---
id: fund-03
title: "Admittance Analysis of the Spark Circuit"
section: "Fundamentals"
difficulty: "intermediate"
estimated_time: 30
prerequisites: ["fund-01", "fund-02"]
objectives:
  - Understand why admittance is preferred over impedance for parallel circuits
  - Derive the total admittance formula for the spark circuit
  - Calculate real and imaginary parts of admittance
  - Convert between admittance and impedance representations
  - Apply formulas to practical Tesla coil examples
tags: ["admittance", "circuit-analysis", "complex-algebra", "formulas"]
---

# Admittance Analysis of the Spark Circuit

## Introduction

The spark circuit topology (R || C_mut in series with C_sh) requires careful analysis. While we could work entirely with impedances, using admittance simplifies the parallel combination and provides clearer insight into circuit behavior.

## Why Use Admittance?

For the spark circuit topology (parallel R||C_mut, in series with C_sh), admittance simplifies calculations.

**Parallel elements:** Add admittances directly
```
Y_total = Y₁ + Y₂ + Y₃ + ...
vs impedances: 1/Z_total = 1/Z₁ + 1/Z₂ + ... (messy!)
```

**Our circuit:**
```
Y_mut_R = Y_Cmut + Y_R  (parallel: C_mut || R)
Then series with C_sh requires impedance: Z = Z_mut_R + Z_Csh
Then convert back: Y_total = 1/Z_total
```

Admittance makes the first step (parallel combination) trivial, and we only need to handle the series combination once.

## Deriving the Total Admittance Formula

Let's work through the complete derivation step by step.

**Step 1:** Admittance of R and C_mut in parallel

```
Y_R = G = 1/R
Y_Cmut = jωC_mut = jB₁  (where B₁ = ωC_mut)

Y_mut_R = G + jB₁
```

**Step 2:** Convert to impedance for series combination

```
Z_mut_R = 1/(G + jB₁)
```

**Step 3:** Add impedance of C_sh in series

```
Z_Csh = 1/(jωC_sh) = -j/(ωC_sh) = 1/(jB₂)  (where B₂ = ωC_sh)

Z_total = Z_mut_R + Z_Csh
Z_total = 1/(G + jB₁) + 1/(jB₂)
```

**Step 4:** Find common denominator

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

**Step 5:** Invert to get admittance

```
Y_total = 1/Z_total = [jB₂(G + jB₁)] / [G + j(B₁ + B₂)]

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

This is the **fundamental admittance equation** for the spark circuit.

## Extracting Real and Imaginary Parts

To use this formula, we need to separate it into Re{Y} and Im{Y}.

Multiply numerator:
```
(G + jB₁) × jB₂ = jGB₂ + j²B₁B₂ = jGB₂ - B₁B₂
                = -B₁B₂ + jGB₂
```

So:
```
Y = [-B₁B₂ + jGB₂] / [G + j(B₁ + B₂)]
```

To separate real and imaginary parts, multiply numerator and denominator by complex conjugate of denominator:

```
Denominator conjugate: G - j(B₁ + B₂)
Denominator magnitude squared: G² + (B₁ + B₂)²
```

After algebra (multiply out and simplify):

```
Re{Y} = GB₂² / [G² + (B₁ + B₂)²]

Im{Y} = B₂[G² + B₁(B₁ + B₂)] / [G² + (B₁ + B₂)²]
```

These are the **working formulas** for calculating admittance from R, C_mut, C_sh.

### Formula Summary

Given R, C_mut, C_sh, and frequency f:

**Step 1:** Calculate component values
```
ω = 2πf
G = 1/R
B₁ = ωC_mut
B₂ = ωC_sh
```

**Step 2:** Calculate admittance
```
Re{Y} = GB₂² / [G² + (B₁ + B₂)²]

Im{Y} = B₂[G² + B₁(B₁ + B₂)] / [G² + (B₁ + B₂)²]

Y = Re{Y} + j×Im{Y}
```

**Step 3:** Magnitude and phase
```
|Y| = √[Re{Y}² + Im{Y}²]
φ_Y = atan(Im{Y}/Re{Y})
```

## Converting to Impedance

From Y = G_total + jB_total:

```
Z = 1/Y = 1/(G_total + jB_total)

Multiply by conjugate:
Z = (G_total - jB_total) / (G_total² + B_total²)

R_total = G_total / (G_total² + B_total²)
X_total = -B_total / (G_total² + B_total²)

