Understanding the Transformer-Coupled Voltage-Switching Class D Amplifier – Technical Articles

Previous articles in this series examined the complementary voltage-switching Class D amplifier and some of the non-idealities that affect its performance. In this article, we’ll discuss a different Class D configuration: the transformer-coupled voltage-switching (TCVS) amplifier. The schematic for the TCVS amplifier appears in Figure 1.

Figure 1. Transformer-coupled voltage-switching Class D amplifier.

Over the course of the article, we’ll explore this amplifier’s operation, analyze its performance, and compare it to the basic Class B configuration. To help cement the concepts we discuss, we’ll also determine the circuit parameters for a TCVS amplifier in two example problems. Before any of that, however, let’s examine the circuit itself.

The TCVS Amplifier Schematic

You may have noticed some similarities between the schematic in Figure 1 and the transformer-coupled push-pull Class B amplifier we learned about in an earlier article. For ease of comparison, the simplified diagram for the push-pull Class B stage is reproduced in Figure 2.

Figure 2. Transformer-coupled push-pull Class B amplifier.

In both of the above configurations, two transistors (Q₁ and Q₂) operate during alternating half-cycles of the input signal. Only one transistor is driven ON at any given time. To avoid both transistors being turned on at the same time, a center-tapped input transformer (T₂) produces opposite-polarity drive signals from a single-ended input signal.

The center tap of T₂ is connected to a fixed bias voltage. In Figure 2, this voltage is labeled as V_bias. For the push-pull Class B stage, V_bias is chosen to appropriately bias the transistors just below their turn-on point. For the TCVS configuration, the bias voltage is ground.

Both types of power amplifier use an output transformer (T₁) to combine the collector currents. In both Figure 1 and Figure 2, the center tap of the output transformer is connected to the power supply (V_CC). One major difference is that the push-pull Class B stage is directly connected to the load. The TCVS configuration, on the other hand, is connected to the load through a series LC circuit.

Despite the above similarities, the two amplifiers operate in completely different ways. In the push-pull Class B amplifier, the transistors operate as current sources, producing a sinusoidal voltage at the secondary of T₁. With the TCVS amplifier, however, the transistors are driven hard enough to act as switches and the voltage signal at the secondary of T₁ is a square wave.

How Does the TCVS Amplifier Operate?

First, let’s consider each half-cycle of the TCVS amplifier’s operation. The simplified diagram in Figure 3 shows the amplifier when the upper switch (S₁) is closed and the bottom switch (S₂) is open. We’re assuming the transistors act as ideal switches, which is why S₁ and S₂ take the place of Q₁ and Q₂ in this and following figures.

Figure 3. The TCVS amplifier when S₁ is ON and S₂ is OFF.

In this half-cycle, node C is clearly at ground potential. A voltage drop of V_CC appears across the upper segment of the primary winding. Its polarity is shown in purple in the figure.

Due to the transformer action, the same voltage is also induced across the lower winding. Node D is therefore at 2V_CC while node C is at ground. Note that we’re assuming an ideal transformer as well as ideal switches.

Figure 4 illustrates the next half-cycle, when S₁ is open and S₂ is closed.

Figure 4. The TCVS amplifier when S₁ is OFF and S₂ is ON.

A voltage of V_CC now appears across the lower segment of the primary winding. The magnetic coupling between the windings produces the same voltage across the upper segment of the primary. Once again, the polarity is shown in purple. In this half-cycle, node D is at ground and node C is at 2V_CC.

Understanding the Voltage Waveforms

To review:

• When S₁ is OFF and S₂ is ON, node C is at 2V_CC and node D is at ground.
• When S₁ is ON and S₂ is OFF, node C is at ground and node D is at 2V_CC.

Therefore, the voltages at node C and node D (V_C and V_D, respectively) are square waves switching between zero and 2V_CC. Figure 5 shows these voltage waveforms over two full cycles, assuming that S₁ is OFF and S₂ is ON in the first half-cycle.

Figure 5. The voltages at node C (top) and node D (bottom) over two full cycles of operation.

A key takeaway here is that the maximum collector-emitter voltage of each transistor is twice the supply voltage (2V_CC). This should be taken into account when choosing transistors for a TCVS amplifier.

Next, let’s determine the voltage at the input of the tuned circuit (node E). From Figures 3 and 4, we observe that a voltage of V_CC with alternating polarity drops across each segment of the primary winding. Since each segment in the primary winding has m turns and the secondary winding has n turns, the voltage at node E has an amplitude of (n/m)V_CC. Because the polarity is positive in one half-cycle and negative in the other, the voltage at node E switches between +(n/m)V_CC and –(n/m)V_CC.

This is actually an advantage of the TCVS circuit. By changing the turns ratio of the output transformer, we can scale the amplitude of the output square wave based on design specifications. The voltage waveform for node E (V_E) is illustrated in Figure 6.

