A push-pull Class AB amplifier is, at its core, a symmetry machine. Two complementary transistors, one NPN and one PNP, share the task of reproducing the signal: the NPN handles the positive half-cycle, the PNP handles the negative. When the topology is truly symmetrical, the harmonic distortion each half contributes to the output is also symmetrical, and a remarkable cancellation follows from that symmetry. Even-order harmonics, the second, fourth, and sixth, vanish from the output because the two halves generate them in equal magnitude and identical phase, so they subtract to zero at the load. What the designer measures on the spectrum analyzer is a spectrum dominated by odd harmonics, the third and fifth especially, which is the characteristic signature of a well-balanced push-pull stage.
That cancellation, clean as it appears on paper, is fragile in practice. It depends on both halves of the circuit being truly identical in every parameter that matters. And one of those parameters, parasitic inductance in the power supply rails, is determined not by the schematic but by the PCB layout. A positive rail trace that runs five centimeters longer than its negative counterpart carries five additional nanohenries of series inductance. That inductance does not appear anywhere in the design file. It is not reviewed in simulation. It does nothing at DC or at low frequencies where it is simply a wire. But the moment the output transistors begin switching at high frequencies, that inductance becomes a circuit element, and its asymmetry becomes a distortion mechanism that breaks exactly the cancellation the topology was designed to provide.
Why Push-Pull Symmetry Cancels Even Harmonics in Theory
To understand what asymmetry destroys, it is worth being precise about what symmetry creates. A nonlinear device driven by a signal x(t) can be described by a Taylor series expansion of its transfer function:
y(t) = a₁·x + a₂·x² + a₃·x³ + a₄·x⁴ + ...
The even-power terms, a₂·x² and a₄·x⁴, generate even harmonics. For a sinusoidal input x = A·sin(ωt), the squared term produces a₂·A²·sin²(ωt) = (a₂·A²/2)·(1 − cos 2ωt), which includes a DC offset and a component at 2ω. In a single-ended stage, this component appears in the output unimpeded.
In a push-pull configuration, the second half of the output stage processes the inverted signal, −x(t). Its even-power contribution becomes a₂·(−x)² = a₂·x², which is identical in sign and magnitude to the contribution from the first half. When the two halves drive the load in opposition, as they do in a complementary emitter-follower topology, their even-order outputs are subtracted, and a₂·x² − a₂·x² = 0. The odd-order terms, a₃·x³ from the first half and a₃·(−x)³ = −a₃·x³ from the second, add rather than cancel: −a₃·x³ − (−a₃·x³) = 2a₃·x³. This is why symmetrical push-pull amplifiers contain only odd harmonics as expected.
This cancellation is conditional on the two halves being governed by the same coefficients a₁, a₂, a₃, and so on. The instant any parameter differs between the positive and negative signal paths, the second-order terms no longer cancel perfectly, and a residual even-harmonic component survives in the output.
How Class AB Rail Currents Become Half-Wave Pulses
Before arriving at the asymmetry mechanism, it is important to understand the nature of the current that flows in each supply rail during Class AB operation. At quiescent conditions, a small standing current flows through both transistors. As the output swings positive, the NPN transistor conducts increasingly hard, drawing current from the positive rail. The PNP transistor simultaneously reduces its conduction. During the positive half-cycle, essentially all of the large-signal load current flows from the positive rail through the NPN device to the load. During the negative half-cycle, the PNP transistor carries the load current, returning it to the negative rail. The supply current drawn from each rail is therefore a half-wave rectified version of the load current, not a full sine wave.
Using the Fourier series for a half-wave rectified sinusoid of amplitude I_p, the current in each rail is:
i_rail(t) = I_p/π + I_p/2 · sin(ωt) − 2I_p/(3π) · cos(2ωt) − 2I_p/(15π) · cos(4ωt) − ...
This expansion makes the problem visible in mathematical form. The half-wave rectified current contains a DC component, a fundamental at ω, even harmonics at 2ω and 4ω, and no odd harmonics beyond the fundamental. These are large, high-amplitude current pulses surging through the supply traces with every audio cycle, and at high output levels their amplitudes are measured in amperes. Each of these pulses interacts with every impedance it encounters in the supply path, including the parasitic inductance of the PCB traces.
