![Kelvin Double Bridge: 1 to 0.00001 (STUDY THE DERIVATION FROM
THE CLASS NOTES)
G
R1
R2
Rx
R3
m
n
Ry
o
k
l
V
I
Rb
p
Ra
2 ratio arms: R1-R2 and Ra-Rb
the connecting lead between m and n: yoke
balance conditions: Vlk = Vlmp or Vok = V onp
lk
R2
V
R1 R2
here V IRlo I[R3 Rx (Ra Rb ) // Ry ]
3lmp
Ry
V I R
Ra Rb Ry
b
V (1)
R (2)
Eq. (1) = (2) and rearrange: 1 a
Rb Ry R R
Rx R3
R1
R2 Ra Rb Ry R2 Rb
If we set R1/R2 = Ra/Rb, the second term of the right hand side will be zero, the relation
reduce to the well known relation. In summary, The resistance of the yoke has no effect
on the measurement, if the two sets of ratio arms have equal resistance ratios.
2
x
R13
R
R R](https://image.slidesharecdn.com/bridgeppt-1-200329073604/85/Bridge-ppt-1-12-320.jpg)
Bridge ppt 1
- 1. AC & DC BRIDGES
- 2. RReessiissttaanncceeMMeeaassuurreemmeennttTTeecchhnniiqquueess A Supply Unknow resistance A Rx Supply Decade resistance box substituted in place of the unknown Substitution Substitution
- 3. DC Bridge (Resistance) AC Bridge Inductance Capacitance Frequency Schering Bridge Wien BridgeMaxwell Bridge Hay Bridge Owen Bridge Etc. Wheatstone Bridge Kelvin Bridge Megaohm Bridge Bridge Circuit Bridge Circuit is a null method, operates on the principle of comparison. That is a known (standard) value is adjusted until it is equal to the unknown value. Bridge Circuit
- 4. Wheatstone Bridge and Balance Condition V R1 R3 R2 R4 I1 I2 I3 I4 Suitable for moderate resistance values: 1 to 10 M A B C D Balance condition: No potential difference across the galvanometer (there is no current through the galvanometer) Under this condition: VAD = VAB I1R1 I2R2 And also VDC = VBC I3R3 I4R4 where I1, I2, I3, and I4 are current in resistance arms respectively, since I1 = I3 and I2 = I4 R3 R4 R1 R2 or 1 x 4 3 R R R R R2
- 5. 12 V 12 V 12 V 12 V (a) Equal resistance (b) Proportional resistance (c) Proportional resistance (d) 2-Volt unbalance Examp le
- 6. Application of Wheatstone Bridge Murray/Varrley Loop Short Circuit Fault (Loop Test) •Loop test can be carried out for the location of either a ground or a short circuit fault. Power or communication cable Murray Loop Test Short circuit fault X1 X2 R1 R2 R3 R4 ground fault short circuit fault Let R = R1+R2 At balance condition: R R R4 R2 3 1 3 1 R R R R R3 4 4 2 R R R R R3 4 The value of R1 and R2 are used to calculate back into distance. Assume: earth is a good conductor
- 7. X1 X2 R1 R2R5 R4 Short circuit fault R3 Varley Loop Test Let R = R1+R2 and define Ratio = R4/R5 5 2 3 R1 At balance condition: Ratio R4 R R R 1 3 R R R Ratio Ratio1 2 R R-RatioR3 Ratio1 Murray/Varrley Loop Short Circuit Fault (Loop Test) Examples of commonly used cables (Approx. R at 20oC) Remark The resistance of copper increases 0.4% for 1oC rise in Temp. Wire dia. In mm Ohms per km. Meter per ohm 0.32 218.0 4.59 0.40 136.0 7.35 0.50 84.0 11.90 0.63 54.5 18.35 0.90 27.2 36.76
- 8. Application of Wheatstone Bridge Unbalance bridge G A B C D R R R R+R V RTH = R G D VTH=V R 4R Consider a bridge circuit which have identical resistors, R in three arms, and the last arm has the resistance of R +R. if R/R << 1 Small unbalance occur by the external environment C Thévenin Voltage (VTH) TH CD V V V R 4R Thévenin Resistance (RTH) RTH R This kind of bridge circuit can be found in sensor applications, where the resistance in one arm is sensitive to a physical quantity such as pressure, temperature, strain etc.