Figure 6. Voltage at node E over two full cycles of operation.

The TCVS circuit operation is otherwise quite similar to that of the basic Class D amplifier in Figure 7.

Figure 7. The basic Class D amplifier.

This is the same simple circuit we used to introduce the concepts of Class D operation. Here, as in the TCVS amplifier, S₁ and S₂ turn ON and OFF alternately to produce a square wave at node E. However, the square wave switches between ground and V_CC rather than +(n/m)V_CC and –(n/m)V_CC.

Now that we understand how the TCVS amplifier operates, let’s examine its performance.

Finding the Output Power of the TCVS Amplifier

The amplifier’s series RLC circuit presents a very large impedance to all but the fundamental frequency component of the input voltage. Consequently, the tuned circuit imposes a sinusoidal current at the fundamental frequency (Figure 8).

Figure 8. A sinusoidal current at the fundamental frequency flows through the RLC circuit.

To find the amplitude of the TCVS amplifier’s output current, we need to find the fundamental component of the square wave at node E. Using the Fourier series representation, we can express the square wave current in Figure 6 in terms of its constituent frequency components:

$$v_E~=~frac{4}{pi} frac{n}{m}V_{CC} sum_{p=1}^{infty}frac{sin((2p~-~1)omega_{0}t)}{2p~-~1}$$

Equation 1.

where:

n = the number of turns in the output transformer’s secondary winding

m = the number of turns in each segment of the output transformer’s primary winding.

Therefore, the fundamental component of the square wave has a peak value of:

$$v_{E, fund}~=~ frac{4}{pi}frac{n}{m} V_{CC}$$

Equation 2.

Dividing by R_L, we find the peak value of the output current:

$$I_p ~=~ frac{4}{pi}frac{n}{m} frac{V_{CC}}{R_L}$$

Equation 3.

Finally, the output power is:

$$P_{L} ~=~ R_Li_{rms}^2 ~=~ frac{8}{pi^2} (frac{n}{m})^2 frac{V_{CC}^2}{R_L}$$

Equation 4.

where i_rms = I_p/(sqrt{2}).

Equations 3 and 4 are the key relationships for designing a TCVS amplifier, as we’ll see shortly. In the mean time, let’s find the TCVS amplifier’s theoretical efficiency.

Finding the Efficiency of the TCVS Amplifier

To calculate the amplifier’s efficiency, we need to know both the output power (Equation 4) and the input power. The input power is equal to the supply voltage multiplied by the average value of the current drawn from the supply.

Though the output current (i_RF) is a sinusoid, the currents passing through the switches (i₁ and i₂ in Figure 1) are half-wave rectified sinusoids. Therefore, the overall current drawn from the supply (i_cc) is a full-wave rectified sinusoid. This current waveform can be seen in Figure 9.

Figure 9. The total current drawn from the supply is a full-wave rectified sinusoid.

Though the peak value of i_RF is I_p, the peak value of i_cc is (n/m)I_p. This is due to the current-scaling function of the transformer. You can easily verify that a full-wave rectified sinusoid of amplitude I_p has a DC component of 2I_p/π. The average value of the waveform in Figure 9, which has an amplitude of (n/m)I_p, is:

$$I_{dc} ~=~ frac{2}{pi} big ( frac{n}{m}I_p big )$$

Equation 5.

Upon multiplying by V_CC and substituting I_p from Equation 3, the power delivered by the supply is found to be:

$$P_{CC} ~=~ frac{8}{pi^2} Big ( frac{n}{m} Big )^2 frac{V_{CC}^2}{R_L}$$

Equation 6.

This is equal to the output power in Equation 4, meaning that the TCVS amplifier—like the complementary voltage-switching amplifier—has an ideal efficiency of 100%.

Example 1: Choosing Maximum Transistor Voltage and Current for a TCVS Amplifier

In a previous article, we found the supply voltage and maximum switch current for a basic Class D amplifier (Figure 7) that delivered 20 W to a 50 Ω load. Let’s repeat this example for an ideal TCVS amplifier that delivers 20 W to a 50 Ω load. Assume that the turns ratio (n/m) is unity.

We’ll start with the supply voltage. Substituting our example values into Equation 4, we have:

$$P_{L} ~=~ frac{8}{pi^2} (frac{n}{m})^2 frac{V_{CC}^2}{R_L} ~~rightarrow~~20 ~=~ frac{8}{pi^2} ~times~ 1 ~times~ frac{V_{CC}^2}{50}$$

Equation 7.

Solving for V_CC, we obtain:

$$V_{CC}~=~ sqrt{frac{1000}{8} pi^2}~=~35.1$$

Equation 8.

The supply voltage of the TCVS amplifier is 35.1 V.

