RF power amplification (PA) is the key to achieving wireless application requirements, such as in ubiquitous communication technologies and radar. To understand the interplay of tradeoffs with requirements, including efficiency and power consumption, range, and linearity, the first in this series of articles on RF PA essentials dealt with basic waveform shapes and the formulae for calculating power supplied, RF output power, and efficiency. That article assumed idealities in waveform shapes — no over- or undershoots — and an ideal transistor.
We continue with these assumptions in this, the second part, as we look at the amplification mechanism, how amplifiers can be classified by circuits needed to generate various waveform shapes, and how the output waveform affects efficiency.
The purpose of the RF PA is to increase the power of the RF input signal. This is achieved by applying the signal to the gate, let us say a sinusoid with a voltage swing of 3 V, which affects the drain-to-source voltage (Vds) and current (Ids) such that their voltage swing is far greater, say >60 V (Figure 1).
The relationship between the signal at the output to the signal at the input is a measure of amplification, called gain. It may be expressed as the ratio of input-to-output voltage, current, and power to express voltage, current, and power gain, respectively.
Note that in Figure 1, the input and output signals are different only in terms of gain, with the output waveform shape essentially unaltered. However, this is not always the case.
One way to categorize amplifiers is by their modes of operation, or classes that represent the portion or conduction angle of a full cycle of a sinusoidal input signal for which the amplifier remains active. These amplifier classes differ in the method of operation, efficiency, and linearity.
This article compares some of the common amplifier classes, starting with the most-linear but least-efficient Class A to the still-linear (due to the ideal transistor) but more-efficient Class F and inverse-F.
Based on Figure 1 plot of Ids vs Vgs, keeping a linear amplification means that the sinusoidal voltage and current swing must stay within the linear slope of the transconductance curve and avoid the abrupt clipping at the edges. Biasing the transistor at the center of this linear region allows the amplifier to deliver the largest swing (output power) while staying linear (Figure 2). This is the Class A amplifier that is used when high linearity and gain are required.
The Class A amplifier conducts through a full cycle, or 360º (2π), of the input signal and is therefore equivalent to a current source. The drain voltage and current waveforms are ideally both sinusoidal.
Because the quiescent bias point is at the center of the current range, 0.5 A in this example, Class A amplifiers are always conducting DC drain current even when there’s no RF swing. This means higher DC power consumption and it affects their efficiency. At their largest RF swing, the efficiency is given by:
With parameters shown in Figure 2 as example,
DC power = 50 V × 0.5 A = 25 W,
RF power = 1⁄2 (50 V × 0.5 A) = 12.5 W, and
Efficiency (%) = (12.5/50) × 100 = 50%
Even with an ideal device and full RF swing, this is the theoretical best performance we can get from Class A amplifiers.
Class B amplifiers address the efficiency problem of Class A. They do this by biasing the gate at the pinch-off point on the characteristic curve, where the transistor just turns off and the quiescent drain current is zero — it is an ideal transistor, after all.
The transistor is therefore conducting half of the time during the RF swing and the drain current is now a half-rectified sine wave. Recall that a half-rectified sine wave consists of even harmonics. Therefore, a very low impedance (short circuit) is needed at these harmonics to maintain the half-rectified shape of the output drain current. The output voltage is still sinusoidal like in Class A.
Since we are assuming an ideal transistor with an abrupt turn-on characteristic, the amplitude of the drain current is proportional to the drive amplitude, producing a linear amplification.
To achieve the same amplitude as the full-sine–wave output of the Class A amplifier, or to occupy the same linear region of the transistor characteristics, the input wave swing must be twice as large as that for Class A. This means that Class-B amplifiers have lower gain than Class-A.
Recall that the DC current of a half-sine wave is given by:
Plugging this into our efficiency calculations,
DC power = 50 V × 1/π A = 15.9 W,
RF power = 1⁄2 (50 V × 0.5 A) = 12.5 W,
and
Efficiency (%) = (12.5/15.9) × 100 = 78.5%
While the RF power has remained the same as before, the DC power has decreased because the Class B amplifier has lower DC drain current during full RF swing because of the half-rectified sine wave shape and no quiescent current. This leads to the significant boost to efficiency over Class A’s limit, but the price for this is gain. We will see in the next article that linearity is also traded-off compared to Class-A.
