Lecture 281 Loop Analysis

Lecture 282 Loop Analysis Nodal analysis was developed by applying KCL at each non-reference node. Loop analysis is developed by applying KVL around loops in the circuit. Loop analysis results in a system of linear equations which must be solved for unknown currents.

Lecture 283 Example: A Summing Circuit The output voltage V of this circuit is proportional to the sum of the two input voltages V 1 and V 2. This circuit could be useful in audio applications or in instrumentation. The output of this circuit would probably be connected to an amplifier.

Lecture 284 Summing Circuit V out = (V 1 + V 2 )/3 + - V out 1k  V1V1 + - V2V2 + -

Lecture 285 Steps of Mesh Analysis 1.Identify mesh (loops). 2.Assign a current to each mesh. 3.Apply KVL around each loop to get an equation in terms of the loop currents. 4.Solve the resulting system of linear equations.

Lecture 286 Mesh 2 1k  Identifying the Mesh’s V1V1 + - V2V2 + - Mesh 1

Lecture 287 Steps of Mesh Analysis 1.Identify mesh (loops). 2.Assign a current to each mesh. 3.Apply KVL around each loop to get an equation in terms of the loop currents. 4.Solve the resulting system of linear equations.

Lecture 288 1k  Assigning Mesh Currents V1V1 + - V2V2 + - I1I1 I2I2

Lecture 289 Steps of Mesh Analysis 1.Identify mesh (loops). 2.Assign a current to each mesh. 3.Apply KVL around each loop to get an equation in terms of the loop currents. 4.Solve the resulting system of linear equations.

Lecture 2810 Voltages from Mesh Currents R I1I1 +- VRVR V R = I 1 R R I1I1 +- VRVR I2I2 V R = (I 1 - I 2 ) R

Lecture k  V1V1 + - V2V2 + - I1I1 I2I2 KVL Around Mesh 1 -V 1 + I 1 1k  + (I 1 - I 2 ) 1k  = 0 I 1 1k  + (I 1 - I 2 ) 1k  = V 1

Lecture k  V1V1 + - V2V2 + - I1I1 I2I2 KVL Around Mesh 2 (I 2 - I 1 ) 1k  + I 2 1k  + V 2 = 0 (I 2 - I 1 ) 1k  + I 2 1k  = -V 2

Lecture 2813 Steps of Mesh Analysis 1.Identify mesh (loops). 2.Assign a current to each mesh. 3.Apply KVL around each loop to get an equation in terms of the loop currents. 4.Solve the resulting system of linear equations.

Lecture 2814 Matrix Notation The two equations can be combined into a single matrix/vector equation.

Lecture 2815 Solving the Equations V 1 = 7V V 2 = 4V

Lecture 2816 Using MATLAB >> A = [1e3+1e3 -1e3; -1e3 1e3+1e3]; >> v = [7; -4]; >> i = inv(A)*v i =

Lecture 2817 Results I 1 = 3.33mA I 2 = -0.33mA V out = (I 1 - I 2 ) 1k  = 3.66V

Lecture 2818 Example: Problem k  2k  12V + - 4mA 2mA I0I0

Lecture 2819 Mesh 2 Mesh 3 Mesh 1 Identify Mesh’s 1k  2k  12V + - 4mA 2mA I0I0

Lecture 2820 Assign Mesh Currents 1k  2k  12V + - 4mA 2mA I0I0 I1I1 I2I2 I3I3

Lecture 2821 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. What to do?

Lecture 2822 Current Sources The 4mA current source sets I 2 : I 2 = -4mA The 2mA current source sets a constraint on I 1 and I 3 : I 1 - I 3 = 2mA We have two equations and three unknowns. Where is the third equation?

Lecture k  2k  12V + - 4mA 2mA I0I0 I1I1 I2I2 I3I3 The Supermesh surrounds this source! The Supermesh does not include this source!

Lecture 2824 KCL Around the Supermesh -12V + I 3 2k  + (I 3 - I 2 )1k  + (I 1 - I 2 )2k  = 0 I 3 2k  + (I 3 - I 2 )1k  + (I 1 - I 2 )2k  = 12V

Lecture 2825 Matrix Notation The three equations can be combined into a single matrix/vector equation.

Lecture 2826 Solve Using MATLAB >> A = [0 1 0; ; 2e3 -1e3-2e3 2e3+1e3]; >> v = [-4e-3; 2e-3; 12]; >> i = inv(A)*v i =

Lecture 2827 Solution I 1 = 1.2mA I 2 = -4mA I 3 = -0.8mA I 0 = I 1 - I 2 = 5.2mA

Lecture 2828 Rules for Mesh Circuit Direct Matrix All mesh currents in same direction (CW) Diagonal terms are sum of impedance in the respective loop Off-diagonal terms are negative of common impedance Voltages are positive if voltage supports mesh current. Negative otherwise