Transcript D2.5
Circuit Theorems Discussion D2.5 Sections 2-9, 2-11 September 2006 1 Circuit Theorems • • • • Linear Circuits and Superposition Thevenin's Theorem Norton's Theorem Maximum Power Transfer 2 Linear Circuits • A linear circuit is one whose output is directly proportional to its input. • Linear circuits obey both the properties of homogeneity (scaling) and additivity. 3 Superposition Principle Because the circuit is linear we can find the response of the circuit to each source acting alone, and then add them up to find the response of the circuit to all sources acting together. This is known as the superposition principle. The superposition principle states that the voltage across (or the current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone. 4 Turning sources off Current source: a i is is We replace it by a current source where is 0 b An open-circuit Voltage source: + DC vs v vs - We replace it by a voltage source where vs 0 i An short-circuit 5 Steps in Applying the Superposition Principle 1. Turn off all independent sources except one. Find the output (voltage or current) due to the active source. 2. Repeat step 1 for each of the other independent sources. 3. Find the total output by adding algebraically all of the results found in steps 1 & 2 above. In some cases, but certainly not all, superposition can simplify the analysis. 6 3 k Superposition Example 9 mA 6 k DC 18 V I0 I0 I1 I 2 2 3 5 mA 3 k 6 k 3 k DC 18 V 9 mA 6 k I2 I1 18 I1 2 mA 63 I2 3 9 3 mA 3 6 7 Example: In the circuit below, find the current i by superposition 24V DC DC 12V Turn off the two voltage sources (replace by short circuits). 3A i 12V v1 v2 3A i1 8 1 4 v1 0 1 4 1 3 1 4 v 1 4 1 4 1 8 2 3 5 1 v1 v2 0 6 4 12V v1 i1 i1 1 v2 1 3 v1 v2 3 4 8 3A 10 v2 v1 3 10 2 v1 3 8 8 v1 3 9 Example: In the circuit below, find the current i by superposition 24V DC DC 12V Turn off the 24V & 3A sources: 3A i i1 DC 12V i2 O.C. i2 10 DC O.C. 12V DC 12V O.C. i2 i2 12 4 3 16 12 i2 2 6 DC 12V O.C. i2 11 Example: In the circuit below, find the current i by superposition 24V DC DC 12V Turn off the 3A & 12V sources: 3A 24V i DC i2 i3 O.C. i3 12 4 8 4 4 i2 24 i 4 3 3 0 4 24V 16i2 4i3 24 DC i2 i3 i3 4i2 7i3 0 O.C. 7 i2 i3 4 i3 28 4 24 i3 1 13 24V DC DC i i1 i2 i3 1A 2A 1A 2A 12V 3A i 24V DC i1 12V v1 i1 i1 1 i2 v2 DC 3A 12V i2 i2 i2 2 O.C. i3 O.C. i3 i3 1 14 Circuit Theorems • • • • Linear Circuits and Superposition Thevenin's Theorem Norton's Theorem Maximum Power Transfer 15 Thevenin's Theorem In many applications we want to find the response to a particular element which may, at least at the design stage, be variable. a + Linear Circuit V b Variable R Each time the variable element changes we have to re-analyze the entire circuit. To avoid this we would like to have a technique that replaces the linear circuit by something simple that facilitates the analysis. A good approach would be to have a simple equivalent circuit to replace everything in the circuit except for the variable part (the load). 16 Thevenin's Theorem Thevenin’s theorem states that a linear two-terminal resistive circuit can be replaced by an equivalent circuit consisting of a voltage source VTh in series with a resistor RTh, where VTh is the open-circuit voltage at the terminals, and RTh is the input or equivalent resistance at the terminals when the independent sources are all turned off. i i a a RTh Linear Circuit RL DC RL VTh b b Rin Rin 17 Thevenin's Theorem Thevenin’s theorem states that the two circuits given below are equivalent as seen from the load RL that is the same in both cases. i i a a RTh Linear Circuit RL DC RL VTh b b Rin Rin VTh = Thevenin’s voltage = Vab with RL disconnected (= ) = the open-circuit voltage = VOC 18 Thevenin's Theorem i i a a RTh Linear Circuit RL DC RL VTh b b Rin Rin RTh = Thevenin’s resistance = the input resistance with all independent sources turned off (voltage sources replaced by short circuits and current sources replaced by open circuits). This is the resistance seen at the terminals ab when all independent sources are turned off. 19 3 k Thevenin Example 9 mA DC 6 k 18 V Rth 3 I0 Voc 45 V 45 V DC 3 k + 27 V - 9 mA Voc 45 V DC 18 V Rth 3 3 k Voc 45 V DC 6 k I0 Rth 3 45 5 mA 9 20 Example DC 10V VTh 5 RTh 2 iSC 2.5 a vOC RTh 2 2 10V 5V VTh 22 DC a VTh 5V b DC 10V a iSC b 10 2 10 2.5A 23 4 2 3 b a RTh 1 2 2 2 22 21 b Circuit Theorems • • • • Linear Circuits and Superposition Thevenin's Theorem Norton's Theorem Maximum Power Transfer 22 Norton's Theorem Norton’s equivalent circuit can be found by transforming the Thevenin equivalent into a current source in parallel with the Thevenin resistance. Thus, the Norton equivalent circuit is given below. i IN VTh RTh a RN RTh RL b Formally, Norton’s Theorem states that a linear two terminal resistive circuit can be replaced by an equivalent circuit consisting of a current source IN in parallel with a resistor RN, where IN is the short-circuit current through the terminals, and RN is the input or equivalent resistance at the terminals when all independent sources 23 are all turned off. Circuit Theorems • • • • Linear Circuits and Superposition Thevenin's Theorem Norton's Theorem Maximum Power Transfer 24 Maximum Power Transfer In all practical cases, energy sources have non-zero internal resistance. Thus, there are losses inherent in any real source. Also, in most cases the aim of an energy source is to provide power to a load. Given a circuit with a known internal resistance, what is the resistance of the load that will result in the maximum power being delivered to the load? Consider the source to be modeled by its Thevenin equivalent. i a RTh DC RL VTh b 25 i a RTh DC RL VTh b The power delivered to the load (absorbed by RL) is p i RL VTh 2 RTh RL 2 RL This power is maximum when p RL 0 p 2 3 2 VTh RTh RL 2 RL RTh RL 0 RL 26 p 2 3 2 VTh RTh RL 2 RL RTh RL 0 RL RTh RL 2RL RL RTh Thus, maximum power transfer takes place when the resistance of the load equals the Thevenin resistance RTh. Note also that pmax VTh RTh RL 2 RL RL RTh pmax VTh 2 RTh RTh VTh 2 4 RTh 2 Thus, at best, one-half of the power is dissipated in the internal resistance and one-half in the load. 27