Bridge Method. Bridge Method

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1 ridge Method EIE 240 Electrical and Electronic Measurement Class 7, March 13, ridge Method Diode bridge is an arrangement of four or more diodes for AC/DC full-wave rectifier. Component bridge methods are used for measurement of resistance, capacitance, inductance, etc. The network will be balanced when the detector reading becomes zero. Component eing Measured ridge Network Detector 2 1

2 Wheatstone ridge Wheatstone bridge was invented by Samuel Hunter Christie in 1833 and improved and popularized by Sir Charles Wheatstone in DC supply, V s R 1 R 2 Output voltage, V o I 1 V o I 2 A C Sir Charles Wheatstone ( ) R 3 R D 4 + V s 3 Wheatstone ridge (Cont d) When V o = 0 V, the potential at must equal to the potential at D I 1 R 1 = I 2 R 3 I 1 R 2 = I 2 R 4 Hence I 1 R 1 = I 2 R 3 = (I 1 R 2 /R 4 ) R 3 R 1 /R 2 = R 3 /R 4 The balance condition is independent of V s 4 2

3 Wheatstone ridge (Cont d) R 1 is the input resistance to be measured by comparing to accurately known resistors (standard). R 2 and R 4 are known-fixed resistances. R 3 can be adjusted to give the zero potential difference condition. A R 1 Adjust R 3 V o = 0 V G D Wheatstone ridge 5 Wheatstone ridge (Cont d) Change in R 1, change R 3 The precision is about 1 to 1 M. The accuracy is mainly up to the known resistors and the sensitivity of the null detector ( 0.1 to 0.2%). Error comes from changes in resistances of the bridge arms by changes in temperatures or thermoelectric EMF in contacts. 6 3

4 Sensitivity of the ridge If no galvanometer at the output, V A = V s R 1 /(R 1 +R 2 ) V AD =V s R 3 /(R 3 +R 4 ) Thus, V o = V A V AD V o = V s ( R 1 /(R 1 +R 2 ) R 3 /(R 3 +R 4 ) ) The relationship between V o and R 1 is non-linear V o = 0 V when R 1 /R 2 = R 3 /R 4 7 Sensitivity of the ridge (Cont d) Changing R 1 to R 1 + R 1 gives a change of V o to V o + V o V o+ V o=v s s(( ((R 1+ R 1 1) )/((R 1+ R 1 1) )+R 2 2) R 3 /(R 3+R 4 4)) Then (V o + V o ) V o = V s R 1 + R 1 R 3 R 1 + R 1 +R 2 R 3 +R 4 V s R 1 R 3 R 1 +R 2 R 3 +R 4 V o = V s R 1 + R 1 R 1 R 1 + R 1 +R 2 R 1 +R 2 8 4

5 Sensitivity of the ridge (Cont d) If small changes R 1 << R 1 then the sensitivity of Wheatstone bridge can be computed from, V o R 1 V s /(R 1 +R 2 ) V o / R 1 V s /(R 1 +R 2 ) Higher R 1 to be measured, lower sensitivity Amplifier can be used to amplify V o 9 Unbalanced ridge If there is a galvanometer, R g, between the two output terminals, the current I g can be determined by Thévenin equivalent circuit. V th = V o = V s ( R 1 /(R 1 +R 2 ) R 3 /(R 3 +R 4 ) ) Short voltage source, then Thévenin resistance is R 1 //R 2 + R 3 //R 4 R th = R 1 R 2 /(R 1 +R 2 ) + R 3 R 4 /(R 3 +R 4 ) 10 5

6 R 1 A Unbalanced ridge (Cont d) R 2 C G R 3 R 4 D R th I g Rg V th V s For unbalanced bridge, I g = V th / (R th + R g ) and V g = I g R g = V th R g /(R th +R g ) If balanced or V g = 0 V like no movement 11 Slightly Unbalanced ridge If R 2 = R 3 = R 4 = R and R 1 = R+ R V A = V s (R+ R)/(R+ R+R) = V s (R+ R)/(2R+ R) V AD = V s R/(R+R) = V s /2 R+ R V o = V A V AD = V s (R+ R) 1 I 1 A (2R+ R) 2 = V s R/ (4R+2 R) I 2 If R < 5% of R R V o V s R/ 4R D V o + V s R R 12 C 6

7 Slightly Unbalanced ridge (Cont d) For Thévenin s equivalent circuit, V th = V o = V s R/ 4R R th = (R+ R)//R + R//R = R(R+ R)/(2R+ R) + R/2 R/2 +R/2 = R I g = V th / (R th + R g ) V g = I g R g 13 Kelvin Double ridge A modification of Wheatstone bridge for low resistance measurement (R 1 < 1 ) ecause non-perfect wire resistances affect the measurement. r R 1 r R 2 3 r 4 V o R 3 R 4 V s 14 7

8 Using four-terminal resistors (two for voltage supply and 2 for current supply) 15 The yoke r is connected to R 1 and R 2 The relationship between r 3, r 4, R 3 and R 4 R 3 /R 4 = r 3 /r 4 Using the delta-star transformation, the equivalent circuit r a r b R 1 r R 2 c r a = r 3 r /( (r 3 +r 4 +r) r b = r 4 r / (r 3 +r 4 +r) R 3 R

9 Note -Y Transformation R a = R 1 R 2 / (R 1 +R 2 +R 3 ) R b = R 2 R 3 / (R 1 +R 2 +R 3 ) R c = R 1 R 3 / (R 1 +R 2 +R 3 ) C R 1 R 3 C R c A R 2 A R a R b 17 The balance condition is the same as Wheatstone bridge (Null V o = 0 V) (R 1 +r a ) / (R 2 +r b ) = R 3 / R 4 R 1 = R 3 3( (R 2+r b b) )/R 4 r a = R 3 R 2 /R 4 + R 3 r b /R 4 r a = R 3 R 2 /R 4 + R 3 r 4 r/(r 3 +r 4 +r)r 4 r 3 r/(r 3 +r 4 +r) = R 3 R 2 /R 4 + r 4 r(r 3 /R 4 r 3 /r 4 ) / (r 3 +r 4 +r) R 1 = R 3 R 2 /R 4 Therefore R 1 /R 2 = R 3 /R 4 = r 3 /r

10 High Resistance ridge For very high resistance, e.g. 1,000 M, there is leakage currents over the surface of the insulated post. Using three-terminal resistors (parallel with 2 R s2 R s1 leakage resistances) R s1 >>R 3 and R s1 //R 3 to R 3 R 4 R s2 may affect the avoid the leakage effect R 1 R 2 detector sensitivity 19 ridge Compensation The resistance of long leads will be affected by changes in temperaturest To avoid this, 3 leads are required to connect to the coils They are all the same length and resistance 20 10

11 ridge Compensation (Cont d) Any changes in lead resistance will affect all 3 leads equally and occur in 2 arms of bridge and will cancel out. 3 R 1 R 2 1 V o 2 R 3 R 4 V s 21 ridge Controlled Circuits The bridge can be used as an error detector in a control circuit, using the potential difference at the output of the bridge that is sensitive to any physical parameters. Passive circuit elements such as strain gauges, temperature sensitive resistors (thermistors) or photo resistors are used as one arm of Wheatstone bridge. A change in the elements (pressure, heat or light) causes the bridge to be unbalanced

12 References Hotek Technologies, Inc webpage : hotektech com/ Yokogawa webpage: MAGNET LA Wheatstone ridge webpage: a/wheatstonebridge/index.html 23 12