Over the decades, a variety of power amplifier topologies have been developed to meet the objectives of different applications. Some power amplifier classes, including those we’ll be discussing, are defined based on their conduction angle (θc). The conduction angle represents the fraction of an input cycle for which the amplifier’s RF transistor is on.
Previous articles in this series discussed Class A and Class B amplifiers. With a Class A amplifier, the transistor is always on. The amplifier therefore has a conduction angle of 360 degrees. In a Class B amplifier, the transistor conducts for only half of the signal cycle, so the conduction angle is 180 degrees.
We learned that reducing the conduction angle allows us to increase the efficiency from 50% in a Class A stage to 78.5% in a Class B amplifier. But what happens to efficiency if we reduce the conduction angle even further?
Power amplifiers with a conduction angle lower than 180 degrees are known as Class C amplifiers. In this article, we’ll examine the operation of Class C amplifiers and explore the impact of a lower conduction angle reduction on various power amplifier performance parameters. We’ll conclude with a classic unifying analysis that compares the performance of Class A, B, and C amplifiers.
Current and Voltage Waveforms in a Class C Amplifier
In a Class C amplifier, the transistor is on for less than half of the input cycle. Being stimulated by a narrow pulse, the transistor produces a short current pulse at the output. The orange curve in Figure 1 shows one period of the collector current for a Class C amplifier with conduction angle θc.
Figure 1. Current waveform for a Class C stage with conduction angle θc. Image used courtesy of Steve Arar
We can see above that the output current is a piece of a sinusoid when the transistor is active and zero when the transistor is in cutoff. The full sinusoid (the blue curve above) has a negative offset of IQ and amplitude of IRF. The DC offset IQ is analogous to the bias current in a linear amplifier, but is negative in the Class C amplifier. The output current can therefore be described by the following expression:
$$i_c ~=~ max (I_Q ~+~ I_{RF} sin(theta), ~0 )$$
Equation 1.
where θ corresponds to the position on the horizontal axis.
By changing IQ, we can produce the waveforms of Class A and B amplifiers as well. For example, IQ = 0 leads to a conduction angle of 180 degrees (Class B). We can therefore use the above waveform to examine the performance of all three amplifier classes—A, B, and C—under discussion.
Figure 2 compares the transistor current and output voltage waveforms for near-perfect Class A, B, and C amplifiers.
Figure 2. Current (a) and voltage (b) waveforms for Class A, B, and C amplifiers. Image used courtesy of George Vendelin
Both the transistor current and output voltage waveforms of the Class A amplifier are sinusoids. Though the transistor current is only a piece of a sinusoid in the Class B and C stages, the output voltage can be approximated as a sinusoid for these modes of operation as well. This is due to the presence of a high-Q tank at the output of the Class B and C amplifiers, as we’ll see in the next section.
The Class C Amplifier Schematic
Figure 3 shows the basic circuit schematic of a Class C amplifier. The high-Q tank at the amplifier’s output is labeled in green.
Figure 3. The basic schematic of a Class C amplifier. Image used courtesy of Steve Arar
Depending on the quiescent bias point we choose for the transistor, the above schematic could also be used to build a Class A stage or a single-transistor Class B stage. The Class A amplifier, which is the most linear of the three types, may use a tank circuit with a relatively lower Q factor.
At the other end of the linearity spectrum, a Class C amplifier produces a train of short current pulses at the output. The high-Q tank circuit shorts the output current harmonics and reduces the out-of-band emissions that result from the inevitable nonlinearities. Note that a high-Q tank circuit necessarily implies narrowband operation.
Waveform Analysis
When analyzing Class C stages, we assume the following:
- The output voltage can be approximated by a sinusoidal waveform. This requires an ideal tank circuit that can short out all the higher harmonics of the output current.
- The collector current waveform is a portion of a sine wave. This is actually true only at low frequencies.
Though not necessarily true in practice, these assumptions allow us to simplify the circuit analysis. With that in mind, let’s examine the Class C amplifier’s performance.
