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Lesson 10: Thevenin’s Theorem and Max Power Transfer 1 Learning Objectives • State and explain Thèvenin's theorem. • List the procedure for determining the Thèvenin equivalence of an actual circuit from the standpoint of two terminals. • Apply Thèvenin's Theorem to simplify a circuit for analysis. • Analyze complex series-parallel circuits using Thèvenin's theorem. • Apply the Maximum Power Transfer theorem to solve appropriate problems. 2 Thévenin’s Theorem • Thévenin’s theorem greatly simplifies analysis of complex circuits by allowing us to replace all of the elements with a combination of just one voltage source and one resistor. • Thévenin’s theorem provides a simplified circuit that provides the same response (voltage and current) at the load terminals. − This allows the response to be easily determined for various load values. 3 Thévenin’s Theorem • Any complex two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source VTh and a series resistor RTh. Original Circuit Thévenin Equivalent Circuit • The Thévenin equivalent circuit provides an equivalence at the terminals only. − The internal construction and characteristics of the original network and the Thévenin equivalent are usually quite different. 4 Thévenin’s Theorem • ETh is the open circuit voltage at the terminals. • RTh is the input or equivalent resistance at the terminals when the sources are turned off. Original Circuit Thévenin Equivalent Circuit 5 Thévenin’s Theorem 5 Steps: 1. Remove the load and 2. Label the terminals a and b. 3. Solve for RTH by setting all sources to zero. 4. Solve for VTH by returning all sources to their original position and finding the open-circuit voltage between a and b. 5. Draw the new equivalent circuit. 6 Thévenin’s Theorem Steps 1 & 2 Convert to a Thévenin circuit: 1. Identify and remove the load from the circuit. 2. Label the resulting open terminals. 7 Thévenin’s Theorem Step 3 3. Solve for RTH and isolate the resistance from the source. Set all sources to zero: Replace voltage sources with shorts. Replace current sources with opens. 8 Zeroing Sources 3. “Zeroing” a source means setting its value equal to zero. • Voltage sources – 0 V is equivalent to a short-circuit. • Current sources – 0 A is equivalent to a open-circuit. Voltage Sources Become Short-Circuits Current Sources Become Open-Circuits 9 Thévenin’s Theorem Step 3 • With the load disconnected, turn off all source. • RTh is the equivalent resistance looking into the “dead” circuit through terminals a-b. Rth 10 Thévenin’s Theorem Step 3 3. Set all sources to zero, and calculate RTh . 1 RTH 1 1 Rab 31 80 60 40 Remember, calculate RTH from the a and b perspective! 11 Thévenin’s Theorem Step 4 4. Solve for VTH and then, as needed: • • Calculate the voltage (VLD) across the RLD. Calculate the current (ILD) through RLD. (VDR) ETH VTH Vab V40 R 40 VTH E * 40 20V * 4.44V RT 40 80 60 VLD I LD 12 RLD ETh RTh RLD ETh RTh RLD Thévenin’s Theorem Step 5 5. REDRAW the circuit showing the Thèvenin equivalents (VTH and RTH) with the load installed. 13 Applying Thévenin Equivalent • Repetitive solutions for various load resistances now becomes easy with the transformed circuit. 