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Unformatted text preview: EIE209 Basic Electronics Basic circuit analysis Prof. C.K. Tse: Basic Circuit Analysis 1 Fundamental quantities Voltage  potential difference bet. 2 points "across" quantity analogous to `pressure' between two points Current  flow of charge through a material "through" quantity analogous to fluid flowing along a pipe Prof. C.K. Tse: Basic Circuit Analysis 2 Units of measurement
n n Voltage: volt (V) Current: ampere (A) NOT Volt, Ampere!! n Prof. C.K. Tse: Basic Circuit Analysis 3 Power and energy
Work done in moving a charge dq from A to B having a potential difference of V is W = V dq A dq
B Power is work done per unit time, i.e., Prof. C.K. Tse: Basic Circuit Analysis 4 Direction and polarity
n n Current direction indicates the direction of flow of positive charge Voltage polarity indicates the relative potential between 2 points: + assigned to a higher potential point; and assigned to a lower potential point. n NOTE: Direction and polarity are arbitrarily assigned on circuit diagrams. Actual direction and polarity will be governed by the sign of the value. Prof. C.K. Tse: Basic Circuit Analysis 5 Independent sources
n n Voltage sources Current sources Independent  stubborn! never change! Maintains a voltage/current (fixed or varying) which is not affected by any other quantities. An independent voltage source can never be shorted. An independent current source can never be opened.
Prof. C.K. Tse: Basic Circuit Analysis 6 Dependent sources
n Dependent sources  values depend on some other variables Prof. C.K. Tse: Basic Circuit Analysis 7 Circuit
n Collection of devices such as sources and resistors in which terminals are connected together by conducting wires.
n n These wires converge in NODES The devices are called BRANCHES of the circuit Circuit Analysis Problem: To find all currents and voltages in the branches of the circuit when the intensities of the sources are known. Prof. C.K. Tse: Basic Circuit Analysis 8 Kirchhoff's laws
n Kirchhoff's current law (KCL)
n The algebraic sum of the currents in all branches which converge to a common node is equal to zero. n Kirchhoff's voltage law (KVL)
n The algebraic sum of all voltages between successive nodes in a closed path in the circuit is equal to zero.
Prof. C.K. Tse: Basic Circuit Analysis 9 Overview of analysis
n Ad hoc methods (not general)
n n n n Series/parallel reduction Ladder circuit Voltage/current division Stardelta conversion n More general
n } Done in Basic Electronics! Mesh and nodal methods n Completely general
n Loop and cutset approach (requires graph theory) NEW Prof. C.K. Tse: Basic Circuit Analysis 10 Series/parallel reduction
n Series circuit each node is incident to just two branches of the circuit KVL gives =
Hence, the equivalent resistance is: Prof. C.K. Tse: Basic Circuit Analysis 11 Series/parallel reduction
n Parallel circuit one terminal of each element is connected to a node of the circuit while other terminals of the elements are connected to another node of the circuit KCL gives Hence, the equivalent resistance is: Prof. C.K. Tse: Basic Circuit Analysis 12 Note on algebra
n For algebraic brevity and simplicity:
n n For series circuits, R is preferably used. For parallel circuits, G is preferably used. For example, if we use R for the parallel circuit, we get the equivalent resistance as which is more complex than the formula in terms of G: G = G1 + G2 + ... + Gn
Prof. C.K. Tse: Basic Circuit Analysis 13 Ladder circuit
n We can find the resistance looking into the terminals 0 and 1, by apply the series/ parallel reduction successively. First, lumping everything beyond node 2 as G2, we have Then, we focus on this G2, which is just G20 in parallel with another subcircuit, i.e., We continue to focus on the remaining subcircuit. Eventually we get
Prof. C.K. Tse: Basic Circuit Analysis 14 Voltage/current division
For the series circuit, we can find the voltage across each resistor by the formula: For the parallel circuit, we can find the voltage across each resistor by the formula: Note the choice of R and G in the formulae!