Due to the high-Q tank, the power at the fundamental frequency is what’s actually delivered to the load. To find the output power, we therefore need to analyze the frequency content of the output current waveform. We can make this analysis simpler by changing the time origin of the waveform. Figure 4, which is symmetric around the vertical axis, is the result.
Figure 4. One period of collector current in a Class C amplifier. Image used courtesy of Steve Arar
The above waveform can be described as a cosine function:
$$i_c ~=~ max (I_Q ~+~ I_{RF} cos(theta), ~0 )$$
Equation 2.
Using the Fourier series, we can write the output current in terms of its constituent frequency components:
$$i_c ~=~ sum_{n=0}^{infty} a_n cos(n theta)$$
Equation 3.
where an denotes the Fourier coefficient of the n-th harmonic. To find the efficiency and output power of the Class C stage, we only need the average value (a0) and the fundamental component (a1) of the current waveform. Without going through all of the math, the results are provided below in Equations 4 and 5:
$$a_0 ~=~ frac{I_M}{pi} ~times~ frac{ sin (phi) ~-~ phi cos (phi) }{ 1 ~-~ cos (phi)}$$
Equation 4.
and:
$$a_1 ~=~ frac{I_M}{pi} ~times~ frac{ phi ~-~ sin(phi) cos (phi) }{ 1 ~-~ cos (phi)}$$
Equation 5.
where ϕ is equal to half the conduction angle ((phi~=~frac{theta_{c}}{2})).
Figure 5 plots the average and fundamental components against the conduction angle.
Figure 5. The average and fundamental components against the conduction angle. Image used courtesy of Steve Arar
This figure shows the plot of the a0 and a1 coefficients normalized to IM (or, equivalently, IM is assumed to be unity). We’ll return to these results shortly. First, though, let’s calculate the efficiency of the Class C amplifier.
Efficiency of Class C Operation
Assuming that the high-Q resonator eliminates the higher harmonic components, the AC output voltage can be computed from Equation 5 as:
$$v_{out} ~=~ R_L a_1 cos(theta)$$
Equation 6.
where RL is the resistance of the load.
Therefore, the average power delivered to the load is:
$$P_L ~=~ frac{(v_{rms})^{2}}{R_L} ~=~ frac{R_L a_1^{2}}{2}$$
Equation 7.
To calculate the power provided by the supply, we multiply the average value of the current drawn from the supply by the supply voltage. The average value of the current is a0, producing:
$$P_{CC} ~=~ V_{CC} a_0$$
Equation 8.
From Equations 7 and 8, we can calculate the efficiency:
$$eta ~=~ frac{P_L}{P_{CC}} ~=~ frac{R_L a_1^2}{2V_{CC} a_0}$$
Equation 9.
With an appropriate load resistance, the fundamental component of the current produces the maximum permissible voltage swing. The maximum voltage swing has an amplitude of VCC. Therefore, from Equation 6, the maximum efficiency is achieved when:
$$R_L a_1 ~=~ V_{CC}$$
Equation 10.
Combining Equations 10 and 9, the maximum efficiency is calculated as:
$$eta_{max} ~=~ frac{a_1}{2 a_0}$$
Equation 11.
Finally, substituting Equations 4 and 5 into the above equation gives us:
$$eta_{max} ~=~ frac{1}{2} ~times~ frac{phi ~-~ sin(phi) cos (phi)}{sin (phi) ~-~ phi cos (phi)}$$
Equation 12.
This equation is plotted in Figure 6.
Figure 6. Maximum efficiency vs. conduction angle. Image used courtesy of Steve Arar
Let’s give these equations some context by working through an example.
Example: Choosing a Maximum Current Specification for a Class C Amplifier
We know that transistors are limited in regard to the maximum voltage and current levels they can handle, as well as in the maximum power they can burn without being damaged. Determine the maximum transistor current for a Class C amplifier that delivers 25 W to a 50 Ω load at a maximum efficiency of 85%. Ignore the saturation effect of the transistor and assume that the supply voltage (VCC) is 12 V.
The output current’s fundamental component (a1) determines the power delivered to the load. From Equation 7, we have:
$$P_L ~=~ frac{R_L a_1^2}{2} ~=~ 25 ~text{W}$$
Equation 13.
