As we know from earlier articles in this series, reducing the conduction angle of a power amplifier allows us to increase its efficiency. A Class A stage has a 360 degree conduction angle and a theoretical efficiency of 50%; a Class B amplifier has a 180 degree conduction angle and an efficiency of 78.5%.
This trend toward increased efficiency continues with Class C amplifiers, which use conduction angles lower than 180 degrees. For example, as we see in Figure 1, a Class C amplifier with a conduction angle of 90 degrees has a theoretical efficiency of 94%.
Figure 1. Plot of maximum efficiency vs. conduction angle.
Unfortunately, Class C power amplifiers have several drawbacks that restrict their application in modern solid-state RF circuits. For one thing, they produce considerably less output power than a Class A or B amplifier for the same input power. If we examine the design of a Class C amplifier with the same output power as a Class A or B amplifier, more limitations become apparent.
In this article, we’ll explore these limitations by comparing the performance of Class A, B, and C amplifiers that must deliver a certain RF power to a given load impedance. To keep things simple, we’ll use a 90 degree conduction angle for the Class C amplifier throughout the article. Let’s start by comparing the maximum current that the output transistor of each amplifier carries.
Transistor Current Requirements
In the previous article, we learned how the conduction angle changes the frequency content of the output current. You may recognize Figure 2 from that discussion. It plots the average (a0) and fundamental (a1) components against the conduction angle for Class A, B, and C amplifiers. In order for these three designs to deliver the same power, their respective output currents should have identical fundamental components.
Figure 2. The average and fundamental components against the conduction angle.
Note that the a0 and a1 coefficients in this figure are normalized to IM, where IM is the peak value of the transistor current. Put a different way, IM is assumed to be unity.
Class A Transistor Current
Let the maximum transistor current in the Class A stage be IMA. From Figure 1, we see that the fundamental component of a Class A stage normalized to IMA is 0.5. We thus have:
$$frac{a_{1}}{I_{MA}} ~=~ 0.5~ rightarrow ~a_{1} ~=~ 0.5 I_{MA}$$
Equation 1.
If you can’t immediately figure out why the fundamental component is half the maximum current of the Class A stage, recall that this type of amplifier biases the transistor at the midpoint of its load line. The maximum collector current of an amplifier is the sum of the bias current and the AC current.
To have the maximum symmetrical swing, the DC current through the transistor should be equal to the amplitude of the AC current delivered to the load. Figure 3 shows how the collector current changes in a Class A amplifier when the transistor reaches the boundaries of its load line.
Figure 3. Minimum (a) and maximum (b) collector current In a Class A stage with maximum symmetrical swing.
As you can see, the collector current goes from zero to twice the maximum RF current. This is consistent with Equation 1.
Class B Transistor Current
What about a Class B amplifier? Let the maximum collector current of the Class B stage be IMB. Figure 1 shows that the normalized fundamental component of the Class B stage is also 0.5, giving us:
$$frac{a_{1}}{I_{MB}} ~=~ 0.5 ~rightarrow~ a_{1} ~=~ 0.5 I_{MB}$$
Equation 2.
From this, we see that the Class A and Class B amplifiers produce identical fundamental components for a given maximum transistor current.
Class C Transistor Current
The fundamental component of a Class C amplifier depends on its conduction angle. We’re assuming that the conduction angle is θc = 90 degrees, which corresponds to a normalized fundamental component of 0.31 (Figure 1). We therefore have:
$$frac{a_{1}}{I_{MC}} ~=~ 0.31 ~rightarrow~ a_{1} ~=~ 0.31 I_{MC}$$
Equation 3.
where IMC is the maximum collector current of the Class C stage.
Comparing Transistor Currents
To summarize:
- For the Class A amplifier, a1 = 0.5IMA.
- For the Class B amplifier, a1 = 0.5IMB.
- For the Class C amplifier a1 = 0.31IMC
For these three designs to produce the same fundamental component, the following must be true:
$$0.31 I_{MC} ~=~ 0.5 I_{MA} ~=~ 0.5 I_{MB}$$
Equation 4.
Solving for IMC, we obtain:
$$I_{MC} ~=~ 1.6 I_{MA} ~=~ 1.6 I_{MB}$$
Equation 5.
The Class C amplifier’s peak current is 1.6 times that of the Class A or Class B amplifier. Keep in mind that we based our calculations on a Class C amplifier with a 90 degree conduction angle. A Class C amplifier with a conduction angle of less than 90 degrees would have an even higher peak current relative to the Class A or B stage.
The key takeaway here is that the transistor in a Class C amplifier must handle larger currents to produce the same output power. In practical terms, this means that the amplifier needs a larger transistor. The further we reduce the conduction angle, the higher the peak current gets, and the larger the transistor needs to be. This can be problematic—for one thing, a larger transistor leads to a lower matching bandwidth.
The transistor size requirement is a major drawback. There are other efficient power amplifier configurations that don’t need such big transistors, and these have taken over Class C operation in modern solid-state RF design.
When picking a transistor, we also need to consider the amplifier’s maximum voltage. As we’ll see in the next section, Class C operation can run into problems here as well.