Prof. C.K. Tse: Basic Circuit Analysis 15 Example (something that can be done with series/parallel reduction)
Consider this circuit, which is created deliberately so that you can solve it using series/parallel reduction technique. Find V2. Solution: Resistance seen by the voltage source is Hence, Current division gives: Then, using V2=I4R4, we get Prof. C.K. Tse: Basic Circuit Analysis 16 Oops!
Series/parallel reduction fails for this bridge circuit! Is there some ad hoc solution? Prof. C.K. Tse: Basic Circuit Analysis 17 Equivalence of star and delta Y (star) D (delta) Problems: 1. Given a star circuit, find the delta equivalence. That means, suppose you have all the G's in the star. Find the G's in the delta such that the two circuits are "equivalent" from the external viewpoint. 2. The reverse problem.
Prof. C.K. Tse: Basic Circuit Analysis 18 Startodelta conversion Y (star) D (delta) For the Y circuit, we consider summing up all currents into the centre node: I1+I2+I3=0, where Thus, , and Prof. C.K. Tse: Basic Circuit Analysis 19 Startodelta conversion Y (star) D (delta) For the D circuit, we have Prof. C.K. Tse: Basic Circuit Analysis 20 Startodelta conversion
Now, equating the two sets of I1, I2 and I3, we get The first problem is solved. Prof. C.K. Tse: Basic Circuit Analysis 21 Deltatostar conversion This problem is more conveniently handled in terms of R. The answer is: Prof. C.K. Tse: Basic Circuit Analysis 22 Example  the bridge circuit again
We know that the series/parallel reduction method is not useful for this circuit! The stardelta transformation may solve this problem. The question is how to apply the transformation so that the circuit can become solvable using the series/parallel reduction or other ac hoc methods. Prof. C.K. Tse: Basic Circuit Analysis 23 Example  the bridge circuit again After we do the conversion from Y to D, we can easily solve the circuit with parallel/series reduction. Prof. C.K. Tse: Basic Circuit Analysis 24 Useful/important theorems Thvenin Theorem Norton Theorem Maximum Power Transfer Theorem Prof. C.K. Tse: Basic Circuit Analysis 25 Thvenin and Norton theorems
Circuit in question External apparatus (another circuit) Problem: Find the simplest equivalent circuit model for N, such that the external circuit N* would not feel any difference if N is replaced by that equivalent model. The solution is contained in two theorems due to Thvenin and Norton.
Prof. C.K. Tse: Basic Circuit Analysis 26 Thvenin and Norton theorems
Let's look at the logic behind these theorems (quite simple really). If we write down KVL, KCL, and Ohm's law equations correctly, we will have a number of equations with the same number of unknowns. Then, we can try to solve them to get what we want. Now suppose everything is linear. We are sure that we can get the following equation after elimination/substitution (some high school algebra): Case 1: a0 Case 2: b0
Prof. C.K. Tse: Basic Circuit Analysis Thvenin Norton
27 Equivalent models
Thvenin equiv. ckt Voltage source in series with a resistor i.e., V + IRT = VT which is consistent with case 1 equation Norton equiv. ckt Current source in parallel with a resistor i.e., I = IN + V/RN which is consistent with case 2 equation
Prof. C.K. Tse: Basic Circuit Analysis 28 How to find VT and IN
Thvenin equiv. ckt Opencircuit the terminals (I=0), we get VT as the observed value of V. Easy! VT is just the opencircuit voltage! Norton equiv. ckt Shortcircuit the terminals (V=0), we get IN as the observed current I. Easy! IN is just the shortcircuit current!