14 Example Problem 1 Find the Thévenin equivalent circuit external to RLD. Determine ILD and VLD when RLD = 2.5 Ω. a 15Ω 9Ω R 9 (VDR) ETH Vab V9 E * 9 10V * 6V RT 6 9 I LD RTH = 18.6Ω = 2.5Ω I LD 1/2. Remove the load label the terminals a and b. 3. Solve for RTH. 4. Solve for VTH. 5. Draw the new equivalent circuit. RTH b 1 Rab 15 18.6 1 1 6 9 ETh RTh RLD 6V 284mA 18.16 2.5 RLD RTh RLD 2.5 6V 0.71V 18.6 2.5 VLD ETh VLD OR VLD I LD * RLD 284mA * 2.5 0.71V 15 Example Problem 2 Find the Thévenin equivalent circuit external to RLD and determine ILD. RTH Rab 20k 50k 70k ETH Vab V50k I S * R50 k 100 A * 50k 5V RTH = 70kΩ ETH = 5V I LD ETh RTh RLD I LD 5V 20 A 70k 180k = 180kΩ 16 Maximum Power Transfer In some applications, the purpose of a circuit is to provide maximum power to a load. Some examples: • Stereo amplifiers • Radio transmitters • Communications equipment The question is: If you have a system, what load should you connect to the system in order for the load to receive the maximum power that the system can deliver? 17 Maximizing PLD • How might we determine RLD such that PLD is maximized? 2 PLD I RLD 2 LD 18 VTh RLD RTh RLD Maximum Power Transfer Theorem • Maximum power is transferred to the load when the load resistance equals the Thévenin resistance as seen from the load (RLD = RTh). − When RLD = RTh, the source and load are said to be matched. 19 Maximizing PLD • As RLD increases, a higher percentage of the total power is dissipated in the load resistor. • But since the total resistance is increasing, the total current is dropping, and a point is reached where the total power dissipated by the entire circuit starts dropping. 2 PLD I RLD 2 LD 20 VTh RLD RTh RLD Maximum Power • The max power happens when RLD = RTh , but what is the level of power at this point? − Showing the derivation, we get: 2 2 PLD I LD RLD 2 V Th 2 VTh VTh RTh RLD RTh RLD RTh RTh Rth 4R 2 Th V 2 Th 4 RTh PMAX • BE CAREFUL!!! Note that this is not true if RLD RTh. − If RLD RTh then use: PLD I LD RLD 2 2 VTh RLD RTh RLD 21 Maximum Power Transfer Theorem • The total power delivered by a supply such as ETh is absorbed by both the Thévenin equivalent resistance and the load resistance. • Any power delivered by the source that does not get to the load is lost to the Thévenin resistance. 22 Example Problem 3 a) Find the Thévenin equivalent circuit to the left of terminals a-b. b) Calculate the maximum power transfer to the load if RLD = RTH. c) Determine the power dissipated by RLD for load resistances of 2 and 6. b) When R LD = R TH V 2 Th 4 RTh PMAX PMAX a 4Ω 302 V 4*4 56.25W c) When R LD R TH 1Ω 2 VTh PL I L 2 * RL OR PL RLD R R Th LD For the 2 load and because we already calculated VTH and R TH let's use: 12Ω b 1 1 1 a) RTH Rab 1 4 4 12 R 12 (VDR) ETH Vab V12 E * 12 40V * 30V RT 4 12 2 VTh 30V PL RLD * 2 50W 4 2 RTh RLD Now, for the 6 load, and just to show it works let's use: PL I L * RLD 2 23 2 2 30V *6 54W 4 6 Example Problem 4 A stereo is rated for max output power of 150W per channel when RLD = 8Ω a) Sketch the Thévenin Equivalent circuit. b) What would the output power be with two 8Ω speakers as the load and are connected in parallel to one of the channels? b) We know that PMAX RTH = 8Ω = 8Ω V 2 Th 4 RTh Rearrange the above and solve for VTH : V 2TH PMAX * 4 RTh VTH PMAX * 4 RTh 150W * 4*8 69.28V Now, to caculate output power for the two 8Ω resistors in parallel, calculate R L (8//8 = 4), RTH = 8Ω and because we already know VTH and R TH let's use: = 8Ω//8 Ω =4Ω 2 VTh 69.28V PL RLD * 4 133.3W 8 4 RTh RLD You should note, PL for this case is less than PMAX . 2 24 Efficiency • When maximum power is delivered to RLD, the efficiency is a mere 50%. − Remember that max power occurs when RLD = RTH. pout PL I L 2 RLD 2 100% 2 pin PS I L RTh I L RLD When R L =R TH i 2 RTh 2 100% 50% 2 i RTh i RTh 25 Efficiency • Communication Circuits and Amplifiers: − Max Power Transfer Is More Desirable Than High Efficiency. • Power Transmission (115 VAC 60 Hz Power ): − High Efficiency Is More Desirable Than Max Power Transfer. − Load Resistance Kept Much Larger Than Internal Resistance Of Voltage Source. 26 QUESTIONS? 27