How Crossover Distortion Occurs in Class B Push-Pull Amplifiers

This article provides a clear explanation of how crossover distortion occurs at the zero-crossing point in Class B push-pull amplifier electronics. It covers the fundamental architecture of the push-pull configuration, the role of transistor threshold voltages in creating a non-conducting dead zone, and the resulting impact on the output signal waveform. Finally, it briefly addresses how engineers resolve this issue by transitioning to Class AB biasing.

Understanding the Class B Push-Pull Configuration

A Class B push-pull amplifier uses two complementary transistors—typically one NPN and one PNP bipolar junction transistor (BJT), or an N-channel and P-channel MOSFET pair—to amplify an alternating current (AC) input signal.

In this setup, each transistor is responsible for amplifying strictly half of the input waveform:

  • The "Push" Transistor (NPN): Conducts and amplifies the positive half-cycle of the input signal (from \(0\text{V}\) to the positive peak).
  • The "Pull" Transistor (PNP): Conducts and amplifies the negative half-cycle of the input signal (from \(0\text{V}\) to the negative peak).

At idle, when no input signal is present, both transistors are biased right at their cutoff points. This means neither transistor conducts current at \(0\text{V}\), making the Class B amplifier highly power-efficient compared to Class A designs. However, this exact biasing strategy is the direct cause of crossover distortion.

The Physics of the Zero-Crossing Dead Zone

To understand why distortion happens as the signal transitions from positive to negative (the zero-crossing point), you must look at the physical characteristics of silicon semiconductors.

A standard silicon BJT does not turn on the exact moment a voltage above \(0\text{V}\) is applied to its base. Instead, it requires a specific forward bias voltage between the base and the emitter (\(V_{BE}\)) to overcome the internal depletion region of the p-n junction. For silicon transistors, this threshold voltage is approximately \(0.6\text{V}\) to \(0.7\text{V}\).

Conversely, the complementary PNP transistor requires the input signal to drop to approximately \(-0.6\text{V}\) to \(-0.7\text{V}\) before it begins to conduct.

Step-by-Step: How the Distortion Happens

When an AC input signal passes through a Class B amplifier, the following sequence occurs around the \(0\text{V}\) mark:

  1. Approaching Zero: As the positive half of the input sine wave drops toward \(0\text{V}\), it eventually falls below \(+0.7\text{V}\). At this exact moment, the NPN transistor turns off completely, even though the input signal has not yet reached \(0\text{V}\).
  2. The Zero-Crossing Point: The input signal crosses \(0\text{V}\) and begins its negative swing. Both transistors are currently turned off.
  3. Leaving Zero: The signal continues to drop into the negative region. Between \(0\text{V}\) and \(-0.7\text{V}\), the PNP transistor still does not have enough base-emitter voltage to turn on.
  4. Resuming Conduction: Once the input signal exceeds \(-0.7\text{V}\), the PNP transistor finally turns on and amplifies the remainder of the negative half-cycle.

Because neither transistor conducts when the input signal is between \(-0.7\text{V}\) and \(+0.7\text{V}\), a \(1.4\text{V}\) "dead zone" is created right at the zero-crossing point.

During this dead zone, the output voltage remains flat at \(0\text{V}\). When viewing the output waveform on an oscilloscope, the smooth transition of the original sine wave is replaced by a flat, horizontal step every time the wave crosses the zero axis. In audio electronics, this clipping at the zero-crossing point introduces severe, harsh-sounding high-frequency harmonics that become especially noticeable at low volume levels.

Eliminating the Distortion

To fix crossover distortion without sacrificing the efficiency of the push-pull topology, engineers modify the circuit into a Class AB amplifier. By adding a biasing network—usually two diodes or a variable voltage multiplier placed between the bases of the two transistors—a small, continuous quiescent current is applied. This pre-biases both transistors right at their \(\pm0.7\text{V}\) turn-on thresholds, completely eliminating the dead zone at the zero-crossing point.