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Unit 5 – DC Bridges Objectives: Describe the Operation of a Wheatstone Bridge

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Unit 5 – DC Bridges Objectives: Describe the Operation of a Wheatstone Bridge RICHLAND COLLEGE INTC 1307 School of Instrumentation Test Equipment Engineering Business & Technology Teaching Unit 5 Rev. 0 – W. Slonecker DC Bridges Rev. 1 – (8/26/2012)– J. Bradbury Unit 5 – DC Bridges Objectives: Describe the operation of a Wheatstone Bridge. Describe and calculate conditions for a balanced bridge. Use a bridge to determine unknown resistance. Thevenize an unbalanced bridge. Use a bridge as an error detector. Describe and use a Varley Loop Test Circuit. Bridge Configuration R1 R3 R1 R3 U1 + - + DET 0.000 A X Y ≡ V1 X Y V R2 R4 R2 R4 Classic Bridge Circuit Equivalent Bridge Circuit Bridge circuits have been widely used in the past for comparative measurements of resistance, capacitance, inductance, impedance, and admittance. The two configurations shown above are equivalent, just drawn differently. The DET (detector) shown may be a galvanometer or microammeter, or it could be a milli-voltmeter. Usually, of the four components (shown as resistors above) two are high precision fixed values, one is a high precision adjustable value and the fourth component is of unknown value. The variable component is adjusted to null the detector reading. When the detector reads zero, the bridge is said to be balanced. Bridges can also be used as inputs to control circuits. In this case, one component is made to vary resistance in response to some physical parameter such as pressure. Wheatstone Bridge The classic bridge schematic above using resistors as components is called a Wheatstone bridge. Ignoring the detector, resistors R1 and R2 are in series as are resistors R3 and R4. These two series circuits are parallel to each other. Since they are parallel, both series circuits have the same voltage across them. So long as the series circuits connect directly across a voltage source, the voltage across each circuit is equal to the source voltage. The bridge is balanced when the voltage at point X is equal to the voltage at point Y. Balance assures that the ratio of voltages on V V the X side is equal to the ratio of voltages on the Y side: R1 = R3 V R2 V R4 Page 1 of 8 Since the current in R1 is the same as the current in R2 at balance, and the current in R3 and R4 is the equal (although probably different from the R1_R2 current) then: R1 R3 = R2 R4 , called The Bridge Equation. Wheatstone bridges are more accurate for resistance measurements than the type of ohmmeter presented in Unit 1. With precision components, they can have typical accuracies of 0.1% compared to about 3% or so for D’Arsonval movement ohmmeters. Remember the non- linearity of the D’Arsonval ohmmeter scale. A bridge is always read at the null point, which can be quite sensitive. Wheatstone Bridge Calculations R1 R3 = Unknown The bridge equation, R2 R4 , is true only at R1 R3 balance, or null. In the bridge circuit on the right, R3 U1 + - is an unknown resistance and R1 is a precision 0.000 A variable resistor. R2 and R4 are both precision fixed V1 resistors. The bridge equation is rewritten as: R2 R4 R1 R3 R4 = ⇒ R1⋅ =R3 R2 R4 R2 Since R4 and R2 are fixed, their ratio is like a multiplying factor to R1. If R4 were equal to R2, then for the bridge to balance, R1 must be exactly Wheatstone Bridge equal to the unknown R3. However, since the multiplying constant, R4/R2 can be greater or less than 1, a single precision potentiometer can provide accurate nulls for a large range of unknown resistors. Examples: Let R1 be a precision 10kΩ potentiometer. With R2 = R4 = 10kΩ, and the unknown equal to 500Ω, the null would be near the bottom of the pot adjustment range. However, if R4 = 1kΩ and R2 = 10kΩ, then with R1 adjusted mid-range to 5kΩ the modified bridge equation becomes R3 = 5kΩ×(1kΩ/10kΩ) = 500Ω. So the bridge is nulled at the midpoint of R1, where the adjustment is less sensitive. Similarly, if the unknown R3 were 200kΩ, it would be impossible to null the bridge for R2 = R4 = 10kΩ. But you could tune a null if R4/R2 were 100. So leaving R2 at 10k and setting R4 = 1000kΩ (1MΩ) makes R4/R2 = 100. This makes R1 = 2kΩ to null the bridge for R3 = 200kΩ. As a practical matter, 1MΩ is probably an upper limit for R4 because precision resistors above 1MΩ have problems. However, if a ratio greater than 100 were needed, the value of R2 could be decreased. In