# 1 Analysis and Simulation Exercises ~DC Circuit Analysis (1) Basic Circuit Laws (Kirchhoff ’ s Voltage/ Current Law) Thévenin’s Theorem Norton ’ s Theorem.

## Presentation on theme: "1 Analysis and Simulation Exercises ~DC Circuit Analysis (1) Basic Circuit Laws (Kirchhoff ’ s Voltage/ Current Law) Thévenin’s Theorem Norton ’ s Theorem."— Presentation transcript:

1 Analysis and Simulation Exercises ~DC Circuit Analysis (1) Basic Circuit Laws (Kirchhoff ’ s Voltage/ Current Law) Thévenin’s Theorem Norton ’ s Theorem

2 Circuit with three meshes 1

3 Spice file Bridge Circuit for use with Basic Circuit Laws V 3 0 25v R1 1 2 100 R2 1 0 75 R3 2 3 50 R4 4 0 60 R5 2 4 150 R6 1 4 200.OP.PRINT DC I(R1) I(R2) I(R3).PRINT DC I(R4) I(R5) I(R6).DC V 25V 25V 25V.OPT nopage.END

4 Output file (1) **** 09/19/02 11:39:09 ************** PSpice Lite (Mar 2000) ***************** Bridge Circuit for use with Basic Circuit Laws **** CIRCUIT DESCRIPTION******************************************** V 3 0 25v R1 1 2 100 R2 1 0 75 R3 2 3 50 R4 4 0 60 R5 2 4 150 R6 1 4 200.OP.PRINT DC I(R1) I(R2) I(R3).PRINT DC I(R4) I(R5) I(R6).DC V 25V 25V 25V.OPT nopage.END

5 Output file (2) **** DC TRANSFER CURVES TEMPERATURE = 27.000 DEG C V I(R1) I(R2) I(R3) 2.500E+01 -9.704E-02 8.885E-02 -1.726E-01 **** DC TRANSFER CURVES TEMPERATURE = 27.000 DEG C V I(R4) I(R5) I(R6) 2.500E+01 8.379E-02 7.560E-02 8.184E-03 * SMALL SIGNAL BIAS SOLUTION TEMPERATURE =27.000 DEG C NODE VOLTAGE NODE VOLTAGE ( 1) 6.6641 ( 2) 16.3680 ( 3) 25.0000 ( 4) 5.0273

6 Output file (3) VOLTAGE SOURCE CURRENTS NAME CURRENT V -1.726E-01 TOTAL POWER DISSIPATION 4.32E+00 WATTS **** OPERATING POINT INFORMATION TEMPERATURE = 27.000 DEG C JOB CONCLUDED TOTAL JOB TIME.07

7 Verify Kirchhoff ’ s Voltage Law Check V 12 +V 23 +V 30 +V 01 =0 ?? V 1 = 6.6641 V 2 =16.3680 V 3 = 25.0000 V 4 = 5.0273 V 12 =V 1 -V 2 =-9.7039, V 23 =V 2 -V 3 =-8.632, V 30 =V 3 =25.000, V 01 =-V 1 =-6.6641 V 12 +V 23 +V 30 +V 01 = -9.7039 -8.632+ 25.000 -6.6641=0

8 Verify Kirchhoff ’ s Current Law Finding the sum of the currents entering node 1. For node 1, I 21 +I 01 +I 41 =0. I 21 = -I(R 1 )= 97.04mA, I 01 = -I(R 2 )=-88.85mA, I 41 = -I(R 6 )=-8.184mA I 21 +I 01 +I 41 =97.04 - 88.85 - 8.184 = 0.0

9 THÉVENIN’S Theorem Circuit to illustrate Thévenin’s theorem

10 Thévenin voltage and series

11 Spice file Thevenin Circuit for spice V 1 0 75v R1 1 2 100 R2 2 3 150 R3 2 0 200 RL 3 0 1E12.OP.OPT nopage.TF V(3) V.END

12 Simulation Output File (Partial) NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE ( 1) 75.0000 ( 2) 50.0000 ( 3) 50.0000 **** SMALL-SIGNAL CHARACTERISTICS V(3)/V = 6.667E-01 INPUT RESISTANCE AT V = 3.000E+02 OUTPUT RESISTANCE AT V(3) = 2.167E+02

13 Norton ’ s Theorem T Circuit for Norton analysis

14 Spice File Find the short-current for R 4 Norton's Theorem; Find Isc V 1 0 48V R1 1 2 20K R2 2 0 20K R3 2 3 5K R4 3 0 0.001.DC V 48V 48V 48V.OP.OPT nopage.PRINT DC I(R4) V(1,2).END

15 Simulation Output File **** DC TRANSFER CURVES TEMPERATURE = 27.000 DEG C V I(R4) V(1,2) 4.800E+01 1.600E-03 4.000E+01 **** SMALL SIGNAL BIAS SOLUTION TEMPERATURE = 27.000 DEG C NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE ( 1) 48.0000 ( 2) 8.0000 ( 3) 1.600E-06