Or directly:
|Z| = 1/|Y|
φ_Z = -φ_Y  (opposite sign!)
```

## Worked Example: Complete Y and Z Calculation

**Given:**
- Frequency: f = 200 kHz → ω = 2π × 200×10³ = 1.257×10⁶ rad/s
- C_mut = 8 pF = 8×10⁻¹² F
- C_sh = 6 pF = 6×10⁻¹² F
- R = 100 kΩ = 10⁵ Ω

**Find:** Y_total (rectangular), Z_total (rectangular and polar)

**Solution:**

Step 1: Calculate component values
```
G = 1/R = 1/(10⁵) = 10⁻⁵ S = 10 μS
B₁ = ωC_mut = 1.257×10⁶ × 8×10⁻¹² = 10.06×10⁻⁶ S = 10.06 μS
B₂ = ωC_sh = 1.257×10⁶ × 6×10⁻¹² = 7.54×10⁻⁶ S = 7.54 μS
```

Step 2: Calculate Re{Y}
```
Re{Y} = GB₂² / [G² + (B₁ + B₂)²]

Numerator: 10 × (7.54)² = 10 × 56.85 = 568.5 μS²
Denominator: (10)² + (10.06 + 7.54)² = 100 + (17.6)² = 100 + 309.8 = 409.8 μS²

Re{Y} = 568.5 / 409.8 = 1.387 μS
```

Step 3: Calculate Im{Y}
```
Im{Y} = B₂[G² + B₁(B₁ + B₂)] / [G² + (B₁ + B₂)²]

Numerator inner: G² + B₁(B₁ + B₂) = 100 + 10.06×17.6 = 100 + 177.1 = 277.1 μS²
Numerator: 7.54 × 277.1 = 2089.3 μS³
Denominator: 409.8 μS² (same as before)

Im{Y} = 2089.3 / 409.8 = 5.10 μS
```

Step 4: Admittance result
```
Y_total = 1.387 + j5.10 μS
|Y| = √(1.387² + 5.10²) = √(1.92 + 26.01) = √27.93 = 5.28 μS
φ_Y = atan(5.10/1.387) = atan(3.68) = 74.8°
```

Step 5: Convert to impedance
```
|Z| = 1/|Y| = 1/(5.28×10⁻⁶) = 189 kΩ
φ_Z = -φ_Y = -74.8°

In rectangular:
R_total = |Z| × cos(φ_Z) = 189 × cos(-74.8°) = 189 × 0.263 = 49.7 kΩ
X_total = |Z| × sin(φ_Z) = 189 × sin(-74.8°) = 189 × (-0.965) = -182 kΩ

Z_total = 49.7 - j182 kΩ = 189 kΩ ∠-74.8°
```

**Interpretation:**
- Impedance is strongly capacitive (φ_Z = -74.8°)
- Equivalent resistance ≈ 50 kΩ (half of actual R due to capacitive divider)
- Large capacitive reactance dominates

![Complex plane plots showing Y and Z](assets/complex-plane-admittance.png)

**Visualization notes:**
- LEFT: Admittance plane (Y = G + jB)
  - Point at (1.387, 5.10) μS
  - Angle φ_Y = 74.8° from horizontal
  - Positive B means capacitive in admittance

- RIGHT: Impedance plane (Z = R + jX)
  - Point at (49.7, -182) kΩ
  - Angle φ_Z = -74.8° below horizontal
  - Negative X means capacitive in impedance

- Connection: Angles are opposite (φ_Z = -φ_Y), magnitudes invert (|Z| = 1/|Y|)

## Key Takeaways

- **Admittance simplifies parallel combinations:** Y_parallel = Y₁ + Y₂ + ...
- **Fundamental formula:** Y = [(G + jB₁) × jB₂] / [G + j(B₁ + B₂)]
- **Working formulas:**
  - Re{Y} = GB₂² / [G² + (B₁ + B₂)²]
  - Im{Y} = B₂[G² + B₁(B₁ + B₂)] / [G² + (B₁ + B₂)²]
- **Conversion:** |Z| = 1/|Y| and φ_Z = -φ_Y
- Typical spark: strongly capacitive with large |Im{Y}| compared to Re{Y}

## Practice

{exercise:fund-ex-03}

**Problem 1:** For f = 150 kHz, C_mut = 10 pF, C_sh = 8 pF, R = 80 kΩ, calculate Y_total (real and imaginary parts).

**Problem 2:** An admittance Y = 2.0 + j4.5 μS. Convert to impedance Z in both rectangular and polar forms.

**Problem 3:** Show algebraically that if R → ∞ (open circuit), the formula reduces to Y = jωC_mut × C_sh/(C_mut + C_sh), which is two capacitors in series.

---

**Next Lesson:** [Phase Angles and Their Meaning](04-phase-angles.md)