The Asymmetric Inductance and How It Breaks the Cancellation
A PCB trace of width w and length l over a ground plane has a parasitic inductance that can be estimated as:
L_trace ≈ (µ₀/2π) · l · ln(2l/w) [H]
For a typical power trace of 3 mm width and 50 mm length on standard FR4, this yields roughly 20 to 30 nanohenries. A trace that runs 10 mm longer than its counterpart adds approximately 5 to 7 nanohenries of additional inductance. At audio frequencies below 20 kHz, the impedance of 5 nanohenries is ωL = 2π · 20000 · 5 × 10⁻⁹ ≈ 0.6 milliohms, which is negligible. However, the half-wave rectified rail current contains harmonic components at 2ω, 4ω, and beyond. At 100 kHz, the second harmonic of a 50 kHz switching event, the impedance of 5 nH becomes 2π · 10⁵ · 5 × 10⁻⁹ ≈ 3.1 milliohms. Against the milliohm-level source impedance of a well-designed Class AB output stage, this is no longer negligible. And against a load of 8 ohms, it seems trivial, but the voltage drop it produces appears directly in the supply rail, which is exactly what the output transistors see as their collector or emitter supply voltage.
Let the parasitic inductance of the positive rail trace be L₊ and that of the negative rail trace be L₋. If these are equal, L₊ = L₋ = L, then the voltage drop across each rail during the corresponding half-cycle is symmetric: ΔV₊ = L · dI₊/dt and ΔV₋ = L · dI₋/dt. Since I₊ and I₋ are mirror images of each other, ΔV₊ and ΔV₋ are also mirror images, and the net effect on the output is symmetric. The even-harmonic content of ΔV₊ and ΔV₋ cancels at the load for the same reason the transistors' own even-order nonlinearity cancels.
Now introduce an asymmetry ΔL = L₊ − L₋. The voltage drop on the positive rail during the positive half-cycle becomes L₊ · dI₊/dt, while the negative rail drop during the negative half-cycle is L₋ · dI₋/dt. These drops are no longer equal. The effective supply voltage seen by the NPN transistor during positive peaks differs from the effective supply voltage seen by the PNP transistor during negative peaks. The two halves of the output stage are now operating from asymmetric instantaneous supply voltages, and their gain and operating point differ on each half-cycle. This is exactly the asymmetry that reintroduces even-order terms into the transfer function. The residual second harmonic voltage at the output is approximately:
V₂ₙd ≈ ΔL · dI_peak/dt · (a₂/a₁)
where a₂ and a₁ are the second- and first-order coefficients of the output stage transfer function. The factor dI_peak/dt grows with both signal amplitude and frequency, explaining why rail inductance asymmetry is a high-frequency, high-level distortion mechanism that is essentially invisible in routine bench testing at moderate levels and audio frequencies, but measurable and audible under dynamic conditions with complex program material.
Why High Frequencies Amplify the Effect
The inductive voltage drop V = L · dI/dt grows linearly with the rate of current change. For a sinusoidal load current I(t) = I_peak · sin(ωt), the derivative is dI/dt = ω · I_peak · cos(ωt), so the voltage drop amplitude across a parasitic inductance is:
V_L(peak) = ω · L · I_peak = 2πf · L · I_peak
At 1 kHz with I_peak = 5 A and ΔL = 10 nH, the differential inductive drop between rails is 2π · 10³ · 10⁻⁸ · 5 ≈ 0.31 mV. Against a 40 V peak output swing this is 0.00078%, undetectable by any practical measurement. At 20 kHz, the same calculation gives 6.3 mV, which is still small but now within range of measurable THD in a precision amplifier. At switching-frequency-related events or with reactive loads that produce fast current transients, dI/dt can be orders of magnitude higher than the audio-frequency calculation suggests, and the inductive drop rises proportionally.
In Class B and Class AB operation, supply rail currents are half-wave rectified sine pulses with strong harmonic content, and if they contaminate the signal, distortion degrades badly. At high frequencies the harmonic content of these pulses concentrates energy in exactly the range where even a few nanohenries of inductance imbalance produces meaningful voltage differentials between the two rails. The extra distortion component from this mechanism rises at 6 dB per octave with frequency, making it progressively more visible as test frequency increases and most damaging to the high-frequency accuracy of the amplifier.
The Role of Decoupling Capacitors and Why They Do Not Simply Solve the Problem
The standard response to rail impedance concerns is to add local decoupling capacitors close to the output devices. A capacitor of value C placed at the transistor's supply pin presents an impedance of 1/(2πfC) at frequency f, and for a 100 µF capacitor this is 0.016 ohms at 100 Hz, falling to 0.00016 ohms at 10 kHz. The capacitor looks like a local charge reservoir, replenishing current to the transistor faster than the full rail trace inductance would allow and effectively bypassing much of the trace inductance at high frequencies.
