Survival Skills for Circuit Analysis

P. R. Nelson Fall 2010 WhatToKnow - p. 1/46 Survival Skills for Circuit Analysis What you need to know from ECE 109 Phyllis R. Nelson prnelson@csupomona.edu Professor, Department of Electrical and Computer Engineering California State Polytechnic University, Pomona

P. R. Nelson Fall 2010 WhatToKnow - p. 2/46 Basic Circuit Concepts All circuits can be analyzed by many methods.

P. R. Nelson Fall 2010 WhatToKnow - p. 2/46 Basic Circuit Concepts All circuits can be analyzed by many methods. Some methods have specific advantages for calculating a particular result in a specific circuit. Some methods yield useful intuition about circuit behavior.

P. R. Nelson Fall 2010 WhatToKnow - p. 3/46 Kirchhoff s Laws These two methods are the most general tools of circuit analysis. Sign errors in writing Kirchhoff s laws are the most common error I observe in student work in higher-level classes. Sign errors are not trivial errors!

P. R. Nelson Fall 2010 WhatToKnow - p. 4/46 Kirchhoff s current law is equivalent to stating that electrical charge cannot be stored, created, or destroyed locally in a conductor. The algebraic sum of all currents entering a node is zero. Alternate statements: The algebraic sum of the currents leaving a node is zero. The total current entering a node is equal to the total current leaving the node.

P. R. Nelson Fall 2010 WhatToKnow - p. 5/46 Kirchhoff s voltage law is derived by considering the forces on an electrical charge that travels around a closed loop in a circuit and has the same starting and ending velocity. The total force on the charge must be zero, so the total change in electrical potential energy (voltage) around the loop must be zero. The algebraic sum of the voltage drops around any closed loop in a circuit is zero.

P. R. Nelson Fall 2010 WhatToKnow - p. 6/46 Solving K*L equations Kirchhoff s laws applied directly result in I i = 0 (KCL) i i V i = 0 (KVL)

P. R. Nelson Fall 2010 WhatToKnow - p. 7/46 Solving K*L... Solving KCL means finding the node voltages. Solving KVL means finding the branch currents. need relationships between current and voltage for each circuit element. These relationships describe the operation of the circuit elements.

P. R. Nelson Fall 2010 WhatToKnow - p. 8/46 Ohm s law + V I V = IR Ohm s law is an example of a current-voltage (I-V ) relation. To use K*L, you must know the I-V relation for every circuit element. I-V relations include definition of the relationship between the direction of the current and the sign of the voltage.

P. R. Nelson Fall 2010 WhatToKnow - p. 9/46 Series resistors, voltage divider I + V T + V1 R 2 + V2 R i + Vi R n + Vn V T = I R eq Prove that these equations are true: R eq = n j=1 R j V i = R i n j=1 R j V T

P. R. Nelson Fall 2010 WhatToKnow - p. 10/46 n n n V T = V j = I R j = I R j = I R eq j=1 j=1 j=1 n R eq = R j j=1 ( VT ) ( ) V i = I R i = R eq R i = R i n j=1 R j V T

P. R. Nelson Fall 2010 WhatToKnow - p. 11/46 Parallel resistors, current divider I + V I 1 R 2 I 2 V = I R eq Prove that these equations are true: R 1 eq = 2 i=1 R 1 i R eq = R 2 + R 2 I x = ( Ry R x + R y Which can be extended to more resistors? ) I

P. R. Nelson Fall 2010 WhatToKnow - p. 12/46 ( 2 ) I = I 1 + I 2 = V + V R 2 = V i=1 R 1 i R 1 eq = = V R eq 2 i=1 R 1 i R eq = 1 + 1 R 2 = R 2 + R 2

P. R. Nelson Fall 2010 WhatToKnow - p. 13/46 Let x = 1 and y = 2. Then ( I 1 = V = I R eq = ) R 2 +R 2 I = ( R2 + R 2 ) I If x = 2 and y = 1 then I 2 = ( R1 + R 2 ) I

P. R. Nelson Fall 2010 WhatToKnow - p. 14/46 For n resistors in parallel, ( n n n V I = I j = = V R j j=1 j=1 j=1 R 1 j ) R 1 eq = = V R eq n j=1 R 1 j This is the only formula for parallel resistors that extends easily to more than two resistors.

P. R. Nelson Fall 2010 WhatToKnow - p. 15/46 Illegal circuits The following configurations are not allowed: a short-circuited voltage source an open-circuited current source Why?

