1. UCSD CSE245 Notes -- Spring 2006 CSE245: Computer-Aided Circuit Simulation and Verification Lecture Notes Spring 2006 Prof. Chung-Kuan Cheng

2. UCSD CSE245 Notes -- Spring 2006 Motivation Why –Whole Circuit Analysis –Interconnect Dominance Wires smaller  R increase Separation smaller  C increase What –Power Net, Clock, Interconnect Coupling, Parallel Processing Where –Matrix Solvers, Integration For Dynamic System –RLC Reduction, Transmission Lines, S Parameters –Whole Chip Analysis –Thermal, Mechanical, Biological Analysis

3. UCSD CSE245 Notes -- Spring 2006 Formulation General Equation (a.k.a. state equations) Equation Formulation –Conservation Laws KCL (Kirchhoff’s Current Law) –n-1 equations, n is number of nodes in the circuit KVL (Kirchhoff’s Voltage Law) –m-(n-1) equations, m is number of branches in the circuit. –Branch Constitutive Equations m equations

4. UCSD CSE245 Notes -- Spring 2006 Conservation Laws KCL KVL n-1 independent cutsets m-(n-1) independent loops

5. UCSD CSE245 Notes -- Spring 2006 Branch Constitutive Laws Each branch has a circuit element –Resistor –Capacitor Forward Euler (FE) Approximation Backward Euler (BE) Approximation Trapezoidal (TR) Approximation –Inductor Similar approximation (FE, BE or TR) can be used for inductor.

6. UCSD CSE245 Notes -- Spring 2006 Formulation - Cutset and Loop Analysis find a cutset for each trunk –write a KCL for each cutset find a loop for each link –write a KVL for each loop Select tree trunks and links cutset matrix loop matrix

7. UCSD CSE245 Notes -- Spring 2006 Formulation - Cutset and Loop Analysis Or we can re-write the equations as: In general, the cutset and loop matrices can be written as

8. UCSD CSE245 Notes -- Spring 2006 Formulation – State Equations From the cutset and loop matrices, we have In general, one should –Select capacitive branches as tree trunks no capacitive loops for each node, there is at least one capacitor (every node actually should have a shunt capacitor) –Select inductive branches as tree links no inductive cutsets Combine above two equations, we have the state equation

9. UCSD CSE245 Notes -- Spring 2006 Formulation – An Example State Equation Output Equation (suppose v 3 is desired output)

10. UCSD CSE245 Notes -- Spring 2006 Responses in Time Domain State Equation The solution to the above differential equation is the time domain response Where

11. UCSD CSE245 Notes -- Spring 2006 Exponential of a Matrix Properties of e A k! can be approximated by Stirling Approximation That is, higher order terms of e A will approach 0 because k! is much larger than A k for large k’s. Calculation of e A is hard if A is large

12. UCSD CSE245 Notes -- Spring 2006 Responses in Frequency Domain: LapLace Transform Definition: Simple Transform Pairs Laplace Transform Property - Derivatives

13. UCSD CSE245 Notes -- Spring 2006 Responses in Frequency Domain Time Domain State Equation Laplace Transform to Frequency Domain Re-write the first equation Solve for X, we have the frequency domain solution

14. UCSD CSE245 Notes -- Spring 2006 Serial Expansion of Matrix Inversion For the case s  0, assuming initial condition x 0 =0, we can express the state response function as For the case s   assuming initial condition x 0 =0, we can express the state response function as

15. UCSD CSE245 Notes -- Spring 2006 Concept of Moments The moments are the coefficients of the Taylor’s expansion about s=0, or Maclaurin Expansion Recall the definition of Laplace Transform Re-Write as Moments

16. UCSD CSE245 Notes -- Spring 2006 Concept of Moments Re-write Maclaurin Expansion of the state response function Moments are

17. UCSD CSE245 Notes -- Spring 2006 Moments Calculation: An Example

18. UCSD CSE245 Notes -- Spring 2006 Moments Calculation: An Example For the state response function, we have A voltage or current can be approximated by

19. UCSD CSE245 Notes -- Spring 2006 Moments Calculation: An Example (Cont’d) (1) Set V s (0) =1 (suppose voltage source is an impulse function) (2) Short all inductors, open all capacitors, derive V c (0), I L (0) (3) Use V c (i), I L (i) as sources, i.e. I c (i+1) =CV c (i) and V L (i+1) =LI L (i), derive V c (i+1), I L (i+1) (4) i++, repeat (3)