Ch2 Basic Analysis Methods to Circuits Engineering Circuit Analysis Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits 2.2 Basic Nodal and Mesh Analysis 2.3 Useful Circuit Analysis Techniques References: Hayt-Ch3, 4; Gao-Ch2;

Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Key Words: Equivalent Circuits Network Equivalent Resistance, Equivalent Independent Sources

Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Equivalent Circuits Network о Two-terminal Circuits Network о a b c d 6 5 15 о b о N1 I a + _ V о N2 I a b + V

Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Equivalent Resistance How do we find I1 and I2? I1 + I2 = I R1 R2 V + - I1 I2 I

Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Equivalent Resistance i(t) + - v(t) i(t) + - v(t) Req Req is equivalent to the resistor network on the left in the sense that they have the same i-v characteristics.

Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Equivalent Resistance Condition : without knowing V&I . We only know Rs Series and parallel Resistance Method 1 (source-free) Condition : without knowing Rs . We only know V&I Method 2 I V R ab o =  a b source -free  source Voc Method 3 Isc

Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Equivalent Resistance In practice , Vs-source voltage ≠ VL- Local Let The IL-VL curve : 10 100 VL(V) IL(A) , Open Circuit (OC) . , Short Circuit (SC) .

Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Equivalent Resistance P2.1  VS R1 R2 R3 a b How do we find Rab? + - Method 1  V R1 R2 R3 a b I Method 2 Method 3

Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Source Transformation Ideally: An ideal current source has the voltage necessary to provide its rated current An ideal voltage source supplies the current necessary to provide its rated voltage Practice: A real voltage source cannot supply arbitrarily large amounts of current A real current source cannot have an arbitrarily large terminal voltage

Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Source Transformation Vs + - Rs Is Rs Note: Consistency between the current source ref. direction and the voltage Source ref. terminals.

Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Equivalent Source How do we find I1 and I2? Is2 V R1 R2 + - I1 I2 Is1 Ieq

Ch2 Basic Analysis Methods to Circuits 2.1 Equivalent Circuits Equivalent Source Series Voltage Source  + - VS1 VS2 VSn  + - VS parallel Current Source  IS1 IS2 ISn IS I RS=RS1// RS2//…// RSn

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Key Words: Branch Analysis, Nodal Analysis, Mesh (Loop) Analysis Why? The analysis techniques previously (voltage divider, equivalent resistance, etc.) provide an intuitive approach to analyzing circuits They are not systematic and cannot be easily automated by a computer Comments: Analysis of circuits using node or loop analysis requires solutions of systems of linear equations. These equations can usually be written by inspection of the circuit.

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Branch Analysis P2.2 How do we find I1 and I2, I3? KVL Mesh 1: I3 I1 I2 Mesh 2: KCL

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Branch Analysis Suppose m branches, n nodals write KCL equation for each independent node. ——(n-1) KCL equations write KVL equation for each independent mesh/loop ——m-(n-1) KVL equations

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Branch Analysis Here’s a quick example of a circuit that we will see later when we model the operation of transistors. For now, let’s assume ideal independent and dependent sources. Ⅰ We can write the following equations: Ⅰ: Ⅱ: Ⅲ : Ⅱ Ⅲ

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis 1) Choose a reference node + - V 500W 1kW I1 I2 The reference node is called the ground node.

. . . Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis 1) Choose a reference node + - V 500W 1kW I1 I2 . . . Ⅰ I4 I5 I6 I7 I8 Ⅱ Ⅲ KCL Ⅰ: Ⅱ: Ⅲ :

V1, V2, and V3 are unknowns for which we solve using KCL. Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis 2) Assign node voltages to the other nodes 500W 1kW I1 I2 1 2 3 V1 V2 V3 I4 I5 I6 I7 I8 V1, V2, and V3 are unknowns for which we solve using KCL.

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis 3) Apply KCL to each node other than the reference-express currents in terms of node voltages. 500W 1kW I1 I2 1 2 3 V1 V2 V3 I4 I5 I6 I7 I8 Node ① : Node ② : Node ③ : KCL Node ① : Node ② : Node ③ : ,

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis 4) Solve the resulting system of linear equations. Node 1: Node 2: Node 3: 500W 1kW I1 I2 1 2 3 V1 V2 V3 The left hand side of the equation: The node voltage is multiplied by the sum of conductances of all resistors connected to the node. The neighbourly node voltages are multiplied by the conductance of the resistor(s) connecting to the two nodes and to be subtracted. The right hand side of the equation: The right side of the equation is the sum of currents from sources entering the node.

