# Basic Principles of the Push-Pull Class B Power Amplifier

## Learn how a push-pull Class B amplifier functions, how to calculate its efficiency, and how its performance compares to the inductively-loaded Class A design.

As we discussed in the preceding article, a single-transistor Class B amplifier (Figure 1) uses a high-Q tank circuit as a load to suppress the higher harmonic components. With the high-Q resonant circuit employed, the output voltage contains only the fundamental component, enabling the amplifier to faithfully reproduce the input signal. Shorting the harmonic components, the high-Q tank makes the output voltage a sinusoid at the fundamental frequency, though the collector current is a half-wave rectified sinusoid.

**Figure 1.** A single-transistor Class B RF amplifier.

**Figure 1.**A single-transistor Class B RF amplifier.

Instead of using a high-Q tank, we could also remove the harmonic components of the Class B stage by forcing two half-sine wave pulses through the load in opposite directions. This is known as a push-pull amplifier. This article presents the basic concepts that underlie the push-pull configuration, along with some example calculations and a comparison of this amplifier type with the inductively-loaded Class A stage.

The push-pull configuration is generally introduced at the undergraduate level, so most EEs have at least a passing familiarity with it. However, RF and microwave implementation of this amplifier can involve some additional intricacies, especially when the goal is to achieve high output power and high efficiency over a wide bandwidth. Still, a push-pull Class B RF power amplifier operates much like a Class B audio-frequency power amplifier.

### The Transformer-Coupled Push-Pull Configuration

There are several different methods for implementing a push-pull amplifier. Figure 2 illustrates a configuration commonly known as the transformer-coupled push-pull power amplifier. It employs two Class B transistors—one that conducts during the positive half-cycle of the waveform, and one that operates during the negative half-cycle.

**Figure 2.** A transformer-coupled push-pull power amplifier.

**Figure 2.**A transformer-coupled push-pull power amplifier.

The two transistors (*Q*_{1} and *Q*_{2}) operate on alternating half-cycles. To drive these two transistors, we need to have both the input signal and its polarity-inverted version. Notice that both transistors are NPN types.

The power supply (*V _{CC}*) is connected to the center tap of the transformer. With the waveforms shown, transistor

*Q*

_{1}is driven into conduction during the first half-cycle. In this half-cycle, transistor

*Q*

_{2}remains off. When

*Q*

_{1}is on and

*Q*

_{2}is off, the circuit can be simplified to the one shown in Figure 3.

**Figure 3.** Transformer-coupled push-pull amplifier when Q_{1}* is on and *Q_{2}* is off.*

**Figure 3.**Transformer-coupled push-pull amplifier when

In Figure 3, transistor *Q*_{1} draws a collector current (*i _{c}*

_{1}) from

*V*. As indicated by the transformer dot convention, the current in the secondary of the transformer flows into

_{CC}*R*. This produces the positive half-cycle of the output voltage.

_{L}Figure 4 illustrates the next half-cycle, when *Q*_{1} turns off and *Q*_{2} turns on.

**Figure 4.** Transformer-coupled push-pull amplifier. Q_{1}* is off and *Q_{2}* is on.*

**Figure 4.**Transformer-coupled push-pull amplifier.

Once again, the collector current (*i _{c}*

_{2}) is drawn from

*V*. However, the direction of the current flowing through the primary winding is reversed. This reverses the output current direction, producing the negative half-cycle of the output voltage waveform. In this way, the transformer combines the collector currents appropriately to produce a sinusoidal output—rather than a rectified sine wave—at the fundamental frequency.

_{CC}

### Power Terms in the Push-Pull and Class A Arrangements

Figure 5 shows how the three power terms (*P _{L}*,

*P*, and

_{CC}*P*) associated with the push-pull circuit change with the amplitude of the collector AC current. It also plots the amplifier’s power efficiency.

_{Tran}

**Figure 5.** Supply power (P_{CC}*), load power (*P_{L}*), transistor power (*P_{Tran}*), and power efficiency versus collector current (*i_{c}*) for a push-pull amplifier.*

**Figure 5.**Supply power (

The parameter *R _{L,c}* used in the above graph is defined as \( (\frac{m}{n})R_{L}^{2}\). It’s the equivalent load resistance seen from the collector of each transistor when the other transistor is off.

