⚙️ ENGINEER LEVEL: Power System Analysis
Electrical System Modeling
Complete system equivalent circuit:
Illustration note: Circuit schematic showing alternator model, battery models, wire impedances, distribution blocks, fuses, and amplifier loads with all parasitic elements
Components:
Alternator model:
V_alt = V_no_load - I_load × R_internal
Typical: Vnoload = 14.4V, R_internal = 0.015-0.030Ω
Battery model (linear approximation):
V_batt = V_oc - I × R_internal - I × R_wire
Non-linear battery model (more accurate):
V_batt = V_oc - I×R_internal - K×log(1 + I/I_0)
Where K = polarization constant, I_0 = exchange current
Wiring impedance:
AC impedance includes resistance and inductance:
Z_wire = R + jωL
For audio frequencies (20-200 Hz):
|Z| ≈ R (inductance effect minimal)
For switching frequencies (tens of kHz):
|Z| = √(R² + (ωL)²)
Distribution block:
Ideally zero impedance, but real blocks have:
R_block = R_connections + R_internal ≈ 1-10 mΩ
Fuse resistance: - ANL fuse: 0.5-2 mΩ (depends on rating and current) - MAXI fuse: 2-5 mΩ - Mini blade: 5-10 mΩ
Total supply impedance:
Z_supply = Z_alt + Z_batt + Z_wire + Z_dist + Z_fuse
Typical: 0.020 - 0.100Ω depending on system
Transient Analysis
Current demand profile:
Music is highly dynamic with rapid transients.
Illustration note: Oscilloscope trace showing actual current draw over time during music playback, highlighting peaks and average
Typical profile: - Average: 30-50% of rated current - Peaks: 100% of rated (brief) - Peak duration: 10-100ms - Repetition rate: 1-10 Hz (music dependent)
System response to transient:
t = 0: Transient begins - Amplifiers demand high current suddenly - Voltage begins to sag
t = 1ms: - Capacitors discharge, supplying current - Voltage drops by V = Q/C - Wiring inductance limits current rate: dI/dt = V/L
t = 10ms: - Batteries begin responding - Chemical reaction rate limits response - High-frequency current still from capacitors
t = 100ms: - Steady-state reached - Alternator picks up load - Batteries and capacitors recharging
Voltage sag calculation:
Capacitor discharge:
ΔV_cap = I × Δt / C
Example: 200A for 50ms, 2F capacitor
ΔV = 200 × 0.05 / 2 = 5V
Supply impedance drop:
ΔV_supply = I × Z_supply
Example: 200A, 0.05Ω supply
ΔV = 200 × 0.05 = 10V
Battery recovery:
After transient ends, battery must recharge capacitors:
Recharge time:
t_recharge = C × ΔV / I_charge
Example: 2F, 5V drop, 50A charge current
t = 2 × 5 / 50 = 0.2 seconds
Energy balance:
Energy delivered to amplifiers:
E = P × t = V × I × t
Example: 2000W for 50ms
E = 2000 × 0.05 = 100 Joules
Energy from capacitor:
E_cap = ½ × C × (V₁² - V₂²)
Example: 2F, 14V to 9V
E = 0.5 × 2 × (14² - 9²) = 1 × (196 - 81) = 115 Joules
Capacitor supplied all transient energy! This is why capacitors work.
Harmonic Analysis
Non-linear loads:
Amplifiers are non-linear loads: - Class AB crossover distortion generates harmonics - Class D switching generates high-frequency components - Transients contain wide frequency spectrum
Current harmonics:
Fourier series:
I(t) = I_dc + Σ(A_n × cos(nωt) + B_n × sin(nωt))
Where n = harmonic number (1, 2, 3...)
Typical harmonic content: - Fundamental (music frequency): 40-200 Hz - 2nd harmonic: 80-400 Hz - 3rd harmonic: 120-600 Hz - Switching frequency: 100-500 kHz (Class D)
Wiring impedance at harmonics:
Fundamental (60 Hz):
Z = R + j×2π×60×L
Z ≈ R (inductance negligible)
3rd harmonic (180 Hz):
Z = R + j×2π×180×L
Switching frequency (300 kHz):
Z = R + j×2π×300,000×L
Z ≈ jωL (resistance negligible, inductance dominates!)
This is why high-frequency bypassing (capacitors) is critical!
Alternator Rectification and Ripple
Three-phase rectification:
Automotive alternator has 3-phase stator, full-wave rectified.
Illustration note: Oscilloscope traces showing 3-phase AC waveforms and resulting DC with ripple after rectification
Ripple frequency:
f_ripple = 3 × (RPM / 60) × (Poles / 2)
Typical alternator: 12 poles
At 2000 RPM:
f = 3 × (2000/60) × 6 = 600 Hz
Ripple magnitude:
Good alternator: 50-100 mV p-p Worn alternator: 500+ mV p-p
Ripple effects:
Audible alternator whine
- 600 Hz tone (varies with RPM)
- Modulates audio signal
- Enters through power supply
Capacitor filtering
V_ripple = I_load / (C × f)Example: 100A load, 10,000 μF, 600 Hz
V_ripple = 100 / (0.01 × 600) = 16.7V (?!)This seems high - actually peak-to-peak current varies, not DC!
More accurate:
V_ripple ≈ I_ripple / (C × f)Where I_ripple ≈ 10-20% of DC load for 3-phase
V_ripple = 10 / (0.01 × 600) = 1.7V p-pStill significant without large capacitance!
Filtering requirements
For <50mV ripple:
C = I_ripple / (f × V_ripple)
C = 10 / (600 × 0.05) = 0.33 Farads
This is why competition systems use massive capacitor banks!