plasma-expert/examples/calculating-ropt.md

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