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Wheatstone Bridge

Wheatstone Bridge

The Wheatstone Bridge is the name given to a combination of four resistances connected to give a null center value

The Wheatstone Bridge diamond shaped circuit who’s concept was developed by Charles Wheatstone can be used to accurately measure unknown resistance values, or as a means of calibrating measuring instruments, voltmeters, ammeters, etc, by the use of a variable resistance and a simple mathematical formula.

Although today digital multimeters provide the simplest way to measure a resistance. The Wheatstone Bridge can
be used to compare an unknown resistance to that of a known resistance to determine its value allowing very low values of resistances down in the milli-Ohms () range to be measured.

The Wheatstone bridge (or resistance bridge) circuit can be used in a number of applications and today, with modern operational amplifiers we can use the Wheatstone Bridge Circuit to interface various transducers and sensors to these amplifier circuits.

The Wheatstone Bridge circuit is nothing more than two simple series-parallel arrangements of resistances connected between a voltage supply terminal and ground producing zero voltage difference between the two parallel branches when balanced. A Wheatstone bridge circuit has two input terminals and two output terminals consisting of four resistors configured in a familiar diamond-like arrangement as shown. This is typical of how the Wheatstone bridge is drawn.

The Wheatstone Bridge

wheatstone bridge

When balanced, the Wheatstone bridge can be analysed simply as two series strings in parallel. In our tutorial about Resistors in Series, we saw that each resistor within the series chain produces an IR drop, or voltage drop across itself as a consequence of the current flowing through it as defined by Ohms Law. Consider the series circuit below.

series resistor circuit

As the two resistors are in series, the same current ( i ) flows through both of them. Therefore the current flowing through these two resistors in series is given as: V/RT.

I = V ÷ R = 12V ÷ (10Ω + 20Ω) = 0.4A

The voltage at point C, which is also the voltage drop across the lower resistor, R2 is calculated as:

VR2 = I × R2 = 0.4A × 20Ω = 8 volts

Then we can see that the source voltage VS is divided among the two series resistors in direct proportion to their resistances as VR1 = 4V and VR2 = 8V. This is the principle of voltage division, producing what is commonly called a potential divider circuit or voltage divider network.

Now if we add another series resistor circuit using the same resistor values in parallel with the first we would have the following circuit.

series parallel resistor circuit

As the second series circuit has the same resistive values of the first, the voltage at point D, which is also the voltage drop across resistor, R4 will be the same at 8 volts, with respect to zero (battery negative), as the voltage is common and the two resistive networks are the same.

But something else equally as important is that the voltage difference between point C and point D will be zero volts as both points are at the same value of 8 volts as: C = D = 8 volts, then the voltage difference is: 0 volts

When this happens, both sides of the parallel bridge network are said to be balanced because the voltage at point C is the same value as the voltage at point D with their difference being zero.

Now let’s consider what would happen if we reversed the position of the two resistors, R3 and R4 in the second parallel branch with respect to R1 and R2.

reversed resistor circuit

With resistors, R3 and R4 reversed, the same current flows through the series combination and the voltage at point D, which is also the voltage drop across resistor, R4 will be:

VR4 = 0.4A × 10Ω = 4 volts

Now with VR4 having 4 volts dropped across it, the voltage difference between points C and D will be 4 volts as: C = 8 volts and D = 4 volts. Then the difference this time is: 8 – 4 = 4 volts

The result of swapping the two resistors is that both sides or “arms” of the parallel network are different as they produce different voltage drops. When this happens the parallel network is said to be unbalanced as the voltage at point C is at a different value to the voltage at point D.

Then we can see that the resistance ratio of these two parallel arms, ACB and ADB, results in a voltage difference between 0 volts (balanced) and the maximum supply voltage (unbalanced), and this is the basic principal of the Wheatstone Bridge Circuit.

So we can see that a Wheatstone bridge circuit can be used to compare an unknown resistance RX with others of a known value, for example, R1 and R2, have fixed values, and R3 could be variable. If we connected a voltmeter, ammeter or classically a galvanometer between points C and D, and then varied resistor, R3 until the meters read zero, would result in the two arms being balanced and the value of RX, (substituting R4) known as shown.

