Lecture 6
Nodal/Mesh Analysi4.2,4.3, 4.5 &4.6
Nodes
Node: a point where two or more circuit elements join
Essential Node
Essential Node: a node where three or more circuit elements join
Step 1: Number of Nodal Voltage Equations
• ne is essential nodes
• Number of nodal voltage Equations is ne-1
ne = 3ne-1 =2 nodal voltage equations are required.
Step 2: Designate a node as the reference node.
• Suggestion: Select the node with the most branches.
(3 branches)(3 branches)
(4 branches)
Step 3:Define the Node Voltage on the Diagram
Step 4: Apply KCL
• Apply KCL to essential nodes.
R1 R2
R10R5VS IS
Simulation Results
Solve ia, ib and ic
Hints
ne=2Need ne-1=1 equation
Branch
Branch: a path that connects two nodes.
Essential Branch
Essential branch: a path which connects two essential nodes without passing through an essential node.
Mesh
Mesh: a loop that does not enclose any other loops
Step 1: Determine the number of essential nodes
ne=3
Step 2: Determine the number of essential branches
be=5
# of equations: be-(ne-1)=5-(3-1)=3
Step 3: Apply KVL Around Loop b
a. Focus initially on ia. 40-iaR2-(ia )8=0b. Account for ib. 40-iaR2-(ia-ib)8=0
+ -
+
-
Step 4: Apply KVL Around Loop b
a. Focus initially on ib. -(ib )8-ib6-(ib )6=0b. Account for ia. -(ib-ia)8-ib6-(ib )6=0c. Account for ic. -(ib-ia)8-ib6-(ib-ic )6=0
+ -
+
-+
-
Step 5: Apply KVL Around Loop c
a. Focus initially on ic. -(ic )6-ic4-20=0b. Account for ib -(ic- ib)6-ic4-20=0
+ -
+
-
Solve 3 EQ and 3 Unknowns Using Mathematica
3 Unknown equations 3 unknowns
Get Mathematica Through SSU
Step 1
# of essential branch: 6# of essential nodes: 4# of equations: 6-(4-1)=3
Step 2
See in the handout.
Use Mathematica to Solve Equations
Format: Solve[{equations separated by a comma},{list of unknowns}]To solve an equation: Evaluation→Evaluate Cells
Degenerated Common Emitter Amplifier
(Small signal model)
What if we drive the base with a small signal?
Vin, m=1 mV Vout, m=46 mV
Mesh Analysis
ne=2 essential nodesbe=3 essential branches
3-(2-1)=2 equations
Mesh Analysis
1 2
Clockwise around loop 1:+Vin-i1rπ-(i1+i2)RE=0i2=gmi1rπ
Vout=-i2RC
Loop 1: clockwiseLoop 2: counter-clockwise