29 = VT I = IN Prof. C.K. Tse: Basic Circuit Analysis How to find RT and RN (they are equal)
I = Isc
Thvenin equiv. ckt Shortcircuit the terminals (V=0), find I which is equal to VT/RT. Thus, RT = VT / Isc Norton equiv. ckt Opencircuit the terminals (I=0), find V which is equal to INRN. Thus, RN = Voc / IN. = Voc For both cases, RT = RN = Voc / Isc
Prof. C.K. Tse: Basic Circuit Analysis 30 Simple example
Step 1: opencircuit The o/c terminal voltage is Step 2: shortcircuit The s/c current is Step 3: Thvenin or Norton resistance Hence, the equiv. ckts are: Prof. C.K. Tse: Basic Circuit Analysis 31 Example  the bridge again
Problem: Find the current flowing in R5.
One solution is by deltastar conversion (as done before). Another simpler method is to find the Thvenin equivalent circuit seen from R5. Prof. C.K. Tse: Basic Circuit Analysis 32 Example  the bridge again
Step 1: open circuit The o/c voltage across A and B is = VT
Step 2: short circuit The s/c current is Step 3: RT Prof. C.K. Tse: Basic Circuit Analysis 33 Example  the bridge again = Current in R5 = VT R5 + RT Prof. C.K. Tse: Basic Circuit Analysis 34 Maximum power transfer theorem
We consider the power dissipated by RL. The current in RL is Thus, the power is This power has a maximum, when plotted against RL. = 0 gives R L = R T. Prof. C.K. Tse: Basic Circuit Analysis 35 A misleading interpretation
It seems counterintuitive that the MPT theorem suggests a maximum power at RL = RT. Shouldn't maximum power occur when we have all power go to the load? That is, when RT = 0! Is the MPT theorem wrong? Discussion: what is the condition required by the theorem? Prof. C.K. Tse: Basic Circuit Analysis 36 Systematic analysis techniques
So far, we have solved circuits on an ad hoc manner. We are able to treat circuits with parallel/series reduction, stardelta conversion, with the help of some theorems. How about very general arbitrary circuit styles? In Basic Electronics, you have learnt the use of MESH and NODAL methods. MESH  planar circuits only; solution in terms of mesh currents. NODAL  any circuit; solution in terms of nodal voltages. BUT THEY ARE NOT EFFICIENT!
Prof. C.K. Tse: Basic Circuit Analysis 37 Mesh analysis (for planar circuits only)
Planar or not? Meshes  windows Prof. C.K. Tse: Basic Circuit Analysis 38 Mesh analysis
Step 1: Define meshes and unknowns Each window is a mesh. Here, we have two meshes. For each one, we "imagine" a current circulating around it. So, we have two such currents, I1 and I2  unknowns to be found. Step 2: Set up KVL equations Step 3: Simplify and solve Once we know the mesh currents, we can find anything in the circuit! e.g., current flowing down the 3 resistor in the middle is equal to I1 I2 ; current flowing up the 42V source is I1 ; current flowing down the 10V source is I2 ; and voltages can be found via Ohm's law.
39 which gives I1 = 6 A and I2 = 4 A. Prof. C.K. Tse: Basic Circuit Analysis Mesh analysis
In general, we formulate the solution in terms of unknown mesh currents: [ R ] [ I ] = [ V ]  mesh equation where [ R ] is the resistance matrix [ I ] is the unknown mesh current vector [ V ] is the source vector For a short cut in setting up the above matrix equation, see Sec. 3.2.1.2 of the textbook. This may be picked up in the tutorial. Prof. C.K. Tse: Basic Circuit Analysis 40 Mesh analysis  observing superposition
Consider the previous example. The mesh equation is given by: or Thus, the solution can be written as Remember what 42 and 10 are? They are the sources! The above solution can also be written as or SUPERPOSITION of two sources
Prof. C.K. Tse: Basic Circuit Analysis 41 Problem with current sources
The mesh method may run into trouble if the circuit has current source(s). Suppose we define the unknowns in the same way, i.e., I1, I2 and I3 . The trouble is that we don't know what voltage is dropped across the 14A source! How can we set up the KVL equation for meshes 1 and 3? One solution is to ignore meshes 1 and 3. Instead we look at the supermesh containing 1 and 3. So, we set up KVL equations for mesh 2 and the supermesh: Mesh 2: Supermesh: One more equation: I1 I3 = 14
Prof. C.K. Tse: Basic Circuit Analysis 42 Finally, solve the equations. Complexity of mesh method
In all cases, we see that the mesh method ends up with N equations and N unknowns, where N is the number of meshes (windows) of the circuit. One important point: The mesh method is overcomplex when applied to circuits with current source(s). WHY? We don't need N equations for circuits with current source(s) because the currents are partly known! In the previous example, it seems unnecessary to solve for both I1 and I3 because their difference is known to be 14! This is a waste of efforts! Can we improve it?
