Introduction:
The Wheatstone Bridge is an electrical bridge circuit for accurately measuring a constant or changing electrical resistance. This circuit can be used to measure physical quantities such as temperature and pressure, as seen in the mass air flow sensor (temperature of the hot wire) and MAP sensor (pressure in the intake manifold).
The Wheatstone Bridge contains four resistors, with three known and one unknown resistance. This essentially makes the bridge two voltage dividers connected in parallel.
The image shows resistors R1 to R3 (known resistance values) and Rx (unknown), with a voltmeter in the middle of the two voltage dividers and a power source to the left of the bridge.
The Wheatstone Bridge is balanced or in equilibrium when the output voltage between points b and c is equal to 0 volts. The following paragraphs show different situations.
Balanced Wheatstone Bridge:
The Wheatstone Bridge is balanced when the output voltage is equal to 0 volts, because the resistance values on the left and right are proportionate.
The circuit in this paragraph is drawn differently than in the previous paragraph but is based on the same operation.
- resistors R1 and R2 have resistances of 270 and 330 a9. Together, this totals 600 a9;
- resistors R3 and Rx have resistances of 540 and 660 a9. Together, this totals 1200 a9.
The proportions of the resistances on the left and right are equal. Therefore, the resistance ratios and the voltage drops are equal between R1 and R3, as well as R2 and Rx.
The formulas below show the equal resistance ratios and voltage drops:
a0 a0 a0anda0 a0a0
With a known supply voltage and resistance values, we can determine the voltage drops across the resistors and thus the voltage difference between points b and c. In the example below, we calculate the voltage difference between points b and c for a balanced Wheatstone Bridge. Knowledge of Ohm’s Law and calculating with series & parallel circuits is required.
1. calculate the currents through resistors R1 and R2a0(RV = equivalent resistance):
2. calculate the voltage drop across resistors R1 and R2:
a0 a0 a0 a0 a0
3. calculate the currents through resistors R1 and R2:
4. calculate the voltage drop across resistors R3 and Rx:
a0 a0 a0 a0 
The voltage at point b and c is 5.4 volts. The potential difference is 0 volts.
Unbalanced Wheatstone Bridge (known resistances):
Due to a change in resistance of Rx, the Wheatstone Bridge will be out of balance. The resistance change can occur, for example, due to a changing temperature, where Rx is a thermistor. The voltage divider between R1 and R2 will remain the same, but not between R3 and Rx. Because the voltage divider changes there, we get a different voltage at point c. In this example, the resistance value of Rx has dropped from 600 a9 to 460 a9.
1. calculate the currents through resistors R1 and R2:
2. calculate the voltage drop across resistors R1 and R2:
a0 a0 a0 a0
4. calculate the voltage drop across resistors R3 and Rx:
a0 a0 a0 a0 a0
In the two examples, the resistance value of Rx has changed from 660 a9 to 460 a9. Due to this change in resistance, the voltage between b-c has changed from 0 volts to 1.08 volts. When this Wheatstone Bridge is embedded in the sensor electronics, the 1.08 volts is seen as a signal voltage. This signal voltage is sent to the ECU via a signal wire. The A/D converter in the ECU converts the analog voltage to a digital message, which can be read by the microprocessor.
Wheatstone Bridge with unknown resistance:
In the previous paragraphs, we assumed a known resistance value for Rx. Since this resistance value is variable, we can go a step further by calculating this resistance value to bring the Wheatstone Bridge in balance.
In this circuit, R1 and R2 are again 270 and 330 a9. The resistance of R3 is reduced to 100 a9 and Rx is unknown. If both the resistance value and the voltages and currents are unknown, we can calculate the resistance value Rx in two ways:
Method 1:
1. First, we look at the general formula and then fill in the resistance values:
a0 a0 a0 –>a0 a0 a0 
2. There is a factor of 2.7 between 270 and 100, as well as between 330 and the unknown value.
By dividing 330 by 2.7, we obtain a resistance of 122.2 a9.a0
Method 2:
1. through the general formula in which we cross-multiply the resistors:
2. we rearrange the formula to bring Rx to the left of the equals sign and divide by R1. Hereby, we also find the resistance value of 122.2 a9.
We will of course check if we have a balanced bridge with the previously calculated resistance of 122 a9.
The resistors R1 and R2 with the currents and partial voltages are the same as in the examples in paragraph 1 and 2, so they are considered known. We focus on the right side of the bridge.
1. calculate the current through R3 and Rx:
2. calculate the voltage drop across resistors R3 and Rx:
The voltage difference between points b and c is 0 volts because resistors R1 and R3 both absorb 5.4, thus the bridge is now balanced.
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