plasma-expert/examples/thevenin-extraction.md

Thévenin Equivalent Extraction and Power Calculations

Overview

This worked example demonstrates the complete procedure for extracting the Thévenin equivalent of a Tesla coil (V_th and Z_th), then using it to calculate power delivery to various spark loads. This method allows you to characterize a coil once and predict performance with any load without re-simulation.

Given Parameters

Tesla Coil Specifications:

• Operating frequency: f = 185 kHz
• Coil type: Medium DRSSTC
• Primary tank: L_primary = 15 μH, C_MMC = 0.8 μF
• Secondary: 800 turns, 6" diameter, 24" height
• Topload: Toroid 12"×3"
• Drive voltage: Variable, 340V bus

Part 1: Extracting Z_th (Output Impedance)

Step 1: SPICE Setup for Z_th Measurement

Circuit configuration:

- Primary drive: Set to 0V AC (short circuit the voltage source)
- Tank components: Keep ALL in place (L_primary, C_MMC, damping resistors)
- Magnetic coupling: k = 0.23 (remains in model)
- Secondary coil: Full distributed model or lumped
- Topload: C_top = 28 pF
- Test source: 1V AC @ 185 kHz applied at topload-to-ground

Why keep tank components? The tank circuit affects output impedance even when not driven. Removing components would give incorrect Z_th that doesn't represent actual coil behavior.

Step 2: Run AC Analysis

Simulation command (SPICE):

.ac lin 1 185k 185k
V_test topload 0 AC 1 0

Measure test current:

I_test = I(V_test)

Step 3: Simulation Results

Raw output:

Frequency: 185,000 Hz
V_test: 1.000 ∠0° V
I_test: 0.000412 ∠87.3° A

Convert to standard units:

I_test_magnitude = 0.412 mA
I_test_phase = 87.3°

Step 4: Calculate Z_th Magnitude

|Z_th| = |V_test| / |I_test|
       = 1.000 V / 0.000412 A
       = 2427 Ω
       ≈ 2.43 kΩ

Physical check: This is reasonable for a medium Tesla coil at RF frequencies (typically 0.5-5 kΩ).

Step 5: Calculate Z_th Phase

Phase of impedance:

φ_Z_th = φ_V - φ_I
       = 0° - 87.3°
       = -87.3°

Polar form:

Z_th = 2427 Ω ∠-87.3°

Physical interpretation: Nearly capacitive (-90° would be pure capacitor). The small difference from -90° is due to resistive losses.

Step 6: Convert to Rectangular Form

Calculate components:

R_th = |Z_th| × cos(φ_Z_th)
     = 2427 × cos(-87.3°)
     = 2427 × 0.0471
     = 114.3 Ω
     ≈ 114 Ω

X_th = |Z_th| × sin(φ_Z_th)
     = 2427 × sin(-87.3°)
     = 2427 × (-0.9989)
     = -2424 Ω

Rectangular form:

Z_th = 114 - j2424 Ω

Step 7: Physical Interpretation of Z_th Components

Resistance (R_th = 114 Ω):

• Represents all resistive losses in system
• Includes secondary winding resistance (~30-50 Ω)
• Includes reflected primary losses (damping resistors, ESR)
• Includes dielectric losses in coil form
• This is the "tax" paid to extract power from the coil

Reactance (X_th = -2424 Ω):

• Negative sign indicates capacitive reactance
• Topload capacitance dominates
• Calculate equivalent capacitance at 185 kHz:

C_eq = 1 / (ω|X_th|)
     = 1 / (2π × 185,000 × 2424)
     = 1 / (2.819×10⁶)
     = 3.548 × 10⁻⁷ F
     = 354.8 pF

But wait, topload is only 28 pF!