P. R. Nelson Fall 2010 WhatToKnow - p. 16/46 KCL Example V x R 2 R 4 I y Find V R2 and V R4.

P. R. Nelson Fall 2010 WhatToKnow - p. 16/46 KCL Example V x R 2 R 4 I y Find V R2 and V R4. Hint: Since KCL is applied at nodes, how many KCL equations will be required?

P. R. Nelson Fall 2010 WhatToKnow - p. 17/46 KCL Example - 1 V x R 2 R 4 I y Find V R2 and V R4.

P. R. Nelson Fall 2010 WhatToKnow - p. 17/46 KCL Example - 1 V x R 2 R 4 I y Find V R2 and V R4. 1. Since KCL is applied at nodes, count the nodes.

P. R. Nelson Fall 2010 WhatToKnow - p. 17/46 KCL Example - 1 V x R 2 R 4 I y Find V R2 and V R4. 1. Since KCL is applied at nodes, count the nodes. (4)

P. R. Nelson Fall 2010 WhatToKnow - p. 18/46 KCL Example - 2, 3 V x R 2 R 4 I y Find V R2 and V R4.

P. R. Nelson Fall 2010 WhatToKnow - p. 18/46 KCL Example - 2, 3 V x R 2 R 4 I y Find V R2 and V R4. 2. Choose one node as ground. Choose for convenience! 3. Label node voltages.

P. R. Nelson Fall 2010 WhatToKnow - p. 19/46 V 2 V 4 V x R 2 R 4 I y

P. R. Nelson Fall 2010 WhatToKnow - p. 19/46 V 2 V 4 V x R 2 R 4 I y Both V R2 and V R4 are connected to ground. The voltage source is connected to ground one less KCL equation.

P. R. Nelson Fall 2010 WhatToKnow - p. 20/46 KCL Example - 4 V 2 V 4 V x R 2 R 4 I y 2 unknown node voltages 2 KCL equations. 4. Define currents for all elements connected to nodes 2 and 4.

P. R. Nelson Fall 2010 WhatToKnow - p. 21/46 KCL Example - Eqn s V 2 V 4 V x I 1R2 I 3R4 I 2 I 4 I y KCL at 2:

P. R. Nelson Fall 2010 WhatToKnow - p. 21/46 KCL Example - Eqn s V 2 V 4 V x I 1R2 I 3R4 I 2 I 4 I y KCL at 2: I 1 I 2 I 3 = 0

P. R. Nelson Fall 2010 WhatToKnow - p. 21/46 KCL Example - Eqn s V 2 V 4 V x I 1R2 I 3R4 I 2 I 4 I y KCL at 2: KCL at 4: I 1 I 2 I 3 = 0

P. R. Nelson Fall 2010 WhatToKnow - p. 21/46 KCL Example - Eqn s V 2 V 4 V x I 1R2 I 3R4 I 2 I 4 I y KCL at 2: KCL at 4: I 1 I 2 I 3 = 0 I 3 I 4 I y = 0

P. R. Nelson Fall 2010 WhatToKnow - p. 22/46 KCL Example - I-V s V 2 V 4 V x I 1R2 I 3R4 I 2 I 4 I y Write unknown I s in terms of node V s:

P. R. Nelson Fall 2010 WhatToKnow - p. 22/46 KCL Example - I-V s V 2 V 4 V x I 1R2 I 3R4 I 2 I 4 I y Write unknown I s in terms of node V s: I 1 = V x V 2 I 2 = V 2 R 2 I 3 = V 2 V 4 I 4 = V 4 R 4

P. R. Nelson Fall 2010 WhatToKnow - p. 23/46 KCL Example - Substitution V x V 2 V 2 R 2 V 2 V 4 = 0 V 2 V 4 V 4 R 4 I y = 0 Check! 2 equations in 2 unknowns This pair of equations can be solved, but it s messy unless resistor values are given.

P. R. Nelson Fall 2010 WhatToKnow - p. 24/46 KVL Example I 3 V A R 2 I B R 4 Find I 3.

P. R. Nelson Fall 2010 WhatToKnow - p. 24/46 KVL Example I 3 V A R 2 I B R 4 Find I 3. Hint: Since KVL is applied around closed loops, how many KVL equations will be required?

P. R. Nelson Fall 2010 WhatToKnow - p. 25/46 KVL Example - 1 I 3 V A R 2 I B R 4 Find I 3.

P. R. Nelson Fall 2010 WhatToKnow - p. 25/46 KVL Example - 1 I 3 V A R 2 I B R 4 Find I 3. 1. Count the number of windowpanes in the circuit.