Class C amplification is what results if the transistor is biased below threshold so that it is active for less than half of the cycle (<π).
In this case, the current waveform starts to look like it comprises a series of pulses (Figure 4). The DC component is thus further lowered compared to Class-B. The output voltage is still sinusoidal like that of Class A and Class B.
Class C efficiency can be increased from 78.5% like that of Class B toward 100% by decreasing the conduction angle with increasingly negative gate bias. However, it comes at the expense of a much lower and non-linear gain. The output power does not increase linearly with input power. Typically, Class C designs settle for tradeoffs that result in 85% efficiency.
The Class F amplifier utilizes the mathematical relationship of a composite square wave to its fundamental sinusoid — the amplitude of the fundamental is higher than that of the composite — as discussed in the first part of this series.
In Class F, a half-wave rectified sinewave for RF current reduces the DC component (same as Class B) and a square wave for its RF voltage increases Vmag of the fundamental, both contributing to an increase in efficiency (Figure 5).
Using values shown in Figure 5,
DC power = 50 V × 1/π (1 A) = 15.9 W,
RF power = 1⁄2 ((4/π × 50 V) × 0.5 A) = 15.9 W,
and
Efficiency (%) = 15.9 W / 15.9 W = 100%
The squaring of the drain voltage thus results in an incredible 100% ideal-case efficiency. An efficiency of 100% is only achievable if you have an infinite number of harmonics. If you are limited to 3rd harmonic only, then the maximum efficiency attainable is 88%.
A significant flattening of the voltage waveform can be achieved by adding the third harmonic (and higher-order odd harmonics) of the wave. But because the input voltage is a pure sine wave, it has no harmonics.
To generate the harmonics, the input voltage is increased such that the output current is clipped by driving the transistor into saturation (Figure 6). Then the third-harmonic voltage is built by having the RF load present an infinitely high impedance (open circuit) to the third harmonic current.
In an inversed F-class amplifier, the voltage is a half-rectified sine wave while the RF current is a square wave.
Consider the current generator plane, such as the one shown in Figure 6 for Class F. Inversed F operation is achieved by presenting, instead of a short circuit, an open circuit at the second harmonic. This develops the second-harmonic voltage and builds the half-rectified voltage shape. Additionally, a short circuit is presented to the third harmonic to keep the third-harmonic current and build a square current waveform.
This has no impact on the efficiency, as the DC and RF values for voltage and current are simply swapped, as shown below:
Class A | Class B | Class F | |
|---|---|---|---|
RF Voltage | 0.71 | 0.71 | 0.9 |
DC Voltage | 1 | 1 | 1 |
RF Current | 0.71 | 0.71 | 0.71 |
DC Current | 1 | 0.64 | 0.64 |
RF Power | 0.5 | 0.5 | 0.64 |
DC Power | 1 | 0.64 | 0.64 |
Efficiency | 50% | 78.5% | 100% |
DC power = (1/π × 100 V) × 0.5 A = 15.9 W,
RF power = 1⁄2 ((4/π × 0.5 A) × 50 V) = 15.9 W,
and
Efficiency (%) = 15.9 W / 15.9 W = 100 %
The F classes thus utilize the fact that the presence of a third-harmonic component allows the fundamental component amplitude to be increased. Because the transistor peak voltage has a maximum allowable swing, Vmax, Class F offers a useful way to increase power within that constraint.
There are many classes of amplifiers, some of which have been discussed here (Table 1). We calculated that Class B is more efficient than Class A because of the lower DC current component. Class F holds on to that advantage and increases RF power to achieve the highest efficiency of the three classes.
However, remember that this discussion has assumed an ideal transistor with characteristics that include a linear region surrounded by strongly nonlinear limits at cutoff and saturation. There is no quiescent current to worry about. There is no “knee” near saturation or soft turn-on.
The next part spells an end to those assumed idealities to consider efficiency degradation.
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