---
id: fund-01
title: "Introduction to Tesla Coil Spark Modeling"
section: "Fundamentals"
difficulty: "beginner"
estimated_time: 20
prerequisites: []
objectives:
  - Understand the scope and goals of Tesla coil spark modeling
  - Review essential AC circuit fundamentals including peak vs RMS values
  - Master complex number notation and phasor representation
  - Learn power calculations using peak phasors
  - Understand impedance and admittance concepts
tags: ["introduction", "ac-circuits", "phasors", "complex-numbers", "power"]
---

# Introduction to Tesla Coil Spark Modeling

## Overview

This lesson plan is designed to take you from basic circuit concepts through advanced Tesla coil spark modeling. Tesla coil sparks are complex plasma phenomena that require understanding of AC circuits, electromagnetic fields, and plasma physics. By the end of this series, you'll be able to predict spark behavior and optimize coil performance.

### What You'll Learn

The complete course is divided into four parts:

1. **Part 1: Fundamentals** - Circuits, impedance, and basic spark behavior
2. **Part 2: Optimization** - Power transfer and efficiency
3. **Part 3: Growth Physics** - FEMM modeling and energy requirements
4. **Part 4: Advanced Topics** - Distributed models and real-world application

This lesson begins Part 1 by establishing the circuit theory foundation you'll need throughout.

## AC Circuit Fundamentals Review

### Peak vs RMS Values

In AC circuits, voltage and current vary sinusoidally with time:

**Time domain:**
```
v(t) = V_peak × cos(ωt + φ)
```

**Two amplitude conventions:**
- **Peak value:** The maximum value reached (V_peak)
- **RMS value:** Root-Mean-Square, V_RMS = V_peak/√2 ≈ 0.707 × V_peak

**For this entire framework, we use PEAK VALUES exclusively.**

**Why peak values?**
1. Tesla coils are concerned with maximum voltage (breakdown, field stress)
2. Consistent with phasor notation in engineering
3. Power formula becomes: P = 0.5 × V_peak × I_peak × cos(θ)

**Example:** If your oscilloscope shows a 100 kV peak-to-peak waveform:
- V_peak-to-peak = 100 kV
- V_peak = 50 kV (one-sided amplitude)
- V_RMS = 50 kV / √2 ≈ 35.4 kV

### Complex Numbers and Phasors

AC circuit analysis uses complex numbers to represent magnitude and phase simultaneously.

**Rectangular form:**
```
Z = R + jX
where j = √(-1) (imaginary unit, engineers use 'j' instead of 'i')
R = real part (resistance)
X = imaginary part (reactance)
```

**Polar form:**
```
Z = |Z| ∠φ = |Z| × e^(jφ)
where |Z| = √(R² + X²) (magnitude)
      φ = atan(X/R) (phase angle)
```

**Conversion:**
```
R = |Z| × cos(φ)
X = |Z| × sin(φ)
```

**Phasor notation:** A complex number representing sinusoidal amplitude and phase:
```
V = V_peak ∠φ_v
I = I_peak ∠φ_i
```

**Complex conjugate:** Used in power calculations
```
If I = a + jb, then I* = a - jb (flip sign of imaginary part)
```

### Resistance, Reactance, Impedance

**Resistance (R):** Opposition to current that dissipates energy as heat
- Units: Ω (ohms)
- Always real and positive
- V = I × R (Ohm's law)

**Reactance (X):** Opposition to current that stores energy (no dissipation)
- Units: Ω (ohms)
- Can be positive (inductive) or negative (capacitive)
- **Capacitive reactance:** X_C = -1/(ωC) where ω = 2πf
- **Inductive reactance:** X_L = ωL

**Impedance (Z):** Total opposition to AC current
```
Z = R + jX (complex)
|Z| = √(R² + X²)
φ_Z = atan(X/R)
```

**Sign conventions:**
- X > 0: inductive (current lags voltage)
- X < 0: capacitive (current leads voltage)
- φ_Z > 0: inductive
- φ_Z < 0: capacitive

### Conductance, Susceptance, Admittance

For parallel circuits, **admittance (Y)** is more convenient than impedance.

