# 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