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id	title	section	difficulty	estimated_time	prerequisites	objectives	tags
opt-05	Direct Power Measurement Method	Optimization & Simulation	intermediate	25	[opt-04 opt-01]	[Understand the direct measurement alternative to Thévenin Set up simulations for direct power measurement Extract spark resistance through power optimization Compare advantages and disadvantages of each method]	[power-measurement simulation optimization methodology]

Direct Power Measurement Method

While the Thévenin equivalent method is powerful and elegant, there's an alternative approach: directly measure power delivered to the spark in a full simulation. Each method has advantages and trade-offs.

The Direct Measurement Approach

Concept

Instead of extracting a simplified equivalent circuit, keep the full coupled model with the spark load present and directly measure power flow.

Setup:

Build complete simulation (primary, secondary, coupling, spark load)
Drive primary at operating frequency and amplitude
Run AC analysis (or transient with post-processing)
Measure power dissipated in spark resistance
Repeat for different spark resistance values

Goal: Find the spark resistance R that maximizes measured power

Procedure

Step 1: Build Full Model

Primary tank circuit (L_primary, C_MMC)
Secondary coil (distributed or lumped model)
Topload capacitance
Magnetic coupling k
Spark load modeled as R||C_mut in series with C_sh

Step 2: Set Operating Point

Drive frequency: f_drive (initially at unloaded resonance)
Drive amplitude: V_drive or I_drive
Spark parameters: Choose initial R, C_mut, C_sh

Step 3: Run AC Analysis

Solve circuit at drive frequency
Extract voltage and current at spark resistor
Calculate power: P = 0.5 × Re{V_spark × I_spark*}

Or more directly:

P = 0.5 × |I_R|² × R

where I_R is current through the resistance R

Step 4: Sweep R Values

Vary R from 10 kΩ to 200 kΩ (typical range)
For each R, measure P
Plot P vs R
Find R that gives maximum P → this is R_opt_power

Step 5: Validate

Compare numerical R_opt_power to analytical formula
Check that it matches: R_opt = 1/[ω(C_mut + C_sh)]

Power Measurement in SPICE

Method 1: Using Current Through Resistor

.param Rspark = 50k
Rspark topload node2 {Rspark}
Cmut node2 0 8p
Csh topload 0 6p

.ac lin 1 185k 185k
.step param Rspark list 10k 30k 50k 70k 100k 150k

.meas ac Ispark_mag find mag(I(Rspark))
.meas ac Pspark param '0.5 * Ispark_mag^2 * Rspark'

This sweeps Rspark and calculates power for each value.

Method 2: Direct Power Function

Some SPICE variants support direct power measurement:

.meas ac Pspark_real find Re(V(topload)*conj(I(Rspark)))

This directly computes complex power and extracts the real part.

Method 3: Voltage and Current

.meas ac Vtop_mag find mag(V(topload))
.meas ac Ispark_mag find mag(I(Rspark))
.meas ac phase_diff param 'ph(V(topload)) - ph(I(Rspark))'
.meas ac Pspark param '0.5 * Vtop_mag * Ispark_mag * cos(phase_diff)'

This accounts for phase difference in power calculation.

Worked Example: Direct Optimization

Given:

DRSSTC simulation at f = 185 kHz
Primary drive: V_drive produces V_top ≈ 350 kV (unloaded)
Spark model: C_mut = 8 pF, C_sh = 6 pF, R = variable

Goal: Find R_opt_power

Analytical Prediction

First, predict what we should find:

C_total = C_mut + C_sh = 8 + 6 = 14 pF
ω = 2π × 185×10³ = 1.162×10⁶ rad/s

R_opt_power = 1/(ωC_total)
            = 1/(1.162×10⁶ × 14×10⁻¹²)
            = 61.5 kΩ

We expect maximum power near 61.5 kΩ.

Simulation Sweep

Run AC analysis with R values:

R = 20 kΩ → P = 85 kW
R = 40 kΩ → P = 115 kW
R = 60 kΩ → P = 125 kW ← Maximum
R = 80 kΩ → P = 118 kW
R = 100 kΩ → P = 105 kW

Result: Maximum power at R ≈ 60 kΩ

Validation: Simulation (60 kΩ) matches theory (61.5 kΩ) within rounding!

Advantages of Direct Measurement

1. No Approximations

Full coupled model captures all interactions
No linearization assumptions
Includes all nonlinear effects (if using transient analysis)

2. Intuitive

Directly see what you care about: power to spark
No intermediate steps
Easy to visualize results

3. Flexibility

Can use any circuit simulator
Works with complex topologies
Easy to add additional elements (damping, protection, etc.)