Resolution: The 354.8 pF is the equivalent capacitance looking into the topload port, which includes:

• Topload capacitance (28 pF)
• Reflected impedances through coupling
• Distributed capacitances in secondary
• Effective value is higher due to resonant enhancement

Step 8: Calculate Quality Factor

Q = |X_th| / R_th
  = 2424 / 114
  = 21.3

Interpretation:

• Q ≈ 21 is relatively low for a Tesla coil
• This is the system Q (including all damping)
• Unloaded secondary Q might be 100-300
• Primary circuit damping reduces effective Q at topload port
• This Q represents the loaded, coupled system behavior

Part 2: Extracting V_th (Open-Circuit Voltage)

Step 1: SPICE Setup for V_th Measurement

Circuit configuration:

- Remove test source
- Primary drive: ON at normal operating conditions
- Drive voltage: 340V DC bus → ~240V AC RMS to half-bridge
- Frequency: 185 kHz (operating frequency)
- Spark load: REMOVED (open circuit at topload)
- All other components: Normal

Step 2: Run AC Analysis

Simulation command:

.ac lin 1 185k 185k
V_drive primary 0 AC 240 0

Step 3: Simulation Results

Raw output:

V(topload) = 350,000 ∠-15° V peak

Convert to standard form:

V_th = 350 kV ∠-15°
|V_th| = 350 kV
Phase = -15° (relative to drive)

Step 4: Sanity Check - Voltage Gain

Calculate voltage gain:

Primary voltage: V_pri = 240 V RMS × √2 = 339 V peak
Secondary voltage: V_th = 350,000 V peak

Voltage gain = V_th / V_pri
             = 350,000 / 339
             = 1032

Turns ratio = N_sec / N_pri
            = 800 / (assumed ~10 turns)
            = 80

Effective gain ratio = 1032 / 80 = 12.9

Physical interpretation:

• Gain exceeds turns ratio due to resonant enhancement
• Q ≈ 21 is consistent with gain boost
• With coupling k = 0.23, some energy transfers efficiently
• Result is physically reasonable

Step 5: V_th at Different Drive Levels

The Thévenin voltage scales linearly with drive (assuming linearity):

Drive V_bus	V_pri (peak)	V_th (estimated)
240V	240V	248 kV
340V	340V	350 kV
400V	400V	412 kV

Linearity assumption valid when:

• No core saturation
• No component heating effects
• No frequency shifting (small spark)

Part 3: Power Calculations for Various Loads

Thévenin Equivalent Summary

Z_th = 114 - j2424 Ω
V_th = 350 kV ∠0° (using 0° reference for simplicity)

Load Case 1: Typical Spark (Lumped Model)

Spark parameters:

C_mut = 9 pF
C_sh = 7 pF (3.5 ft spark)
f = 185 kHz
ω = 1.162 × 10⁶ rad/s

Calculate R_opt_power:
R_spark = 1/(ω(C_mut + C_sh))
        = 1/(1.162×10⁶ × 16×10⁻¹²)
        = 53,800 Ω
        = 53.8 kΩ

Spark impedance (lumped model):

Z_mut = R_spark || (1/jωC_mut)
X_mut = -1/(ωC_mut) = -1/(1.162×10⁶ × 9×10⁻¹²) = -95.6 kΩ

Parallel combination (R || Xc):
Y_mut = 1/R + jωC
      = 1/53800 + j×1.162×10⁶×9×10⁻¹²
      = 1.859×10⁻⁵ + j1.046×10⁻⁵

|Y_mut| = √((1.859×10⁻⁵)² + (1.046×10⁻⁵)²)
        = 2.134×10⁻⁵

Z_mut = 1/Y_mut = 46,860 Ω

Phase of Z_mut:
φ_mut = atan(Im/Re) = atan(-1.046/1.859) = -29.4°

Z_mut ≈ 46.9 kΩ ∠-29.4°

Shunt capacitor:

X_sh = -1/(ωC_sh) = -1/(1.162×10⁶ × 7×10⁻¹²) = -123.4 kΩ
Z_sh = -j123.4 kΩ

Total spark impedance:

Z_spark = Z_mut + Z_sh (series combination)

Convert to rectangular:
Z_mut = 46.9k × cos(-29.4°) - j×46.9k × sin(29.4°)
      = 40.9k - j23.0k

Z_spark = (40.9k - j23.0k) + (0 - j123.4k)
        = 40.9k - j146.4k

|Z_spark| = √(40.9² + 146.4²) = 152.0 kΩ
φ_spark = atan(-146.4/40.9) = -74.4°

Total impedance (coil + spark):

Z_total = Z_th + Z_spark
        = (114 - j2424) + (40900 - j146400)
        = (114 + 40900) - j(2424 + 146400)
        = 41014 - j148824