- 9. 5 k 6 V Rv Output signal5 k 5 k 6 5 4 3 2 1 0 0 20 40 60 80 100 120 Temp (o C) Rv(k (b) Example Circuit in Figure (a) below consists of a resistor Rv which is sensitive to the temperature change. The plot of R VS Temp. is also shown in Figure (b). Find (a) the temperature at which the bridge is balance and (b) The output signal at Temperature of 60oC. (a) SOLUTION (a) at bridge balance, we have v R1 R R3 R2 5 k5 k 5 k 5 k The value of Rv = 5 k corresponding to the temperature of 80oC in the given plot. (b) at temperature of 60oC, Rv is read as 4.5 k thus R = 5 - 4.5 = 0.5 k We will use Thévenin equivalent circuit to solve the above problem. THV V R 6 V 0.5 k 0.15 V 4R 45 k It should be noted that R = 0.5 k in the problem does not satisfy the assumption R/R << 1, the exact calculation gives VTH = 0.158 V. However, the above calculation still gives an acceptable solution. 4.5 k
- 10. G R3 R1 R2 Rx V m p n Ry Low resistance Bridge: Rx < 1 The effects of the connecting lead and the connecting terminals are prominent when the value of Rx decreases to a few Ohms Effect of connecting lead At point m: Ry is added to the unknown Rx, resulting in too high and indication of Rx At point n: Ry is added to R3, therefore the measurement of Rx will be lower than it should be. Ry = the resistance of the connecting lead from R3 to Rx At point p: R1 R2 Rx Rnp R3 Rmp x 3 mp np R2 R2 R R R1 R R1 Rrearrange Where Rmp and Rnp are the lead resistance from m to p and n to p, respectively. The effect of the connecting lead will be canceled out, if the sum of 2nd and 3rd term is zero. 1 2 2 np mp np mp RR R R R R R 1 R 0 or 2 x R13 R R R
- 11. Kelvin Double Bridge: 1 to 0.00001 Four-Terminal Resistor Current terminals Voltage terminals Current terminals Voltage terminals Four-terminal resistors have current terminals and potential terminals. The resistance is defined as that between the potential terminals, so that contact voltage drops at the current terminals do not introduce errors. r4 R3 R1 R2 r1 r2 r3 Rx Ra Rb G Four-Terminal Resistor and Kelvin Double Bridge •r1 causes no effect on the balance condition. •The effects of r2 and r3 could be minimized, if R1 >> r2 and Ra >> r3. •The main error comes from r4, even though this value is very small.
- 12. Kelvin Double Bridge: 1 to 0.00001 (STUDY THE DERIVATION FROM THE CLASS NOTES) G R1 R2 Rx R3 m n Ry o k l V I Rb p Ra 2 ratio arms: R1-R2 and Ra-Rb the connecting lead between m and n: yoke balance conditions: Vlk = Vlmp or Vok = V onp lk R2 V R1 R2 here V IRlo I[R3 Rx (Ra Rb ) // Ry ] 3lmp Ry V I R Ra Rb Ry b V (1) R (2) Eq. (1) = (2) and rearrange: 1 a Rb Ry R R Rx R3 R1 R2 Ra Rb Ry R2 Rb If we set R1/R2 = Ra/Rb, the second term of the right hand side will be zero, the relation reduce to the well known relation. In summary, The resistance of the yoke has no effect on the measurement, if the two sets of ratio arms have equal resistance ratios. 2 x R13 R R R
- 13. Capacitor Capacitance – the ability of a dielectric to store electrical charge per unit voltage conductor A 0 r d C Area, A Dielectric, r thickness, d Typical values pF, nF or F Dielectric Construction Capacitance Breakdown,V Air Meshed plates 10-400 pF 100 (0.02-in air gap) Ceramic Tubular 0.5-1600 pF 500-20,000 Disk 1pF to 1 F Electrolytic Aluminum 1-6800 F 10-450 Tantalum 0.047 to 330 F 6-50 Mica Stacked sheets 10-5000 pF 500-20,000 Paper Rolled foil 0.001-1 F 200-1,600 Plastic film Foil or Metallized 100 pF to 100 F 50-600
- 14. Inductor l A l 2 L o r N A o = 410-7 H/m r – relative permeability of core material Ni ferrite: Mn ferrite: r > 200 r > 2,000 Distributed capacitance Cd between turns Cd Equivalent circuit of an RF coil L Re Air core inductor Iron core inductor Inductance – the ability of a conductor to produce induced voltage when the current varies. N turns
- 15. Quality Factor of Inductor and Capacitor Inductance series circuit: Q Xs Ls Quality factor of a coil: the ratio of reactance to resistance (frequency dependent and circuit configuration) Typical D ~ 10-4 – 0.1 Typical Q ~ 5 – 1000 Rs Rs Rp X p Lp R Inductance parallel circuit: Q p Dissipation factor of a capacitor: the ratio of reactance to resistance (frequency dependent and circuit configuration) Capacitance parallel circuit: Capacitance series circuit: 1X D p s s Xs Rp Cp Rp D Rs C R
- 16. AC Bridge: Balance Condition D Z1 Z2 Z4 Z3 A C D B I1 I2 all four arms are considered as impedance (frequency dependent components) The detector is an ac responding device: headphone, ac meter Source: an ac voltage at desired frequency Z1, Z2, Z3 and Z4 are the impedance of bridge arms At balance point: EBA = EBC or I1Z1 = I2Z2 V V I1 = and I2 = Z1 + Z3 Z2 + Z4 V Z1Z4 = Z2Z3 Z1Z4 1 4 =Z2Z3 2 3 General Form of the ac Bridge Complex Form: Polar Form: Magnitude balance: Phase balance: Z1Z4 =Z2Z3 1 4 =2 3
- 17. Example The impedance of the basic ac bridge are given as follows: Z 100 80o (inductive impedance)1 Z2 250 (pure resistance) Determine the constants of the unknown arm. SOLUTION The first condition for bridge balance requires that Z 400 30o (inductive impedance)3 Z4 unknown 4 100 Z Z1 Z2Z3 250 400 1,000 The second condition for bridge balance requires that the sum of the phase angles of opposite arms be equal, therefore o 4 =2 3 1 0 30 80 50 Hence the unknown impedance Z4 can be written in polar form as o Z4 1,000 50 Indicating that we are dealing with a capacitive element, possibly consisting of a series combination of at resistor and a capacitor.
