When the switching frequency of a Class D amplifier isn’t exactly the same as the resonant frequency, the switches need to conduct negative currents for a portion of their ON cycles. As we learned in the previous article, BJT implementations require anti-parallel diodes to provide a path for these negative currents. Some MOSFET implementations use anti-parallel diodes as well, though not necessarily for the same reason.
However, a phenomenon known as diode reverse recovery can cause problems when anti-parallel diodes are used. In this article, we’ll discuss how, and under what circumstances, diode reverse recovery affects the performance of Class D amplifiers. This discussion will cover a lot of ground: we’ll examine the Class D amplifier’s operation both above and below its resonant frequency, and we’ll use both theoretical explanations and LTspice simulations to do so.
Before we dive in, let’s review some necessary general knowledge.
What Is Diode Reverse Recovery?
From basic electronics courses, we know that the I–V characteristic of a P-N diode is an exponential function. Note that this exponential relationship describes the device behavior in equilibrium conditions.
During a transition from forward bias to reverse bias, the behavior of a diode deviates from the exponential characteristic due to changes in minority charges stored in the device. We won’t go through the physical details of operation; the key point here is that some reverse current passes through the diode before it can actually turn off. This reverse current is required to remove the minority charge stored in the vicinity of the diode P-N junction.
The time interval between the instant the forward current drops to zero and the instant the reverse current reduces to 25% of its maximum value is known as the reverse recovery time. Depending on the type of material, doping level, and structure, different diodes exhibit different reverse recovery characteristics. To help you visualize this, Figure 1 compares the reverse recovery characteristics of two different diodes:
- A slow, snappy diode (the red curve).
- A fast, soft-recovery diode (the blue dotted curve).
Figure 1. Different diodes exhibit different reverse recovery characteristics. Image used courtesy of Ichiro Omura, et al
How Reverse Recovery Affects Switching-Mode Circuits
The reverse recovery current can have a significant impact on the switching loss of switch-mode power converters and switch-mode power amplifiers. For example, consider the buck converter in Figure 2.
Figure 2. Circuit diagram for a buck converter. Image used courtesy of Steve Arar
A buck converter takes a DC input voltage and steps it down to a lower DC voltage. During the first half-cycle, the switch (S1) is closed and the rail voltage is applied to the inductor. In this half-cycle, the diode (D1) is reverse-biased.
When S1 opens in the second half-cycle, the inductor tries to keep the current flowing. This turns on the diode, which provides the required current (iD in the above figure). In this case, node A in the circuit is at about –0.7 V, as determined by the forward drop of the diode.
The diode transitions from forward to reverse bias when S1 closes. However, as we discussed earlier, it can’t do so instantly. Before it turns off, the diode is capable of conducting current in the reverse direction.
This leads to a short from power supply to ground when S1 closes. The brief, large current spikes flowing through this short result in lost energy and EMI. The longer the diode reverse recovery time, the more the power loss we’ll have.
Let’s turn our attention back to the Class D amplifier. Figure 3 shows the complementary voltage-switching configuration we examined in the preceding article.
Figure 3. Complementary voltage-switching Class D configuration. Image used courtesy of Steve Arar
You’ve probably noticed some similarities between Figure 3 and the buck converter in Figure 2. But is the effect of reverse recovery similar as well? We’ll find out in the next two sections, which explain how diode reverse recovery affects the Class D amplifier when it’s operated either above or below the resonant frequency.
Operating the Class D Amplifier Above the Resonant Frequency
Consider what happens if the amplifier in Figure 3 operates above the resonant frequency of its tuned circuit. When the switching frequency is above the resonant frequency, the series LC circuit acts as an inductive load. The current flowing through the load (iRF) therefore lags behind the fundamental component of the square wave applied to node A of the amplifier (VA). Figure 4 compares the two currents.
Figure 4. Above the resonant frequency, the current lags the fundamental component of the voltage. Image used courtesy of Steve Arar
Each switch conducts negative currents during part of its ON cycle. Since NPN transistors can’t conduct currents in the reverse (emitter to collector) direction, the anti-parallel diodes provide a path for the negative currents.
Figure 5(a) and Figure 5(c) show the currents passing through the transistors Q1 and Q2, respectively. Likewise, Figure 5(b) and Figure 5(d) show the currents that pass through the diodes D1 and D2.