From Figure 5, we can see that the following occurs as the collector AC current (*i _{c}*) increases from zero to its maximum value:

- The power delivered to the load (
*P*) increases as the square of_{L}*i*._{c} - The power drawn from the supply (
*P*) increases linearly._{CC} - The power dissipated in the transistors (
*P*) reaches its maximum at the maximum value of_{Tran}*i*._{c}

The maximum power dissipated in each transistor is one-fifth the maximum power that the push-pull stage can deliver to the load.

Let’s contrast this with Figure 6, which reproduces the power terms of the inductively-loaded Class A configuration.

**Figure 6.** Supply power (P_{CC}*), load power (*P_{L}*), transistor power (*P_{Tran}*), and power efficiency versus collector current (*i_{c}*) for an inductively-loaded Class A amplifier.*

**Figure 6.**Supply power (

In a Class A stage, the transistor is always biased on. A constant DC current is drawn from the supply even when no AC signal is applied. As a result, *P _{CC}* is always constant, and remains non-zero even in the absence of an AC signal. Because the transistor is always biased on, a significant amount of power is wasted in the transistor as heat.

With a Class B amplifier, *P _{CC}* increases linearly with the collector current. No power is burned in the transistors when the amplitude of the AC signal is very small, leading to a much higher efficiency.

### Calculating the Push-Pull Amplifier’s Efficiency

To calculate the efficiency of the push-pull configuration, we need to find the DC power drawn from the supply (*P _{CC}*) and the AC power delivered to the load (

*P*). We’ll start with

_{L}*P*.

_{CC}The current drawn by each transistor is a half-wave rectified sinusoid. However, the overall current drawn from the supply—labeled as *i _{cc}* in the diagrams above—is a full-wave, rectified sine wave. Below, Figure 7 shows the

*i*waveform.

_{cc}*T*is the period of the input sine wave, and

*I*denotes the maximum current flowing through the transistors.

_{p}

**Figure 7. **Overall current flowing through the push-pull amplifier’s transistors over time.

**Figure 7.**Overall current flowing through the push-pull amplifier’s transistors over time.

You can easily verify that a full-wave, rectified sinusoid of amplitude *I _{p}* has a DC component of \(\frac{2I_{p}}{\pi}\). Knowing that, we can calculate the average power provided by the supply as:

$$P_{CC}~=~2\frac{I_pV_{CC}}{\pi}$$

**Equation 1.**

**Equation 1.**

Now we compute the power delivered to the load. Considering the current scaling of the transformer, if *i _{cc}* has a peak value of

*I*, the current flowing through the load is a sinusoidal waveform of amplitude \((\frac{m}{n})I_{p}\). Therefore, the voltage across

_{p}*R*can be written as:

_{L}$$v_{out}~=~(\frac{m}{n})I_p R_L \sin(\omega_0 t)$$

**Equation 2. **

**Equation 2.**

The average power delivered to the load is:

$$P_{L}~=~\big (\frac{m}{n} \big )^2 \frac{I_p^2 R_L}{2}$$

**Equation 3.**

**Equation 3.**

Equation 3, together with Equation 1, gives us the efficiency of the circuit:

$$\eta ~=~ \frac{P_{L}}{P_{CC}}~=~\big (\frac{m}{n} \big ) ^2 \frac{\pi}{4} \frac{I_p R_L}{V_{CC}}$$

**Equation 4.**

**Equation 4.**

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

*V*. We can find this relationship by noting that the collector voltage has a maximum swing

_{CC}*amplitude*of

*V*. In other words, the collector voltage can swing from zero volts to 2

_{CC}*V*, assuming that the transistors’ saturation voltage is zero (

_{CC}*V*= 0).

_{CE(sat)}Taking into account the voltage scaling of the transformer, we observe that the maximum amplitude of the output voltage swing is \((\frac{n}{m}) V_{CC}\). On the other hand, from Equation 2, we know that the output swing is \((\frac{m}{n}) I_{p} R_{L} \). Equating these two values, we find the current amplitude that corresponds to maximum voltage swing:

$$(\frac{n}{m})V_{CC}~=~(\frac{m}{n})I_{p}R_{L} \; ~~\Rightarrow~~ \; I_p~=~ (\frac{n}{m})^2 \frac{V_{CC}}{R_L}$$

**Equation 5.**

**Equation 5.**

Using this in the efficiency equation (Equation 4), we find the maximum efficiency:

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

**Equation 6.**

**Equation 6.**

which is the same as that of the single-transistor Class B stage.

### Maximum Power Dissipated in the Transistors

Transistors have limitations as to the maximum power they can burn without being damaged. For that reason, it’s important to know how much power is going to be dissipated in the transistor of a given power amplifier.

