output voltage
What the equation gives you. Everything else feeds it.
Supply voltage across the bridge
The supply voltage across the whole bridge, in V. Scales the bridge output in direct proportion. Raising it by 10 % (to 11) moves the result by +10 %, from −1 to −1.1 V.
Upper left resistor
The upper resistor of the left leg, in Ω. A larger R₁ lowers the left midpoint voltage, so the output falls. Raising it by 10 % (to 1,100) moves the result by +23.8 %, from −1 to −1.238 V.
Lower left resistor
The lower resistor of the left leg, in Ω. A larger R₂ raises the left midpoint voltage, so the output rises. Raising it by 10 % (to 1,100) moves the result by −23.8 %, from −1 to −0.7619 V.
Upper right resistor
The upper resistor of the right leg, in Ω. A larger R₃ lowers the right midpoint voltage, so the output rises. Raising it by 10 % (to 2,200) moves the result by −23.1 %, from −1 to −0.7692 V.
Lower right resistor
The lower resistor of the right leg, in Ω. A larger R₄ raises the right midpoint voltage, so the output falls. Raising it by 10 % (to 3,300) moves the result by +22.6 %, from −1 to −1.226 V.
Worked example with the values above
- Left leg: R₂/(R₁ + R₂) = 0.5; right leg: R₄/(R₃ + R₄) = 0.6.
- Difference of ratios: −0.1.
- Vₐ − Vᵦ = 10 × that = −1 V.
Assumptions and limits
- Ideal sources and linear resistors with positive values, all in base SI units.
- The circuit is exactly as described: no extra loads, and instruments that draw no current.
Common mistakes
- Forgetting a load across a divider, which lowers the output.
- Using the resistor on the wrong side of the divider or the wrong branch for the current share.
- Mixing kΩ and Ω in one calculation.
Statistics and uncertainty
Resistor values are only known to their tolerance, typically 1 % to 5 %, so a result built from them carries that scatter. Ratios such as dividers and bridges are less sensitive to a common drift than to differences between the resistors.
Where it is used
- Sensor interfaces, bias networks, bridge circuits and mixers.
- Reducing a network to simpler series and parallel parts.
Maths you need
- Ohm’s law: V = I R
- Kirchhoff’s current law: currents into a node add to zero
- Series and parallel resistors: R₁ + R₂ and R₁R₂/(R₁ + R₂)