Efficiency Analysis of Class F/Inverse Class F Power Amplifiers

With the rapid development of wireless communication systems, the demand for high-efficiency RF power amplifiers has gradually increased. How to improve the working efficiency of power amplifiers has become an important topic. In order to improve efficiency, researchers have focused a lot of energy on the working modes of amplifiers, such as class D, class E, class F and inverse class F power amplifiers. High-efficiency power amplifiers in class F and inverse class F have become the focus of research in recent years. The efficiency of class F working mode is 75% in the literature, and the efficiency of inverse class F working mode is 80% in the literature. Both use GaN power tube CGH400 10, whose 1 dB compression point output power is about 10 W, which cannot meet the requirements of large output power.

Aiming at the power tube CGH40045 with an output power of 45 W at 1 dB compression point, two power amplifiers, Class F and inverse Class F, are designed respectively. The drain efficiencies of the two modes of power amplifiers are compared, and the efficiency difference between the two amplifiers is quantitatively analyzed from both theoretical and simulation aspects.

1 Theoretical Analysis

Figure 1 shows the time domain waveforms of the ideal voltage and current at the drain output of the Class F and Inverse Class F power amplifiers. The drain current of the Class F power amplifier is a half-sine wave, and the voltage is a square wave; on the contrary, the drain current of the Inverse Class F power amplifier is a square wave, and the voltage is a half-sine wave. The opposite drain voltage and current waveforms of the two will have different effects on the efficiency of the amplifier. The following is the calculation of the working efficiency of the two amplifiers under ideal conditions. Both waveforms can be expanded through the Fourier series expansion, so that the DC signal, fundamental signal and each harmonic signal can be easily analyzed.

First, the class F power amplifier is analyzed, and its voltage and current waveforms are expanded through Fourier series. The variables are shown in Figure 1.

The power of the DC signal and the power of the fundamental wave signal are calculated respectively according to the Fourier expansion, thereby obtaining the efficiency of the class F power amplifier, where Ron is the on-state internal resistance of the transistor.

Analyze the inverse class F power amplifier, expand its voltage and current waveforms through Fourier series, and the variables are shown in Figure 1.

According to the method of calculating the efficiency of Class F, the efficiency of the inverse Class F power amplifier can be obtained by the same logic:

It can be seen from equations (5) and (10) that when V0=V0', the efficiency of the power amplifiers in the two modes will differ due to the difference in current imax and i'max. If the on-resistance Ron of the transistor is zero, the efficiency of the power amplifiers in both modes will reach 100%. However, in reality, power tubes all have a certain on-resistance. When Ron exists, the efficiency of the power amplifiers in the two modes will differ.

To study this difference, we need to have the same output power and the same bias point, that is, P1=P1', V0=V0'. From equations (4) and (9), we can list an equation about i'max. Solving this equation, we can get

Substituting this result into equation (10), we can get a function expression of η' with respect to imax. Here, assuming V0 = 28 V, imax = 6 A, substituting the values ​​into equations (5) and (10), and using Matlab to plot the variation curves of η and η' with respect to Ron, as shown in Figure 2. It can be seen from the figure that with the increase of Ron, the inverse class F power amplifier has better efficiency than the class F power amplifier, and for each value of Ron, the quantitative calculation results of the difference in efficiency between the two modes are given.

2 Solution Design

In order to obtain the required waveforms of the two-mode amplifier shown in Figure 1 at the drain end of the power tube, it is necessary to control all harmonic impedances at the drain output end. It can be seen from equations (1) and (2) that to achieve the square wave voltage and half-sine wave current required by the class F power amplifier, it is necessary to achieve the odd harmonic superposition of the voltage and the even harmonic superposition of the current at the drain end. Similarly, it can be seen from equations (6) and (7) that to achieve the half-sine wave voltage and square wave current required by the inverse class F power amplifier, it is necessary to achieve the even harmonic superposition of the voltage and the odd harmonic superposition of the current at the drain end, which can be achieved by adding a shaping circuit at the output end. The output circuit of the class F power amplifier must satisfy equation (12), and the output circuit of the inverse class F power amplifier must satisfy equation (13).

However, it is obviously impossible to control all harmonics in real applications. Generally, in engineering applications, it is sufficient to control the second and third harmonics. Controlling more harmonics will increase the complexity of the circuit and will not significantly improve the performance. The F-class output shaping circuit is shown in Figure 3, and the inverse F-class output shaping circuit is shown in Figure 4. The impedance requirements of the F-class and inverse F-class power amplifiers are met by shaping the second and third harmonic impedances. Electromagnetic simulation software is used to perform load traction design on the power tube and the shaping circuit as a whole to find the optimal output impedance value of the fundamental wave at the output end. The overall circuit is matched to the standard impedance of 50Ω through an additional matching network at the back end.

3 Simulation Results Analysis

According to the above circuit structure, the class F and inverse class F power amplifiers are designed respectively. The power tube adopts the 45 W GaN power tube CGH40045 of Cree Company, and its on-resistance Ron=0.3 Ω. In order to make an objective and true comparison of the efficiency of the two power amplifiers, all parameters should be kept consistent as much as possible. The substrate adopts the R05870 of Rogers Company, and the thickness of the substrate is 0.79 mm. The working bias point is selected as VDS=28 V, VGS=-2.5 V, and the working frequency is 1.5 GHz.

Figures 5 and 6 are curves showing the gain and output power of the class F power amplifier and the inverse class F power amplifier as a function of input power. Figure 5 is a gain diagram of the two. It can be seen from the figure that the gains of the two are basically equal, and the gain in the linear region is about 16.5 dBm. Figure 6 is the output power of the two. It can be seen from the figure that the output power change curves of the two are basically consistent, with a maximum output power of about 55 W and an output power of about 45 W at the P1dB point.

Under the premise of equal gain and output power, the drain efficiency is compared. Figure 7 shows the curve of the drain efficiency of the class F power amplifier and the inverse class F power amplifier when working at 1.5 GHz versus input power. It can be seen from the curve in the figure that the drain efficiency of the inverse class F power amplifier reaches a maximum value of 91.8% when the input is 40 dBm, and the drain efficiency of the class F power amplifier is 89.3%. It can be seen from Figure 2 that when Ron is 0.3 Ω, the inverse class F efficiency is 96% and the class F efficiency is 93.6%. Since ideal square wave and half-sine wave signals cannot be achieved in reality, there are certain differences between the simulation results and the calculation results, but the efficiency difference between the two modes is basically equal, which confirms that the theoretical calculation and simulation are consistent.

4 Conclusion

Through theoretical calculation, it is found that under the same working bias point and output power, the inverse class F power amplifier has higher efficiency than the class F power amplifier due to the existence of the power tube on-resistance, and this difference increases with the increase of the on-resistance Ron. For the GaN power tube CGH40045, an inverse class F power amplifier and a class F power amplifier with an operating frequency of 1.5 GHz were designed respectively, and the simulation results are consistent with the theoretical analysis.