Introduction: Wheatstone Bridge Implementation for a Light-Dependent Resistor
The Wheatstone Bridge is a basic resistive network that can be utilized in both the alternating current (AC) and direct current (DC) regimes. This document serves as a guideline for beginner Electrical Engineers to understand the basic operation and implementation of the bridge under DC conditions. This guide does not rigorously cover many other topics and skills required (like how to bread board, use a multimeter, use Ohm's Law, read data-sheets, etc.), but it does link to some useful sites along the way.
The primary use-case for the Wheatstone Bridge is usually to determine the value of a resistor that is dependent on any number of physical or pseudo-physical quantities such as light, strain, or temperature. In this article, a Wheatstone bridge will be implemented for a photoresistor/light-dependent resistor (LDR).
Using the Wheatstone Bridge, LDR, Operational Amplifier, and some standard resistors, a circuit can be constructed to output a high or low level signal which is dependent on a selectable threshold. An LED will be used to determine the state of the system.
This document assumes a basic understanding of circuit analysis (familiarity with passive components and Ohm’s Law).
- V = IR
- V: Voltage
- R: Resistor
- I: Current
Step 1: Gather Supplies
If you will be building the circuit in real life, you should have the following materials:
- Photoresistor (LDR)
- 4 Standard Resistors
- Op-Amp (LM324)
- Power Supply
- Wire Stripper
If you do not have access to these physical materials, download a free circuit analysis software, like LTSpice, in order simulate the circuit.
Step 2: Learn the Theory (Wheatstone Bridge)
Observe the resistive network pictures. Points A and B have corresponding voltages Va and Vb. Using voltage division, we can solve for both:
Va = Vin*(R1)/(R1+R2)
Vb = Vin*(Rx)/(R3+Rx)
The voltage between the points can be defined as V = Va-Vb. The bridge is said to be balanced when Va = Vb. So,
Vin*(R1)/(R1+R2) = Vin*(Rx)/(R3+Rx)
(R1)/(R1+R2) = (Rx)/(R3+Rx)
(R1)*(R3+Rx) = (Rx)(R1+R2)
R1*R3 + R1*Rx = Rx*R1 + Rx*R2
R1*R3 = Rx*R1 + Rx*R2 - R1*Rx
Rx*R2 = R1*R3
This shows that when the bridge is balanced, the product of the resistance in the top of the first branch and the bottom of the second branch is equal to the product of the resistance in the bottom of the first branch and the top of the second branch.
This information is beneficial. A base state can be defined such that there is no voltage difference between Va and Vb. A useful base state (as a function of Rx) can be derived from the resistance of the sensor under a specified “threshold” condition. The ratio of R2:R1 = R3:Rx should be the same at the threshold condition (creates equal voltage division, which should be intuitive). Fulfilling the ratio, rather than exact values, allow for standard resistor values to be used.
Something to keep in mind is the overall resistance values as well. The higher the resistance of your network, the lower the amount of power dissipated (compared to a similar network with lower resistance).
Say a resistive device is at the threshold when its value is 10kOhms. Since 10kOhms is a standard resistor, the entire bridge can be made with 10kOhm resistors for simplicity. Currently:
V = Va - Vb = Vin/2 - Vin/2 = 0, The bridge is balanced.
If the resistance increases to some arbitrary value like 20kOhms
V = Va - Vb = Vin/2 - 2Vin/3 = -Vin/6
If the resistance decreases to some arbitrary value like 5kOhms
V = Va - Vb = Vin/2 - Vin/3 = Vin/6
It is useful to observe that Va never changes. Since it is fixed, it can be referred to as the reference voltage. As resistance of the device increases, Vb becomes greater than the reference voltage. As resistance decreases, Vb becomes less than the reference voltage.
Say a resistive device has a non-ideal resistance at the threshold, 13.6 kOhm.
Rx = 13.6 kOhms
Rx*R2 = R1*R3 R2*13.6 = R1*R3
There are many solutions that will fulfill this. Start by assuming a resistance value for one of the other resistors, then calculating the others.
R2 = 1 kOhm
13.6*1 = R1*R3
13.6 = ~1.3*10
R1 = 1.3 kOhm
R3 = 10 kOhm<br>
Step 3: Learn the Theory (Comparator)
A comparator is one of the most basic Op-Amp configurations. If the reader is unfamiliar with op-amps, a whole host of literature is available on the internet.
The basic idea of the comparator is to compare a voltage to a particular reference. In the last step, it was determined that the voltage on one side of the bridge was constant, which will serve nicely as the reference voltage. The other voltage, which changes, tells us the behavior of the LDR.
In the inverting terminal of the op-amp will be the reference voltage. The non-inverting terminal will host the varying signal.
- If Vin > Vref, the op-amp will output a high signal.
- If Vin < Vref, the op-amp will output a low signal.
Think back to the behavior of the LDR. As light increases, resistance decreases. If resistance decreases, the voltage decreases. The Vin will then drop below Vref (as more light is introduced past the threshold) and the output of the op-amp will go low. The opposite is true, that as light decreases past the threshold, the op-amp will go high.
Step 4: Obtain Real Values
Use a light, lamp, or any other light source to adjust the brightness of your surroundings. Use your multimeter to measure the resistance of the LDR. Determine at what brightness (or lack thereof) you want the light to turn on at. In my case, the value of the LDR was 10 kOhms at the brightness I wanted to define my threshold at.
R = 10 kOhm
Since 10 kOhms is a common value, it is easiest to make all the resistance values 10 kOhms. This is suppported by the previous step where the sample calculations were performed.
R1, R2, R3 = 10 kOhm
The output of the system is driven by the output of the Op-Amp. The output of the voltage is about 7.5 Volts. Look up the datasheet for the LED to determine the diode voltage and maximum current. Use (Vout-Vbias)/Imax = R. That is the output resistor needed in order to avoid damaging the LED.
For the components I am using:
- Vout = 7.5 V,
- Vbias = 2 V,
- Imax = 10 mA
R = (7.5 - 2)/(.010) = 550 ohms, The closest standard resistor is 560 ohms
Step 5: Build the Circuit
Following the schematic above, build the circuit on a breadboard. Since a comparator uses the input rails of the opamp to determine the output voltage (saturated to positive rail or saturated to negative rail), the negative supply port should be grounded for this particular case.
Step 6: Test the System
In order to test your system, simply change the brightness of the surrounding area to cross the threshold “brightness.” That change should cause your LED to either turn on or off. This image shows the output of the system (by way of output voltage).