R_total = 41,014 Ω ≈ 41.0 kΩ
X_total = -148,824 Ω ≈ -148.8 kΩ

|Z_total| = √(41.0² + 148.8²) = 154.3 kΩ

Current through spark:

I = V_th / Z_total
|I| = 350,000 V / 154,300 Ω
    = 2.268 A peak
    = 1.604 A RMS

Voltage across spark:

|V_spark| = |I| × |Z_spark|
          = 2.268 A × 152,000 Ω
          = 344,700 V
          ≈ 345 kV peak

Voltage check:

V_spark / V_th = 345 / 350 = 0.986 = 98.6%

Most voltage appears across spark (excellent!)
This is because Z_spark >> Z_th

Power in spark:

P_spark = 0.5 × |I|² × Re{Z_spark}
        = 0.5 × (2.268)² × 40,900
        = 0.5 × 5.144 × 40,900
        = 105,200 W
        ≈ 105 kW

Validation check:

Alternative calculation:
I_RMS = 2.268 / √2 = 1.604 A
P = I_RMS² × R = 1.604² × 40,900 = 105,100 W ✓

Load Case 2: Theoretical Maximum Power (Conjugate Match)

Conjugate match condition:

For maximum power: Z_load = Z_th*

Z_th = 114 - j2424
Z_th* = 114 + j2424  (conjugate)

Optimal load would be R = 114 Ω, X = +2424 Ω (inductive)

Total impedance with conjugate match:

Z_total = Z_th + Z_th*
        = (114 - j2424) + (114 + j2424)
        = 228 + j0
        = 228 Ω (purely resistive!)

Current at conjugate match:

I_max = V_th / Z_total
      = 350,000 / 228
      = 1535 A peak (!)

Maximum power:

P_max = 0.5 × |I_max|² × R_th
      = 0.5 × (1535)² × 114
      = 0.5 × 2,356,225 × 114
      = 134,305,000 W
      = 134.3 MW (!)

Alternative formula:
P_max = |V_th|² / (8 × R_th)
      = (350,000)² / (8 × 114)
      = 1.225×10¹¹ / 912
      = 134.3 MW ✓

Load Case 3: Efficiency Comparison

Actual spark (Case 1):

P_actual = 105 kW

Theoretical maximum (Case 2):

P_max = 134.3 MW

Power transfer efficiency:

η = P_actual / P_max
  = 105,000 / 134,300,000
  = 0.000782
  = 0.0782%

Less than 0.1% of theoretical maximum!

Why such low efficiency?

1. Impedance mismatch:

Z_spark = 40.9k - j146.4k Ω
Z_th* = 114 + j2424 Ω

Resistance ratio: 40,900 / 114 = 359×  (way too high!)
Reactance wrong sign: Need +2424 Ω, have -146,400 Ω

2. Topological constraints:

• Spark structure (R || C_mut in series with C_sh) is inherently capacitive
• Cannot produce positive (inductive) reactance
• Cannot achieve R as low as 114 Ω with realistic plasma
• The 0.1% is NOT a design flaw - it's fundamental physics!

3. Different optimization goal:

• We optimize for high voltage (field at tip)
• Power efficiency is secondary
• 98.6% voltage transfer is excellent (what matters for sparks)

Load Case 4: Shorter Spark (2 feet)

Spark parameters:

C_mut = 9 pF (same topload)
C_sh = 4 pF (2 ft spark, lower shunt capacitance)

R_spark = 1/(ω × 13 pF) = 66.3 kΩ

Following same procedure:

Z_spark ≈ 53 - j175 kΩ
Z_total ≈ 53.1 - j177.4 kΩ
|Z_total| ≈ 185.2 kΩ

I = 350kV / 185.2kΩ = 1.89 A peak

P_spark = 0.5 × (1.89)² × 53,000 = 94.8 kW

Comparison:

• Shorter spark: 95 kW
• Original 3.5 ft spark: 105 kW
• Longer spark gets more power (better matched at this voltage)

Load Case 5: Resistive Load (Theoretical)

Pure resistor: R_load = 50 kΩ, no reactance

Z_total = Z_th + Z_load
        = (114 - j2424) + 50,000
        = 50,114 - j2424

|Z_total| = √(50114² + 2424²) = 50,173 Ω

I = 350,000 / 50,173 = 6.98 A

P_load = 0.5 × (6.98)² × 50,000 = 1,219 kW

Wow! 1.2 MW into resistor vs 105 kW into spark.