Wheatstone Bridge Circuit

wheatstone bridge circuit

By replacing R4 above with a resistance of known or unknown value in the sensing arm of the Wheatstone bridge corresponding to RX and adjusting the opposing resistor, R3 to “balance” the bridge network, will result in a zero voltage output. Then we can see that balance occurs when:

wheatstone bridge ratio

The Wheatstone Bridge equation required to give the value of the unknown resistance, RX at balance is given as:

wheatstone bridge equation

Where resistors, R1 and R2 are known or preset values.

Example No1

The following unbalanced Wheatstone Bridge is constructed. Calculate the output voltage across points C and D and the value of resistor R4 required to balance the bridge circuit.

wheatstone bridge example

For the first series arm, ACB

wheatstone bridge arm acb

For the second series arm, ADB

wheatstone bridge arm adb

The voltage across points C-D is given as:

wheatstone bridge voltage

The value of resistor, R4 required to balance the bridge is given as:

balance resistor

We have seen above that the Wheatstone Bridge has two input terminals (A-B) and two output terminals (C-D). When the bridge is balanced, the voltage across the output terminals is 0 volts. When the bridge is unbalanced, however, the output voltage may be either positive or negative depending upon the direction of unbalance.

Wheatstone Bridge Light Detector

Balanced bridge circuits find many useful electronics applications such as being used to measure changes in light intensity, pressure or strain. The types of resistive sensors that can be used within a wheatstone bridge circuit include: photoresistive sensors (LDR’s), positional sensors (potentiometers), piezoresistive sensors (strain gauges) and temperature sensors (thermistor’s), etc.

There are many wheatstone bridge applications for sensing a whole range of mechanical and electrical quantities, but one very simple wheatstone bridge application is in the measurement of light by using a photoresistive device. One of the resistors within the bridge network is replaced by a light dependent resistor, or LDR.

An LDR, also known as a cadmium-sulphide (Cds) photocell, is a passive resistive sensor which converts changes in visible light levels into a change in resistance and hence a voltage. Light dependent resistors can be used for monitoring and measuring the level of light intensity, or whether a light source is ON or OFF.

A typical Cadmium Sulphide (CdS) cell such as the ORP12 light dependent resistor typically has a resistance of about one Megaohm (MΩ) in dark or dim light, about 900Ω at a light intensity of 100 Lux (typical of a well lit room), down to about 30Ω in bright sunlight. Then as the light intensity increases the resistance reduces. By connecting a light dependant resistor to the Wheatstone bridge circuit above, we can monitor and measure any changes in the light levels as shown.

Light Detection

wheatstone bridge light detector

The LDR photocell is connected into the Wheatstone Bridge circuit as shown to produce a light sensitive switch that activates when the light level being sensed goes above or below the pre-set value determined by VR1. In this example VR1 either a 22k or 47kΩ potentiometer.

The op-amp is connected as a voltage comparator with the reference voltage VD applied to the non-inverting pin. In this example, as both R3 and R4 are of the same 10kΩ value, the reference voltage set at point D will therefore be equal to half of Vcc. That is Vcc/2.

The potentiometer, VR1 sets the trip point voltage VC, applied to the inverting input and is set to the required nominal light level. The relay turns “ON” when the voltage at point C is less than the voltage at point D.

Adjusting VR1 sets the voltage at point C to balance the bridge circuit at the required light level or intensity. The LDR can be any cadmium sulphide device that has a high impedance at low light levels and a low impedance at high light levels.

Note that the circuit can be used to act as a “light-activated” switching circuit or a “dark-activated” switching circuit simply by transposing the LDR and R3 positions within the design.

The Wheatstone Bridge has many uses in electronic circuits other than comparing an unknown resistance with a known resistance. When used with Operational Amplifiers, the Wheatstone bridge circuit can be used to measure and amplify small changes in resistance, RX due, for example, to changes in light intensity as we have seen above.

But the bridge circuit is also suitable for measuring the resistance change of other changing quantities, so by replacing the above photo-resistive LDR light sensor for a thermistor, pressure sensor, strain gauge, and other such transducers, as well as swapping the positions of the LDR and VR1, we can use them in a variety of other Wheatstone bridge applications.

