8.4 KiB

Raw Blame History

id	title	section	difficulty	estimated_time	prerequisites	objectives	tags
opt-01	The Two Critical Resistances	Optimization & Simulation	intermediate	35	[fund-08]	[Derive and understand R_opt_phase for minimum phase angle Derive and understand R_opt_power for maximum power transfer Compare the two resistances and their physical meanings Calculate phase angles at different operating points]	[optimization impedance power-transfer phase-angle]

The Two Critical Resistances

In spark gap modeling, we encounter two fundamentally different optimization criteria that lead to two different "optimal" resistance values. Understanding the distinction between these is critical for both analysis and practical coil operation.

The Topological Phase Constraint

Before we dive into the two resistances, we need to understand a fundamental limitation imposed by circuit topology.

What is a Topological Constraint?

Definition: A limitation imposed by the structure of the circuit itself, independent of component values.

Example: A series RLC circuit can only have impedance phase between -90° (pure capacitive) and +90° (pure inductive). You cannot achieve φ_Z = +120° no matter what component values you choose. This is a topological constraint.

For spark circuits: The specific arrangement (R||C_mut) in series with C_sh creates a fundamental limit on how resistive the impedance can appear.

Deriving the Minimum Phase Angle

From Part 1 fundamentals, we have the spark admittance:

Y = [(G + jB₁) × jB₂] / [G + j(B₁ + B₂)]

where:
  G = 1/R (conductance)
  B₁ = ωC_mut (mutual capacitance susceptance)
  B₂ = ωC_sh (sheath capacitance susceptance)

The impedance phase is:

φ_Z = atan(-Im{Y}/Re{Y})

Question: For fixed C_mut and C_sh, which R value minimizes |φ_Z| (makes the impedance most resistive)?

Mathematical result: Taking the derivative ∂φ_Z/∂G = 0 and solving:

G_opt = ω√[C_mut(C_mut + C_sh)]

Therefore:
R_opt_phase = 1 / [ω√(C_mut(C_mut + C_sh))]

At this resistance, the phase angle magnitude is minimized to:

φ_Z,min = -atan(2√[r(1 + r)])

where r = C_mut/C_sh (capacitance ratio)

The Critical Ratio r = 0.207

Let's find when φ_Z,min = -45° is achievable:

-45° = -atan(2√[r(1 + r)])
tan(45°) = 1 = 2√[r(1 + r)]
0.5 = √[r(1 + r)]
0.25 = r(1 + r) = r + r²
r² + r - 0.25 = 0

Using quadratic formula:
r = [-1 ± √(1 + 1)] / 2 = [-1 ± √2] / 2

Taking positive root:
r = (√2 - 1) / 2 ≈ 0.207

Critical insight:

If r < 0.207: Can achieve φ_Z = -45° (with appropriate R)
If r > 0.207: Cannot achieve φ_Z = -45° no matter what R you choose!
If r ≥ 0.207: φ_Z,min is more negative than -45°

Typical Tesla Coil Values

Large topload, short spark:

C_mut = 10 pF, C_sh = 4 pF (2 feet)
r = 10/4 = 2.5

φ_Z,min = -atan(2√[2.5 × 3.5]) = -atan(2 × 2.96) = -atan(5.92) = -80.4°

Small topload, long spark:

C_mut = 6 pF, C_sh = 12 pF (6 feet)
r = 6/12 = 0.5

φ_Z,min = -atan(2√[0.5 × 1.5]) = -atan(2 × 0.866) = -atan(1.732) = -60.0°

Common range: r = 0.5 to 2.0, giving φ_Z,min ≈ -60° to -80°

Conclusion: For most Tesla coil geometries, achieving -45° is mathematically impossible!

R_opt_phase: Closest to Resistive

Formula:

R_opt_phase = 1 / [ω√(C_mut(C_mut + C_sh))]

Purpose: Minimizes |φ_Z| to achieve φ_Z,min

Use case: If you want the "most resistive-looking" impedance possible for your given capacitances.

Physical meaning: This is the geometric mean of the capacitive reactances, representing the resistance that balances the phase contributions from C_mut and C_sh.

R_opt_power: Maximum Power Transfer

Different question: Which R maximizes real power delivered to the spark for a given topload voltage?

Setup: Fixed voltage source V_top, variable load resistance R

Power to load:

P = 0.5 × |V_top|² × Re{Y(R)}

where Y(R) depends on R through G = 1/R.

Mathematical derivation: Take ∂P/∂G = 0 and solve for G:

After applying calculus (expanding Re{Y} and differentiating):

R_opt_power = 1 / [ω(C_mut + C_sh)]

Simpler formula! Just the total capacitance reactance, not a geometric mean.

