Class D Power Amplifier

This article intends to cover basic concepts of switching mode power amplifier, and the design of class D power amplifier.

Simple Switching Amplifier

Schematic of switching amplifier

Current and voltage waveform

where we have

$$V_{dc}=\frac{1}{2\pi}\int_{-\pi}^{\pi} v_{sw}(\theta)\ d\theta=\frac{\pi-\alpha}{\pi}V_{pk}\\ I_{dc}=\frac{1}{2\pi}\int_{-\pi}^{\pi} i_{sw}(\theta)\ d\theta=\frac{\alpha}{\pi}I_{pk}$$

As for the fundamental components

$$V_1=\frac{1}{\pi}\int_{-\pi}^{\pi}v_{sw}(\theta)\cos\theta\ d\theta=-\frac{2\sin\alpha}{\pi}V_{pk}\\ I_1=\frac{1}{\pi}\int_{-\pi}^{\pi}i_{sw}(\theta)\cos\theta\ d\theta=\frac{2\sin\alpha}{\pi}I_{pk}\\$$

Mathematica code for Fourier series

f[x_] := Piecewise[{{0, -a <= x <= a}, {1, x < -a}, {1, x > a}}]
FourierCosCoefficient[f[x], x, 1]

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Hence, we can calculate the efficiency as

$$\eta=\frac{P_{rf}}{P_{dc}}=\frac{-\frac{1}{2}V_1I_1}{V_{dc}I_{dc}}=\frac{2\sin^2\alpha}{\alpha(\pi-\alpha)}$$

To compare with the class A PA, we write the output power as

$$PUF=\frac{P_1}{P_{lin}}=\frac{V_{dc}I_{pk}\frac{2\sin^2\alpha}{\pi(\pi-\alpha)}}{\frac{1}{4}V_{dc}I_{pk}}=\frac{8\sin^2\alpha}{\pi(\pi-\alpha)}$$

We can plot the efficiency and PUF vs. conduction angle

Efficiency and PUF vs conduction angle

Eff[a_] := 2 Sin[a]^2/a/(Pi - a);
PUF[a_] := 8 Sin[a]^2/Pi/(Pi - a);
Plot[{Eff[a], PUF[a]}, {a, 0, Pi},
 PlotLegends -> "Expressions"]

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The peak power occurs at a conduction angle of around 113 degrees. Though the structure dissipate no heat, the peak efficiency is about 81%, as some of the power is wasted in harmonic components.

Switching Amplifier with Harmonic Short

To remove harmonic components, we can place a harmonic short across the load.

Switching amplifier with harmonic short

In this case, the voltage waveform must assume a sinusoidal form.

Waveform of the switching amplifier with harmonic short

Repeat the analysis, we have

$$\eta=\frac{P_{rf}}{P_{dc}}=\frac{\sin\alpha}{\alpha}\\ PUF=\frac{P_1}{P_{lin}}=\frac{4\sin\alpha}{\pi}$$

It can be plotted as

Efficiency and PUF vs conduction angle of switching amplifier with harmonic short

Eff[a_] := Sin[a]/a;
PUF[a_] := 4 Sin[a]/Pi;
Plot[{Eff[a], PUF[a]}, {a, 0, Pi},
 PlotLegends -> "Expressions"]

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The peak power occurs at a conduction angle of 90 degrees with an efficiency of 63 %.

Class D Power Amplifier

Class D power amplifier

The key point is that the current in the LCR branch is constrained to remain sinusoidal. With a sufficiently large bypass capacitor, the voltage $v_{sw}$ would remain $V_{dc}$ when the switch is turned to A, and remain zero when turned to B.

Waveform of class D power amplifier

We have

$$V_1=\frac{2}{\pi}V_{dc}\\ I_1=\frac{1}{2}I_{pk}\\ I_{pk}=\pi I_{dc}\\ \eta=\frac{P_{rf}}{P_{dc}}=\frac{\frac{V_{dc}I_{pk}}{\pi}}{V_{dc}I_{dc}}=1\\ PUF=\frac{\frac{V_{dc}I_{pk}}{\pi}}{\frac{1}{4}V_{dc}I_{pk}}=\frac{4}{\pi}$$

It can achieve theoretically 100 % efficiency with 1 dB higher power than class A amplifier.

A realization

Schematic

Simulated result

Reference

1. S. C. Cripps. RF Power Amplifier for Wireless Communications. Artech House, 2014.