Figure 5. The currents passing through Q1 (a), D1 (b), Q2 (c) and D2 (d) when the amplifier is operated above the resonant frequency. Image used courtesy of Steve Arar
As we see from these waveforms, each transistor turns on after the corresponding anti-parallel diode when we operate above the resonant frequency. The device turn-on sequence is:
$$D_1~rightarrow~Q_1~rightarrow~D_2~rightarrow~Q_2$$
Consider the time instant t = t1, which is marked in Figure 5(d) and Figure 5(c). At t = t1, the diode D2 goes from forward to reverse bias and the transistor Q2 turns on to conduct the output current. Because it can’t instantly turn off, D2 will sink some current from node A to ground. Figure 6 shows the direction of the reverse recovery current that flows through D2 at t = t1 as it transitions from forward to reverse bias.
Figure 6. At t = t1, the reverse recovery current of the diode contributes to the output current. Image used courtesy of Steve Arar
The reverse recovery current through D2 flows in the same direction as the positive current conducted by Q2. Essentially, the reverse recovery current becomes part of the positive current that the Class D amplifier’s lower switch is supposed to sink.
This situation is completely different from what we saw with the buck converter. There, the reverse recovery current caused a short circuit, producing a large current spike flowing from the supply to ground. Here, when the reverse current through D2 is insufficient to provide the output current, Q2 turns on to provide a path for the output current instead.
But what if we operate the amplifier below the resonant frequency of the tuned circuit? Let’s find out.
Operating the Class D Amplifier Below Its Resonant Frequency
When the Class D amplifier’s switching frequency is below the resonant frequency, the series LC circuit acts as a capacitive load. The load current (iRF) therefore leads the fundamental component of the square wave at node A (VA). This is illustrated in Figure 7.
Figure 7. Below the resonant frequency, the current leads the fundamental component of the voltage. Image used courtesy of Steve Arar
Each switch still conducts negative currents during part of its ON cycle, and the anti-parallel diodes once again provide a path for the negative currents. Figure 8 shows the currents passing through Q1 (a), D1 (b), Q2 (c), and D2 (d).
Figure 8. The currents passing through Q1 (a), D1 (b), Q2 (c), and D2 (d) when the amplifier is operated below the resonant frequency. Image used courtesy of Steve Arar
As we’ll shortly see, the above waveforms need some adjustments to take the reverse recovery effect into account.
For operation below the resonant frequency, the device turn-on sequence is:
$$Q_1~rightarrow~D_1~rightarrow~Q_2~rightarrow~D_2$$
Each transistor is turned on before its own anti-parallel diode and after the anti-parallel diode of the other switch. This leads to a situation similar to what we saw in the buck converter—at the moment the current diverts from a diode to the transistor of the other switch, a current spike flows from the power supply to ground due to the reverse recovery of the diode. Figure 9 shows the reverse recovery current path when diode D1 transitions from forward to reverse bias.
Figure 9. The current spike flowing through D1 and Q2 when D1 transitions from forward to reverse bias is shown in green. Image used courtesy of Steve Arar
To better understand how this phenomenon impacts performance, let’s run some simulations.
Simulating a Class D Amplifier in LTspice
Figure 10 shows the LTspice diagram that I’ll use to explore the switching dynamics of the complementary voltage-switching Class D amplifier.
Figure 10. LTspice schematic of the Class D amplifier from Figure 3. Image used courtesy of Steve Arar
Please note that the above schematic is for simulation purposes only—it doesn’t aim to recommend components for a practical amplifier. The components and parameters are only chosen to show the switching behavior of the output devices, such as the reverse recovery effect.
The output LC circuit is tuned to 1 MHz. A transformer is simulated using three 10 H inductors (L2, L3, and L4) along with the following k statement:
k1 L2 L3 L4 1
The k statement specifies the coupling coefficient (k) between the windings of the transformer. The coupling coefficient can vary between 0 and 1, where 1 represents the ideal situation with no leakage inductance. As you can see above, k = 1 for this simulation.
The .ic directive can be found below the k statement in the schematic. It specifies the initial conditions—in this case, the initial currents—for the inductors.
The multi-pane plot in Figure 11 shows the specified currents when the switching frequency is the same as the resonant frequency. It can be interpreted as follows:
- Top pane: Load current.
- Middle pane: Upper switch current, comprising the currents through Q1 (the blue waveform) and D1 (cyan).
- Bottom pane: Lower switch current, comprising the currents through Q2 (magenta) and D2 (yellow).
