INTC 1307 Instrumentation Test Equipment Teaching Unit 6 AC Bridges

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1 IHLAN OLLEGE chool of Engineering & Technology ev. 0 W. lonecker ev. (8/6/0) J. Bradbury INT 307 Instrumentation Test Equipment Teaching Unit 6 A Bridges Unit 6: A Bridges OBJETIVE:. To explain the operation of an A Bridge.. To review A concepts of frequency and effects on L circuits. 3. To explain the requirements of balancing an A Bridge. 4. To null an A Bridge. A Bridge onfiguration When the four resistive arms of the basic Wheatstone bridge are replaced by impedances and the bridge is excited by an A source, the result is an A Bridge. Now, to balance the bridge, two 3 conditions must be satisfied, the resistive () and the reactive components ( or L ). Once balanced, the A Bridge indicates a null. A bridge circuits are also used for shifting phase, 4 providing feedback paths for oscillators and amplifiers, filtering out undesired signals, and measuring the frequency of audio and radio frequency (rf) signals. At balance: indicates the magnitude of the impedance θ = 3 θ 3 θ 4 θ 4 ( θ )( 4 θ 4 )=( θ )( 3 θ 3 ) 4 3 and θ θ 4 = θ 3 θ 3 The null or balanced condition occurs when detector current becomes ero (no voltage difference occurs from Y to ). This means that the impedance () of the detector circuit appears infinite ( ), or as an 3 apparent open circuit, so that each leg of the bridge is isolated from the other leg. I A I B Y For this balance condition, 4 V V3 V V4 I A = I B 3 I A = I B 4 I A I B3 so 3 I I and 4 = 3 A B 4 4 and 4 = 3 indicates both magnitude and phase Page of 8

2 This is the general bridge equation and applies to any four arm bridge circuit whether branches are pure resistances or combinations of,, and L. Most of the time, the balance equation is not dependent upon frequency. When the bridge is not balanced, these equations are not correct. The circuit is complex and must be written in complex form = θ. A Bridge alculations 00Ω 60 50Ω 3 For example, the A Bridge at right is balanced. In order to use the bridge equation, complex forms must be multiplied. When multiplying or dividing, the polar complex form is easiest to use. When adding or subtracting, the rectangular complex form is easiest to use j 300Ω 4 UNKNOWN ince and 3 are already in polar form, change to polar form and solve for 4. onversion of, 00 + j 300Ω, to polar form: tan j L = 300 L 3.0 o Arc tan = 00 36Ω7. 6 in L L 300Ω = 36Ω sin 7.6 sin = 3 The bridge equation 3 (36.6 )(50) 4 4 = = =37 Ω.6 4 = 37 Ω. 6 onversion of 37Ω.6 to rectangular form j 37Ω. 6 in L L in 37Ω(in.6 ) os = os 37Ω(os.6 ) 4 = 3 + j47.7 Ω Page of 8

3 eview of Basic A oncepts esistance: esistors limit current and dissipate power. There is no phase shift across pure resistance. hanges in frequency cause no change in the value of the resistance. V I = 0 freq. no change with frequency = Page 3 of 8

4 Inductance: Inductance opposes any change in current. is the symbol for Inductive eactance which is the opposition to the flow of A current measured in ohms. An inductor causes the current through the inductor to lag the voltage across the inductor by 90 degrees. N A L VL L L 90 I L j L =πƒl frequency apacitance: apacitance opposes any change in voltage. is the symbol for apacitive eactance which is the opposition to the flow of A current measured in ohms. apacitance causes the voltage across the capacitor to lag the current through the capacitor by 90 degrees. Q Q V= V Q = charge j L ircuits: = j I is the same everywhere in a series circuit. I T V j I T L j L L L = + j L 0 ƒ ƒ ƒ = + jl at, L = 0 so = 0 θ=0 +j90 at mid freq. = + j L, when L =, 45 at high freq. L = = L θ=90 If V g is constant as ƒ increases, remains constant, both L and increase, and I T decreases. Page 4 of 8

