plasma-expert/spark-lessons/lessons/02-optimization/04-thevenin-calculations.md

id	title	section	difficulty	estimated_time	prerequisites	objectives	tags
opt-04	Using the Thévenin Equivalent - Power Calculations	Optimization & Simulation	intermediate	45	[opt-03 opt-01]	[Calculate load voltage and current using Thévenin equivalent Compute power delivered to arbitrary loads Determine maximum theoretical power (conjugate match) Understand why conjugate match is usually unachievable]	[thevenin power-calculation impedance-matching circuit-analysis]

Using the Thévenin Equivalent - Power Calculations

Now that we've extracted the Thévenin equivalent (V_th and Z_th), we can use it to predict coil performance with any load without re-running full simulations. This lesson shows how to perform these calculations and interpret the results.

Predicting Behavior with Any Load

Once you have V_th and Z_th, the Tesla coil looks like this simple circuit:

     ┌────┐
     │V_th├───[Z_th]───┬─── Output
     └────┘             │
                     [Z_load]
                        │
                       GND

This is just a voltage divider! We can apply basic circuit analysis.

Voltage Across Load

Using voltage divider rule:

V_load = V_th × [Z_load / (Z_th + Z_load)]

Complex arithmetic: Both numerator and denominator are complex numbers, so you need to handle magnitude and phase carefully.

Current Through Load

Using Ohm's law on the series circuit:

I = V_th / (Z_th + Z_load)

This current flows through both Z_th and Z_load since they're in series.

Power Delivered to Load

Power dissipated in the load (real power only):

P_load = 0.5 × |I|² × Re{Z_load}

Or equivalently:

P_load = 0.5 × Re{V_load × I*}

where I* is the complex conjugate of I.

Direct formula combining everything:

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

This formula is gold! It lets you sweep different Z_load values and calculate power without any additional simulation.

Step-by-Step Calculation Process

Given Information

• V_th (complex voltage phasor)
• Z_th = R_th + jX_th (complex impedance)
• Z_load = R_load + jX_load (spark impedance from model)

Step 1: Calculate Total Impedance

Z_total = Z_th + Z_load
        = (R_th + R_load) + j(X_th + X_load)

R_total = R_th + R_load
X_total = X_th + X_load

|Z_total| = √(R_total² + X_total²)

Step 2: Calculate Current

I = V_th / Z_total

|I| = |V_th| / |Z_total|

φ_I = φ_V_th - φ_Z_total

where φ_Z_total = atan(X_total / R_total)

Step 3: Calculate Load Voltage

V_load = I × Z_load

|V_load| = |I| × |Z_load|

φ_V_load = φ_I + φ_Z_load

Or use voltage divider directly (often simpler):

|V_load| = |V_th| × |Z_load| / |Z_total|

Step 4: Calculate Power in Load

P_load = 0.5 × |I|² × R_load

P_load = 0.5 × |I|² × Re{Z_load}

The factor of 0.5 accounts for peak phasor to RMS conversion in AC power.

Worked Example: Complete Thévenin Analysis

Given:

• Z_th = 114 - j2424 Ω (from previous lesson)
• V_th = 350 kV ∠0° (measured with drive on, no load)
• Spark load: Z_spark = 60 kΩ - j160 kΩ (from lumped model)

Find: (a) Current through spark (b) Voltage across spark (c) Power dissipated in spark (d) Theoretical maximum power (conjugate match)

Part (a): Current Through Spark

Calculate total impedance:

Z_total = Z_th + Z_spark
        = (114 - j2424) + (60000 - j160000)
        = (60114 - j162424) Ω

R_total = 60114 Ω
X_total = -162424 Ω

|Z_total| = √(60114² + 162424²)
         = √(3.614×10⁹ + 2.638×10¹⁰)
         = √(3.000×10¹⁰)
         = 173.2 kΩ

Calculate current:

I = V_th / Z_total
|I| = 350 kV / 173.2 kΩ = 2.02 A peak

Part (b): Voltage Across Spark

Method 1: Voltage divider

|Z_spark| = √(60000² + 160000²)
         = √(3.6×10⁹ + 2.56×10¹⁰)
         = √(2.92×10¹⁰)
         = 171 kΩ

|V_spark| = |V_th| × |Z_spark| / |Z_total|
          = 350 kV × (171 kΩ / 173.2 kΩ)
          = 350 kV × 0.987
          = 345 kV

Method 2: Using current

|V_spark| = |I| × |Z_spark|
          = 2.02 A × 171 kΩ
          = 345 kV

Observation: Most voltage appears across the spark! This makes sense because Z_spark >> Z_th.

Part (c): Power in Spark

P_spark = 0.5 × |I|² × Re{Z_spark}
        = 0.5 × (2.02)² × 60000
        = 0.5 × 4.08 × 60000
        = 122 kW

This is the real power dissipated in heating, ionization, radiation, and sound in the spark.

Part (d): Theoretical Maximum Power

The maximum power transfer theorem states that power is maximized when the load impedance is the complex conjugate of the source impedance.

Conjugate match condition:

Z_load = Z_th*  (complex conjugate)

If Z_th = R_th + jX_th
Then Z_load = R_th - jX_th

For our case:
Z_th = 114 - j2424 Ω
Z_load_optimal = 114 + j2424 Ω

Why this maximizes power:

• Reactive components cancel: Z_total = Z_th + Z_th* = 2R_th (purely real)
• No reactive power circulation
• All delivered power is real

Maximum power formula:

P_max = |V_th|² / (8 × R_th)

Calculate:

P_max = (350×10³)² / (8 × 114)
      = 1.225×10¹¹ / 912
      = 134.3 MW

Wait, this seems enormous!