**Conductance (G):** Inverse of resistance
```
G = 1/R
Units: S (siemens)
```

**Susceptance (B):** Inverse of reactance (BUT with opposite sign convention!)
```
For capacitor: B_C = ωC (positive!)
For inductor: B_L = -1/(ωL) (negative)
```

**Important:** Susceptance sign convention is OPPOSITE of reactance:
- Capacitor: X_C < 0, but B_C > 0
- Inductor: X_L > 0, but B_L < 0

**Admittance (Y):** Inverse of impedance
```
Y = G + jB = 1/Z
|Y| = 1/|Z|
φ_Y = -φ_Z (opposite sign!)
```

**Conversion between Z and Y:**
```
Y = 1/Z = 1/(R + jX) = R/(R² + X²) - jX/(R² + X²)

Therefore:
G = R/(R² + X²)
B = -X/(R² + X²)
```

### Power in AC Circuits

**Using peak phasors:**
```
P = 0.5 × Re{V × I*}

where V and I are complex peak phasors
      I* is the complex conjugate of I
      Re{·} means "real part of"
```

**Why the 0.5 factor?**
- Average power over a full AC cycle
- Comes from time-averaging cos²(ωt), which equals 0.5
- If you used RMS values, formula would be P = V_RMS × I_RMS × cos(θ), NO 0.5

**Expanded form:**
```
If V = V_peak ∠φ_v and I = I_peak ∠φ_i, then:
P = 0.5 × V_peak × I_peak × cos(φ_v - φ_i)
```

The angle difference (φ_v - φ_i) is the power factor angle.

## Worked Example: Power Calculation with Peak Phasors

**Given:**
- Voltage: V = 50 kV ∠0° (peak, using 0° as reference)
- Impedance: Z = 100 kΩ ∠-60° (capacitive load)

**Find:** Real power dissipated

**Solution:**

Step 1: Calculate current using Ohm's law
```
I = V/Z = (50 kV ∠0°)/(100 kΩ ∠-60°)
I = 0.5 A ∠(0° - (-60°)) = 0.5 A ∠60°
```

Step 2: Calculate power
```
P = 0.5 × Re{V × I*}
P = 0.5 × Re{(50 kV ∠0°) × (0.5 A ∠-60°)}
P = 0.5 × Re{25 kW ∠-60°}
```

Step 3: Convert to rectangular to get real part
```
25 kW ∠-60° = 25 kW × (cos(-60°) + j×sin(-60°))
            = 25 kW × (0.5 - j×0.866)
            = 12.5 kW - j×21.65 kW
```

Step 4: Extract real part and apply 0.5 factor
```
P = 0.5 × 12.5 kW = 6.25 kW
```

**Alternative method:** Using power factor angle
```
P = 0.5 × V_peak × I_peak × cos(φ_v - φ_i)
P = 0.5 × 50 kV × 0.5 A × cos(0° - 60°)
P = 0.5 × 25 kW × cos(-60°)
P = 0.5 × 25 kW × 0.5
P = 6.25 kW
```

## Key Takeaways

- Always use **peak values** for Tesla coil analysis
- Complex numbers combine magnitude and phase: Z = R + jX = |Z|∠φ
- Power calculation: **P = 0.5 × Re{V × I*}** with peak phasors
- Admittance (Y = G + jB) is the inverse of impedance
- **Sign convention critical:** X < 0 for capacitors, but B > 0
- Phase angles are opposite: φ_Y = -φ_Z

## Practice

{exercise:fund-ex-01}

**Problem 1:** A capacitor has reactance X_C = -80 kΩ at 200 kHz. What is its capacitance? What is its susceptance?

**Problem 2:** An impedance Z = 50 kΩ - j75 kΩ has current I = 0.2 A ∠30° (peak). Calculate: (a) Voltage magnitude and phase, (b) Real power

**Problem 3:** An admittance Y = 0.00001 + j0.00002 S. Convert to impedance Z = R + jX.

---

**Next Lesson:** [Basic Circuit Model](02-basic-circuit-model.md)