Also more than one resistive sensor can be used within the four arms (or branches) of the bridge formed by the resistors R1 to R4 to produce “full-bridge”, “half-bridge” or “quarter-bridge circuit arrangements providing thermal compensation or automatic balancing of the Wheatstone bridge.

204 Comments

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  • kenneth gokgana

    indeed a very pleasing information

  • Sandeep

    All method clear in this page

  • IMBAL JAYSON

    I like to know more about physics so thanks much for your help

  • Emmanuel John

    It’s easy to understand

  • Charity Jeffrey

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  • Charity Jeffrey

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  • Abdulbasi

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  • Shahzad p s

    Metro bridge

  • Fasil

    It is the best of best

  • Stella

    Nyc explanation.

  • Mike

    I feel like this has sort of missed the point of the Wheatstone bridge. It’s most useful because you can remove or reduce transducer drift, unwanted thermal effects etc by adding a passive version of the transducer to the other arm of the bridge and not exposing it to light, weight, sound (whatever). Both transducers will remain balanced in the bridge even when environmental properties of the transducers shift – the only way to get an imbalance is by exposing one of the transducers to the thing that you are really interested in measuring (light, sound etc). This is a very specific and valuable property of the bridge.

  • NiCK

    1. “first, the voltage at point D, which is also the voltage drop across resistor, R4
    2. “The voltage at point C, which is also the voltage drop across the lower resistor, R2 is calculated as:”
    Can you please elaborate on these statements?
    I think of it this way-> “Voltage at point D should be the voltage after the voltage drop has occurred in R3”, or, “The voltage at point C should be the voltage after the voltage drop has occurred in R1”

    Although in this case, the voltage at point D is also the voltage drop across R4 because there is no other resistor in the series left, so all the voltage has to drop beyond point D.

    • Wayne Storr

      With regards to the example given, the I*R voltage drops and potential differences between any two points in a circuit are generally referenced to a common point (usually earth, or 0V). Resistors R1 and R2 form one series branch, while resistors R3 and R4 form the second series branch. As point “B” is common to both branches, this is taken as the reference point.

      The voltage at point (or Node) A is 12 volts as it is connected to the 12 volt source. The I*R voltage drop across resistor R1 is the potential difference between point A and C, i.e., voltage at point A minus the voltage at point C. The voltage at point C can also be found from the voltage at point C minus the reference 0V at point B. Thus using the voltage divider formula:

      VR1 = V x (R1/(R1 + R2))
      VR1 = 12 x (10/(10 + 20)) = 4 volts

      VR2 = V x (R2/(R1 + R2))
      VR2 = 12 x (20/(10 + 20)) = 8 volts

      We know that the current going through resistors R1 and R2 is the same (series circuit rule). Their voltage drops can therefore be calculated using Ohms Law’s I*R rule and some basic maths to find the voltage at point C as shown in the tutorial.

      The same rules apply for the second series branch of R3 and R4.

  • Kelvin John

    How can you calculate the seven s voltage on unbalanced wheatstome bridge

  • Kelvin Jackson

    Nice teacher

  • Odia pius

    How does weighbridge work with the principle of wheatstone bridge?

    • Wayne Storr

      It depends on the size and weight limit of the weighbridge. There could be one load cell in each corner and a mean average value taken, or one single pivoted load cell. Internally, one or more strain gauges whose electrical resistance varies with applied force are arranged to form the resistive arms of a Wheatstone Bridge Circuit. An excitation voltage (in Volts) is applied to the load cell with the low output voltage (in milli-volts) indicating the weight.

  • Adnan dawar

    👍👍👍👍👍

  • nadeem ahmed

    good explanation

  • sayon John

    I love the side pls I will like to be notify when u post next
    Thanks

  • Jude Onyekachi

    Help me out with this question from principles of Electronics.
    👉 Using Thevenin’s theorem , find the current through the galvanometer in the wheatstone bridge with R1=1000 ohms, R2=100 ohms, R3=1000 ohms , R4=99 ohms and a voltage of 10 volts.

  • Annu Sharma

    I agree