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis 4) Solve the resulting system of linear equations. Node 1: Node 2: Node 3: 500W 1kW I1 I2 1 2 3 V1 V2 V3 Matrix Notation(Symmetric)

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis 4) Solve the resulting system of linear equations. Node 1: Node 2: Node 3: 500W 1kW I1 I2 1 2 3 V1 V2 V3 G11V1+G12V2 +G13V3 =I11 G21V2+G22V2 +G23V3 =I22 G31V1+G32V2+G33V3=I33

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis What if there are dependent sources? 1kW 5mA 100Ib + - Vo 50W Ib 2 1 V2 V1 Example: Node ①： Node ② ： Matrix is not symmetric due to the dependent source.

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis What if there are voltage sources? 0.7V R1 Ib V2 V3 R3 V4 + - 1 2 3 4 + 1kW 50W + Vo 100Ib R4 3kW V1 - R2 1kW - Difficulty: We do not know Ib – the current through the voltage source? Equations: KCL at node 2, node 3, node 4, and Unknowns: Ib, V1, V2 (V3),V4 Node 2: Node 3: Independent Voltage Source: Node 4:

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis What if there are voltage sources? CURRENT CONTROLLED VOLTAGE SOURCE Io=? KCL AT SUPERNODE

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Nodal Analysis Advantages of Nodal Analysis Solves directly for node voltages. Current sources are easy. Voltage sources are either very easy or somewhat difficult. Works best for circuits with few nodes. Works for any circuit.

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh(Loop) Analysis 1) Identifying the Meshes 1kW 1kW V1 + - + - Mesh 1 Mesh 2 V2 1kW Mesh: A special kind of loop that doesn’t contain any loops within it.

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh(Loop) Analysis 2) Assigning Mesh Currents 1kW + - V2 I1 I2 V1 3) Apply KVL around each loop to get an equation in terms of the loop currents. For Mesh 1: I1 ( 1kW + 1kW) - I2 1kW = V1 - I1 1kW + I2 ( 1kW + 1kW) = -V2 -V1 + I1 1kW + (I1 - I2) 1kW = 0 For Mesh 2: (I2 - I1) 1kW + I2 1kW + V2 = 0

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh(Loop) Analysis 3) Apply KVL around each loop to get an equation in terms of the loop currents. 1kW + - V2 I1 I2 V1 I1 ( 1kW + 1kW) - I2 1kW = V1 - I1 1kW + I2 ( 1kW + 1kW) = -V2 4) Solve the resulting system of linear equations.

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh (Loop) Analysis I1 I2 I3 I5 I6 I4 Im1 Mesh 1 Mesh 1: Mesh 2: Mesh 3: Im2 Mesh 2 Im3 Mesh 3

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh (Loop) Analysis P2.4(P2.2) Mesh 1: Im1 Im2 Mesh 2: -12

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh (Loop) Analysis What if there are current sources? The current sources in this circuit will have whatever voltage is necessary to make the current correct. We can’t use KVL around the loop because we don’t know the voltage. P2.6 Im1 Super Mesh Im2 Mesh 1 P2.5 I = ? + - V Super Mesh: Im1 Im2 Im3 Mesh 2 Mesh 2: Mesh 1:

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh (Loop) Analysis What if there are current sources? The Supermesh does not include this source! 2kW The Supermesh surrounds this source! 2mA I3 1kW + 2kW I1 12V 4mA I2 - I0

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh (Loop) Analysis What if there are current sources? 1kW 2kW 12V + - 4mA 2mA I0 I1 I2 I3 The 4mA current source sets I2: I2 = -4mA The 2mA current source sets a constraint on I1 and I3: I1 - I3 = 2mA We have two equations and three unknowns. Where is the third equation? -

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh (Loop) Analysis What if there are current sources? Mesh 2 Mesh 1 P2.6 Im1 Im2 Mesh 1: Mesh 2: Node 3: + - V Im3 Node 3

Ch2 Basic Analysis Methods to Circuits 2.2 Useful Circuit Analysis Techniques Mesh(Loop) Analysis Dependent current source. Current sources not shared by meshes. We treat the dependent source as a conventional source. Equations for meshes with current sources We are asked for Vo. We only need to solve for I3 . Replace and rearrange Then KVL on the remaining loop(s) And express the controlling variable, Vx, in terms of loop currents

Ch2 Basic Analysis Methods to Circuits 2.2 Basic Nodal and Mesh Analysis Mesh (Loop) Analysis Advantages of Loop Analysis Solves directly for some currents. Voltage sources are easy. Current sources are either very easy or somewhat difficult. Works best for circuits with few loops. Disadvantages of Loop Analysis Some currents must be computed from loop currents. Choosing the supermesh may be difficult.

