Wheatstone Bridge PDF

Wheatstone Bridge Known resistors (ratio arms) Normally used to measure resistance change in the range of 1 Ω to 1 M Ω. Variable The bridge is balanced...

Wheatstone Bridge Known resistors (ratio arms) Normally used to measure resistance change in the range of 1 Ω to 1 M Ω. Variable The bridge is balanced resistance (by varying R1) ➔ Unknown potential difference resistance between points B and D (transducer) is zero ➔ voltage drop across R2 = voltage drop across R1. Gage (G) indicates no current flow. 1 When Balanced Bridge Method balanced Balanced Bridge Null Bridge Leads connecting Rx to the bridge must have resistance that is small (compared) to the unknown Rx. Wheatstone bridge is widely used for measuring the output resistance of many transducers that show the change in physical variable as a change in output resistance. 2 Unbalanced Bridge = Deflection Bridge Method Normally used for measuring dynamic signals. When slightly unbalanced, Rb does not affect the effective bridge resistance. R is the effective resistance of the bridge as presented to the galvanometer G; 3 Unbalanced Bridge = Deflection Bridge Method Using voltage divider relations  R1  ib VA = E    R1 + R4   R2  VC = E    R2 + R3   R1 R2  E g = VA − VC  E g = E  −   R1 + R4 R2 + R3  R2 R3 For small unbalance, the effective B D resistance of the bridge as presented to the battery is; R1 R4 E Ro Voltage impressed on bridge = E = ib.Ro = Eb 4 Ro + Rb   R1  VA = E    R1 + R4  Voltage divider relation  R2  VC = E    2 R + R3   R1 R2  Eg = VA − VC  Eg = E  −   R1 + R4 R2 + R3   R1 R3 − R2 R4   ( R1 + R1 ) R3 − R2 R4  Eg = E   = E   ( R1 + R4 )( R2 + R3 )   ( R1 + R1 + R4 )( R2 + R3 )  Assume that the resistance R1 changes by the amount R1  5   ( R1 + R1 ) R3 − R2 R4  Eg = E    ( R1 + R1 + R4 )( R2 + R3 )   Eg = E  ( R1 + R1 ) R3 − R2 R4    R1 (1 + R1 / R1 + R4 / R1 ) R3 ( R2 / R3 + 1)   1 + ( R1 / R1 ) − ( R2 R4 / R1R3 )  Eg = E    1 + ( R1 / R1 ) + ( R4 / R1 )  1 + ( R2 / R3 )      Simplify by assuming that all the resistances are initially equal to R ➔ Eg = 0 initially Eg / E = ( R1 / R ) 4 + 2 ( R1 / R ) 6  Eg / E = ( R1 / R ) 4 + 2 ( R1 / R ) Eg / E ( R1 / R ) Curve shows nonlinear characteristic of Wheatstone bridge 7 Effects of long leadwires: Figure (a) illustrates a two-wire connection from a sensor to a bridge circuit. With this configuration, each of the leadwire resistances R' adds to the resistance of the strain gage branch of the bridge. If the leadwire temperature changes, it causes changes in the resistance of the bridge branch (i.e., long leadwires extending through environments where the temperature changes).  R1 R2  Eg = E  −   R1 + R4 R2 + R3  Figure (b) illustrates a three-wire connection that solves this problem. With this configuration, equal leadwire resistances are added to adjacent branches in the bridge so the effects of changes in the leadwire resistances offset each other. The third leadwire is connected to the high-input impedance voltage measuring circuit, and its resistance has a negligible effect since it carries negligible current. 8 Assuming resistor’s material is homogeneous and has a constant cross-sectional area. ρL Resistance = R = A where; ρ = resistivity (specific resistance) L = wire length A = cross - sectional area. Ohm’s law: V = IR 9 A capacitor is a passive element that stores energy in the form of an electric field (electric charge). DC does not flow through a capacitor, rather, charges are displaced from one side of the capacitor through the circuit to the other side, establishing the electric field. Capacitor’s voltage-current relationship is defined as: t 1 q(t) V(t) = C 0 I(τ)d(τ) = C , dV or, I(t) = C dt where : q(t) = accumulated charge in coulumbs C = capacitance in farads (F = coulumbs/volts) I(t) = displacement current Typical capacitor range (1pF to 1000  F) 10  Inductor: is a passive element (wire coil) that stores energy in the form of a magnetic field. dλ dI Faraday's law of indcution :V(t)= =L dt dt where; λ = LI = magnetic flux through the coil windings due to the current. measured in webers (Wb). L = coil coil inductance, incuctance,henry (H = Wb / A) or, t 1 I(t)=  V(τ)dτ L0 τ = dummy variable of integration. Typical inductors range(1μH to 100 mH) 11 Kirchhoff’s Laws: Are used in circuits analysis. Kirchhoff’s voltage law: The sum of voltages around a closed loop or path is 0. N V = 0 i =1 i −V1 − V2 + V3 +.... − VN = 0 Kirchhoff’s current law: The sum of the currents flowing into a closed surface or node is 0. N I i =1 i =0 I1 + I 2 − I 3 = 0 12 Series Circuit N Req = R1 + R2 =  Ri i=1 C1C2 Ceq = C1 +C2 Leq = L1 + L2 R1 R2 Voltage divider ; VR1 = Vs , VR 2 = Vs R1 + R2 R1 + R2 Ri Ri generalized voltage divider = VRi = Vs = N Vs R Req j j=1 Voltage dividers allowus to create different reference voltages in a circuit energized by a single voltage supply. 13  Parallel Circuit N 1 1 1 = + =  1/Ri Req R1 R2 i=1 R1 R2 1 Req = = N  1/R R1 + R2 i i=1 Ceq = C1 +C2 L1 L2 Leq = L1 + L2 R2 R1 Current divider = I1 = I , I2 = I R1 + R2 R1 + R2 14 Impedance Matching: Power Rs RL + + By voltage divider ; VL = Vs RL + Rs Vs RL Power transmitted toload ; - - VL2 RL PL = = V 2 RL ( RL + Rs ) 2 s load source RL that maximizes power canbe found by setting the derivative of the power = 0. 2 ( RL + Rs ) − 2 RL ( RL + Rs ) 2 dPL = Vs =0 dRL ( RL + Rs ) 4 or , ( RL + Rs ) 2 = 2 RL ( RL + Rs )  RL = Rs To maximize power transmission to a load, the load’s impedance should match the source’s impedance. 15 Impedance Matching: Signal Integrity RL By voltage divider ; VL = Vs RL + Rs RL  RS  VL = 0 + Rs + Vs RL RL  RS  VL = VS - - load source Proper circuit or signal termination may be required when connecting different devices and circuits. 16