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Method employs PSpice for Thevenin-equivalent circuits

18 Dec 2015  | David Karpaty

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Thevenin- and Norton-equivalent circuits, among the most fundamental circuit-analysis theorems, can be useful for determining a load resistance for maximum power transfer, simplifying circuit models, and a variety of other analysis techniques. Unfortunately, calculating the Thevenin voltage and resistance can become difficult as circuit complexity increases. Figure 1, figure 2, and figure 3 illustrate a simple method for obtaining the Thevenin voltage and resistance—and, subsequently, the Norton equivalence—with the aid of simulation. First, you choose an arbitrary load resistance, RLOAD—2 kΩ in this example—and run the simulation to get the current through the load resistance. Next, you remove the load resistance and simulate the open-circuit voltage across nodes A and B to obtain the Thevenin voltage. You obtain the Thevenin resistance from those two values.


Figure 1: To calculate Thevenin-equivalent circuits, you first choose a load resistance—2 kV in this circuit.


Figure 2: The simulation for current through the load resistance yields

260.3 mA.


Figure 3: The simulation for the open-circuit voltage yields approximately

2374 mV.


The Thevenin-equivalent circuit must produce the same current through the load. The total resistance in the Thevenin circuit is RTOTAL=(VTH/ILOAD)=(374.095 mV/60.301µA)≈6.203 kΩ, where RTOTAL is the total resistance. Therefore, the Thevenin resistance is simply [(VTH/ILOAD)–RLOAD]=(RTOTAL–RLOAD)=6.203 kΩ –2 kΩ≈4.203 kΩ, where VTH is the Thevenin voltage and ILOAD is the load current.


Figure 4: In the Thevenin-equivalent circuit, current flows to the left, so the VTHEVENIN terminal is grounded.


Figure 5: In the Norton-equivalent circuit, RNORTON is 4.203 kV.


Figure 4 shows the Thevenin-equivalent circuit, and figure 5 shows the Norton-equivalent circuit. Note that, because the net current through the load flows to the left, the positive Thevenin terminal is grounded. Without the aid of simulation, you can calculate VTHEVENIN and RTHEVENIN as follows. The array for the loop currents in figure 2, assuming a clockwise current flow in each loop, gives the current through the load resistance (Equation 1).


From Equation 1, you can calculate I2 and I5: I2≈217.77µA, and I5≈157.47µA. Thus, I2–I5≈60.3µA, assuming a leftward flow through the load resistor.


You calculate the array for the loop currents in figure 3 without the load resistance, as Equation 2 shows. From Equation 2, you can calculate the following currents: I1≈807.92µA, I2≈1.744 mA, I3≈179.87µA, I4≈53.64µA, and I5≈148.27µA. Thus, VA= –V4+[(I2–I3)×R7]≈–1.8719V, where the net current flows downward. Further, VB= [(–V4+(I3×R9))+((I3–I5)×R10)+V2]≈–1.498V, where the net current in R10 flows downward. Thus, VTHEVENIN=VA–VB≈–374 mV, and you can calculate RTHEVENIN according to the previous description.


About the author
David Karpaty contributed this article.


This article is a Design Idea selected for re-publication by the editors. It was first published on April 23, 2009 in EDN.com.




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