5 ircuits: = j = f 0 -j 0 0 ƒ ƒ ƒ at, =, = 90 at mid freq. =, θ= 45 = - j 45 at high freq. = 0Ω = 0 L ircuits = + j ( L ) L at low freq. (dc) L = 0, = j 90 The circuit is open! at ƒco min, > L, = j( L) and = j, when =, = -45 at ƒ 0, = 0, L = so they cancel each other leaving = at ƒco max, < L, = + j( L ) = + j when =, = 45 at high freq. L =, = 0, = + j L, = 90 ƒo min ƒo max ƒo + - : I ƒo 0 ƒ ƒ Impedance vs. freq. urrent vs. freq. ƒo is the resonant frequency when L = Practical problem: Find the value of 4 to balance this A bridge. 400Ω 00Ω in series with 0.059H L = πƒl = L3 = 00Ω 7 f = Ω 3 = 400Ω0 = 300 j398ω = 500Ω = 00 + j00ω = 3.6Ω Ω in series with 0.4F UNKNOWN = 4 = 3 ( ) ( ) = = Ω.8f Page 5 of 8

6 apacitor Equivalent ircuits: Except for electrolytic capacitors, capacitors have almost no intrinsic resistance. Electrolytic leakage and dissipation factors are modeled using parallel/series resistances. P is called leakage resistance and is in parallel with the actual capacitance. The P / P parallel circuit shown below is equivalent to the series / circuit. Electrolytic capacitors have a relatively low P ; other types of capacitors have extremely high P. The higher the P the less the leakage current so the capacitor can better maintain its charge and voltage. If P decreases, the capacitor no longer acts as a capacitor; the leakage increases to the point that the capacitor appears as a resistor. P P = P = P( + ) Bridges measure the ratio of reactance to resistance for the component under test. This ratio is called the dissipation factor,, which is directly proportional to the power loss per cycle. A capacitor with a low leakage loss (keeps its charge when disconnected) is a high quality capacitor. P P f PP f P P f f f Inductor Equivalent ircuits: Inductors are formed by winding a coil of wire. This wire has resistance. Each inductor then has a series resistance along with its inductance. This ratio of Inductive eactance to its coil wire resistance is the storage factor or quality factor, Q. This Q factor is directly proportional to the energy stored per cycle. The greater the dissipation factor, the lower the Q of the coil. L Q = fl P L P Q = P P f L Q = P P Page 6 of 8

7 apacitance Bridges: If the bridge is balanced, an apparent open circuit occurs from Y to. At that balance point, the ratio of each leg to the other is such that the voltage drops and the phase angles are the same, just as in a balanced Impedance Bridge. The standard frequency of the voltage source for these A bridges is 000 Hz. The following formulae show that frequency is not necessary for a capacitance bridge to operate at balance. V V khz 0eg U V f f f f f f f (ivide out πf) apacitance Bridge Example: For unknown, = 50kΩ, = 0.0F, and = 300kΩ for balance, then = 0.06F ince reactance is a function of frequency, it is possible with capacitors to obtain a very low reactance for (the unknown) for a frequency of khz. For example, if = 0F then its reactance is 5.9Ω. For an fixed at 0kΩ, the source voltage divider ratio for - would be only.5% or 5mV for a v M source. The sharpest nulls are obtained for divider ratios of 50%, so a.5% ratio would give a very weak null. However, if the source frequency were lowered to 50 Hz and lowered to kω, the reactance would be 38.3Ω and the divider ratio would be 30%. This ratio gives a much better null. o although frequency divides out of the balance equation, it can and does impact bridge accuracy. Other Bridges: This discussion has covered the basic concepts of bridges. Other bridges are Inductance, imilar Angle, Opposite Angle, Wien, adio Frequency, chering, and Owen bridges. A Maxwell-Wien Bridge is shown below (often called the Maxwell bridge), and is used to measure unknown inductances in terms of calibrated resistance and capacitance. alibration-grade inductors are more difficult to manufacture than capacitors of similar precision, and so the use of a simple "symmetrical" inductance bridge is not always practical. Because the phase shifts of inductors and capacitors are exactly opposite each other, a capacitive impedance can balance out an inductive impedance if they are located in opposite legs of a bridge, as they are here. Page 7 of 8

8 V V khz 0eg L V U 3 50% Maxwell-Wien Bridge The balance of the Maxwell-Wien bridge is independent of source frequency, (with the understanding of divider ratios already discussed) and in some cases this bridge can be made to balance in the presence of mixed frequencies from the A voltage source, the limiting factor being the inductor's stability over a wide frequency range. Bridge Equations: L = = /3 Page 8 of 8

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