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

The previous article in this series introduced the transformer-coupled voltage-switching (TCVS) Class D amplifier. In this article, we’ll learn about another transformer-coupled Class D configuration—the transformer-coupled current-switching (TCCS) amplifier. We’ll go over the principles of its operation, analyze its performance, and work through an example where we determine its circuit parameters for a given load and output power.

The TCCS circuit is the final Class D amplifier we’ll cover in this series. For that reason, we’ll wrap up the article by comparing the requirements of the different Class D configurations we’ve discussed, including the TCCS amplifier.

The TCCS Configuration

Figure 1 shows the schematic of the TCCS amplifier.

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

The TCVS schematic from the previous article is reproduced in Figure 2. Let’s briefly go over the similarities and differences between the two amplifiers.

Figure 2. Transformer-coupled voltage-switching (TCVS) Class D amplifier.

We’ll start with the similarities. Both amplifiers use a center-tapped input transformer (T₂) to produce opposite-polarity signals from the input signal. The center tap of T₂ is connected to ground; the opposite-polarity signals at the secondary of T₂ are used to drive two transistors (Q₁ and Q₂) hard enough that they act as switches. Q₁ and Q₂ operate on alternating half-cycles of the input signal.

Both circuits also use an output transformer (T₁) to combine the collector currents. In the TCVS configuration, however, the center tap of T₁ is directly connected to the power supply (V_CC). The TCCS amplifier connects the center tap to V_CC through an RF choke (RFC). Having a reactance of infinity at the switching frequency, the RF choke blocks AC currents and forces a constant current into the transformer center tap.

The center tap of T₁ is one of the two major differences between the configurations. The second difference is the load network. Unlike the voltage-switching arrangement, which uses a series RLC circuit as its load, the current-switching amplifier has a parallel RLC network at the output.

Operation of the TCCS Amplifier

During each half-cycle of operation, a current (I_dc) is forced by the RF choke into the center tap of the output transformer (T₁). It then passes through one of the two switches (the transistors Q₁ and Q₂) before reaching ground.

The simplified diagrams in Figures 3 and 4 illustrate the TCCS amplifier’s operation over two consecutive half-cycles. The ideal switches S₁ and S₂ take the place of Q₁ and Q₂, respectively.

Figure 3. The TCCS amplifier when S₁ is closed and S₂ is open (Q₁ is ON and Q₂ is OFF).

Figure 4. The TCCS amplifier when S₁ is open and S₂ is closed (Q₁ is OFF and Q₂ is ON).

In Figure 3, S₁ is closed and S₂ is open. I_dc therefore passes through S₁ (i₁ = I_dc). Since S₂ is open, no current can flow through the lower path (i₂ = 0). In the next half-cycle (Figure 4), S₁ is open and S₂ is closed, leading to i₁ = 0 and i₂ = I_dc. Assuming that S₁ is open and S₂ is closed in the first half-cycle, we obtain the current waveforms in Figure 5.

Figure 5. Current waveforms for i₁ and i₂ over two full cycles of operation.

As we can see, the collector currents (i₁ and i₂) are square waves switching between 0 and I_dc. Note that I_dc is still an unknown parameter—we won’t be able to find its value until the end of our analysis. For now, we need to determine the current flowing through the secondary of T₁. This current is denoted by i₃ in the diagrams above.

Depending on the half-cycle, the current delivered by the choke passes through either the upper or lower half of the primary winding. Therefore, due to the transformer action, the amplitude of the current in the secondary winding is (m/n)I_dc. However, the direction of this current alternates between half-cycles. Figure 6 shows the current waveform for i₃ over two full cycles of operation.

Figure 6. The secondary current waveform over two cycles of operation.

To understand why i₃ switches between (m/n)I_dc and –(m/n)I_dc, we need to consider Figure 5’s collector current waveforms in context of the circuit diagrams. During the first half-cycle, the current delivered by the choke passes through the lower half of the primary winding. Since it enters the dotted end of the primary winding, the secondary current exits the dotted end, leading to i₃ = (m/n)I_dc.

In the next half-cycle, however, the choke’s current flows through the upper half of the primary winding. Since the primary current exits the dotted end of that winding, the output current enters the dotted end of the secondary winding. We therefore have i₃ = –(m/n)I_dc.

