RICHLAND COLLEGE School of Engineering Business & Technology Rev. 0 W. Slonecker Rev. 1 (8/26/2012) J. Bradbury INTC 1307 Instrumentation Test Equipment Teaching Unit 5 DC Bridges 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 + X DET Y X Y V 0.000 A 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 and are in series as are resistors and. 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 the X side is equal to the ratio of voltages on the Y side: V V = V V Page 1 of 8
Since the current in is the same as the current in at balance, and the current in and is the equal (although probably different from the _ current) then: =, 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 nonlinearity of the D Arsonval ohmmeter scale. A bridge is always read at the null point, which can be quite sensitive. Wheatstone Bridge Calculations The bridge equation, =, is true only at balance, or null. In the bridge circuit on the right, is an unknown resistance and is a precision variable resistor. and are both precision fixed resistors. The bridge equation is rewritten as: = = Since and are fixed, their ratio is like a multiplying factor to. If were equal to, Wheatstone Bridge then for the bridge to balance, must be exactly equal to the unknown. However, since the multiplying constant, / can be greater or less than 1, a single precision potentiometer can provide accurate nulls for a large range of unknown resistors. Examples: Let be a precision 10kΩ potentiometer. With = = 10kΩ, and the unknown equal to 500Ω, the null would be near the bottom of the pot adjustment range. However, if = 1kΩ and = 10kΩ, then with adjusted mid-range to 5kΩ the modified bridge equation becomes = 5kΩ (1kΩ/10kΩ) = 500Ω. So the bridge is nulled at the midpoint of, where the adjustment is less sensitive. Similarly, if the unknown were 200kΩ, it would be impossible to null the bridge for = = 10kΩ. But you could tune a null if / were 100. So leaving at 10k and setting = 1000kΩ (1MΩ) makes / = 100. This makes = 2kΩ to null the bridge for = 200kΩ. As a practical matter, 1MΩ is probably an upper limit for because precision resistors above 1MΩ have problems. However, if a ratio greater than 100 were needed, the value of could be decreased. In the second example above, = 100kΩ and = 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 and 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. and should be selected for the null to occur at a readable point on the scale. 0.000 A Unknown Page 2 of 8
Wheatstone Bridge Analysis When and are equal, the two series circuits of the bridge are identical when =. When = 10, the total resistance in the _ side is high, so the current is in that side is lower than the current in the _ side. The need to make larger than is because is much larger than can be adjusted. If is a 10kΩ pot and = 10kΩ, then the voltage at Vx can be changed from a maximum of when = 0, to a minimum of one half when = 10kΩ. To obtain a null, Vx must equal to Vy. Vy= ( + ) is the divider equation for Vy. It should be obvious that if is larger than then Vy will less than one half and the bridge will never balance. For examipe if = 50kΩ, so no null is possible. By changing to 100kΩ you get: and a null is simple to obtain. With = 100kΩ and = 10kΩ then = 10, giving a null when 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, V X and V Y 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, V TH. 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, R TH. The Thevenin equivalent circuit consists of V TH in series with R TH. 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 Vx Wheatstone Bridge Vy= ( 10 k Ω 50 k Ω+10 k Ω ) = ( 1 6 Vy= ( 100 k Ω 50 k Ω+100 kω ) = ( 0.000 A Unknown Vy 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. VB 30V 6 OHM D X D 4 OHM 12 OHM Y First, remove D, and redraw the circuit. Calculate the voltage at X and the voltage at Y. Then calculate the open circuit voltage from Y to X V X = V S x R 2 R 1 +R 2 = 30 x 12 18 = 20v. 12 OHM 8 OHM V Y = V S x R 4 R 3 +R 4 = 30 x 8 20 = 12v.. VB 30 20V X V OC = V Y V X = 12 20 = -8v. V OC = -8v and Y Remove 12V V B and replace with its Rint (0Ω for a DC po(0ω for the DC power supply). 6 OHM 12 OHM X 0 OHMS 12 OHM Y 8 OHM Then calculate the Thevenin Resistance of the circuit. Look carefully, one side of connects to X and the other side connects to ground. Likewise for. This is shown in the circuit below 6 ohm 12 ohm X Y 12 ohm 8 ohms R X to gnd = R 1 R 2 = R Y to gnd = R 3 R 4 = R 1 ( R 2 ) R 1 +R 2 = 6 Ω(12Ω ) 6 Ω+12Ω = 72 18 = 4Ω R 3 ( R 4 ) 12 Ω(8Ω ) = R 3 +R 4 12 Ω+8 Ω =96 Ω 20 Ω = 4.8Ω X Y 4 ohms 4.8 ohms The two equivalent resistors are effectively in series so that RTH = 4 + 4.8 = 8.8Ω Voc - 8v Rint 8.8 ohms RD 4 ohms The Thevenin equivalent circuit is shown at the left with the 4Ω detector inserted across its output The detector current is given by Ohm s Law as: I D = 8v = 0.625 A 8.8 Ω+4 Ω V D = 0.625 A 4 Ω= 2.5v Page 4 of 8