Let's double-check:

With Z_load = 114 + j2424 Ω:

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

I = 350 kV / 228 Ω = 1535 A

P = 0.5 × (1535)² × 114 = 134.3 MW ✓

Part (e): Reality Check - Why Such a Huge Difference?

Actual spark power: 122 kW Theoretical maximum: 134.3 MW Efficiency: 122 / 134,300 = 0.09% of theoretical maximum

Why such a huge discrepancy?

1. Conjugate match requires Z_load = 114 + j2424 Ω

   • This means R_load = 114 Ω (extremely low!)
   • This means X_load = +2424 Ω (inductive, not capacitive)

2. Actual spark: Z_spark = 60 kΩ - j160 kΩ

   • R_spark = 60 kΩ (525× too high!)
   • X_spark = -160 kΩ (capacitive, wrong sign, 66× too large)

3. Topological constraints prevent achieving conjugate match:

   • Spark structure (R||C_mut in series with C_sh) is inherently capacitive
   • Cannot produce positive (inductive) reactance
   • Cannot achieve R_load as low as 114 Ω with realistic plasma

This is normal for Tesla coils! The impedance mismatch is fundamental to the physics of spark discharges. We cannot achieve conjugate match in practice.

Understanding Efficiency

What Does 0.09% Mean?

It does NOT mean the coil is "inefficient" in the usual sense. Rather:

• The coil has very low output impedance (114 Ω)
• The spark has very high impedance (171 kΩ)
• This is a 1500:1 impedance mismatch
• The voltage divider heavily favors the spark (good!)
• Most voltage appears at the spark, but current is limited

Voltage Transfer Efficiency

Voltage across spark / Total voltage:
345 kV / 350 kV = 98.6%

We achieve excellent voltage transfer! This is what matters for spark length (field at tip).

Why Not Match Impedances?

In conventional circuits: Match impedances for maximum power transfer

In Tesla coils: We WANT high spark impedance because:

• High voltage at spark tip drives field
• High resistance means controlled current (safety)
• Mismatch is unavoidable due to plasma physics
• Optimization focuses on maximizing power given the constraints

Practical Use: Sweeping Spark Parameters

The real power of Thévenin analysis is rapid parameter sweeps:

Given: V_th = 350 kV, Z_th = 114 - j2424 Ω

Sweep: Spark resistance R from 10 kΩ to 200 kΩ

For each R value:

1. Construct Z_spark from R and capacitances (using lumped model)
2. Calculate Z_total = Z_th + Z_spark
3. Calculate I = V_th / Z_total
4. Calculate P = 0.5 × |I|² × R
5. Plot P vs R

Result: You find P_max at R ≈ R_opt_power without any new simulations!

When Thévenin Analysis Fails

Nonlinearity

Assumption: Coil behaves linearly (impedances don't change with voltage/current)

Breaks down when:

• Magnetic cores saturate
• Component heating changes parameters
• Very large sparks significantly load the coil

Solution: Iterate - use results to update model, re-extract Thévenin

Frequency Dependence

Assumption: Operating at a single frequency

Breaks down when:

• Spark loading shifts resonant frequency
• Comparing different loads at fixed frequency (detuning varies)

Solution: Extract Z_th(ω) and V_th(ω), account for frequency shift (next lessons)

Coupled Modes

Assumption: Single-mode operation

Breaks down when:

• Operating between two coupled poles
• Mode hopping as spark changes loading

Solution: Full coupled-mode analysis or stay clearly in one mode

Key Takeaways

• Thévenin circuit: Simple series combination of V_th and Z_th
• Load voltage: V_load = V_th × Z_load/(Z_th + Z_load)
• Load current: I = V_th / (Z_th + Z_load)
• Load power: P = 0.5 × |I|² × Re{Z_load} or P = 0.5 × |V_th|² × Re{Z_load}/|Z_th + Z_load|²
• Maximum power: Requires conjugate match Z_load = Z_th*
• P_max = |V_th|²/(8R_th) but usually unachievable
• Tesla coils operate far from conjugate match due to physics constraints
• High voltage transfer efficiency matters more than impedance matching
• Parameter sweeps become trivial with Thévenin equivalent

Practice

{exercise:opt-ex-04}

Problem 1: Given Z_th = 95 - j1850 Ω, V_th = 280 kV, and a spark model with Z_spark = 50 kΩ - j140 kΩ: (a) Calculate total impedance (b) Calculate current through spark (c) Calculate power delivered to spark (d) Calculate theoretical maximum power (conjugate match) (e) What percentage of theoretical maximum is achieved?

Problem 2: A load Z_load = 200 + j200 Ω is connected to a coil with Z_th = 100 - j2000 Ω and V_th = 300 kV. (a) Calculate the load voltage (b) Calculate power delivered (c) Is this load inductive or capacitive? (d) Is this load closer to conjugate match than a typical spark?

Problem 3: For Z_th = 120 - j2200 Ω: (a) What load impedance gives conjugate match? (b) Calculate P_max if V_th = 400 kV (c) If actual spark has R = 70 kΩ, X = -180 kΩ, calculate actual power (d) Calculate the power transfer efficiency ratio

Problem 4: A coil has V_th = 350 kV and Z_th = 110 - j2500 Ω. You want to deliver 100 kW to a purely resistive load. What resistance value is required? (Hint: Set P = 100 kW in power formula and solve for R)

Problem 5: Explain physically why Tesla coils operate so far from conjugate match. Why can't we just add inductance to the spark to cancel its capacitive reactance?

---

Next Lesson: Direct Power Measurement Method