Why the difference?

• Resistor has no large capacitive reactance
• Better impedance match
• But this is theoretical - no such "plasma resistor" exists
• Spark inherently has large capacitance (physics limitation)

Summary Table: Power Delivery to Various Loads

| Load Type | Z_load | |Z_total| | Current | Power | Efficiency | |-----------|---------|-----------|---------|-------|------------| | 3.5 ft spark | 40.9k-j146k | 154 kΩ | 2.27 A | 105 kW | 0.078% | | 2 ft spark | 53k-j175k | 185 kΩ | 1.89 A | 95 kW | 0.071% | | Pure 50k resistor | 50k+j0 | 50 kΩ | 6.98 A | 1219 kW | 0.91% | | Conjugate match | 114+j2424 | 228 Ω | 1535 A | 134.3 MW | 100% |

Practical Implications

Why Thévenin Method is Powerful

Once you have V_th and Z_th, you can:

1. Instantly calculate power for any load (no new simulations!)
2. Sweep resistance values to find optimal R
3. Compare different spark lengths quickly
4. Validate lumped vs distributed models
5. Predict behavior with varying drive levels (V_th scales linearly)

Example: Sweep R_spark from 10 kΩ to 200 kΩ

For each R value:

• Construct Z_spark from R and capacitances
• Calculate Z_total = Z_th + Z_spark
• Calculate I = V_th / Z_total
• Calculate P = 0.5 × I² × R
• Plot P vs R

Find peak: This is R_opt_power, typically 40-80 kΩ for this coil.

Time savings:

• Full simulation: 10-30 seconds each × 100 points = 16-50 minutes
• Thévenin method: <1 second total (simple formula)

Voltage Transfer vs Power Transfer

Two different goals:

Power transfer efficiency:

η_power = P_load / P_max = 0.078% (poor)

Voltage transfer efficiency:

η_voltage = V_spark / V_th = 345kV / 350kV = 98.6% (excellent!)

For Tesla coils:

• Voltage transfer is critical (drives E-field at tip)
• Power efficiency is secondary
• The mismatch is fundamental, not a flaw
• High Z_spark is desirable (safety, controlled current)

Key Insights

Extraction Process

1. Z_th measurement: Drive OFF, apply 1V test, measure current
2. V_th measurement: Drive ON, no load, measure voltage
3. Both are complex (magnitude and phase matter)
4. Frequency-specific: Extract at operating frequency

Physical Meaning

Z_th = 114 - j2424 Ω:

• R_th: System losses (copper, dielectric, damping)
• X_th: Predominantly topload capacitance with resonant enhancement
• Q = 21: Coupled system quality factor

V_th = 350 kV:

• Open-circuit voltage achievable
• Scales with drive voltage
• Voltage gain ~1000× due to resonance and turns ratio

Power Calculations

Simple formula once Z_th known:

P_load = 0.5 × |V_th|² × Re{Z_load} / |Z_th + Z_load|²

Maximum possible:

P_max = |V_th|² / (8 × R_th) = 134 MW (unachievable)

Typical spark:

P ≈ 50-150 kW for medium coil (0.05-0.1% of theoretical max)

Common Mistakes to Avoid

1. Removing tank components when measuring Z_th (changes network!)
2. Using magnitude only (phase information is critical)
3. Comparing to wrong maximum (conjugate match is unachievable)
4. Expecting high power efficiency (voltage efficiency matters, not power)
5. Forgetting factor of 0.5 in power calculation (peak vs RMS)
6. Wrong current measurement point (use port current, not base current)
7. Assuming linearity at all levels (valid for small signals, breaks at saturation)

See Also

• Related Lessons:

  • Module 2, Lesson 3: Thévenin Extraction (theory)
  • Module 2, Lesson 4: Thévenin Calculations (applications)
  • Module 2, Lesson 5: Direct Measurement Method (alternative)

• Related Worked Examples:

  • calculating-ropt.md: Finding optimal spark resistance
  • distributed-model-complete.md: Multi-segment analysis

• Related Exercises:

  • Exercise opt-ex-03: Thévenin extraction practice
  • Exercise opt-ex-04: Power calculation problems