Remember, electrons flow from - to +. In this example, current flow is from Y to X. By adjusting R 3, the bridge will be balanced and no current flows. If R 1 is adjusted past the null point, the current through D will change direction (change the polarity). The amount of unbalance in the bridge is indicated by the amount of current through D. Keep in mind that if the resistance is changed, the Thevenin equivalent circuit must be recalculated. You need to be able to solve this type of analysis problem. Error Signals from a Bridge If a balanced bridge is upset, an output voltage or error signal is generated. For example, a thermostat is set for a certain temperature. A heat sensitive device such as a thermistor is calibrated to balance the bridge when the temperature is 60 degrees. At that temperature, the bridge is balanced and no error signal is developed from X to Y. The thermistor has a negative temperature coefficient. When the temperature decreases, the resistance of the thermistor increases, unbalancing the bridge and developing an error signal that is fed to a field effect transistor amplifier (input impedance is infinity). This error signal activates the furnace which heats the air and the thermistor until the temperature reaches 60 degrees again. At this point the error signal is zero and the furnace is turned off. If the temperature rises, the resistance of the thermistor decreases. This changes the polarity of the error signal. The furnace remains off but the FET amplifier turns on the air conditioner to cool the room. When the temperature returns to 60 degrees, the error voltage is again Zero and the air conditioner is turned off. The value of and the polarity of the error signal controls other systems or pieces of equipment. Carefully read the problem statement above, then solve the following problem. V error = V Y V X defines the polarity of the error voltage. The bridge is balanced when the temperature is at 60 degrees. Error Voltage = v. (1) At this temperature, the resistance of the 5kOhm 5kOhm thermistor leg = ohms. (2) 18 V The temperature drops to 40 degrees and the Det resistance of the thermistor leg reads 5K. The voltage at X = v, (3) voltage at Y = 4.5kOhm Thermistor v. (4) The error voltage at 40 degrees = v. (5) The furnace is activated until the room temperature return to degrees Equivalent Bridge Circuit (zero error voltage). (6) The temperature rises to 80 degrees and the thermistor leg reads 4K ohms. The voltage at X = v; (7) the voltage at Y = v. (8) The error voltage is now v. (9) The air conditioner is turned on and operates until the thermistor leg of the bridge again balances the circuit. Graph the error vs. temperature on the chart below 40 60 80 100 Bridge Error Voltage 0.75v 0.5v 0.25v 0.0-0.25v -0.5v -0.75v Page 5 of 8
(1)=0V (2) = 4.5kΩ (3) = V X = 8.53v (4) = V Y = 9.0v (5) = +0.47v (6) = 60 (7) = = 8.53v (8) = V Y = 8.0v (9) = -0.53v Murry Loop The Murray Loop is a modification of the Wheatstone bridge where the resistance of a cable or wire is used for two legs of the bridge. This loop can be used to locate faults such as a shorting of a conductor to ground or a shorting of a conductor to another conductor in the same cable. Since many cables such as telephone cables are buried underground, repair can be costly, time consuming, and labor intensive. With the Murray Loop the fault can be located precisely on a long run of cable. It is helpful to know were to dig to make repairs. Because of long lengths of cable and the high resistance offered to the testing, a Wheatstone bridge can seldom be used because its limited voltage source to drive current through the long run of wire. 0.000 V 0.000 V Faulted Good Line2 Line1 Good LX LX Fault to gnd LB LA RA ohms Fault Line to line Fault LB LA RA ohms = 2RA RX = RX Short = 2RA RX = RX Short Fault-to-Ground Murry Loop Line-to-line Fault Murry Loop Both Murry Loop configurations have the same equation derived from the schematic on the right. Assume LA = LB: (2 RA RX ) = RX RX= 2 RA + = 2 RA RX RX 0.000 A 2RA - RX The last equation solves for the resistance to the fault (hence the distance to the fault). It is assumed that RA is known precisely, as are and (after nulling the detector). It is even possible to substitute the actual distance for cable resistance instead of RA. Page 6 of 8
LX= 2LA + If LA = 3000 feet, = 100Ω, and adjusted to 300Ω for the null, then 2 3000 ' 100 Ω LX= 100Ω+300Ω =1500' I think that is clever! Varley Loop The Varley Loop is another piece of test equipment to accurately locate ground faults and short circuits in a multi-conductor cable such as a telephone or wiring cable. From the diagram the Varley Loop is a Murray Loop with an added resistor, R. Configuration A Cable Length Resistance = R A V R X Varley loop for measuring R A In the configuration A above, the ground fault does not affect the null since it is the only ground in the circuit. The null simply measures the total roundtrip cable length resistance, 2R A as: 2R A =, the bridge equation that solves for 2R A in terms of. When the same components are reconfigured to the B configuration shown on page 8, there is a new bridge equation. Page 7 of 8
Configuration B = 2R A - R X R A R X + R X Varley loop for measuring R X = 2R A R X +R X, rearranges to + R X =2R A R X, then collecting terms gives R X ( +1 ) =2R A or R X ( + ) = 2R A finally isolating R X gives R X = 2R A +. As you see, is only a part of the total. The A configuration gave a value for 2 R A which plugs into the R X equation above, along with the new null value of to solve for R X. A practical example follows: A fault is located somewhere on a 256 line. When the A loop was nulled, = 100Ω, = 100Ω, and the bridge nulled at = 36.5Ω. 2R A = 100 100 36.5=36.5Ω or R A = 18.25Ω 36.5 100 100 20. 4 In the B configuration, the null is obtained for = 20.4Ω. R X = =8. 05 Ω 200 We conclude that the ground fault is located 8.05 18.25 256 '=112. 9' from the measurement end of the line. Varley Loop Bridges are made so that the reconfiguration from A to B is made by a single switch for convenience. The two nulls can be quickly obtained. Time Domain Reflectometry In more recent times, faults on coaxial cables are pinpointed with Time Domain Reflectometry or TDR. This works like a fish finder, by sending a short pulse down a line and monitoring the type of reflection that is returned on an instrument like an oscilloscope. The appearance of the reflection gives information about the type of fault, and the timing of the reflection gives the distance to the fault. We will look at TDR later in this course. Page 8 of 8