P. R. Nelson Fall 2010 WhatToKnow - p. 25/46 KVL Example - 1 I 3 V A R 2 I B R 4 Find I 3. 1. Count the number of windowpanes in the circuit. (3)

P. R. Nelson Fall 2010 WhatToKnow - p. 26/46 KVL Example - 2, 3 I 3 V A R 2 I B R 4 Find I 3.

P. R. Nelson Fall 2010 WhatToKnow - p. 26/46 KVL Example - 2, 3 I 3 V A R 2 I B R 4 Find I 3. 2. Choose one closed loop for each windowpane. Choose for convenience! 3. Choose a direction for each loop.

P. R. Nelson Fall 2010 WhatToKnow - p. 27/46 KVL Example I 3 V A R 2 I B R 4 Only one loop passes through, ensuring that the unknown can be used as a loop current.

P. R. Nelson Fall 2010 WhatToKnow - p. 28/46 KVL Example - 4, 5 I 3 V A R 2 I B R 4 4. Define loop current labels for each loop. Be careful not to give different currents the same label! 5. Define voltages for all branch elements.

P. R. Nelson Fall 2010 WhatToKnow - p. 29/46 KVL Example - I & V V A R 2 I B R 4 I 1 I 3 I 4 V A + V 1 V 3 R 2 V 2 + I B V 4 R 4

P. R. Nelson Fall 2010 WhatToKnow - p. 30/46 KVL Example - Eqn s KVL for loop 1:

P. R. Nelson Fall 2010 WhatToKnow - p. 30/46 KVL Example - Eqn s KVL for loop 1: V A + V 1 + V 2 = 0

P. R. Nelson Fall 2010 WhatToKnow - p. 30/46 KVL Example - Eqn s KVL for loop 1: KVL for loop 3: V A + V 1 + V 2 = 0

P. R. Nelson Fall 2010 WhatToKnow - p. 30/46 KVL Example - Eqn s KVL for loop 1: V A + V 1 + V 2 = 0 KVL for loop 3: V 2 + V 3 + V 4 = 0

P. R. Nelson Fall 2010 WhatToKnow - p. 30/46 KVL Example - Eqn s KVL for loop 1: V A + V 1 + V 2 = 0 KVL for loop 3: V 2 + V 3 + V 4 = 0 KVL for loop 4:??!?

P. R. Nelson Fall 2010 WhatToKnow - p. 30/46 KVL Example - Eqn s KVL for loop 1: KVL for loop 3: V A + V 1 + V 2 = 0 V 2 + V 3 + V 4 = 0 KVL for loop 4:??!? V 4 + V 4 = 0 is a tautology...

P. R. Nelson Fall 2010 WhatToKnow - p. 30/46 KVL Example - Eqn s KVL for loop 1: KVL for loop 3: V A + V 1 + V 2 = 0 V 2 + V 3 + V 4 = 0 KVL for loop 4:??!? V 4 + V 4 = 0 is a tautology...... but the current source current is I B = I 4 I 3

P. R. Nelson Fall 2010 WhatToKnow - p. 31/46 KVL Example - I-V s V A + V 1 V 3 R 2 V 2 I B V 4 R 4 I 1 I 3 I 4 Write unknown V s in terms of loop I s:

P. R. Nelson Fall 2010 WhatToKnow - p. 31/46 KVL Example - I-V s V A + V 1 V 3 R 2 V 2 I B V 4 R 4 I 1 I 3 I 4 Write unknown V s in terms of loop I s: V 1 = I 1 V 2 = (I 1 I 3 ) R 2 V 3 = I 3 V 4 = I 4 R 4

P. R. Nelson Fall 2010 WhatToKnow - p. 32/46 KVL Example - Substitution V A + I 1 + (I 1 I 3 ) R 2 = 0 (I 1 I 3 ) R 2 + I 3 + I 4 R 4 = 0 I 4 I 3 = I B Check! 3 equations in 3 unknowns Collect terms, then solve this set of equations by Kramer s rule.

P. R. Nelson Fall 2010 WhatToKnow - p. 33/46 KVL Example - Matrix + R 2 R 2 0 R 2 R 2 + R 4 0 1 1 I 1 I 3 I 4 = V A 0 I B = + R 2 R 2 0 R 2 R 2 + R 4 0 1 1 = ( + R 2 ) (R 2 + ) + ( + R 2 )R 4 R 2 2 = R 2 + + R 4 + R 2 + R 2 R 4

P. R. Nelson Fall 2010 WhatToKnow - p. 34/46 I 3 = + R 2 V A 0 R 2 0 R 4 0 I B 1 = R 2 V A ( + R 2 )R 4 V A I 3 = R 2 V A ( + R 2 )R 4 I B R 2 + + R 4 + R 2 + R 2 R 4

Equivalent circuits Equivalent circuits enable us to treat a portion of a circuit like a black box in that we can give it the simplest possible model. There are two possible models that account for both power sources and impedance: a voltage source in series with a resistor a current source in parallel with a resistor Equivalent circuits can be used to simplify a circuit before solving for an unknown! P. R. Nelson Fall 2010 WhatToKnow - p. 35/46

P. R. Nelson Fall 2010 WhatToKnow - p. 36/46 Thèvenin equivalent R eq V oc I + V V = V oc R eq I I = V oc R eq V R eq If the terminals are connected to an open circuit, I = 0 and V = V oc. If the terminals are shorted, V = 0 and I = V oc /R eq = I sc.

