1. EE 201C Homework 1 Solution Fang Gong gongfang@ucla.edu

2. Given three wires, each modeled by at least 2 filaments, find the 3x3 matrix for (frequency-independent) inductance between the 3 wires. We assume that the ground plane has infinite size and is 10 um away for the purpose of capacitance calculation. Homework wire width: W=6um, wire thickness: T=4um, wire length: l=6000um, wire spacing: S = 10um, distance to ground: H=10um, Copper electrical resistivity 0.0175 Ωmm 2 /m (room temperature), µ =1.256×10 −6 H/m, free space  0 = 8.85×10 -12 F/m H T l S S W WW l l

3. (1) Inductance Calculation 1.1 discretize the wires into filaments, and calculate the self- inductance for each filament using formula. T l S S W WW l l

4. (1) Inductance Calculation (cont.) 1.2 Calculate the mutual inductance for each pair of filaments 1.3 calculate the self-inductance and mutual-inductance for wires: L1_self=L2_self=L3_self=3.66e-8 H L12_mutual=L21_mutual=2.7e-8 H L23_mutual=L32_mutual=2.7e-8 H L13_mutual=L31_mutual =2.37e-8 H Problem? the inductance of wire will increase along with the number of filaments.

5. (1) Inductance Calculation (cont.) L_self and L_mutual of wire should be calculated as follows: n m is the number of filaments in wire k n n is the number of filaments in wire l Reason: current in source filament and magnetic flux though target filament are reduced. n For our problem, all inductance should divided by 2*2=4 as Field solver (fasthenry): L matrix =[ 9.1040e-09 6.7526e-095.9179e-09 6.7527e-09 9.1037e-09 6.7523e-09 5.9182e-09 6.7525e-09 9.1034e-09] T l S W W l L11=L22=L33=9.15e-9 H; L12=L21=L23=L32=6.75e-9 H; L13=L31 =5.925e-9 H

6. (2) Capacitance 2.1 Using formula [Sakurai-Tamaru, ED’83] to calculate ground capacitance and coupling capacitance. For a single wire above ground, the ground capacitance per unit length is For three wires above ground, the total capacitance of middle wire (=C 20 +2*C 21 ) per unit length is Ground cap. for single wire Coupling cap.

7. (2) Capacitance (cont.) For our problem: C1 C2 C3 C4 C5 (1) Ground capacitance of middle wire: C3=e*(1.15*(width/h)+2.8*(thickness/h)^0.222)*length =1.579e-13 F; (2) Coupling capacitance of middle wire C2=C4=e*(0.03*(width/h)+0.83*(thickness/h)-0.07*(thickness/h)^0.222)*(s/h)^- 1.34*length = 1.5552e-014 F; (3) Ground capacitance of edge wires: C1=C5=(single wire + three wire)/2 = 1.7350e-013 F; CAPACITANCE MATRIX, (1e-15 F) 1 2 3 1%GROUP1 1 166.7 -19.76 -3.139 2%GROUP1 2 -19.76 157.7 -19.85 3%GROUP1 3 -3.139 -19.85 166.8 C3=1.577e-13 F C1=C5=1.667e-13 F C2=C4=1.976e-14 F

8. (3) Resistance Copper electrical resistivity 0.0175 Ωmm 2 /m (room temperature), R1 = R2 = R3 = 4.375 Ω

9. (4) Simulation Generate net-list for Spice. K12 = L12/sqrt(L11*L22) Different rising time lead to different waveforms.

10. (4) Simulation RCL model

11. Thanks.