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Key Words: Linearity Superposition Thevenin’s and Norton’s theorems

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Linearity Linearity is a mathematical property of circuits that makes very powerful analysis techniques possible. Linearity leads to many useful properties of circuits: Superposition: the effect of each source can be considered separately. Equivalent circuits: Any linear network can be represented by an equivalent source and resistance (Thevenin’s and Norton’s theorems)

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Linearity Linearity leads to simple solutions: Nodal analysis for linear circuits results in systems of linear equations that can be solved by matrices 500W 1kW I1 I2 1 2 3 V1 V2 V3

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Linearity The relationship between current and voltage for a linear element satisfies two properties: Homogeneity Additivity *Real circuit elements are not linear, but can be approximated as linear

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Linearity Homogeneity: Let v(t) be the voltage across an element with current i(t) flowing through it. In an element satisfying homogeneity, if the current is increased by a factor of K, the voltage increases by a factor of K. Additivity Let v1(t) be the voltage across an element with current i1(t) flowing through it, and let v2(t) be the voltage across an element with current i2 (t) flowing through it In an element satisfying additivity, if the current is the sum of i1 (t) and i2 (t), then the voltage is the sum of v1 (t) and v2 (t). Example: Resistor: V = R I If current is KI, then voltage is R KI = KV If current is I1 + I2, then voltage is R(I1 + I2) = RI1 + RI2 = V1 + V2

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Superposition I2 I Superposition is a direct consequence of linearity It states that “in any linear circuit containing multiple independent sources, the current or voltage at any point in the circuit may be calculated as the algebraic sum of the individual contributions of each source acting alone.”

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Superposition How to Apply Superposition? To find the contribution due to an individual independent source, zero out the other independent sources in the circuit. Voltage source  short circuit. Current source  open circuit. Solve the resulting circuit using your favorite techniques. Nodal analysis Loop analysis

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Superposition For the above case: Zero out Vs, we have : Zero out E2, we have : R1 R3 E2 R2 I2’ I + _ Vs R1 R3 E2 R2 I2’’

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Superposition 2kW 1kW 12V + - I0 2mA 4mA P2.7

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Superposition P2.7 2kW 1kW Io’ 2mA KVL for mesh 2: I1 I2 Mesh 2

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Superposition P2.7 2kW 1kW I’’0 4mA I1 KVL for mesh 2: I2 Mesh 2

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Superposition P2.7 2kW 1kW 12V + - I’’’0 KVL for mesh 2: I2 Mesh 2

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Superposition 2kW 1kW 12V + - I0 2mA 4mA 2kW 1kW 12V + - I0 2mA 4mA P2.7 I0 = I’0 +I’’0+ I’’’0 = -16/3 mA

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Thevenin’s theorem Any circuit with sources (dependent and/or independent) and resistors can be replaced by an equivalent circuit containing a single voltage source and a single resistor Thevenin’s theorem implies that we can replace arbitrarily complicated networks with simple networks for purposes of analysis

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Thevenin’s theorem Independent Sources RTh Voc + - Circuit with independent sources Thevenin equivalent circuit

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Thevenin’s theorem No Independent Sources RTh Circuit without independent sources Thevenin equivalent circuit

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Norton’s theorem Very similar to Thevenin’s theorem It simply states that any circuit with sources (dependent and/or independent) and resistors can be replaced by an equivalent circuit containing a single current source and a single resistor

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Norton’s theorem Norton Equivalent: Independent Sources Isc RTh Circuit with one or more independent sources Norton equivalent circuit

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Norton’s theorem Norton Equivalent: No Independent Sources RTh Circuit without independent sources Norton equivalent circuit

Ch2 Basic Analysis Methods to Circuits Motivation of applying the Thevenin’s theorem and Norton’s theorem: Sometimes, in a complex circuit, we are only interested in working out the voltage /current or power being consumed by a single load (resistor); We can then treat the rest of the circuit (excluding the interested load) as a voltage (current) source concatenated with a source resistor; Simplify our analysis. The Thevenin’s theorem: Given a linear circuit, rearrange it in the form of two networks of A and B connected by two wires. Define Voc as the open-circuit voltage which appears across the terminals of A when B is disconnected. Then all currents and voltage in B will remain unchanged, if we replace all the independent current or voltage source in A by an independent voltage source which is in series with a resistor (RTh). The Norton’s Theorem: Given a linear circuit, rearrange it in the form of two networks of A and B connected by two wires. Define isc as the short-circuit current which appears across the terminals of A when B is disconnected. Then all currents and voltage in B will remain unchanged, if we replace all the independent current or voltage source in A by an independent current source isc which is in parallel with a resistor (RN).