However, decoupling capacitors have their own parasitic inductance, their equivalent series inductance (ESL), typically 5 to 20 nH for a standard electrolytic and 1 to 3 nH for a film or ceramic type. Below the self-resonant frequency f_res = 1/(2π · √(L_ESL · C)), the capacitor behaves as intended. Above it, the capacitor's own inductance dominates and the device presents rising rather than falling impedance. For a 100 µF electrolytic with 10 nH ESL, f_res = 1/(2π · √(10⁻⁸ · 10⁻⁴)) ≈ 159 kHz. Above this frequency, the decoupling capacitor is no longer a capacitor; it is an inductor. At the switching-related frequencies where Class AB rail inductance asymmetry does its most significant damage, poorly chosen or poorly placed decoupling capacitors can worsen the problem rather than cure it.
More critically, even a perfect decoupling capacitor placed identically close to both transistors does not cure an asymmetric rail trace layout. The inductance between the reservoir capacitor on the PSU board and the local decoupler on the amplifier board remains asymmetric if the traces have different lengths. The local decoupler shields the transistor from the high-frequency transients that originate downstream of it, but the inductance of the trace between the reservoir and the decoupler is still part of the circuit and still contributes to the differential voltage between the two rails during the switching transient.
Practical Consequences for PCB Layout
The corrective implication is straightforward but demands discipline in execution. The positive and negative rail traces from the reservoir capacitors to the output transistors must be routed with matched electrical lengths, which in practice means matched physical lengths, matched widths, and matched geometry relative to any ground or reference plane. A difference of 10 mm in trace length between the positive and negative rails, which is a modest and easily committed error when routing around a heatsink mounting hole or an output inductor, translates directly into an asymmetric inductance of 5 to 7 nH and an amplifier whose harmonic spectrum contains a second harmonic component that grows with frequency.
Several layout practices enforce the required symmetry.
- Rail traces should be routed as mirror images of each other, with both lengths measured from the reservoir capacitor ground reference to the output transistor's supply pin.
- Where the positive rail must route around an obstacle that the negative rail does not encounter, the negative rail should be deliberately lengthened by an equivalent trace section to restore the inductance balance.
- Local decoupling capacitors of identical type, value, and placement relative to the transistor supply pins should be used on both rails, since even a few millimeters of difference in capacitor placement introduces an asymmetric ESL contribution.
- Via placement matters: each via adds approximately 0.5 to 1 nH of inductance. If the positive rail trace requires an additional via that the negative trace does not, an equivalent via should be added to the negative trace at a corresponding location.
- Twisting the positive and negative rail supply wires from the PSU to the amplifier board reduces the loop area of the supply current path and limits both radiated EMI and the inductive voltage drop that depends on the loop area enclosed.
What the Spectrum Analyzer Reveals
The experimental signature of rail inductance asymmetry is distinctive enough to allow diagnosis without a simulation model. When a well-designed Class AB amplifier with matched transistors and careful biasing produces a THD residual that rises at 6 dB per octave and contains a prominent second harmonic component that worsens with output level and frequency, but the transistors themselves measure well in isolation, the cause is almost always in the supply path rather than the active devices.
The distortion signal may intrude into the input circuitry, the feedback path, or even the cables to the output terminals, resulting in a kind of sawtooth on the distortion residual that is very distinctive, an extra distortion component which rises at 6 dB per octave with frequency. This rise rate matches exactly the ω · L · I_peak scaling of inductive voltage drop, confirming the mechanism. A spectrum showing H2 at -60 dB relative to fundamental at 1 kHz, rising to -48 dB at 4 kHz and -40 dB at 10 kHz, with H3 remaining flat across the same range, is almost diagnostic of this problem. The odd harmonics, anchored to the transistors' own nonlinearity, do not scale with frequency in this way. The even harmonics, fed by the rail asymmetry mechanism, do.
The fix does not require new components, revised topology, or additional feedback. It requires re-laying the supply traces with matched geometry. Five millimeters of extra copper on one rail is the origin of the problem. Five millimeters removed or equalized is the complete cure. Few other distortion mechanisms in analog amplifier design respond so cleanly to such a physically simple intervention.
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