Prof. C.K. Tse: Basic Circuit Analysis 43 Nodal analysis
Step 1: Define unknowns Each node is assigned a number. Choose a reference node which has zero potential. Then, each node has a voltage w.r.t. the reference node. Here, we have V1 and V2  unknowns to be found. Step 2: Set up KCL equation for each node
Node 1: Node 2: Step 3: Simplify and solve Once we know the nodal voltages, we can find anything in the circuit! e.g., voltage across the 5 resistor in the middle is equal to V1 V2 ; voltage across the 3A source is V1 ; voltage across the 2A source is V2 ; and currents can be found via Ohm's law. which gives V1 = 5 V and V2 = 2.5 V. Prof. C.K. Tse: Basic Circuit Analysis 44 Nodal analysis
In general, we formulate the solution in terms of unknown nodal voltages: [ G ] [ V ] = [ I ]  nodal equation where [ G ] is the conductance matrix [ V ] is the unknown node voltage vector [ I ] is the source vector For a short cut in setting up the above matrix equation, see Sec. 3.3.1.2 of the textbook. This may be picked up in the tutorial. Prof. C.K. Tse: Basic Circuit Analysis 45 Nodal analysis  observing superposition
Consider the previous example. The nodal equation is given by: Thus, the solution can be written as Remember what 3 and 2 are? They are the sources! The above solution can also be written as or SUPERPOSITION of two sources
Prof. C.K. Tse: Basic Circuit Analysis 46 Problem with voltage sources
The nodal method may run into trouble if the circuit has voltage source(s). Suppose we define the unknowns in the same way, i.e., V1, V2 and V3 . The trouble is that we don't know what current is flowing through the 2V source! How can we set up the KCL equation for nodes 2 and 3? One solution is to ignore nodes 1 and 3. Instead we look at the supernode merging 2 and 3. So, we set up KCL equations for node 1 and the supernode: One more equation: V3 V 2 = 2
Prof. C.K. Tse: Basic Circuit Analysis 47 Finally, solve the equations. Complexity of nodal method
In all cases, we see that the mesh method ends up with N equations and N unknowns, where N is the number of nodes of the circuit minus 1. One important point: The nodal method is overcomplex when applied to circuits with voltage source(s). WHY? We don't need N equations for circuits with voltage source(s) because the node voltages are partly known! In the previous example, it seems unnecessary to solve for both V2 and V3 because their difference is known to be 2! This is a waste of efforts! Can we improve it?
Prof. C.K. Tse: Basic Circuit Analysis 48 Final note on superposition
Superposition is a consequence of linearity. We may conclude that for any linear circuit, any voltage or current can be written as linear combination of the sources. Suppose we have a circuit which contains two voltage sources V1, V2 and I3. And, suppose we wish to find Ix. Without doing anything, we know for sure that the following is correct: Ix = a V1 + b V 2 + c I3
where a, b and c are some constants. Is this property useful? Can we use this property for analysis? We may pick this up in the tutorial. V2 V1 Ix I3 Prof. C.K. Tse: Basic Circuit Analysis 49 ...
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This note was uploaded on 05/04/2010 for the course BEX 404 taught by Professor Arunpaudyal during the Spring '10 term at Institute of Engineering.
 Spring '10
 ArunPaudyal

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