Analyzing the Performance of the TCCS Amplifier

A 50% duty cycle square wave contains all odd harmonics (1st, 3rd, 5th, etc.). Using the Fourier series representation, we can express the square wave current in Figure 6 in terms of its constituent frequency components:

$$i_3~=~frac{4}{pi} frac{m}{n}I_{dc} sum_{p=1}^{infty}frac{sin((2p~-~1)omega_{0}t)}{2p~-~1}$$

Equation 1.

The current-switching circuit applies the sum of all these harmonic components to the load network. However, a high-Q parallel RLC circuit tuned to the switching frequency presents a very small impedance to all of the current’s frequency components except for the fundamental component. At the fundamental frequency, the RLC circuit exhibits a resistance of R_L. As a result, all output current harmonics are shorted and only the fundamental component flows into the load resistor (R_L).

Denoting the current through R_L as i_out, we have:

$$i_{out}~=~frac{4}{pi} frac{m}{n}I_{dc} sin(omega_0 t)$$

Equation 2.

The peak value of i_out is equal to (frac{4}{pi} frac{m}{n} I_{dc}). The RMS value of i_out (i_rms) is equal to the peak value divided by the square root of 2. Knowing that, we can now calculate the average power delivered to the load:

$$P_{L} ~=~ R_L i_{rms}^2 ~=~ frac{8}{pi^2} (frac{m}{n})^2 {I_{dc}^2}{R_L}$$

Equation 3.

The snag is that we still don’t know I_dc, the current passing through the choke. Until we do, Equation 3 isn’t useful to us. To find I_dc, we need to investigate the circuit further.

Calculating the Current Through the RF Choke

Assume that the RF choke is ideal. Since that’s the case, no DC voltage drops across it. The DC component of the voltage at the output transformer’s center tap is therefore equal to V_CC. This piece of information is crucial to finding I_dc.

To find the voltage waveform for the transformer’s center tap, we first use Equation 2 to determine the voltage across the transformer’s secondary winding:

$$v_{out}~=~frac{4}{pi} frac{m}{n}I_{dc} R_L sin(omega_0 t)$$

Equation 4.

Because of the voltage-scaling function of the transformer, the voltage across each segment of the primary winding is therefore:

$$v_{p}~=~frac{4}{pi} (frac{m}{n})^2 I_{dc} R_L sin(omega_0 t)$$

Equation 5.

Figure 7 provides a simplified diagram of the circuit during the first half-cycle (Q₁: OFF, and Q₂: ON). The relevant voltage quantities (v_out and v_p) are shown in purple.

Figure 7. Voltages when Q₁ is OFF and Q₂ is ON.

Consider nodes C, D, and E in the figure above. Nodes C and D are located at the collectors of Q₁ and Q₂, respectively. Node E appears at the center tap of T₁. In this half-cycle, the node voltages are as follows:

v_C = 2v_p

v_D = 0

v_E = v_p

Note that nodes E and C have sinusoidal voltages.

Figure 8 shows the next half-cycle (Q₁: ON and Q₂: OFF).

Figure 8. Voltages when Q₁ is ON and Q₂ is OFF.

For this half-cycle, the node voltages are:

v_C = 0

v_D = –2v_p

v_E = –v_p

The voltages at nodes E and D are sinusoidal during this half-cycle. However, they are inverted with respect to v_p and v_out.

Figure 9 shows voltage waveforms for these nodes, along with the output, over one full cycle of operation.

Figure 9. From top to bottom: voltage waveforms at the output, node D, node C, and node E.

In the above figure, A₁ and A₂ are the amplitudes of v_out and v_E, respectively. A₁ is simply the peak value of v_out in Equation 4:

$$A_{1}~=~frac{4}{pi} frac{m}{n}I_{dc} R_L$$

Equation 6.

Depending on the half-cycle, v_E is equal to either v_p or –v_p. Taking the peak value of v_p from Equation 5, we obtain:

$$A_{2}~=~ frac{4}{pi} (frac{m}{n})^2 I_{dc} R_L$$

Equation 7.

A full-wave rectified sinusoid of amplitude A₂ has a DC component of 2A₂/π. Therefore, the average component of v_E is:

$$v_{E, ave} ~=~ frac{2}{pi} ~times~ Big ( frac{4}{pi} (frac{m}{n})^2 I_{dc} R_L Big ) ~=~ frac{8}{pi^2} (frac{m}{n})^2 I_{dc} R_L$$

Equation 8.