- 18. Example an ac bridge is in balance with the following constants: arm AB, R = 200 in series with L = 15.9 mH R; arm BC, R = 300 in series with C = 0.265 F; arm CD, unknown; arm DA, = 450 . The oscillator frequency is 1 kHz. Find the constants of arm CD. SOLUTION This result indicates that Z4 is a pure inductance with an inductive reactance of 150 at at frequency of 1kHz. Since the inductive reactance XL = 2fL, we solve for L and obtain L = 23.9 mH D Z1 Z2 Z4 Z3 A C D The general equation for bridge balance states that B I1 I2 V Z1 R jL 200 j100 Z2 R 1/ jC 300 j600 Z3 R 450 Z4 unknown Z1Z4 = Z2Z3 (300 j600) 4Z = Z2Z3 450(200 j100) j150 Z1
- 19. Maxwell Bridge Measure an unknown inductance in terms of a known capacitanceR2 R1 R3 Rx V C1 D Lx Unknown inductance At balance point: Zx = Z2Z3Y1 2 2 3 3 1 1 R1 where Z = R ; Z R ; and Y = 1 jC 1 R1 jC1 Zx = Rx jLx R2R3 1 x Diagram of Maxwell Bridge Separation of the real and imaginary terms yields: R R R2 R3 Lx R2 R3C1and Frequency independent Suitable for Medium Q coil (1-10), impractical for high Q coil: since R1 will be very large.
- 20. Hay Bridge Similar to Maxwell bridge: but R1 series with C1 Diagram of Hay Bridge V At balance point: Z1Zx = Z2Z3 2 3 jwhere Z1 = R1 ; Z R2; and Z3 R C1 1 Rx jLx R2R3 R1 jC1 which expands to D R2 R1 C1 R3 Rx Lx Unknown inductance Lx jRx C1 C1 R1Rx jLxR1 R2R3 1 x 2 3 L C1 x 1 Rx C1 Solve the above equations simultaneously R R x R R (1) L R (2)
- 21. Hay Bridge: continues 1 1 x R2R3C1 L 12 C2 R2 1 1 xR 2 C2 R R R 1 1 2 3 12 C2 R2 Lx Rx Z L R1 Z C C1 and Phasor diagram of arm 4 and 1 xR R XL Lx tanL Q Ctan R C1R1 XC 1 1 1 L Ctan C R tan or Q 1 Thus, Lx can be rewritten as x R2R3C1 L 1 (1/Q2 ) For high Q coil (> 10), the term (1/Q)2 can be neglected Lx R2R3C1
- 22. Schering Bridge Used extensively for the measurement of capacitance and the quality of capacitor in term of D D R2 R1 C1 C3 Rx V Cx Unknown capacitance Diagram of Schering Bridge At balance point: x 2 3 1 where 1 Z = Z Z Y 1 1 R1jC3 jCZ2 = R2; Z3 ; and Y = x j Cx R j 1 R jC2 1 Cx R1 which expands to x Cx C3 C3R1 R j R2C1 jR2 Separation of the real and imaginary terms yields: 3 x 2 C R R C1 2 x R13 R and C C
- 23. Schering Bridge: continues Dissipation factor of a series RC circuit: x x Xx D Rx R C Dissipation factor tells us about the quality of a capacitor, how close the phase angle of the capacitor is to the ideal value of 90o D RxCx R1C1For Schering Bridge: For Schering Bridge, R1 is a fixed value, the dial of C1 can be calibrated directly in D at one particular frequency
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