From Figure 9, the maximum current passing through the switches is (n/m)I_p. Substituting for I_p from Equation 3, we have:

$$I_{max} ~=~ frac{4}{pi}(frac{n}{m})^2 frac{V_{CC}}{R_L}$$

Equation 9.

We know that V_CC = 35.1 V, (n/m) = 1, and R_L = 50 Ω. Plugging these values into Equation 9, we get:

$$I_{max}~=~(frac{4}{pi})(frac{35.1}{50})~=~0.89$$

Equation 10.

The maximum switch current for the TCVS amplifier is 0.89 A.

The TCVS Amplifier vs. the Basic Class D Amplifier

Recall that we previously worked through Example 1 with a basic Class D amplifier instead of a TCVS amplifier. This gives us a useful jumping-off point for comparing the two designs, as we’ll see in this section.

Supply Voltage and Maximum Switch Current for a Given Output Power

To summarize the results of the previous section, V_CC = 35.1 V and I_max = 0.89 A for a TCVS amplifier delivering 20 W to a 50 Ω resistive load. For the same output power and load resistance, the basic Class D amplifier required a supply voltage of 70.2 V. Like the TCVS amplifier, each of its switches conducted a maximum current of 0.89 A. In other words, the TCVS configuration allowed us to halve the supply voltage while using the same maximum current.

How does the TCVS amplifier accomplish this? Assuming that n/m = 1, the TCVS circuit produces a square wave with a peak-to-peak value of 2V_CC at the input of the tuned circuit. The basic Class D amplifier, on the other hand, produces a peak-to-peak value of V_CC. This is what allows the TCVS circuit to halve the supply voltage for a given output power and load.

Maximum Collector-Emitter Voltage

What about the maximum collector-emitter voltage? In a basic Class D configuration, the maximum collector-emitter current is equal to V_CC, which is 70.2 V. However, the waveforms in Figure 5 show that the maximum collector-emitter voltage in a TCVS circuit is twice the supply voltage (2V_CC). Therefore, while we can use a supply voltage of V_CC = 35.1 V in our TCVS design, the maximum collector-emitter voltage that the transistor should tolerate is 70.2 V—the same as in the basic Class D design.

Maximum Switch Current and Output Power for a Fixed Supply

Finally, let’s say that we keep the supply voltage and load resistance fixed. How do the maximum switch current and output power change from the basic Class D amplifier and the TCVS amplifier?

From the article that first introduced the configuration, we know that the maximum switch current for a basic Class D amplifier is:

$$I_{max} ~=~ frac{2V_{CC}}{pi R_L}$$

Equation 11.

If we compare this with the TCVS amplifier’s maximum switch current (Equation 9), we observe that the TCVS circuit requires double the maximum current of the basic Class D amplifier for a given supply voltage (assuming that n/m = 1). Meanwhile, the output power of the basic Class D amplifier is:

$$P_{L} ~=~ frac{2V_{CC}^2}{pi^2 R_L}$$

Equation 12.

Equations 4 and 12 show that, for the same supply voltage and load impedance, the output power in a TCVS circuit is four times that of the basic Class D amplifier. Once again, we are assuming that n/m = 1.

But what if n/m didn’t equal 1? The turns ratio of the output transformer in a TCVS circuit provides us with an additional design parameter. As our next example will illustrate, this parameter can be used to trade off the supply voltage and the maximum switch current.

Example 2: Understanding The Role of the Output Transformer

Assume that a TCVS amplifier is to deliver 20 W to a 50 Ω load, as in the previous example. However, the turns ratio is now n/m = 2. What would be the required supply voltage and maximum switch current?

Substituting the above values into Equation 4, we have:

$$P_{L} ~=~ frac{8}{pi^2} (frac{n}{m})^2 frac{V_{CC}^2}{R_L} ~~rightarrow ~~20 ~=~ frac{8}{pi^2} ~times~ 2^2 ~times~ frac{V_{CC}^2}{50}$$

Equation 13.

Solving for V_CC, we get:

$$V_{CC}~=~sqrt{frac{1000}{32} pi^2}~=~17.56$$

Equation 14.

The supply voltage is V_CC = 17.56 V, half the value we obtained in the previous example. The maximum switch current is:

$$I_{max} ~=~ frac{4}{pi}(frac{n}{m})^2 frac{V_{CC}}{R_L} ~~rightarrow~~ I_{max} ~=~ frac{4}{pi} ~times~ 2^2~ times~ frac{17.56}{50}~=~1.79$$

Equation 15.

I_max = 1.79 A, which is twice the previous value. In short, doubling the turns ratio reduces the required supply voltage by a factor of two and increases the maximum current by the same factor.

Up Next

This article introduced us to the transformer-coupled voltage-switching Class D amplifier, otherwise known as the TCVS amplifier. In the next article, we’ll compare this configuration with the transformer-coupled current-switching (TCCS) Class D amplifier, which we’ll discuss at length.

All images used courtesy of Steve Arar