Download Thevenin Equivalent Circuit - Pre Lab Notes | ECE 3254 and more Lab Reports Electrical and Electronics Engineering in PDF only on Docsity!
ECE 3254 PreLab 2 notes Edited 08-27-
Thevenin's Theorem: Any two terminal network composed of a combination of voltage sources, current sources, and resistors, is electrically equivalent to a single voltage source (Vth) and a single resistor (Rth). For single frequency AC circuits, Thevenin's theorem can also be applied to complex impedances (with a complex Thevenin impedance Zth).
Thevenin Equivalent Circuit
Any circuit can be represented as a “black box” Thevenin source. Vth = VOC (open circuit voltage) Rth = Vth / ISC (short circuit current) Given VOC, VL, RL: Vth = VOC, IL = VL/RL, VRth = VOC - VL, Rth = VRth / IL Given Vth, Rth, RL: IL = Vth/(Rth+RL), VL = RLIL or VL = VthRL/ Rth+RL), IL = VL/RL The Norton Source is interchangeable with the Thevenin source. In = ISC and Rn = Rth Rn = VOC/In = Vth/ISC = Rth and VOC = Vth = InRn
PreLab 2 Thevenin Circuit Calculations
Step 1 - Reduce the series parallel combination to a single equivalent resistor. 2.2K || 3.9K = 1/(1 / 2.2K + 1 / 3.9K) = 1.4K, 1.4K + 3.3K = 4.7K Prelab 2 Notes - Page 1 of 2
Step 2 - Redraw circuit using the equivalent resistor and remove RL (open circuit), then write node equations to find Voc. Note: You should not include the load resistor in the Thevenin source calculations. Write a Node equation at the V1 terminal. Write a Node equation at the V2 terminal. Solve for V 2 , then Voc = V 2. Step 3 - Place a short circuit across termianls A and B, and repeat to find V 1. Note: The 4.7K resistor at V2 will be shorted and V2 = 0. Isc = V 1 / 1.0K Step 4 - Vth = Voc, Rth = Vth / Isc Alternative method: Step 1 is the same. At step 2, write loop equations and solve for the current through the 4.7K resistor. Now Voc = I4.7k * 4.7k. At step 3, write loop equations and solve for the short circuit current (Isc) at terminals A and B. Note: The 4.7K resistor is shorted and will carry no current. Step 4 is the same.
Maximum Power Transfer
The maximum power transfer from the source to the load occurs when RTH = RL. Efficiency at maximum power transfer is 50% - the load and the source each dissipate half of the power produced. When RL < RTH - power transfer is lower, and efficiency is low because the source dissipates more power than the load. When RL > RTH - power transfer is lower, and efficiency is high because the load dissipates more power then the source. A practical example of the relationship between RL, RS, power transfer, and efficiency can be found in the common stereo system audio amplifier. The output impedance of the amplifier is typically 2Ω or less. For our example amplifier, VTH = 20VRMS and RTH = 2Ω. With an 8Ω speaker, Ptotal = 40W, Pspeaker = 32W, Pamp dissipation = 8W With a 4Ω speaker, Ptotal = 67W, Pspeaker = 45 W, Pamp dissipation = 22W (the amp begins to get warm!). With speakers paralleled for 2Ω, Ptotal = 100W, Pspeaker = 50 W (WooHoo!), Pamp dissipation = 50W The speakers get 50W BUT, the amplifier must dissipate the other 50W and many amplifiers do not have heat sinks capable of dissipating enough heat when a low impedance load is used. With more speakers paralleled for 1Ω, Ptotal = 133W, Pspeaker = 44 W, Pamp dissipation = 89W (OUCH!). Not only has power transfer dropped, but efficiency is terrible and the amplifier will probably fry. In the case of a complex source or load impedance, maximum power transfer occurs when ZL equals the conjugate match of ZTH (RTH = RL and jωXTH = -jωXL). Prelab 2 Notes - Page 2 of 2
