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## DC & AC BRIDGES Part 1 (DC bridge)

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**DC & AC BRIDGES**Part 1 (DC bridge)**Objectives**• Ability to explain operation of Wheatstone Bridge and Kelvin Bridge. • Ability to solve the Thevenin’s equivalent circuit for an unbalance Wheatstone Bridge. • Define terms null or balance. • Define sensitivity of Wheatstone bridge.**Introduction**• DC & AC Bridge are used to measure resistance, inductance, capacitance and impedance. • Operate on a null indication principle. This means the indication is independent of the calibration of the indicating device or any characteristics of it. • Very high degrees of accuracy can be achieved using the bridges**Types of bridges**Two types of bridge are used in measurement: 1) DC bridge: a) Wheatstone Bridge b) Kelvin Bridge 2) AC bridge: a) Similar Angle Bridge b) Opposite Angle Bridge/Hay Bridge c) Maxwell Bridge d) Wein Bridge e) Radio Frequency Bridge f) Schering Bridge**Wheatstone Bridge**AWheatstone bridge is a measuring instrument invented by Samuel Hunter Christie (British scientist & mathematician) in 1833 and improved and popularized by Sir Charles Wheatstone in 1843. It is used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one leg of which includes the unknown component. Its operation is similar to the original potentiometer except that in potentiometer circuits the meter used is a sensitive galvanometer. Sir Charles Wheatstone (1802 – 1875)**Wheatstone Bridge**Definition: Basic circuit configuration consists of two parallel resistance branches with each branch containing two series elements (resistors). To measure instruments or control instruments Basic dc bridge used for accurate measurement of resistance: Fig. 5.1: Wheatstone bridge circuit**How a Wheatstone Bridge works?**• The dc source, E is connected across the resistance network to provide a source of current through the resistance network. • The sensitive current indicating meter or null detector usually a galvanometer is connected between the parallel branches to detect a condition of balance. • When there is no current through the meter, the galvanometer pointer rests at 0 (midscale). • Current in one direction causes the pointer to deflect on one side and current in the opposite direction to otherwise. • The bridge is balanced when there is no current through the galvanometer or the potential across the galvanometer is zero.**Cont.**At balance condition; voltage across R1 and R2 also equal, therefore (1) Voltage drop across R3 and R4 is equal I3R3= I4R4 (2) No current flows through galvanometer G when the bridge is balance, therefore: I1 = I3 and I2=I4 (3)**Cont.**Substitute (3) in Eq (2), I1R3 = I2R4 (4) Eq (4) devide Eq (1) R1/R3 = R2/R4 Then rewritten as R1R4 = R2R3 (5)**Example 5-1**Figure 5.2 consists of the following, R1 = 12k, R2 = 15 k, R3 = 32 k. Find the unknown resistance Rx. Assume a null exists(current through the galvanometer is zero). Fig. 5-2: Circuit For example 5-1**Solution 5-1**RxR1 = R2R3 Rx = R2R3/R1 = (15 x 32)/12 k, Rx = 40 k**Sensitivity of the Wheatstone Bridge**When the bridge is in unbalanced condition, current flows through the galvanometer, causing a deflection of its pointer. The amount of deflection is a function of the sensitivity of the galvanometer.**Cont.**Deflection may be expressed in linear or angular units of measure, and sensitivity can be expressed: Total deflection,**Unbalanced Wheatstone Bridge**Fig. 5-3: Unbalanced Wheatstone Bridge Fig. 5-4: Thevenin’s resistance Vth = Eab Rth = R1//R3 + R2//R4 = R1R3/(R1 + R3) + R2R4(R2+R4)**Thévenin’s Theorem**An analytical tool used to extensively analyze an unbalance bridge. Hermann von Helmholtz (1821 – 1894) Léon Charles Thévenin (1857-1926) German Physicist French Engineer Thévenin's theorem for electrical networks states that any combination of voltage sources and resistors with two terminals is electrically equivalent to a single voltage source V and a single series resistor R. For single frequency AC systems the theorem can also be applied to general impedances, not just resistors. The theorem was first discovered by German physicist Hermann von Helmholtz in 1853, but was then rediscovered in 1883 by French telegraph engineer Léon Charles Thévenin (1857-1926).**Thevenin’s Equivalent Circuit**If a galvanometer is connected to terminal a and b, the deflection current in the galvanometer is where Rg = the internal resistance in the galvanometer**R2 = 1.5 kΩ**R1 = 1.5 kΩ Rg = 150 Ω E= 6 V R3 = 3 kΩ R4 = 7.8 kΩ Figure 5.5: Unbalance Wheatstone Bridge Calculate the current through the galvanometer ? Example 5-2**Slightly Unbalanced Wheatstone Bridge**If three of the four resistors in a bridge are equal to R and the fourth differs by 5% or less, we can developed an approximate but accurate expression for Thevenin’s equivalent voltage and resistance.**Cont..**To find Rth: An approximate Thevenin’s equivalent circuit**500 Ω**500 Ω 10 V 525 Ω 500 Ω Use the approximation equation to calculate the current through the galvanometer in Figure above. The galvanometer resistance, Rg is 125 Ω and is a center zero 200-0-200-μA movement. Example 5-3**Kelvin Bridge**• The Kelvin Bridge is a modified version of the Wheatstone bridge. The purpose of the modification is to eliminate the effects of contact and lead resistance when measuring unknown low resistances. • Used to measure values of resistance below 1 Ω .**Fig. 5-6: Basic Kelvin Bridge showing a second set of ratio**arms**Cont.**It can be shown that, when a null exists, the value for Rx is the same as that for the Wheatstone bridge, which is Therefore when a Kelvin Bridge is balanced**Cont.**Fig. 5-6: Basic Kelvin Bridge showing a second set of ratio arms The resistor Rlc shown in figure represents the lead and contact resistance present in the Wheatstone bridge. The second set of ratio arms (Ra and Rb in figure) compensates for this relatively low lead contact resistance. At balance the ratio of Ra to Rb must be equal to the ratio of R1 to R3**Example 5-4**If in Figure 5-6, the ratio of Ra and Rb is 1000, R1 is 5 and R1 =0.5R2. What is the value of Rx.**Solution**The resistance of Rx can be calculated by using the equation, Rx/R2=R3/5=1/1000 Since R1=0.5R2, the value of R2 is calculated as R2=R1/0.5=5/0.5=10 So, Rx=R2(1/1000)=10 x (1/1000)=0.01