Figure 11. (Click to enlarge) Load current (top); transistor Q1 and diode D1’s currents (middle); and transistor Q2 and diode D2’s currents (bottom). Image used courtesy of Steve Arar
When the switching frequency is equal to the resonant frequency, the upper and lower switches don’t need to pass negative currents. Therefore, the whole current is provided by the transistors. As you can see above, the diodes don’t turn on.
We now have a picture of ideal operation. Let’s see what our simulation results look like when the switching frequency and resonant frequency aren’t equal. Once again, we’ll start by examining operation above the resonant frequency.
LTspice Simulation of Operation Above the Resonant Frequency
By keeping the resonant frequency at 1 MHz and changing the input frequency to 1,010 kHz (1.01 MHz), we obtain the waveforms in Figure 12. From top to bottom, these waveforms are:
- The output voltage.
- The load current.
- Current through Q1 (magenta) and D1 (cyan).
- Current through Q2 (blue) and D2 (yellow).
Figure 12. (Click to enlarge) From top to bottom: output voltage; load current; transistor Q1 and diode D1’s currents; transistor Q2 and diode D2’s currents. Image used courtesy of Steve Arar
As an example, let’s examine the switching dynamics at approximately 966.5 μs. At this instant, the square wave at the output is driven to the positive supply rail. Since the output current is negative at t = 966.5 μs, diode D1 turns on to conduct the current. When the current becomes positive, the transistor conducts the current and the diode turns off.
Figure 13 provides a zoomed-in view of the relevant currents at the instant the diode turns off. The load current is shown in red, D1’s current in turquoise, and Q1’s current in magenta.
Figure 13. (Click to enlarge) The load current (red), diode D1’s current (cyan), and transistor Q1’s current (magenta). Image used courtesy of Steve Arar
From about t = 966.530 μs to t = 966.550 μs, the diode provides the negative output current. After this point, the diode’s current becomes negative for a short interval due to the reverse recovery effect, contributing to the positive output current.
Shortly after t = 966.560 μs, the diode’s reverse current drops to zero and the transistor turns on to supply the positive output current. From Figure 12, you can see that a similar sequence of events occurs when the output current diverts from D2 to Q2.
LTspice Simulation of Operation Below the Resonant Frequency
Our final simulation still uses the Class D amplifier in Figure 10, and still with a resonant frequency of 1 MHz. However, the input frequency is now 990 kHz. Figure 14 shows the simulation results.
Figure 14. (Click to enlarge) From top to bottom: output voltage; load current; transistor Q1 and diode D1’s currents; transistor Q2 and diode D2’s currents. Image used courtesy of Steve Arar
These waveforms are consistent with the theoretical discussion that accompanied Figure 8.
For example, the current is diverted from D1 to Q2 some time before t = 987.6 μs. However, current spikes appear. We can see these spikes more clearly in Figure 15, which provides a zoomed-in view of the relevant currents (as well as the output voltage). As in Figures 12 and 14, the output voltage is brown, D1’s current is cyan, Q2’s current is blue, and the load current is red.
Figure 15. (Click to enlarge) The square wave applied to the tuned circuit (brown), the load current (red), diode D1’s current (cyan), and transistor Q2’s current (blue). Image used courtesy of Steve Arar
Just before t = 987.56 μs, the load current becomes negative. This diverts the current from Q1 (not shown) to D1. Next, VOUT is driven to ground as Q2 turns on. Q2 can then provide the output negative current.
As the transition from D1 to Q2 occurs, the diode goes from forward to reverse bias. Due to the diode’s reverse recovery current, a considerable current flows through D1 and Q2. This causes a current spike in the waveforms. A similar sequence of events takes place when the current diverts from D2 to Q1.
So what does all of this mean for the Class D amplifier’s performance? Let’s conclude this article by examining the key takeaways from these simulations.
The Impact of Diode Reverse Recovery: Key Takeaways
When we operate the Class D amplifier above the resonant frequency, the reverse recovery current becomes part of the output current. Examining Figure 12’s voltage plot (the top pane), we also see that the voltage across the diode during reverse recovery is relatively small. Power loss from reverse recovery is therefore low.
However, reverse recovery becomes a problem if the Class D amplifier’s switching frequency is below its resonant frequency. As Figure 15 illustrates, the anti-parallel diodes generate high reverse-recovery power spikes. These current spikes can increase noise and damage the transistors.
We also see in Figure 15 that the voltage across the diode is relatively large when reverse recovery occurs, leading to a high reverse recovery power loss. For all of these reasons, operating a Class D amplifier below the resonant frequency or with a capacitive load isn’t recommended.
