5. Handy Circuit Analysis Techniques

Transcription

1 1 5. Handy Circuit Analysis Techniques The nodal and mesh analysis require a complete set of equations to describe a particular circuit, even if only one current, voltage, or power quantity is of interest In this chapter, we investigate several different techniques for isolating specific parts of a circuit in order to simplify the analysis This chapter covers the following sections: 1. Linearity and Superposition 2. Source Transformations 3. Thévenin and Norton Equivalent Circuits 4. Maximum Power Transfer 5. Delta-Wye Conversion 6. A Summary of Various Techniques

2 2 5.1 Linearity and Superposition Linear Elements and Linear Circuits A linear circuit is a circuit composed of independent sources, linear dependent sources, and linear elements (resistors) The sources are often called forcing functions, and the nodal voltages are termed response functions The response of linear circuits can be given as a weighted linear combination of the forcing functions Example Using node voltage analysis, we have The response, v1, can be given as Linear circuit

3 3 The Superposition Principle The superposition method uses the linearity property of the linear circuit to simplify the analysis We remove all sources (independent only) expect one, and analyze the circuit We repeat the procedure for another source, and so on Finally, the net result is found by summing all the single-source responses Example (cont.) 1- We reduce a current source to zero (open circuit), we have 2- We reduce a voltage source to zero (short circuit), we get 3- The solution is

4 Use superposition to compute the current ix 1- We reduce a current source to zero (open circuit), we have 2- We reduce a voltage source to zero (short circuit), we get 3- The solution is 4

5 Use superposition to compute the current ix 1- We reduce a current source to zero (open circuit), we have 2- We reduce a voltage source to zero (short circuit), we get ( ) 3- The solution is 5

6 6 5.2 Source Transformation A practical voltage source is defined as an ideal voltage source connected in series with an internal resistance Rs A practical current source is defined as an ideal current source in parallel with an internal resistance Rp

7 7 Equivalent Practical Sources With transformation, we can simplify a complex circuit so that in the transformed circuit, the devices are all connected in series or in parallel Changing the practical voltage source to an equivalent current source (or vice versa) requires the following conditions: The internal resistors must be equal in both sources Rs=Rp The source transformation must be constrained by vs=rsis=rpis The conversion of a voltage source into an equivalent current source can be given as follows: Electrical Circuit Electrical Circuit

8 8 Equivalent Practical Sources The conversion of a current source into an equivalent voltage source can be given as follows: Electrical Circuit Electrical Circuit

9 Compute the current IX through the 47 k resistor after performing a source transformation on the voltage source 1- We replace the voltage source with an equivalent current source 2- The current sources are divided as 9

10 Compute the current the labeled current 10

11 Compute the current the labeled current The current I can now be found using KVL: where. Thus, 11

12 Thévenin and Norton Equivalent Circuits Efficient methods to simplify a complex circuit containing a load resistor into a very simple equivalent circuit A Thévenin equivalent circuit A Norton equivalent circuit Thévenin s theorem replaces the complex part of the circuit, except the load resistor, with an independent voltage source in series with a resistor The voltage and current seen by the load resistor will be unchanged Norton s equivalent circuit consists of an independent current source in parallel with a resistor

13 Using repeated source transformations, determine Thévenin and Norton Equivalent Circuits 13

14 14 The circuit is much simpler, and we can now easily compute the equivalent circuits A Thévenin equivalent circuit A Norton equivalent circuit

15 15 1. Thévenin s equivalent circuit (independent sources) 1. Given any linear circuit, rearrange it in the form of two networks, A and B, connected by two wires 2. Disconnect network B and determine a voltage Voc, which appears across the terminals of network A 3. If the network A contains only independent sources 1. Replace the voltage source with a short circuit and a current source with the open circuit 2. Compute the equivalent resistance Roc seen by the terminals of the network A 4. Finally, Vth=Voc and Rth=Roc + Network A Voc Network B - Network A vc sc cs oc Roc Network B

