In this tutorial, we will learn about Wheatstone Bridge. We will see the Working principle of Wheatstone Bridge, few example circuits and some important applications.

### Introduction to Wheatstone Bridge

In the real world we come across various signals, some of them are measured by changes in resistance and some of them are with inductance and capacitance.

If we consider the resistance, most of the industrial sensors like temperature, strain, humidity, displacement, liquid level, etc. produces the change in value of the resistance for a variable change. Therefore, there is a need for a signal conditioning for every resistance sensor.

Generally the resistance measurement is divided into three types, low resistance measurement, medium resistance measurement and the high resistance measurement. If the resistance measurement is possibly from a few milliohms to micro ohms, then it is considered as a low resistance measurement.

This measurement is actually used for research purpose. If the measurement is from 1 ohm to 100 k is generally referred as a medium resistance measurement. Potentiometer, thermistors, etc. measurement comes under this category.

And very high resistance measurement is considered from 100 kilo ohm to greater than 100 mega ohms. For finding the medium value of the resistance different methods are used, but mostly Wheatstone bridge is used.

**What is Wheatstone Bridge?**

The most common and simplest bridge network to find the resistance is the DC Wheatstone Bridge. This bridge is used where small changes in resistance are to be measured like in sensor applications. This is used to convert a resistance change to a voltage change of a transducer.

The combination of this bridge with operational amplifier is used extensively in industries for various transducers and sensors. A Wheatstone bridge consists of four resistors that are connected in the shape of a diamond with the supply source and indicating instruments as shown in figure.

This bridge is used to find the unknown resistance very precisely by comparing it with a known value of resistances. In this bridge null or balanced condition is used to find the resistance.

For this bridge balanced condition voltage at points C and D must be equal. Hence, no current flows through the galvanometer. For getting the balanced condition one of the resistors must be variable.

From the figure,

The voltage at point **D = V × R _{X }/ (R_{3} + R_{X})**

The voltage at point **C = V × R _{2 }/ (R_{1} + R_{2})**

The voltage (V) across galvanometer or between C and D is,

**V _{CD} = V × R_{X }/ (R_{3} + R_{X}) − V R_{2 }/ (R_{1} + R_{2})**

When the bridge is balanced V_{CD} = 0,

So,

**V × R _{X }/ (R_{3} + R_{X}) = V R_{2 }/ (R_{1} + R_{2})**

**R _{X}R_{1 }+ R_{X}R_{2} = R_{2}R_{3 }+ R_{2}R_{X}**

**R _{1}R_{X}= R_{2}R_{3}**

**R _{2}/R_{1}= R_{X}/R_{3}**

This is the condition to balance the bridge. And for finding the unknown value of resistance

**R _{X} = R_{3} × (R_{2 }/ R_{1}) **

From the above equation R4 or Rx can be computed from the known value of resistance R3 and the ratio of R2/R1. Therefore, most of the cases R2 and R1 values are fixed and the R3 value is variable so that null value is achieved and the bridge gets balanced.

**Working Principle**

Without the galvanometer, the bridge circuit just looks like a voltage divider circuit as shown in the figure below. Consider R_{1}= 20 ohm, R_{2}= 40 ohm for one arm and for the other consider same values of R_{3} and R_{4} respectively.

Current flow in first arm is

**I _{1 }= V/ (R_{1}+R_{2})**

**I _{1} = 12V/ (20+40)**

**I _{1 }= 0.2 A**

And voltage at point C is equal to the voltage drop at resistor R_{2},

**V _{R2} = I_{1} × R_{2}= 0.2 × 40 = 8V**

Similarly the voltage across R_{1} is 4V (0.2 × 20). Due to the same resistance values, voltages at R_{4} and R_{3} will be same as that of R_{1} and R_{2} respectively. Hence at the points A and B voltages are same, therefore the galvanometer shows zero reading as the potential difference is zero. In this case the bridge is said to be in balanced condition.

Suppose if we reverse the resistors in the second arm, current flow is same due to the series circuit. But the voltage across the resistor R_{4} changes, i.e., 0.2 * 20 = 4V. So at this condition voltage across the points A and B are different and exists a potential difference of 8 – 4 = 4V. This is the unbalanced condition of the bridge.

**Example of Wheatstone Bridge**

From above, the Wheatstone bridge is unbalanced when the voltmeter reading is not zero. This reading can be positive or negative depends on the magnitudes of the voltages at the meter terminals. Let us consider the below circuit of Wheatstone bridge which is connected to find the unknown resistance value with use of resistor decade box to get the variable resistance of R_{3}.

We know that the condition for bridge balance is

**R _{4} = R_{3} × R_{2} / R_{1}**

**R _{x }= R_{BOX }× (10 x 10^{3})/ (10 x 10^{3})**

**R _{x} = R_{BOX}**

Here in this case, the Wheatstone bridge is balanced by adjusting the decade resistance box until the voltmeter reads zero value. And the corresponding resistance value in the box is equal to the unknown resistance. Suppose if the voltage null condition occurs at 250 ohms of the resistance decade box, then the unknown resistance is also 250 ohms.

**Wheatstone Bridge for Strain Measurement**

Most commonly for measuring the strain, strain gages are used whose electrical resistance varies with proportionate strain in the device. In practice, the range of strain gauge resistance is from 30 ohms to 3000 ohms. For a given strain, the resistance change may be only a fraction of full range. Therefore, to measure extremely a fraction of resistance changes with high accuracy, Wheatstone bridge configuration is used. The below figure shows the Wheatstone bridge where the unknown resistor is replaced with a strain gauge.

In the above circuit, two resistors R_{1} and R_{2} are equal to each other and R_{3} is the variable resistor. With no force applied to the strain gauge, rheostat is varied and finally positioned in such that the voltmeter will indicate zero deflection. This is called a bridge balancing condition. This condition represents that there is no strain on the gauge.

If the strain gauge is either tensed or compressed, then the resistance can increase or decrease. Therefore, this causes unbalancing of the bridge. This produces a voltage indication on voltmeter corresponds to the strain change. If the strain applied on a strain gauge is more, then the voltage difference across the meter terminals is more. If the strain is zero, then the bridge balances and meter shows zero reading.

This is about the resistance measurement using a Wheatstone bridge for precise measurement. Due to the fractional measurement of resistance, Wheatstone bridges are mostly used in strain gauge and thermometer measurements.

**Applications**

- The Wheatstone bridge is used for measuring the very low resistance values precisely.
- Wheatstone bridge along with operational amplifier is used to measure the physical parameters like temperature, strain, light, etc.
- We can also measure the quantities capacitance, inductance and impedance using the variations on the Wheatstone bridge.

pratik says

Which circuit is used to increase the output voltage of Wheatstone’s bridge?

Syed Arsalan says

You can cascade a differential op amp to amplify the output voltage.

muhlis says

Which bridge would you prefer to use for the measurement of a test resistor having a value around 50 k Ω ?