the second example above, R4 = 100kΩ and R2 = 1kΩ also give a ratio of 100 and would work well. The dial scale for the adjustable 10kΩ pot could look like that on the right. It is important to null the bridge between 1.00 and 10.00 if possible. A poor choice of R4 and R2 could make the null between 0.00 and 1.00 or even between 0.00 and 0.1. In either case you would lose setting precision. R4 and R2 should be selected for the null to occur at a readable point on the scale. Page 2 of 8 Wheatstone Bridge Analysis When R2 and R4 are equal, the two series circuits of the bridge are identical when R1 = R3. When R4 = 10×R2, the total resistance in the R3_R4 side is high, so the current is in that side is Unknown R1 R3 lower than the current in the R1_R2 side. The U1 need to make R4 larger than R2 is because R3 is + - 0.000 A Vy much larger than R1 can be adjusted. V1 Vx If R1 is a 10kΩ pot and R2 = 10kΩ, then R2 R4 the voltage at Vx can be changed from a maximum of V1 when R1 = 0, to a minimum of one half V1 when R1 = 10kΩ. To obtain a null, Vx must equal to Vy. Wheatstone Bridge R4 Vy=V1 ( R3+R4 ) is the divider equation for Vy. It should be obvious that if R3 is larger than R4 then Vy will less than one half V1 and the bridge will never balance. For examipe if R3 = 50kΩ, 10 k Ω 1 Vy=V1 =V1 (50 k Ω+10 k Ω ) ( 6) so no null is possible. 100 k Ω 2 By changing R4 to 100kΩ you get: Vy=V1 =V1 (50 k Ω+100 kΩ ) ( 3) and a null is simple to obtain. With R4 = 100kΩ and R2 = 10kΩ then R3 = 10×R1, giving a null when R1 is set to 5.0kΩ. Thevenizing the Unbalanced Bridge Up to now, we have only considered bridge circuits in a balanced state. The detector current was nulled (no current flow), so the detector appeared as an open circuit. The voltages at X and Y could be calculated by the Voltage Divider Law. This calculation led to the bridge equation. If the bridge is unbalanced, VX and VY are unequal and a potential difference exists across the detector. Current now flows through the detector. It is not as easy to analyze the unbalanced bridge. The Voltage Divider Law cannot be used. The best way to find the voltage drop across the detector as well as the detector current in an unbalanced condition is by using Thevenin’s theorem. If you need to refresh, pull out your CETT 1403 text. Applying Thevenin’s theorem requires the following sequence: 1. Select a circuit component as the “load” of the Thevenin circuit, (in our case it is the detector) and remove it from the circuit. 2. Calculate the open circuit voltage across the terminals left after component removal. This gives the Thevenin voltage, VTH. 3. Short all voltage sources (open all current sources) and calculate the impedance between the terminals left after component removal. This gives the Thevenin resistance, RTH. The Thevenin equivalent circuit consists of VTH in series with RTH. You can reattach the removed component to the equivalent circuit and calculate voltage across it and current through it. Be careful! The equivalent circuit only holds so long as no circuit resistors or voltage sources change Page 3 of 8 values. Since a Wheatstone bridge has a precision variable resistor, the equivalent circuit will only be good for one setting of that resistor. First, remove D , and redraw the circuit. Calculate the voltage at X and the voltage at R1 R3 Y. Then calculate the open circuit voltage 6 OHM D 12 OHM VB 4 OHM from Y to X 30V D X Y R 2 12 R2 R4 VX = VS x = 30 x 18 = 20v. R1+R2 12 OHM 8 OHM R4 8 VY = VS x = 30 x 20 = 12v. R3+R4 VOC = VY – VX = 12 – 20 = -8v. VOC = -8v R1 R3 VB 30 20V X Y 12V and Remove VB and replace with its Rint (0Ω for a DC po(0Ω for the DC power supply). R2 R4 R1 R3 6 OHM 12 OHM X Y 0 OHMS R2 R4 12 OHM 8 OHM Then calculate the Thevenin Resistance of the circuit. Look carefully, one side of R1 connects to X and the other side connects to ground. Likewise for R2. This is shown in the circuit below X Y R1( R2 ) 6Ω(12Ω ) 72 RX to gnd = R1 R2 = = = = 4Ω R1 R2 R3 R4 R1+R2 6Ω+12Ω 18 6 ohm 12 ohm 12 ohm 8 ohms R3( R4 ) 12 Ω(8Ω ) 96 Ω RY to gnd = R3 R4 = = = = 4.8Ω R3+R4 12 Ω+8Ω 20 Ω X Y The two equivalent resistors are effectively in series so that RTH = 4 + 4.8 = 8.8Ω 4 ohms 4.8 ohms The Thevenin equivalent circuit is shown at the left with Rint the 4Ω detector inserted across its output 8.8 ohms Voc RD The detector current is given by Ohm’s Law as: - 8v 4 ohms −8v I D= =−0 .625 A 8.8 Ω+4 Ω V D=−0 .625 A×4Ω=−2.5v Page 4 of 8 Remember, electrons flow from - to +.
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