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 . 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 Equivalent Bridge Circuit

Bridge circuits have been widely used in the past for comparative measurements of resistance, , , 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 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 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 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 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 . 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 . 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 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 = =−0 .625 A D 8.8 Ω+4 Ω V D=−0 .625 A×4Ω=−2.5v

Page 4 of 8 Remember, electrons flow from - to +. In this example, current flow is from Y to X. By adjusting R3, the bridge will be balanced and no current flows. If R1 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 R1 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. Verror = VY – VX defines the polarity of the error voltage.

The bridge is balanced when the temperature is at 60 degrees. Error Voltage = ______v. (1) R1 R2 5kOhm 5kOhm At this temperature, the resistance of the V1 thermistor leg = ______ohms. (2) 18 V The temperature drops to 40 degrees and the Det resistance of the thermistor leg reads 5K. The R3 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 0.75v 0.5v 0.25v 0.0 -0.25v -0.5v -0.75v 40° 60° 80° 100° Bridge Error Voltage Page 5 of 8 (1)=0V (2) = 4.5kΩ (3) = VX = 8.53v

(4) = VY = 9.0v (5) = +0.47v (6) = 60° (7) = = 8.53v (8) = VY = 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.

R1 R3 R1 R3 U1 U1 + - + - 0.000 V 0.000 V V1 V1

Faulted Good Line2 Line1 Good

LX LX

LB LA LB LA Fault RA ohms RA ohms Fault Line to to gnd line Fault

R4 = 2RA – RX R4 = 2RA – RX Short R2 = RX Short R2 = RX

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 R1 R3 right. Assume LA = LB: U1 + - ( 2 RA−RX )⋅R1 2 RA⋅R1 0.000 A R3= = −R1 V1 RX RX 2 RA⋅R1 RX 2RA - RX RX= R1+R3

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 R1 and R3 (after nulling the detector). It is even possible to substitute the actual distance for cable resistance instead of RA.

Page 6 of 8 2LA⋅R1 LX= R1+R3 If LA = 3000 feet, R1 = 100Ω, and R3 adjusted to 300Ω for the null, then 2⋅3000 '⋅100 Ω LX= =1500' 100Ω+300Ω 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 R3 R1 Cable Length Resistance = RA

V RX

R2 R4

Varley loop for measuring RA

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, 2RA as: R1 2R = ⋅R4 A R2 , the bridge equation that solves for 2RA in terms of R4.

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

R3 = 2RA - RX R1 RA

RX

R2 R4

R4 + RX

Varley loop for measuring RX

R1 R2 = R1 R1 , rearranges to ⋅R4+ ⋅RX =2RA −RX , then collecting terms gives 2R A−R X R4+RX R2 R2

R1 R1 R1+R2 2R A⋅R2−R1⋅R4 R ⋅ +1 =2R − ⋅R4 R ⋅ = X ( R2 ) A R2 or X ( R2 ) R2

2R ⋅R2−R1⋅R4 R A X R = finally isolating gives X R1+R2 . As you see, R4 is only a part of the total.

R R The A configuration gave a value for 2 A which plugs into the X equation above, along with

R the new null value of R4 to solve for X . A practical example follows: A fault is located somewhere on a 256’ line. When the A loop was nulled, R1 = 100Ω, R2 = 100Ω, and the bridge nulled at R4 = 36.5Ω. 100 2R = ⋅36 .5=36 .5Ω Ω A 100 or RA = 18.25

36 .5⋅100−100⋅20 .4 Ω R = =8.05 Ω In the B configuration, the null is obtained for R4 = 20.4 . X 200

8.05 ⋅256 '=112. 9' We conclude that the ground fault is located 18.25 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 Reflecto- metry 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.

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