# Introduction to the Class B Power Amplifier

## How does a Class B power amplifier work? What makes it more efficient than a Class A power amplifier? Learn the answers in this article.

High-efficiency RF power amplifiers (PAs) are key in many applications, from handheld communications equipment to large, active-element, phased-array antennas. In the aforementioned handheld devices, for instance, higher efficiency translates into lower power consumption, and hence more talk time.

We learned in a previous article that the inductively-loaded Class A amplifier provides a maximum obtainable efficiency of only 50%. This article introduces the Class B power amplifier, which has a theoretical maximum efficiency of 78.5%. After discussing the Class B amplifier’s basic characteristics, we’ll explore the reasons for that difference in efficiency.

Note that the analytical expressions used in this article are approximate—because they involve large signals and strong nonlinearities, power amplifier stages are difficult to analyze. However, these figures still help us gain a good understanding of the circuit’s behavior.

### Class B Power Amplifier Characteristics

In a Class B amplifier, the transistor is biased just below its turn-on point. When not quite on, the transistor is driven into conduction by the positive half-cycle of the input signal. For the other half-cycle of the signal, where the input swings negative, the transistor remains off. This is illustrated in Figure 1.

**Figure 1.** Operation of a transistor as a Class B power amplifier. Image used courtesy of Steve Arar

**Figure 1.**Operation of a transistor as a Class B power amplifier. Image used courtesy of Steve Arar

For a bipolar transistor, we need to bias the base-emitter junction at roughly 0.7 V. For a FET device, the gate-source is biased at pinch-off. These bias points allow the positive half-cycle of the input to drive the transistor into conduction.

#### The Conduction Angle

We’ll use a concept known as the *conduction angle* to help us describe the operation of the power amplifier’s transistor. The conduction angle is the fraction of an input cycle for which the transistor is on, expressed in angles or radians. In a Class A amplifier, for example, the transistor is always on, and so the conduction angle is 360 degrees. In a Class B amplifier, the transistor conducts for only one half of the signal cycle, so the conduction angle is 180 degrees.

An exact conduction angle of 180 degrees is a mathematical concept—in practice, Class B amplifiers might have conduction angles that differ slightly from this theoretical value. However, the conduction angle concept is still a helpful way of classifying different types of power amplifiers. Later on in the article, we’ll discuss how reducing the conduction angle from 360 to 180 degrees allows a Class B stage to achieve a higher efficiency than the Class A design.

#### Typical Waveforms

Due to the intermittent conduction of the transistor in a Class B stage, the current flowing through the transistor isn’t a faithful reproduction of the input signal (see Figure 1). Being a half-wave, rectified sinusoid, the output current contains different harmonics of the applied signal.

If we pass this distorted current through a resistive load, the output voltage will also be rich in different harmonics. However, we normally have a band-pass or low-pass filter at the output that can sufficiently suppress the higher-harmonic components. Figure 2 shows a single-transistor Class B amplifier.

**Figure 2.** An example Class B amplifier with a single transistor. Image used courtesy of Steve Arar

**Figure 2.**An example Class B amplifier with a single transistor. Image used courtesy of Steve Arar

The RF chokes allow the passage of DC current, but act as open circuits for RF signals. The input RF choke sets the quiescent value of the base-emitter voltage; the RF choke at the collector provides the transistor’s DC current. The load is an AC-coupled resistor (*R _{L}*) with a high-Q resonant circuit at the fundamental frequency.

Shorting the harmonic components, the high-Q tank makes the output voltage a sinusoid at the fundamental frequency. Figure 3 shows the transistor current and output voltage waveforms for near-ideal Class A, B, and C amplifiers.

**Figure 3.** Current (a) and voltage (b) waveforms for Class A, B, and C amplifiers. Image used courtesy of George Vendelin

**Figure 3.**Current (a) and voltage (b) waveforms for Class A, B, and C amplifiers. Image used courtesy of George Vendelin

We can assume that the resonant parallel LC network provides some harmonic filtering at the output. However, due to the intermittent conduction of the transistor, some harmonic distortion is still observable in the Class B and C stages. The output voltage and current waveforms of the Class A stage are almost undistorted.

### Calculating a Class B Amplifier’s Efficiency

Now that we have a good understanding of how a Class B amplifier works, let’s calculate the efficiency of a single-transistor Class B amplifier. Assume that the transistor current is a half-wave rectified sinusoid of amplitude *I _{p}* and period

*T*, as shown in Figure 4.

**Figure 4.** Output current of a Class B amplifier's transistor. Image used courtesy of Steve Arar

**Figure 4.**Output current of a Class B amplifier's transistor. Image used courtesy of Steve Arar

Using the Fourier series representation, we can express the output current in terms of its constituent frequency components:

$$i_{out}(t)~=~ \frac{I_p}{\pi} ~+~ \frac{I_p}{2} \sin(\omega_0 t) ~-~ \frac{2I_p}{3\pi} \cos(2 \omega_0t)~-~ \frac{2I_p}{15\pi} \cos(3 \omega_0t) ~+~~ …$$

