Thevenin Equivalent Circuits: (Material for exam - 3)

Transcription

1 Thevenin Equivalent Circuits: (Material for exam 3) The Thevenin equivalent circuit is a two terminal output circuit that contains only one source called E TH and one series resistors called R TH. This circuit is used to replace a much more complex circuit consisting of one or more sources and resistors connected in various cominations around a set of output terminals. The simple steps to convert this more complex circuit to a Thevenin Equivalent at its output terminals can e expressed as, 1.) n the original circuit, remove any load (resistor, ETC.) From its output terminals so the circuit at those terminals is an open circuit. 2.) Use KCL, KVL and Ohm s Law as needed to calculate the open circuit voltage across the output terminals. This open circuit output voltage is (E TH ), the Thevenin voltage source magnitude. 3.) To find the Thevenin resistance ( R TH ) in the original circuit, follow these steps. a. Replace all independent voltage sources with short circuits. n other words, remove the sources(s) and put a wire in its place.. Replace all independent current sources with open circuits. Remove the source(s) and leave the circuit open where the source(s) were located. c. Now to calculate R TH, use series and parallel resistor cominations as needed to find the total resistance the circuit would have at its output terminals. This total resistance is R TH. Let s Look t Few Sample Circuits: d 8Ω c a 1 V 3Ω Ω Diagram #1 First calculate V which is E OC (E open circuit). Using KVL, Using KVL at output to find V

2 Now to find R TH, we replace the voltage source with a wire, d 8Ω c a Wire 3Ω Ω Diagram #2 a 8.1Ω 2.73 V Diagram #3 This is the Thevenin Equivalent circuit of the original circuit. n conclusion The Thevenin equivalent circuit is a terminal equivalent to the original circuit at its terminals. Check out the Multisim circuit on the wesite.

3 Thevenin Example Circuit: (Used to show different methods) D C 2Ω V 2 2 E 1Ω Diagram # Let s first solve for the open circuit voltage V. Use KCL at node C. So,, where we have two unknowns, 1 and 3. This means we need another equation in terms of these two unknowns. t turns out KVL will give us this expression. What are the possiilities for KVL equations? re these two equations possiilities?

4 However, if we look at equation #1 and #2 we find that oth equations contain V C or V C. f we recognize V C = V C and sustitute it into equation #1, Solving for V C, Now taking this and sustituting this into equation #2, We now have an expression of 1 and 3. f this seems to e a roundaout way of getting an expression of 1 and 3, you re right. Note, these two equations oth contain V C or V C, the voltage across the current which is an unknown. The KVL loops we took oth included the current source which has een defined as 2. Since oth loops contained V C or V C and either 1 or 3, V C = V C could e used to get an expression of 1 and 3. However, a etter solution would e to take a KVL loop around the outside components of the circuit. n other words, do not take the loop that includes a current source. The outside loop does contain 1 and 3 which would immediately give us our expression in terms of 1 and 3. Hence, t this point you are proaly wondering why went through the first method and then the second method which got the results we needed more directly. did this to show that there are in this case, more than one way to derive what we need. also wanted to show that if we follow a KVL loop that does not contain current sources ut does contain the two currents we need 1 and 3, which yields a much simpler solution. Now finalizing the solution to this example. KCL

5 KVL = 2 So, Now calculating R TH in the original circuit, replacing voltage sources with wires (short circuits) and current sources with open circuits. This will yield, D C 2 2Ω 9 E 1Ω Diagram #5 The Thevenin equivalent of the original circuit can e drawn as, 17. Diagram #

6 The next two pages are multisim results showing open circuit, 2 ohm load and when the output was shorted. R1 R3 2Ω 3 J1 8 V1 2 V R2 1Ω 1 2 The Original Circuit R 25. V U1 R 2Ω 5 R5 17. J2 7 V2 25. V U2 R7 2Ω Thevenin Equivalent Circuit

7 R1 R3 2Ω 3 J1 8 V1 2 V R2 1Ω 1 2 The Original Circuit R V U1 R 2Ω 5 R5 17. J2 7 V V U2 R7 2Ω V1 2 V R1 5 V2 R2 1Ω Thevenin Equivalent Circuit R R3 2Ω The Original Circuit.1u V J2 U2 R 8 J1 3 7 U U1.1u V U 1.29 = = = Yields each circuits output short circuit current. Thevenin Equivalent Circuit