The Problem of Reverse Breakdown Voltage
In amplifiers with small conduction angles, swinging the output current to its maximum permissible value requires a large voltage applied to the input of the transistor. To understand why, consider the voltage waveform in Figure 4.
Figure 4. A large input voltage is required when using small conduction angles.
In the above figure:
- θc is the amplifier’s conduction angle.
- VT is the turn-on voltage of the transistor.
- Vmax is the input voltage that produces the maximum output current.
The difference between Vmax and VT should be large enough for the transistor to produce its maximum current when the input is at Vmax. However, the portion of the input that exceeds VT corresponds to only a short fraction of the RF cycle (a small conduction angle). A very large drive signal is therefore required.
Now consider the negative swing of the input. As the input voltage approaches its minimum value, the output voltage swings to its maximum. This produces a large voltage across the collector-base terminals, which can cause breakdown of the collector-base junction. This reverse breakdown can damage the transistor and result in a large, uncontrolled current flow from the collector to the emitter.
Distortion and Nonlinearity
To understand how the conduction angle affects linearity, it’s instructive to compare the output current waveforms of the Class A, B, and C amplifiers when they’re designed for the same output power. These waveforms are illustrated in Figure 5, which can be read as follows:
- The purple curve corresponds to a 360 degree conduction angle (Class A operation).
- The magenta curve corresponds to a 180 degree conduction angle (Class B operation).
- The orange curve corresponds to a 90 degree conduction angle (Class C operation).
Figure 5. The current waveforms for Class A (purple), B (magenta), and C (orange) amplifiers.
The blue dashed curve is a sine wave with an amplitude of IRF and negative offset of IQ. The Class C waveform is a portion of this sine wave.
As the blue curve reminds us, the output current is a sinusoidal segment for only part of the cycle when we use a conduction angle other than 360 degrees. For the remainder of the cycle, the output current is zero. Since the input waveform is a pure sinusoid, the output waveform is obviously distorted with respect to it. By this definition, only the Class A amplifier is linear.
In the context of power amplifiers, however, we can also define linearity based on how the power of the fundamental component at the output changes with the input power. Even if the amplifier involves highly nonlinear processes, its overall input-output characteristics can still be linear. Note that this definition of linearity assumes that all harmonic components of the output are shorted out by the high-Q tank at the output.
Along with the average and fundamental components (a0 and a1), Figure 6 shows how the second and third harmonics (a2 and a3) of an amplifier’s current waveform change with its conduction angle. Using this figure, let’s re-examine the linearity of the Class A, B, and C amplifiers.
Figure 6. The average value and first three harmonic frequency components of the output current against the conduction angle.
Linearity of the Class A and Class B Amplifiers
Class A operation is still assumed to be linear, since there’s no clipping of the signal. In Figure 6, the second and third harmonic components are both zero at a 360 degree conduction angle.
The input signal for a Class B stage is symmetric with respect to the transistor’s turn-on point. There is a second harmonic component at the 180 degree conduction angle, but the third harmonic is zero. Changing the amplitude of the input signal doesn’t affect the conduction angle.
Consequently, the power of the fundamental component at the Class B stage’s output is proportional to the input power. In that sense, a single-transistor Class B amplifier is linear. A reduction in the input drive power of a Class B stage results in an equivalent reduction in the output power.
Linearity of the Class C Amplifier
For conduction angles of less than 180 degrees (Class C operation), we see in Figure 6 that there are both second and third harmonic components. The Class C amplifier uses a narrower pulse to achieve a higher efficiency, and its conduction angle changes with the amplitude of the input signal. Referring back to Figure 5 should help you visualize this.
For a given bias point, reducing the amplitude of the input signal reduces the conduction angle and, consequently, the gain of the stage. This means that Class C amplifiers aren’t linear even if we consider the fundamental component of the output with respect to the input power. Reducing the input power by 3 dB, for example, may lower the output fundamental power by more than 3 dB.
Linearity of the Class AB Amplifier
As an aside, a Class AB amplifier (meaning an amplifier with a conduction angle between 180 and 360 degrees) is nonlinear even when using a perfectly linear transistor. This is because the amplifier’s conduction angle changes with the drive level.
However, it should be noted that real-world transistors are nonlinear devices, so in the real world all classes of amplifier have some nonlinearity. In practice, it turns out that Class AB amplifiers provide the best linearity for amplitude-modulated signals. Since Class AB amplifiers are also advantageous from the efficiency point of view, they’re commonly used in applications involving amplitude modulation.
Suppressing Harmonic Components
Returning to our main topic of discussion, there’s one last problem with the Class C mode of operation that we’ve yet to mention: namely, that the harmonic components of the output current are larger than those of either Class A and B amplifiers. This is especially true as we use smaller and smaller conduction angles (Figure 6). We therefore need a higher-Q resonant circuit to suppress the harmonic components.
Wrapping Up
Class C amplifiers were more common in the vacuum tube era than they are now—in fact, modern vacuum tube applications still benefit from the high efficiency of Class C amplifiers. However, they’re rarely used in modern RF semiconductor circuits. Solid-state designs typically use other amplifier types that provide a high efficiency without causing the problems we enumerated in this article. We’ll discuss some of these configurations in future articles.
All images used courtesy of Steve Arar