In this example, we’re operating at maximum efficiency. We therefore know that RLa1 = VCC (Equation 10). Since VCC =12 V, we have RLa1 = VCC = 12 V. Substituting the 12 V value into Equation 13 gives us the amplitude of the fundamental component:
$$frac{12 ~times~ a_1}{2} ~=~ 25~text{W} ~Rightarrow~ a_1 ~=~4.17 ~text{A}$$
Equation 14.
From Figure 7, which reproduces the plot of maximum efficiency versus conduction angle, we observe that an ηmax = 85% corresponds to θc = 147 degrees.
Figure 7. The maximum efficiency of 85% is achieved at θc = 147 degrees. Image used courtesy of Steve Arar
At a conduction angle of θc = 147 degrees, the normalized value of a1 is 0.45 (Figure 8).
Figure 8. At θc = 147 degrees, the normalized fundamental component is 0.45. Image used courtesy of Steve Arar
In other words, we have:
$$frac{4.17}{I_M} ~=~0.45 ~Rightarrow~ I_M ~=~ 9.27 ~text{A}$$
Equation 15.
The transistor should be able to handle a maximum current of 9.27 A. The maximum voltage that the transistor experiences is 24 V, or twice the supply voltage (2VCC = 2 × 12 V = 24 V).
Comparing Class A, B, and C Operation
Next, let’s broaden our focus a bit and see how changing the conduction angle affects the following performance parameters:
- Supply power.
- Output power.
- Maximum efficiency.
How Does the Power Drawn from the Supply Change with θc?
Figure 5 shows that the DC component of the output current decreases monotonically as the conduction angle is reduced from 360 degrees (Class A operation) to 180 degrees (Class B) to 0 degrees.
This makes sense if we consider the current waveform in Figure 4. A smaller conduction angle means a smaller region where the current is non-zero, which also corresponds to a smaller average value. Reducing the conduction angle thus lowers the average value and the DC power drawn from the supply.
How Does the Power Delivered to the Load Change with θc?
The fundamental component in Figure 5 exhibits a more interesting behavior. At 360 degrees, the fundamental component value is 0.5. As we reduce the conduction angle from 360 degrees toward 180 degrees, the fundamental component rises slightly.
At 180 degrees, however, the fundamental value is once again 0.5. This means that for identical transistor specifications and supply voltages, Class A and B stages produce the same maximum output power.
What about the Class C region of operation? If we substitute Equation 10 into Equation 7, we observe that the maximum output power is proportional to a1. From Figure 5, as θc approaches zero, so does a1. As a result, the output power of the Class C amplifier falls to zero as well. This is a significant drawback of Class C operation.
How Does the Efficiency of the Class C Amplifier Change with θc?
In the Class C region, both the average and fundamental components drop as we decrease the conduction angle. From Equation 11, we know that the maximum efficiency (ηmax) is proportional to the ratio of the fundamental component to the average value. Meanwhile, a visual examination of Figure 6 confirms that ηmax increases as we reduce the conduction angle. As the conduction angle moves toward zero, the efficiency approaches 100%.
Though this result might seem exciting at first, keep in mind that the output power of a Class C amplifier is much lower than that of a Class A or Class B amplifier for the same input power. A configuration that produces half the power of the same device in a Class A amplifier, for example, has little practical use even if its efficiency is as high as 95%.
Drawbacks of Class C Amplifiers
The Class C configuration suffers from several other limitations:
- For a given output power, the transistor used in a Class C amplifier must handle larger currents than the transistor in a Class A or Class B amplifier. This worsens as we reduce the conduction angle to achieve a higher efficiency.
- A larger maximum current means we need a larger device, which leads to a lower matching bandwidth.
- The Class C configuration is more nonlinear than either Class A or Class B amplifiers.
- The Class C stage requires transistors with higher breakdown voltage.
- Compared to a Class A or B amplifier, a Class C stage requires a higher-Q resonant circuit to suppress the harmonic components.
In the next article of this series, we’ll explore these limitations in detail.