The power dissipated in both transistors of the push-pull configuration is equal to the DC power provided by the supply minus the power delivered to the load (*P _{CC}* –

*P*). Half this value is dissipated in each transistor, giving us:

_{L}$$P_{Tran}~=~ \frac{1}{2} \big ( P_{CC}~-~P_L \big )~=~\frac{1}{2} \bigg ( 2\frac{i_cV_{CC}}{\pi} ~-~ \frac{1}{2}R_L \big (\frac{m}{n} \big ) ^2 i_c^2 \bigg )$$

**Equation 7.**

**Equation 7.**

where *i _{c}* denotes the amplitude of the collector AC current. By taking the derivative of this function with respect to

*i*, we can verify that the maximum value of

_{c}*i*occurs at:

_{c}$$i_c ~=~ \frac{2V_{CC}}{\pi} \frac{1}{(m/n)^2 R_L}$$

**Equation 8.**

**Equation 8.**

Substituting this value into Equation 4, we can find the maximum power dissipated in each transistor:

$$P_{Tran, max} ~=~ \frac{V_{CC}^2}{\pi^2} \frac{1}{(m/n)^2 R_L}$$

**Equation 9.**

**Equation 9.**

### Producing Opposite-Polarity Inputs

As we discussed above, the transformer-coupled push-pull configuration needs both the input signal and the signal’s inverse. Figure 8 shows how a center-tapped transformer can be used at the input of the push-pull configuration to produce opposite-polarity signals from a single-ended input signal.

**Figure 8.** A push-pull amplifier configuration with a center-tapped transformer for input signal generation.

**Figure 8.**A push-pull amplifier configuration with a center-tapped transformer for input signal generation.

The primary winding and the two segments of the secondary winding each have *k* turns. With a turns ratio of unity, the input signal (*v _{s}*) appears across both windings of the secondary. However, since the center tap is connected to the bias voltage (

*V*), the voltages at nodes

_{bias}*A*and

*B*swing in opposite directions around

*V*.

_{bias}The voltage at node *A* is in phase with the input, whereas the voltage at node *B* is 180 degrees out of phase. *V _{bias}* is chosen appropriately to bias the transistors just below their turn-on point. With that in mind, let's work through an example.

### Example: Choosing the Turns Ratio For Maximum Output Power

Assume that the transistor for the push-pull amplifier in Figure 8 has the following specifications:

- Maximum collector current (
*i*) = 1 A._{c,max} - Maximum collector-emitter breakdown voltage (
*BV*) = 40 V._{CEO} - Maximum power the transistor can handle without being damaged (
*P*) = 4 W._{C,max}

Let’s find the appropriate turns ratio for the output transformer to deliver maximum power to a 50 Ω load.

First, we’ll find the voltage and current swing at the collector. To find the corresponding output swings, we’ll scale them by the turns ratio of the transformer. We know that the amplitude of the half-sine wave collector current is limited by the maximum permissible current of the transistor: *I _{p}* =

*i*= 1 A. Besides, since the collector voltage swings from 0 to 2

_{c,max}*V*, we should choose the supply voltage to be half

_{CC}*BV*to avoid damaging the transistor.

_{CEO}We therefore have V_{CC} = 20 V, which corresponds to a voltage swing amplitude of 20 V at the collectors. The current and voltage swing at the secondary of the transformer are \((\frac{m}{n})~\times~ 1~\text{A}\) and \((\frac{n}{m})~\times~ 20~\text{V}\), respectively. Using Ohm's Law, we can relate the output voltage and current swing to the load resistance:

$$\frac{(\frac{n}{m}) ~\times~ 20}{(\frac{m}{n}) ~\times~ 1}~=~50 \; ~~\Rightarrow~~ \; \frac{m}{n}~=~0.63$$

**Equation 10.**

**Equation 10.**

Finally, we use Equation 9 to find the maximum power dissipated in each transistor:

$$P_{Tran,max} ~=~ \frac{(V_{CC})^2}{\pi^2} \frac{1}{(m/n)^2 R_L}~=~2.04~\text{W}$$

**Equation 11.**

**Equation 11.**

which is below the maximum specified value for the transistors (*P _{C,max}* = 4 W).

### Wrapping Up

I hope that you now have a basic understanding of the push-pull amplifier. When we return to this configuration in a future article, we’ll look at some of the challenges and limitations that affect the performance of practical RF and microwave push-pull amplifiers.

*Featured image used courtesy of Adobe Stock; all other images used courtesy of Steve Arar*

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