We mentioned earlier that the average component of the voltage at the output transformer’s center tap (node E) is V_CC. Substituting V_CC for v_{E, ave} in the above equation, we can solve for I_dc:

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

Equation 9.

At last, by substituting I_dc from Equation 9 into Equation 3, we obtain the output power:

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

Equation 10.

Now that we know I_dc, we can also find a simple equation for the maximum voltage at the collectors. In Figure 9, we saw that v_C and v_D—the two collector voltages—both have a peak value of 2A₂. Combining Equations 9 and 7 leads to:

$$A_{2}~=~ frac{pi}{2} V_{CC}$$

Equation 11.

The maximum voltage drop across the transistors is 2A₂ = πV_CC.

We’re now ready to find the TCCS amplifier’s efficiency. This is the last performance parameter we’ll examine before we move on.

Calculating the Efficiency

The efficiency is calculated by comparing the output power and the input power. We know the output power from Equation 10. As for the input power, it’s equal to the supply voltage (V_CC) multiplied by the average value of the supply current (I_dc). Therefore, from Equation 9, the input power is:

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

Equation 12.

This is identical to the power delivered to the load, which means that the theoretical efficiency of this amplifier is 100%. Note that this is the ideal efficiency. In practice, the circuit’s efficiency may be degraded by non-idealities such as parasitic inductances in series with the collector leads.

Example: Determining Maximum Transistor Voltage and Current for a TCCS Amplifier

Now that we’ve completed our analysis, let’s find the supply voltage, maximum switch current, and maximum switch voltage for a TCCS circuit that delivers 20 W to a 50 Ω load. For simplicity, we’ll assume that the turns ratio (n/m) is unity.

We know that P_L = 20, R_L = 50, and n/m. To find the supply voltage (V_CC), we first plug those numbers into Equation 10:

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

Equation 13.

We then solve for V_CC:

$$V_{CC}~=~ sqrt{frac{20 ~times~ 50 ~times~ 8}{pi ^2}}~=~28.47~text{V}$$

Equation 14.

Next, let’s calculate the maximum switch current. From Figure 5, this is equal to I_dc. Equation 9 produces:

$$I_{dc} ~=~ frac{pi^2}{8} (frac{n}{m})^2 frac{V_{CC}}{R_L} ~=~ frac{pi^2}{8} ~times~ 1 ~times~ frac{28.47}{50}~=~0.7 ~text{A}$$

Equation 15.

Finally, we know from Figure 9 that the maximum voltage across the collector-emitter of each switch transistor is 2A₂. Applying Equation 11, we obtain:

$$2A_2~=~ pi V_{CC}~=~pi~times~28.47~=~89.44~text{V}$$

Equation 16.

To deliver 20 W to a 50 Ω load, the TCCS amplifier requires a supply voltage of 28.47 V. Its transistors must be able to tolerate a maximum switch current of 0.7 A and a maximum collector-emitter voltage of 89.44 V.

Which Class D Amplifier Provides the Best Performance?

We previously worked through the same example for a TCVS amplifier and a basic voltage-switching Class D amplifier. Table 1 summarizes the results alongside those for the TCCS amplifier. All three circuits were designed to deliver 20 W to a 50 Ω load.

Table 1. Requirements for three different Class D configurations to deliver 20 W to a 50 Ω load.

Configuration	Supply Voltage	Max Switch Voltage	Max Switch Current
Basic voltage-switching	70.20	70.20	0.89
TCVS	35.10	70.20	0.89
TCCS	28.47	89.44	0.70

Based on that data, how do the three configurations stack up against each other?

The basic voltage-switching Class D amplifier requires a significantly higher supply voltage than either of the transformer-coupled configurations. Its transistors must handle the same maximum current and voltage as the TCVS amplifier. Compared to the TCVS circuit, the TCCS arrangement requires a lower supply voltage and maximum switch current. However, it increases the voltage stress across the transistor collector-emitter.

The TCVS and TCCS circuits provide comparable performance—there seems to be no absolute winner here. Depending on the supply voltage and available transistor parameters, either configuration might be the right choice. Note that the transistors’ frequency of operation is one of the parameters that should be taken into account—with a square wave collector current, the transistors need to switch faster.

Wrapping Up

In this article, we examined the operation and performance of the transformer-coupled current-switching Class D amplifier. This is the final Class D configuration we’ll discuss. In the next article, we’ll examine some limitations of Class D power amplifiers and explain how they can be addressed by Class E amplifiers.

All images used courtesy of Steve Arar