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EE 201C Homework 1 Solution

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

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EE 201C Homework 1 Solution

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

  2. Homework • 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. l l l T W W W S S H • 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 Ωmm2/m (room temperature), • µ =1.256×10−6H/m, • free space 0=8.85×10 -12F/m

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

  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. l l T W W S Field solver (fasthenry): L matrix =[ 9.1040e-09 6.7526e-09 5.9179e-09 6.7527e-09 9.1037e-09 6.7523e-09 5.9182e-09 6.7525e-09 9.1034e-09] (1) Inductance Calculation (cont.) • L_self and L_mutual of wire should be calculated as follows: • m is the number of filaments in wire k • n is the number of filaments in wire l • Reason: current in source filament and magnetic flux though target filament are reduced. • For our problem, all inductance should divided by 2*2=4 as 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 (=C20+2*C21) per unit length is Ground cap.for single wire Coupling cap.

  7. C2 C4 C1 C3 C5 (2) Capacitance (cont.) • For our problem: (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 Ωmm2/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.

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