4. Transient Capability

Can extend to time-domain (transient) analysis
Capture burst mode, ramping, dynamics
See energy transfer over time

Disadvantages of Direct Measurement

1. Computational Cost

Must re-run full simulation for each R value
Sweep of 20 points = 20 full simulations
Slow for large parameter spaces

2. Limited Insight

Doesn't reveal underlying equivalent circuit
Harder to understand why maximum occurs where it does
Less portable to different load types

3. Frequency Coupling

Operating frequency may need adjustment for each R (see next lesson!)
Fixed-frequency comparison can be misleading
Must account for resonance shift

4. Sensitivity to Setup

Results depend on drive amplitude, frequency, damping
Harder to isolate spark effects from system effects

Comparison: Thévenin vs Direct

Aspect	Thévenin Method	Direct Method
Speed	Fast (single extraction + algebra)	Slow (simulation per R value)
Insight	High (reveals equivalent circuit)	Moderate
Accuracy	Excellent (if linear)	Excellent (includes nonlinearities)
Flexibility	Any load instantly	One load per simulation
Complexity	Requires understanding of method	Straightforward
Best for	Sweeps, optimization, understanding	Validation, nonlinear cases

When to Use Each Method

Use Thévenin When:

Exploring many different load configurations
Optimizing spark parameters
Building intuition about matching
Preparing design curves
Speed is important

Use Direct Measurement When:

Validating Thévenin results
Dealing with significant nonlinearities
Need transient/time-domain behavior
Checking specific operating points
Learning circuit behavior

Best Practice: Use Both

Start with Thévenin: Fast exploration, find optimal regions
Validate with Direct: Confirm key points, check assumptions
Iterate: If discrepancies exist, understand why

Accounting for Displacement Currents

Both methods can fall victim to the "I_base error" discussed in Module 2.4.

The Problem

Wrong: Measuring total current returning through secondary base

Right: Measuring current specifically through spark resistance

Why It Matters

Total base current includes:

Spark current (what we want)
Displacement currents from secondary to ground
Coupling currents to primary
Environmental coupling

In SPICE: This isn't usually a problem because you can measure specific branch currents. Use I(Rspark) not I(V_secondary_base).

In physical measurements: You must use current probes on the spark return path, not the coil base.

Implementation Tips

Tip 1: Automate Sweeps

Use SPICE .STEP or scripting:

.step param Rspark 10k 200k 5k

This automatically sweeps from 10 kΩ to 200 kΩ in 5 kΩ steps.

Tip 2: Log Scale for Wide Ranges

Spark resistance varies over decades (10 kΩ to 1 MΩ). Use logarithmic stepping:

.step param Rspark list 10k 20k 50k 100k 200k 500k

Tip 3: Extract Peak Directly

Use .MEAS to find maximum automatically:

.meas ac Pmax MAX Pspark
.meas ac Ropt WHEN Pspark=Pmax

Tip 4: Verify Power Components

Separately measure real and reactive power:

P_real = Re{V × I*}
Q_reactive = Im{V × I*}
S_apparent = |V × I*|

Check that Q >> P (highly reactive, as expected).

Key Takeaways

Direct measurement: Keep full model, measure power in spark, sweep R
Advantages: Intuitive, no approximations, handles nonlinearity
Disadvantages: Slow, less insight, multiple simulations required
Power formula: P = 0.5 × |I_R|² × R or P = 0.5 × Re{V × I*}
Find R_opt: Sweep R, plot P vs R, identify maximum
Validation: Should match analytical R_opt = 1/[ω(C_mut + C_sh)]
Best practice: Use Thévenin for exploration, direct measurement for validation
Beware: Measure spark current, not base current (displacement current issue)

Practice

{exercise:opt-ex-05}

Problem 1: You run simulations with the following results:

R (kΩ)	P (kW)
30	92
50	118
70	128
90	125
110	115

(a) Estimate R_opt_power from this data (b) If C_total = 12 pF and f = 200 kHz, what does theory predict? (c) Do they match?

Problem 2: A simulation reports I_R = 2.1 A (peak) through R = 55 kΩ. Calculate the power dissipated.

Problem 3: You measure V_topload = 340 kV ∠0° and I_spark = 1.8 A ∠-72°. (a) Calculate apparent power S = V × I* (b) Extract real power P = Re{S} (c) Extract reactive power Q = Im{S} (d) Is the spark more resistive or reactive?

Problem 4: List two scenarios where direct measurement would be preferred over Thévenin extraction.

Problem 5: Why is it important to measure I(Rspark) rather than I(V_secondary_base) when calculating power? Sketch the circuit showing both current paths.

Next Lesson: Frequency Tracking and Loaded Poles

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