16 16 2. Thévenin s equivalent circuit (dependent/independent sources) 1. Given any linear circuit, rearrange it in the form of two networks, A and B, connected by two wires 2. Disconnect network B and determine a voltage Voc, which appears across the terminals of network A 3. If the network A contains dependent and independent sources 1. Leave the dependent and independent sources unchanged 2. Short circuit the terminals of network A and determine Isc 4. Finally, Vth=Voc and Rth=Voc/Isc + Network A Voc Network B - Network A Isc Network B

17 17 Determine the Thévenin equivalent of the shown circuit + Voc - 1- Since the circuit is linear, we use a superposition method to compute Voc + Voc Voc

18 2- We kill the independent sources to determine the equivalent resistor 18

19 19 Determine the Thévenin equivalent of the shown circuit 1- We note that vx=voc and the current of dependent passes only through the 2 k resistor, since no current can flow through the 3 k resistor. Using KVL, we have Voc Then, 2- To determine Rth, we short-circuit the output terminals. Since vx=0, then Isc is given by 3- Finally, Isc

20 20 1. Norton s equivalent circuit (independent sources) 1. Given any linear circuit, rearrange it in the form of two networks, A and B, connected by two wires 2. Short circuit the terminals of network A and determine a current Isc 3. If the network A contains only independent sources 1. replace the voltage sources with short circuits and current source with open circuits 2. Compute the equivalent resistance Roc seen by the terminals of the network A 4. Finally, IN=Isc and RN=Roc Network A Isc Network B Network A vc sc cs oc Roc Network B

21 21 2. Norton s equivalent circuit (dependent/independent sources) 1. Given any linear circuit, rearrange it in the form of two networks, A and B, connected by two wires 2. Short circuit the terminals of network A and determine a current Isc 3. If the network A contains dependent and independent sources 1. Connect an open circuit between the terminal of network A 2. Determine Voc 4. Finally, IN=Isc and RN=Voc/Isc Network A Isc + Network B Network A Voc Network B -

22 22 Determine the Norton equivalent of the shown circuit Voc 1- We note that 1.5 k resistor is in parallel with the short circuit, Isc flows through the 2.5 k resistor. Using KCL, we have 2500 Isc where 2- To determine RN, we open-circuit the output terminals to determine Voc where Then 3- Finally,

23 23 If there are only dependent sources When there are only dependent sources, the equivalent network is merely Rth, that is, no current or voltage sources 1. Connect 1 A current source flowing from terminal b to a 2. Determine vtest, then RN=Rth=vtest/1 a Network A b

24 24 Determine the Norton equivalent of the shown circuit i 1- Since the circuit has no independent sources, then i=0 2- We next seek the value of RTH represented by this two-terminal network. Applying nodal analysis where 3- Finally, i

25 Maximum Power Transfer In some applications, the purpose of a circuit is to provide maximum power to a load If we have a circuit, what is the value of a load resistor that absorbs the maximum power delivered by the source For the practical voltage source, the power delivered to the load RL is To find the value of RL that absorbs the maximum power Thus, we have The maximum power transfer theorem states that a circuit delivers a maximum power to a load resistance RL when RL is equal to a Thévenin equivalent resistance of the circuit

26 The shown circuit represents a model for the common-emitter bipolar junction transistor amplifier. Choose a load resistance so that maximum power is transferred to it from the amplifier, and calculate the actual power absorbed 1- We first determine the Thévenin equivalent resistance. Thus, we remove RL and shortcircuit the independent source we note that, the dependent current source is an open circuit 26

27 2- To find vth we consider the shown circuit. We may write where the voltage vπ may be found from simple voltage division Then, 3- The maximum power is given by 27

28 Delta-Wye Conversion Sometimes we are not sure in electric circuits that the resistors are neither in parallel nor in series Delta connection Wye connection We can simplify this circuit using three terminal equivalent networks Transformation from Δ to Y connection Transformation from Y to Δ connection

29 Homework Assignments 4 P5.4, P5.5, P5.12, P5.17, P5.17, P5.21, P5.27, P5.31(a), P5.39, P5.42, P5.56, P5.60, P