VL VL Ch2 Basic Analysis Methods to Circuits Thevenin equivalent Equivalent transform between a Thevenin equivalent circuit and a Norton equivalent circuit Thevenin equivalent Norton equivalent VL N If and VL VL =

Ch2 Basic Analysis Methods to Circuits Therefore, for a source transform: Thevenin Norton: Norton Thevenin :

Ch2 Basic Analysis Methods to Circuits Example 45 (P88) Find a Thevenin and Norton equivalent circuit for the following circuit excluding A B

Ch2 Basic Analysis Methods to Circuits Thevenin equivalent Norton equivalent

Ch2 Basic Analysis Methods to Circuits Application of Thevenin’s theorem when there are only independent sources. step : 1) Determine of two connection points between network A and network B. 2) Determine of two connection points by replacing the voltage source by a short- circuit or the current source by a open-circuit. Similarly, for Norton’s Theorem steps : 1) Determine between the two connection points between network A and B. 2) Determine by two connection points by replacing the voltage source by a short- circuit or the current source by an open circuit. Test the above case?

Ch2 Basic Analysis Methods to Circuits Practice 4.6 (P90)

Ch2 Basic Analysis Methods to Circuits Thevenin equivalent :

Ch2 Basic Analysis Methods to Circuits When there are multiple independent source, we shall use “superposition” . Example 4.7 (P91) Find the Thevenin and Norton equivalent for the network excluding the resistor. To determine , when only 4v voltage source is functioning.

Ch2 Basic Analysis Methods to Circuits When only 2mA current source is functioning : Therefore,

Ch2 Basic Analysis Methods to Circuits To determine Thevenin equivalent:

Ch2 Basic Analysis Methods to Circuits Norton equivalent Try to look into this problem from the Norton approach. (Figure out the Norton equivalent circuit first)

Ch2 Basic Analysis Methods to Circuits When there are both independent source and dependent source. —Dependent source cannot be “zero out” as far as its controlling variable is not zero. —Similar as before, —But we cannot determine directly, however, we can use Example 4.8 (P92) — Determine the Thevenin equivalent of the following circuit To determine since ,applying KVL to the supermesh:

Ch2 Basic Analysis Methods to Circuits To determine Therefore its Thevenin equivalent is:

Ch2 Basic Analysis Methods to Circuits When there are only dependent sources: VOC = 0 RTH can be determined by implying a test (imaginary) voltage across the two terminals. 3 Ω 2Ω i 1.5i Example 4.9 (p93) Open circuit : To determine RTh, imagine an independent current source xA. as : i = -xA, Apply KCL: Giving : 0.6Ω 3 Ω 2Ω i 1.5i v xA - - - - v = 0.6x V RTh = 0.6Ω

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Thevenin’s theorem Circuits with independent sources Compute the open circuit voltage, this is Voc Compute the Thevenin resistance (set the sources to zero – short circuit the voltage sources, open circuit the current sources), and find the equivalent resistance, this is RTh Circuits with independent and dependent sources: Compute the open circuit voltage Compute the short circuit current The ratio of the two is RTh Circuits with dependent sources only* Voc is simply 0 RTh is found by applying an independent voltage source (V volts) to the terminals and finding voltage/current ratio * Not required by this course.

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Norton’s theorem Circuits with independent sources, w/o dependent sources Compute the short circuit current, this is Isc Compute the Thevenin resistance (set the sources to zero – short circuit the voltage sources, open circuit the current sources), and find the equivalent resistance, this is RN Circuits with both independent and dependent sources Find Voc and Isc Compute RN= Voc/ Isc Circuits w/o independent sources* Apply a test voltage (current) source Find resulting current (voltage) Compute RN

Ch2 Basic Analysis Methods to Circuits 2.3 Useful Circuit Analysis Techniques Maximum power transfer For every choice of RL we have a different power. How do we find the maximum value? Consider PL as a function of RL and find the maximum of such function The maximum power transfer theorem The load that maximizes the power transfer for a circuit is equal to the Thevenin equivalent resistance of the circuit. The value of the maximum power that can be transferred is

Ch2 Basic Analysis Methods to Circuits Analysis methods Review KVL, KCL, I — V Combination rules Node method Mesh method Superposition Thévenin Norton Any circuits linear circuits