Comparing the Two Resistances

Relationship

R_opt_power = 1 / [ω(C_mut + C_sh)]
R_opt_phase = 1 / [ω√(C_mut(C_mut + C_sh))]

Since √(C_mut(C_mut + C_sh)) < (C_mut + C_sh):

R_opt_power < R_opt_phase  ALWAYS

Numerical relationship: For typical r = 0.5 to 2:

R_opt_power ≈ (0.5 to 0.7) × R_opt_phase

Phase Angle at R_opt_power

Always more negative (more capacitive) than φ_Z,min
Typically φ_Z ≈ -55° to -75° at R_opt_power
More capacitive than R_opt_phase, but delivers more power

Key insight: The impedance that transfers maximum power is NOT the same as the impedance with minimum phase angle!

Worked Example: Calculating Both Critical Resistances

Given:

Frequency: f = 200 kHz → ω = 1.257×10⁶ rad/s
C_mut = 8 pF = 8×10⁻¹² F
C_sh = 6 pF = 6×10⁻¹² F

Find: R_opt_phase, R_opt_power, and compare

Solution

Part 1: R_opt_phase

R_opt_phase = 1 / [ω√(C_mut(C_mut + C_sh))]
            = 1 / [1.257×10⁶ × √(8×10⁻¹² × 14×10⁻¹²)]
            = 1 / [1.257×10⁶ × √(112×10⁻²⁴)]
            = 1 / [1.257×10⁶ × 10.58×10⁻¹²]
            = 1 / (13.30×10⁻⁶)
            = 75.2 kΩ

Part 2: R_opt_power

C_total = C_mut + C_sh = 8 + 6 = 14 pF = 14×10⁻¹² F

R_opt_power = 1 / (ωC_total)
            = 1 / (1.257×10⁶ × 14×10⁻¹²)
            = 1 / (17.60×10⁻⁶)
            = 56.8 kΩ

Part 3: Comparison

Ratio: R_opt_power / R_opt_phase = 56.8 / 75.2 = 0.755

R_opt_power is 75.5% of R_opt_phase

Part 4: Phase angle at R_opt_power

Calculate admittance with R = 56.8 kΩ:

G = 1/56800 = 17.61 μS
B₁ = ωC_mut = 1.257×10⁶ × 8×10⁻¹² = 10.06 μS
B₂ = ωC_sh = 1.257×10⁶ × 6×10⁻¹² = 7.54 μS

Re{Y} = GB₂²/[G² + (B₁+B₂)²]
      = 17.61 × 56.85 / [310 + 309.8]
      = 1001.2 / 619.8
      = 1.615 μS

Im{Y} = 7.54[310 + 176.9] / 619.8
      = 7.54 × 486.9 / 619.8
      = 5.928 μS

φ_Y = atan(5.928/1.615) = atan(3.67) = 74.7°
φ_Z = -74.7°

Summary:

R_opt_phase = 75.2 kΩ gives φ_Z = -74.2° (minimum)
R_opt_power = 56.8 kΩ gives φ_Z = -74.7° (slightly more capacitive)
Power is maximized at R_opt_power despite not having minimum phase
Difference is small: both are strongly capacitive

Visual Aid: Power vs Resistance Curves

Image shows two overlaid plots:

Top: Power vs R (bell curve peaking at R_opt_power = 56.8 kΩ)
Bottom: Phase angle vs R (minimum at R_opt_phase = 75.2 kΩ)
Key insight: The two optimal points do not coincide

Key features:

X-axis: R (kΩ), range 20 to 150, log scale
Power curve: Bell-shaped, peaks at R_opt_power
Phase curve: Rises from -90° (R→0), peaks at R_opt_phase, falls back
Vertical lines show the two different optimum points

Key Takeaways

R_opt_phase = 1/[ω√(C_mut(C_mut + C_sh))] minimizes phase angle magnitude
R_opt_power = 1/[ω(C_mut + C_sh)] maximizes power transfer
R_opt_power < R_opt_phase always (typically 50-75% of R_opt_phase)
Most Tesla coils operate with r > 0.207, making φ_Z = -45° impossible
The impedance must be strongly capacitive due to topological constraints
Power optimization and phase optimization are different goals with different solutions

Practice

{exercise:opt-ex-01}

Problem 1: For f = 150 kHz, C_mut = 10 pF, C_sh = 8 pF: Calculate R_opt_power and R_opt_phase.

Problem 2: At 200 kHz, a spark has C_total = 12 pF. What is R_opt_power? If V_top = 400 kV, estimate the maximum deliverable power (assume R at optimal value).

Problem 3: Prove algebraically that R_opt_power < R_opt_phase always (hint: compare 1/(C_mut+C_sh) with 1/√(C_mut(C_mut+C_sh))).

Problem 4: A measurement shows φ_Z = -68° at the operating point. Is R likely above or below R_opt_phase? Above or below R_opt_power? Explain your reasoning.

Problem 5: Calculate the capacitance ratio r and minimum achievable phase angle φ_Z,min for: (a) C_mut = 12 pF, C_sh = 8 pF (b) Can this circuit achieve -45°?

Next Lesson: The Hungry Streamer - Self-Optimization

8.4 KiB Raw Blame History