# Calculating R_opt_power and R_opt_phase ## Overview This worked example demonstrates the complete calculation procedure for the two critical resistance values in spark modeling: R_opt_power (maximum power transfer) and R_opt_phase (minimum impedance phase angle). We show multiple scenarios with different frequencies and capacitances to build intuition. ## Fundamental Formulas **R_opt_power** (Maximum Power Transfer): ``` R_opt_power = 1 / [ω(C_mut + C_sh)] ``` **R_opt_phase** (Minimum Phase Angle): ``` R_opt_phase = 1 / [ω√(C_mut(C_mut + C_sh))] ``` **Angular frequency:** ``` ω = 2πf [rad/s] ``` ## Scenario 1: Medium Coil at 200 kHz ### Given Parameters - Operating frequency: f = 200 kHz - Mutual capacitance: C_mut = 8 pF = 8×10⁻¹² F - Shunt capacitance: C_sh = 6 pF = 6×10⁻¹² F - Spark length: 2 feet (where C_sh ≈ 2 pF/ft × 2 = 4 pF, but using measured value) ### Step 1: Calculate Angular Frequency ``` ω = 2π × f = 2π × 200,000 Hz = 2 × 3.14159 × 200,000 = 1.257 × 10⁶ rad/s ``` **Validation:** Units are rad/s, positive value as expected. ### Step 2: Calculate Total Capacitance ``` C_total = C_mut + C_sh = 8 pF + 6 pF = 14 pF = 14 × 10⁻¹² F ``` **Physical check:** Total capacitance is reasonable for a 2-foot spark with medium topload. ### Step 3: Calculate R_opt_power ``` R_opt_power = 1 / (ω × C_total) = 1 / (1.257×10⁶ rad/s × 14×10⁻¹² F) = 1 / (1.760×10⁻⁵) = 56,818 Ω ≈ 56.8 kΩ ``` **Dimensional analysis:** ``` [R] = 1 / ([rad/s] × [F]) = 1 / ([1/s] × [C/V]) = [V·s/C] = [Ω] ✓ ``` ### Step 4: Calculate Product for R_opt_phase ``` C_mut × (C_mut + C_sh) = 8×10⁻¹² × 14×10⁻¹² = 112 × 10⁻²⁴ = 1.12 × 10⁻²² F² √(C_mut × (C_mut + C_sh)) = √(1.12 × 10⁻²²) = 1.058 × 10⁻¹¹ F = 10.58 pF ``` ### Step 5: Calculate R_opt_phase ``` R_opt_phase = 1 / (ω × √(C_mut(C_mut + C_sh))) = 1 / (1.257×10⁶ × 1.058×10⁻¹¹) = 1 / (1.330×10⁻⁵) = 75,188 Ω ≈ 75.2 kΩ ``` ### Step 6: Compare the Two Resistances ``` Ratio = R_opt_power / R_opt_phase = 56.8 kΩ / 75.2 kΩ = 0.755 = 75.5% Difference = R_opt_phase - R_opt_power = 75.2 - 56.8 = 18.4 kΩ ``` **Key insight:** R_opt_power is always less than R_opt_phase. For this geometry, it's about 75% of R_opt_phase. ### Step 7: Calculate Phase Angles **Capacitance ratio:** ``` r = C_mut / C_sh = 8 / 6 = 1.333 ``` **Minimum achievable phase angle:** ``` φ_Z,min = -atan(2√[r(1 + r)]) = -atan(2√[1.333 × 2.333]) = -atan(2√3.111) = -atan(2 × 1.764) = -atan(3.528) = -74.2° ``` **Phase angle at R_opt_power:** Calculate admittance components: ``` G = 1/R = 1/56800 = 1.761 × 10⁻⁵ S = 17.61 μS B₁ = ωC_mut = 1.257×10⁶ × 8×10⁻¹² = 1.006 × 10⁻⁵ S = 10.06 μS B₂ = ωC_sh = 1.257×10⁶ × 6×10⁻¹² = 7.542 × 10⁻⁶ S = 7.54 μS ``` Real part of admittance: ``` Re{Y} = G×B₂² / [G² + (B₁+B₂)²] = 17.61 × (7.54)² / [17.61² + (17.60)²] = 17.61 × 56.85 / [310.1 + 309.8] = 1001.2 / 619.9 = 1.615 μS ``` Imaginary part of admittance: ``` Im{Y} = B₂[G² + B₁(B₁+B₂)] / [G² + (B₁+B₂)²] = 7.54[310.1 + 10.06×17.60] / 619.9 = 7.54[310.1 + 177.1] / 619.9 = 7.54 × 487.2 / 619.9 = 5.929 μS ``` Phase angle: ``` φ_Y = atan(Im{Y}/Re{Y}) = atan(5.929/1.615) = atan(3.671) = 74.7° φ_Z = -φ_Y = -74.7° ``` **Observation:** At R_opt_power, phase is -74.7° (slightly more capacitive than the minimum of -74.2°). The difference is small because we're close to the optimal point. ## Scenario 2: Large Coil at 150 kHz ### Given Parameters - Operating frequency: f = 150 kHz - Mutual capacitance: C_mut = 12 pF (larger topload) - Shunt capacitance: C_sh = 10 pF (5 feet spark) ### Step 1: Angular Frequency ``` ω = 2π × 150,000 = 942,478 rad/s ≈ 9.425 × 10⁵ rad/s ``` ### Step 2: Total Capacitance ``` C_total = 12 + 10 = 22 pF = 22 × 10⁻¹² F ``` ### Step 3: R_opt_power ``` R_opt_power = 1 / (9.425×10⁵ × 22×10⁻¹²) = 1 / (2.074×10⁻⁵) = 48,220 Ω ≈ 48.2 kΩ ``` ### Step 4: R_opt_phase ``` C_mut × C_total = 12×10⁻¹² × 22×10⁻¹² = 264×10⁻²⁴ F² √(C_mut × C_total) = 1.625 × 10⁻¹¹ F R_opt_phase = 1 / (9.425×10⁵ × 1.625×10⁻¹¹) = 1 / (1.532×10⁻⁵) = 65,274 Ω ≈ 65.3 kΩ ``` ### Step 5: Comparison ``` Ratio = 48.2 / 65.3 = 0.738 = 73.8% ``` **Observation:** Lower frequency gives lower resistance values. Ratio is similar to Scenario 1. ## Scenario 3: Small Coil at 300 kHz ### Given Parameters - Operating frequency: f = 300 kHz - Mutual capacitance: C_mut = 6 pF (smaller topload) - Shunt capacitance: C_sh = 4 pF (2 feet spark) ### Step 1: Angular Frequency ``` ω = 2π × 300,000 = 1.885 × 10⁶ rad/s ``` ### Step 2: Total Capacitance ``` C_total = 6 + 4 = 10 pF = 10 × 10⁻¹² F ``` ### Step 3: R_opt_power ``` R_opt_power = 1 / (1.885×10⁶ × 10×10⁻¹²) = 1 / (1.885×10⁻⁵) = 53,050 Ω ≈ 53.1 kΩ ``` ### Step 4: R_opt_phase ``` C_mut × C_total = 6×10⁻¹² × 10×10⁻¹² = 60×10⁻²⁴ F² √(C_mut × C_total) = 7.746 × 10⁻¹² F R_opt_phase = 1 / (1.885×10⁶ × 7.746×10⁻¹²) = 1 / (1.460×10⁻⁵) = 68,493 Ω ≈ 68.5 kΩ ``` ### Step 5: Comparison ``` Ratio = 53.1 / 68.5 = 0.775 = 77.5% ``` **Observation:** Higher frequency, but smaller capacitances gives R values similar to Scenario 1. The ratio is slightly higher due to lower capacitance ratio. ## Scenario 4: Effect of Varying C_sh (Spark Length) **Fixed parameters:** f = 200 kHz, C_mut = 8 pF **Variable:** C_sh (representing different spark lengths) ### Short Spark: L = 1 ft → C_sh ≈ 2 pF ``` ω = 1.257 × 10⁶ rad/s C_total = 8 + 2 = 10 pF R_opt_power = 1/(1.257×10⁶ × 10×10⁻¹²) = 79.6 kΩ √(8 × 10) = √80 = 8.944 pF R_opt_phase = 1/(1.257×10⁶ × 8.944×10⁻¹²) = 89.0 kΩ Ratio = 79.6/89.0 = 0.894 = 89.4% ``` ### Medium Spark: L = 3 ft → C_sh ≈ 6 pF ``` (Already calculated in Scenario 1) R_opt_power = 56.8 kΩ R_opt_phase = 75.2 kΩ Ratio = 75.5% ``` ### Long Spark: L = 6 ft → C_sh ≈ 12 pF ``` C_total = 8 + 12 = 20 pF R_opt_power = 1/(1.257×10⁶ × 20×10⁻¹²) = 39.8 kΩ √(8 × 20) = √160 = 12.65 pF R_opt_phase = 1/(1.257×10⁶ × 12.65×10⁻¹²) = 62.9 kΩ Ratio = 39.8/62.9 = 0.633 = 63.3% ``` ### Summary Table: Effect of Spark Length | Length | C_sh | C_total | R_opt_power | R_opt_phase | Ratio | |--------|------|---------|-------------|-------------|-------| | 1 ft | 2 pF | 10 pF | 79.6 kΩ | 89.0 kΩ | 89.4% | | 3 ft | 6 pF | 14 pF | 56.8 kΩ | 75.2 kΩ | 75.5% | | 6 ft | 12 pF | 20 pF | 39.8 kΩ | 62.9 kΩ | 63.3% | **Key insight:** As spark grows longer: - Both R values decrease (higher capacitance) - The ratio R_opt_power/R_opt_phase also decreases - Longer sparks have larger separation between the two optimal points ## Frequency Dependence Study **Fixed parameters:** C_mut = 8 pF, C_sh = 6 pF **Variable:** Frequency from 100 kHz to 400 kHz ### f = 100 kHz ``` ω = 6.283 × 10⁵ rad/s R_opt_power = 1/(6.283×10⁵ × 14×10⁻¹²) = 113.6 kΩ R_opt_phase = 1/(6.283×10⁵ × 10.58×10⁻¹²) = 150.4 kΩ ``` ### f = 200 kHz ``` (Scenario 1 result) R_opt_power = 56.8 kΩ R_opt_phase = 75.2 kΩ ``` ### f = 400 kHz ``` ω = 2.513 × 10⁶ rad/s R_opt_power = 1/(2.513×10⁶ × 14×10⁻¹²) = 28.4 kΩ R_opt_phase = 1/(2.513×10⁶ × 10.58×10⁻¹²) = 37.6 kΩ ``` ### Frequency Scaling Table | Frequency | ω (Mrad/s) | R_opt_power | R_opt_phase | |-----------|------------|-------------|-------------| | 100 kHz | 0.628 | 113.6 kΩ | 150.4 kΩ | | 200 kHz | 1.257 | 56.8 kΩ | 75.2 kΩ | | 400 kHz | 2.513 | 28.4 kΩ | 37.6 kΩ | **Scaling law:** R_opt ∝ 1/f Doubling frequency halves resistance (inverse relationship). ## Final Results Summary ### Scenario Comparison | Scenario | f (kHz) | C_mut | C_sh | C_total | R_opt_power | R_opt_phase | Ratio | |----------|---------|-------|------|---------|-------------|-------------|-------| | 1 (Medium, 200 kHz) | 200 | 8 pF | 6 pF | 14 pF | 56.8 kΩ | 75.2 kΩ | 75.5% | | 2 (Large, 150 kHz) | 150 | 12 pF | 10 pF | 22 pF | 48.2 kΩ | 65.3 kΩ | 73.8% | | 3 (Small, 300 kHz) | 300 | 6 pF | 4 pF | 10 pF | 53.1 kΩ | 68.5 kΩ | 77.5% | ### Physical Bounds Check **All scenarios fall within expected ranges:** - R_opt_power: 28-114 kΩ (reasonable for 100-400 kHz, 1-6 ft sparks) - R_opt_phase: 38-150 kΩ (always higher than R_opt_power) - Both values are well above R_min ≈ 1-10 kΩ (plasma lower limit) - Both values are well below R_max ≈ 1-100 MΩ (plasma upper limit) ## Key Insights ### Power vs Phase Optimization **R_opt_power maximizes power transfer:** - This is what the "hungry streamer" seeks - Plasma adjusts conductivity to approach this value - Results in phase angles of -55° to -75° (typical) **R_opt_phase minimizes phase angle:** - Represents "most resistive-looking" impedance - NOT necessarily maximum power transfer - The -45° target is often unachievable (topological constraint) ### Relationship Between the Two **Mathematical relationship:** ``` R_opt_power = 1/(ω·C_total) R_opt_phase = 1/(ω·√(C_mut·C_total)) Ratio = R_opt_power/R_opt_phase = √(C_mut/C_total) = √(C_mut/(C_mut+C_sh)) ``` **For typical r = C_mut/C_sh ratios:** - r = 0.5: Ratio = 0.816 (82%) - r = 1.0: Ratio = 0.707 (71%) - r = 2.0: Ratio = 0.577 (58%) As capacitance ratio increases, the two optimal points diverge more. ### Common Mistakes to Avoid 1. **Forgetting the square root** in R_opt_phase calculation 2. **Using C_mut instead of C_total** in R_opt_power formula 3. **Mixing units** (pF vs F, kHz vs Hz) 4. **Calculating ω = 2f instead of ω = 2πf** 5. **Assuming -45° is always achievable** (topological constraint!) 6. **Using wrong capacitance in product:** C_mut × (C_mut + C_sh), NOT C_mut × C_sh ### Validation Checks **Always verify:** 1. R_opt_power < R_opt_phase (mathematical certainty) 2. Ratio typically 0.5-0.9 (depends on capacitance ratio) 3. Values in physically reasonable range (5-500 kΩ for typical coils) 4. Dimensional analysis: units are Ohms 5. Scaling: R ∝ 1/f and R ∝ 1/C_total ## See Also - **Related Lessons:** - Module 2, Lesson 1: Two Critical Resistances (theory) - Module 2, Lesson 2: Hungry Streamer (self-optimization physics) - Module 1, Lesson 5: Phase Constraint (topological limitations) - **Related Exercises:** - Exercise opt-ex-01: Practice calculations with different parameters - Exercise fund-ex-05: Phase angle calculations - **Related Worked Examples:** - thevenin-extraction.md: Using R_opt in Thévenin analysis - distributed-model-complete.md: R_opt for distributed segments