Thevenin Equivalent Circuits (EC 4.10)


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1 Thevenin Equivalent Circuits (EC 4.10) Thevenin equivalent Current delivered to any load resistance by a circuit is equal to: Voltage source equal to open circuit voltage V th at load n series with a simple resistor th (the source impedance).
2 Procedure for Finding Thevenin Equivalent (1) emove all elements not included in the circuit, emove all loads at the output. (2) Find the open circuit voltage = V th (3) Do one of the following to get th : (3a) Find the output resistance th : Turn off all sources Voltage sources become shorts Current sources become open Then calculate resistance of circuit as seen from output. (3b) Find short circuit current: Calculate the load resistance by V th = th short
3 Example Thevenin Equivalents Consider the simple voltage divider circuit Finding V th by open circuit voltage This acts a simple voltage divider Vth = Vopen = V = 10 = 7 V Setting sources off (short voltage source) Then input resistance is parallel resistors th 1 = tin 1 1 = + = th = 4.2 KΩ mhos
4 Example Thevenin Equivalents Using alternate method for th Short the output Hence 2 removed Short circuit current is Then V 10 = 6000 short = = ma Vth 7 th = = = 4.2 KΩ short ote often easier to use short output method.
5 orton Equivalent Circuits orton Equivalent Current delivered to any load resistance by a circuit is equal to: Constant current source = to short circuit current at the load Shunt resistor = resistance circuit when all sources are stopped.
6 Procedure for orton Equivalent Circuits (1) emove all elements not included in the circuit emove all loads at the output. (2) Find the short circuit current = (3) Do one of the following: (3a) Find the output resistance: turning off all constant sources Voltage sources become shorts Current sources become open Then calculate resistance of circuit as seen from output. (3b) Find the open circuit voltage V open Calculate the orton resistance by V = open
7 Example orton Equivalent Circuit Consider a current source with 1 in parallel & 2 on output Find the short circuit current = When shorted becomes a simple current divider 2 = = short = = = = Then find the output resistance Setting the current source off (open) ow 1 and 2 in series = = + = = KΩ 6 out 1 2 ma
8 Example orton Equivalent Circuit Con d Alternate way for Finding the open circuit voltage Thus the output resistance is V open = = V 2 = Vopen 24 = = = 6 KΩ Depending on circuit this may be faster then shutting off source
9 elationship Between Thevenin and orton Circuits Can easily change Thevenin into orton or vice versa By definition Thus th = th V = = V th = = th V th th
10 When to Use Thevenin and orton Circuits Thevenin equivalents most useful for: Where the information wanted is a single number Such as the current out of a given line. Circuits similar to nonideal voltage sources Eg Battery or simple power supply (50 ohm input) There th provides current limit & internal resistance orton equivalents are most useful for: Circuits that approximate a nonideal current source. Eg. current limited power supply There creates the voltage limit & internal resistance
11 Superposition of Elements Another way of solving circuits with linear circuit elements With linear circuits the effect of the combined system is linear Obtained by adding together effect of each source on circuit To use superposition analysis: (1) Set all power sources except one to zero Short circuit all voltage sources Open circuit all current sources These can be thought of as simplified circuits (2) Calculate the voltages and currents from that one source (3) epeat 12 for each source individually (4) Final result, at each point in the circuit: Sum the voltages and currents of all the simplified circuits esult is the combined voltage/current of the circuit.
12 Example of Superposition Solve the circuit below with a current and voltage source First turn off (open circuit) the current source Then really only have two resistors in series: V = = = 1 2 Using a voltage divider for the 2 line 1 ma V 1 = V = 6 = 4 V
13 Example of Superposition Continued ow turn on the current source Turn off the voltage source (short circuit it) Then using the current divider formula 2 2 = = = = = Similarly for the 1 branch = = = = The voltage across the 2 is V 2 2 = 2 = = 8 V 2 ma 4 ma
14 Example of Superposition Continued ow add the voltages and currents For the voltage across 2 For the current in 2 V 2 = V2 on + V2V on = = 12 V 2 = 2 on + 2V on = = 3 ma For the current in 1 ote: the V and source only currents are in opposite directions Voltage drop across 1 1 = 1 on + 1V on = 4 1 = 3 ma V 1 1 = 1 = = 6 V With this we have a full analysis of the circuit
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