Class-E功率放大器波形推导详解：Matlab实现与效率优化

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本文档探讨了Class-E类功率放大器的波形推导，这是一种特殊的电路设计，其中负载网络与单个晶体管配合工作，晶体管作为开关在输出信号的载频上进行操作。Class-E放大器的核心是其简单的负载网络，通常包含并联的电容器和串联的谐振输出电路，可能还带有剩余的阻抗。这种放大器的工作原理取决于晶体管的开/关状态，以及晶体管关闭时负载网络的暂态响应。作者弗雷德里克·H·拉巴特在1977年的IEEE Transactions on Acoustics, Speech, and Signal Processing（CAS-24, 第12期）上发表了一篇名为《理想化的Class-E调谐功率放大器操作》的文章。文章中，他利用傅立叶级数技术并假设高Q值，对放大器的基本方程进行了推导，以实现100%效率的理想运行。通过这些方程，可以确定组件的最佳值，以优化Class-E放大器的性能。论文深入分析了集电极电压波形的对称性结构，并讨论了不同组件配置和占空比参数如何影响效率。此外，作者还研究了不同效率水平下可能的组件组合，以及这些配置对放大器性能的影响。值得注意的是，虽然文章的初衷是对Class-E放大器进行理论分析，但提供的MATLAB代码使得读者能够方便地调整参数来观察实际的波形，从而在实际应用中灵活地理解和优化这类功率放大器的设计。这篇论文为理解和设计高效、低失真的Class-E功率放大器提供了重要的理论基础和技术指导，对于电子工程师和音频系统设计者来说，具有很高的参考价值。通过学习和应用这些原理，研究人员和实践者能够更好地理解Class-E类放大器的性能极限，并在实际电路设计中实现高效的信号传输。

RAAB: OPERATION OF CLASS E TUNED PA

727

A. Basic Relationships

The output voltage and current are sinusoidal and have

the forms

u,(~)=c sin (wt+cp)=c sin (0+cp)

(2.1)

and

iO(f3)= $

sin (e+cp)

(2.2)

where 0 is an “angular time” used for mathematical

convenience. The parameters c and rp are to be de-

termined; cp is defined in Fig. 2.

The hypothetical voltage u,(0) is also a sinusoid, but

has a different phase because of reactance X

u1(e)=%(e)+%(e)

(2.3)

sin

(e+q)+x$ cos (e+cp)

(2.4)

=c, sin

(e+q,)

(2.5)

where

c,=c

1,s =pc

and

Since the RF choke forces a dc input current and the

series-tuned output circuit forces a sinusoidal output cur-

rent, the difference between those two currents must flow

into (or out of) the switch-capacitor combination. When

switch S is open, the difference flows into capacitor C;

when switch S is closed, the current difference flows

through switch S. If a capacitor voltage of other than zero

volts is present at the time the switch closes, the switch

discharges that voltage to zero volts, dissipating the stored

energy of $ CV2.

For purposes of this analysis, the discharge of the

capacitor may be assumed to be a current impulse

qS (0 -

e,), where 0, is the time at which the switch closes.

However, in a real amplifier, discharge of the capacitor

requires a nonzero length of time, during which the collec-

tor voltage and current are simultaneously nonzero. In

any case, however, the total dissipation is the energy

stored in the capacitor, and does not depend on the

particulars of the discharge waveforms. The model re-

mains valid as long as the time required to discharge the

capacitor is a relatively small fraction of the RF cycle.

When S is off, the collector voltage is produced by the

charging of capacitor C by the difference current, hence,

where 0, indicates the time at which S opens.

Since the nonzero collector voltage appears when the

transistor is off, it is convenient to describe the waveforms

in terms of the half off-time y, which is in radians. The

center of the off-time is arbitrarily defined as 7r/2 (Fig. 2).

The switching instants are now 0, = a/2-y and 0, = 7r/2

+y. Note that in the event that 0, < 0 and is therefore

outside of the 277 interval corresponding to one cycle, it

can be repaced by &,+27, with appropriate modification

to the integral in (2.8). Equivalently, the interval of a cycle

can be redefined when necessary. The collector voltage at

time 0 can now be evaluated by expanding (2.8)

(2.9)

+$e+&

cOs(e+cp)

where

B=wC.

(2.11)

Since the tuned circuit has zero impedance to funda-

mental frequency current, there can be no fundamental

frequency voltage drop across it. This means that the

fundamental frequency component of the collector volt-

age waveform must be the hypothetical voltage u,(B). The

magnitude of this component can be calculated by a

Fourier integral. Unfortunately, the collector voltage is

not symmetrical around 7r/2, which makes the phase cp an

unknown.

B. Fourier Analysis

The magnitude c, of the fundamental frequency compo-

nent of the collector voltage is then

c,= -

~]02”o(e)

sin

(e+cpl)

(2.12)

4 -$(t-Y+‘P~)+--$

My-d]

.coscp, siny+$

[

-2 sin ‘pi siny

-&[sin(2’p+$) sin2y-2y sin$]. (2.13)

It is now possible to solve for c by substituting pc for c,

and collecting terms

pc+c

sin (2~ + Ir/) sin 2y - 2y sin I+L

2rBR

2 sin (y - cp) cos cpr sin y

?TBR

=$c-

77+2y-29,+7r+2r+7,) cos cp, siny

+(2y cosy-2 siny) sinq,]. (2.14)

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