Experimental Procedure
Wheatstone Bridge
Background: Bridges, a parallel combination of two sets of two components in series, are used frequently in electrical and mechanical engineering. Wheatstone bridges are used to detect small changes in the properties of an element, in this case a resistor, and is used frequently in sensing applications. Typically, the bridge is composed of three resistors with known and equal resistances and one resistor with an unknown value (Fig. 1). The resistance of the unknown resistor is determined by a voltage measurement across the bridge. An alternate measurement is to measure the current that flows between node A to node B, where the current meter has some internal resistance associated with it. A description of the operation of this measurement technique is described in Chapter 4.10.2 of the course textbook.
For simplicity (and the fact that parts cannot be rotated by 45o in PSpice), I have redrawn the circuit, shown in Fig. 2. The operation of the bridge is as follows for the circuit in Fig. 2a. The two branches of the bridge function as a current divider, where the currents flowing through the two branches are equal only when the resistance of the unknown resistor is equal to the resistance of the other three resistors. Within each branch is a voltage divider. When all four resistors have exactly the same resistance, the voltage across each resistor will be half of the input voltage. In this case, the voltage at node A, located between R1 and R3, will also be equal to the voltage at node B, located between R2 and R4. Therefore, the voltage difference between the two nodes VBA will be equal to VB – VA = 0V. The bridge is considered to be balanced.
If the resistor R4 has a larger resistance than the other three resistors, the current division is not equal and more of the current supplied by the voltage source, Vs, will flow through the branch containing R1 and R3. In addition, the total current Is will be smaller than in the case where the bridge is balanced because the equivalent resistance of the four resistor network has increased. Since R1 and R3 have not changed value, the voltage at the node between R1 and R3 is still equal to VA = ½ Vs. However, the voltage at node B is no longer equal to ½ Vs. Thus, the bridge is not balanced and there will be a difference in the two node voltages. It is this difference in node voltages that is measured and used to calculate the value of R4.
Experimental Procedure:
Analysis:
1. For the circuit shown in Figure 2a, derive an analytical expression for the voltages at nodes A and B as a function of the variables R1-R4 and Vs and the expression for the difference in the node voltages, VB – VA. Show that VA is always equal to half of the source voltage for any value of R4 when R1, R2, and R3 are equal. Also derive the equations for the current through R3 and R4 when R1, R2, and R3 are equal.
2. Substitute in a value of 4.7 kW for R1, R2, and R3 into the expressions derived in Step 1. Plot the value of VA, VB, and VB – VA as R4 is varied from 0 to 10 kW using MATLAB.
3. Take the first derivative of the expression found in Step 1for the difference in the node voltages, VB – VA as a function of R4. This is a measure of the sensitivity of the measurement to changes in R4. Is the measurement most sensitive when R4 is smaller, equal to, or larger than the values of R1, R2, and R3?
4. Plot the equation for the first derivative of VB – VA from Step 3 as a function of R4 where R1 = R2 = R3 = 4.7 kW and R4 is varied from 0 to 10 kW using MATLAB.
Modeling:
5. Plot the value of VA, VB, and VB – VA where R1 = R2 = R3 = 4.7 kW as R4 is swept from 0 to 10 kW using PSpice.
6. Run a second simulation on the same circuit and plot the currents through R3 and R4 as R4 is swept from 0 to 10 kW.
Measurements
7. Measure the resistance of three 4.7 kW resistors and record the values. Pair the resistors that have the most similar resistances and assign their positions in the circuit as R1 and R2.
8. Construct the circuit in Fig. 2a.
9. Measure the value of the node voltages, VA and VB, the voltage between nodes B and A (VBA), and the currents flowing through R3 and R4 (IR3 and IR4, respectively) when (a) R4 = 0W and (b) when R4 = 10 kW using your digital multimeter.
10. Remove the trim pot from the circuit. Adjust the resistance of the trim pot to a value close to 2.5 kW. Measure the resistance.
11. Replace the trim pot in the circuit. Measure node voltages, VA and VB, VBA, and the currents IR3 and IR4.
12. Repeat steps 10 through 11 where R4 is set to approximately 4.7 kW.
13. Repeat steps 10 through 11 where R4 is set to approximately 7.5 kW.
14. For the last measurement, adjust the value of the trim pot to its maximum. Replace the trim pot in the circuit. Measure node voltages, VA and VB, VBA, IR3, and IR4..
15. Calculate the differences in voltage VB – VA for each value of R4 using the node voltages VA and VB measured in Step 11. Should the values of the differences between the node voltages be smaller than, equal to, or greater than the measured voltage VBA and why?
16. Plot the experimental values of VA, VB, and VBA as a function of R4.
17. Calculate the difference in VB – VA using the node voltages measured in Step 10 and Step 11 and divide by the difference in the trim pot resistances. Repeat this calculation for the difference in VB – VA obtained in Step 11 and Step 12 and divide by the difference in the trim pot resistances. How close are these values to the magnitude of the first derivative when R4 = 4.7 kW found in Step 4?
18. Calculate the percent difference between the expected value of VB – VA. and the calculated values from Step 15 when (a) R4 = 0W, (b) when R4 = 4.7 kW and (c) when R4 = 10 kW.
19. Calculate the percent difference between the expected values of .the currents through R3 and R4.and the measured values from Step 11 when (a) R4 = 0W, (b) when R4 = 4.7 kW and (c) when R4 = 10 kW.
20. What are some of the reasons why VB – VA is not equal to 0V when R4 = 4.7 kW?
Last Revision: 5/30/2012