**Equation 1.**

**Equation 1.**

where ⍵_{0} is the angular frequency of the signal. Assuming that a high-Q resonator eliminates the higher harmonic components, the AC voltage across the load (*R _{L}*) can be computed as:

$$v_{out}(t)~=~ \frac{R_L I_p}{2} \sin(\omega_0 t)$$

**Equation 2.**

**Equation 2.**

Now that we have the output voltage, we can calculate the average power delivered to the load as:

$$P_{L}~=~ \frac{(v_{rms})^2}{R_L}~=~ \frac{R_L I_p^2}{8}$$

**Equation 3.**

**Equation 3.**

To calculate the power provided by the supply, we find the average value of the current drawn from the supply (the waveform in Figure 4) and multiply it by the supply voltage. As we know from Equation 1, the average value of the half-wave rectified signal is *I _{p}*/π. Therefore, the power delivered by the supply is:

$$P_{CC}~=~ \frac{I_p V_{CC}}{\pi}$$

**Equation 4.**

**Equation 4.**

This equation, along with Equation 3, gives us the efficiency of the amplifier:

$$\eta ~=~ \frac{P_L}{P_{CC}} ~=~ \frac{\pi R_{L} I_p}{8 V_{CC}}$$

**Equation 5.**

**Equation 5.**

In order to find the *maximum* efficiency of the amplifier, we need to express the maximum value of *I _{p}* in terms of

*V*. To find this relationship, note that the maximum amplitude of the output swing—which is given by Equation 2—is equal to

_{CC}*V*. This is the case when, as in the inductively-loaded Class A amplifier we discussed, the collector (or drain) of the transistor is biased through an RF choke. The DC bias of the output is

_{CC}*V*; the output can swing above and below

_{CC}*V*.

_{CC}From Equation 2, we therefore have:

$$I_p ~=~ \frac{2V_{CC}}{R_L}$$

**Equation 6.**

**Equation 6.**

By substituting the above for *I _{p}* in Equation 5, we find the maximum efficiency:

$$\eta_{max} ~=~ \frac{\pi}{4}~=~78.5 \, \%$$

**Equation 7.**

**Equation 7.**

### Power Efficiency Compared to the Class A Amplifier

To better understand what makes Class B amplifiers more efficient than their Class A counterparts, let’s start by reviewing of what makes Class A amplifiers *less* efficient.

An inductively-loaded Class A amplifier provides relatively low efficiency because the transistor is always biased on, and draws a constant DC current from the supply even when no signal is applied. This results in a considerable amount of power being consumed in the transistor rather than in the load.

Figure 5, which first appeared in “Introduction to the Inductively-Loaded Class A Power Amplifier,” demonstrates how three power terms behave in this class of PA as collector current increases:

*P*: Supply power._{CC}*P*: Load power._{L}*P*: Transistor power._{Tran}

The figure also plots the amplifier’s power efficiency (the cyan curve).

**Figure 5.** Supply power, load power, transistor power, and power efficiency vs. collector current for an inductively-loaded Class A amplifier. Image used courtesy of Steve Arar

**Figure 5.**Supply power, load power, transistor power, and power efficiency vs. collector current for an inductively-loaded Class A amplifier. Image used courtesy of Steve Arar

As you can see, no power is delivered to the load in the absence of an AC signal. Instead, the transistor dissipates all the power provided by the supply. Even when the signal swing and efficiency are at their maximum, however, the power dissipated in the transistor equals the power delivered to the load.

So, how does the Class B amplifier address this? We saw in Figure 3 that the output voltages are almost identical for Class A and B stages. When we compare the conduction angles, though, we observe that the Class B transistor’s current is nonzero for a relatively smaller fraction of the signal cycle.

By reducing the fraction of the signal cycle for which *both* the drain current and drain voltage are simultaneously nonzero, the Class B amplifier reduces the power dissipated in the transistor. This is achieved at the cost of making the output highly distorted, which can be addressed either by incorporating a high-Q resonator or by using a push-pull configuration that we’ll discuss in a later article.

### Output Power Compared to the Class A Amplifier

But how does the output power of a Class B stage compare with a Class A amplifier for a given transistor and supply voltage? As we’ve already seen, both Class A and B stages can have a maximum swing amplitude of about *V _{CC}*. In other words, assuming that the transistor’s saturation voltage (

*V*) is zero, the peak-to-peak swing will be from zero to 2

_{CE(sat)}*V*. Therefore, the output swing for a given supply voltage is identical across both types of amplifiers.

_{CC}Now, let’s assume that the maximum current that can flow through the transistor is specified as *I _{max}*. The maximum AC current through the load of an inductively-loaded Class A stage is

*I*/2, with the other half of

_{max}*I*being lost in the transistor. When the transistor in a Class B stage operates at its maximum current, the peak of the half-wave rectified current (

_{max}*I*) shown in Figure 4 is equal to

_{p}*I*.

_{max}With *I _{p}* =

*I*, the amplitude of the fundamental component of the current is

_{max}*I*/2 (Equation 1)—which is also equal to the load current swing of the Class A amplifier. This means that a Class A and Class B stage, given the same transistor specifications and identical supply voltages to one another, can produce the same maximum AC current and output voltage. (We’re ignoring the effect of breakdown voltage on the bias point and voltage swing in this analysis).

_{max}It follows that both Class A and B designs can produce the same maximum output power. Additionally, since the voltage and current swings are the same, the optimum load resistances for both designs are identical.

### Wrapping Up

The Class B amplifier cuts off half of the input signal cycle and uses only the remaining half to produce the output signal, accepting an increase in signal distortion in exchange for significant improvements in efficiency. Because it wastes less power as heat, a Class B amplifier can also run much cooler than an equivalent Class A stage. We’ll go into more detail on Class B power amplifier configurations in the next article—for now, I hope you’ve found this introduction interesting and informative.

*Featured image (modified) used courtesy of Adobe Stock*

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