P. R. Nelson Fall 2010 WhatToKnow - p. 37/46 Norton equivalent I + I sc R eq V I = I sc V R eq V = R eq I sc R eq I If the terminals are shorted, V = 0 and I = I sc. If the terminals are connected to an open circuit, I = 0 and V = R eq I sc.

P. R. Nelson Fall 2010 WhatToKnow - p. 38/46 Graphical representation All possible combinations of pairs of voltage and current values for either equivalent circuit model can be represented as the line through the points (V = 0, I SC ) and (V OC, 0). I SC I V OC V

P. R. Nelson Fall 2010 WhatToKnow - p. 39/46 Superposition A voltage or current in a linear circuit is the superposition (sum) of the results for each source alone. Superposition can be used along with parallel and series resistors, voltage and current dividers, and Thèvenin and Norton equivalent circuits to simplify the analysis of a circuit. Superposition often gives solutions in a mathematical form that enhances intuition.

P. R. Nelson Fall 2010 WhatToKnow - p. 40/46 Superposition Example V x R 2 I 2 I y Find the current I 2 using superposition. Solve for the current I 2 with I y = 0 so that V x is the only source V x = 0 so that I y is the only source then add the results.

P. R. Nelson Fall 2010 WhatToKnow - p. 41/46 Superposition Example - V Set I y = 0 and solve for the current I 2V. V x R 2 I 2V

P. R. Nelson Fall 2010 WhatToKnow - p. 41/46 Superposition Example - V Set I y = 0 and solve for the current I 2V. V x R 2 I 2V I 2V = ( R 2 + + R 2 ) V x

P. R. Nelson Fall 2010 WhatToKnow - p. 42/46 Superposition Example - I, Result Set V x = 0 and solve for the current I 2I. R 2 I 2I I y

P. R. Nelson Fall 2010 WhatToKnow - p. 42/46 Superposition Example - I, Result Set V x = 0 and solve for the current I 2I. R 2 I 2I I y I 2I = ( ) R 2 + + R 2 I y

P. R. Nelson Fall 2010 WhatToKnow - p. 42/46 Superposition Example - I, Result Set V x = 0 and solve for the current I 2I. R 2 I 2I I y I 2I = ( ) R 2 + + R 2 I y I 2 = ( R 2 + + R 2 ) (V x I y )

P. R. Nelson Fall 2010 WhatToKnow - p. 43/46 Dependent Sources Circuits containing dependent sources are solved by the same methods used for other circuits. There is one extra step, because the final result cannot contain the dependent variable! Not eliminating the dependent variable from the final result is another of the most common errors I see in student work!

P. R. Nelson Fall 2010 WhatToKnow - p. 44/46 Dependent Source Example V x R 2 + V 2 BV 2 + + V 3 Find V 3.

P. R. Nelson Fall 2010 WhatToKnow - p. 44/46 Dependent Source Example V x R 2 + V 2 BV 2 + + V 3 Find V 3. Hint: The answer may only contain V x,, R 2,, and B.

P. R. Nelson Fall 2010 WhatToKnow - p. 45/46 Dependent Source Example - Eqn s V x R 2 + V 2 BV 2 + + V 3 Find V 3.

P. R. Nelson Fall 2010 WhatToKnow - p. 45/46 Dependent Source Example - Eqn s V x R 2 + V 2 BV 2 + + V 3 Find V 3. V 2 V x + V 2 R 2 + V 3 = 0

P. R. Nelson Fall 2010 WhatToKnow - p. 45/46 Dependent Source Example - Eqn s V x R 2 + V 2 BV 2 + + V 3 Find V 3. V 2 V x + V 2 R 2 + V 3 = 0 V 2 + B V 2 + V 3 = 0

P. R. Nelson Fall 2010 WhatToKnow - p. 46/46 Dependent Source Example - Result V x R 2 + V 2 BV 2 + + V 3 Find V 3. V 3 = [ V x ] +R 2 (1 B) R 2 + 1