Class D audio amplifiers with single-ended outputs can exhibit a strange behavior, in which the power-supply voltage increases with power delivered to the load. Called “power-supply pumping,” it can puzzle an engineer unfamiliar with the phenomenon. If power-supply pumping is not controlled, the higher supply voltages can cause permanent damage by exceeding absolute-maximum-voltage ratings for the amplifier and power-supply capacitors. The following discussion covers the causes of power-supply pumping and what to do about it.
Understanding power-supply pumping
Pumping usually occurs when a common laboratory supply powers an amplifier that drives low-frequency waveforms into a low-impedance load such as a speaker. Because the outputs of such supplies (linear power supplies) are rectified with diodes, they are not able to sink current at the positive output or source current at the negative output.
Figure 1 shows a linear power supply connected to a stereo Class-D amplifier with differential inputs (a MAX9742 or similar devices from other vendors). When the speaker output swings below ground during a negative half-cycle of the sinewave, current flows through paths shown by the arrows, and into the output of the supply .
Figure 1: The long arrows illustrate current flow into the outputs of a Class D amplifier during a negative half-cycle of the sinewave output.
(Click on images to enlarge)
Because the device is a Class D amplifier, the output at node A swings from VDD to VSS : it pulls up to VDD when MOSFET M1 turns on, and pulls down to VSS when MOSFET M2 turns on. The currents indicated by the arrows alternate, according to whether M1 or M2 is on. With M2 on, current flows through M2 and returns to the power supply through D2. When M1 is on, however, current flows through M1, produces reverse bias across D1, and charges C1. The result is an increase in the magnitude of VDD .
In Figure 2 , power-supply pumping is illustrated by power-supply and speaker waveforms from the single stereo output of Figure 1. VDD and VSS are 15 V, the power-supply capacitors are 1000 μF, the output load is a 4 Ω resistor, and the output signal frequency is 20 Hz.
Figure 2: These waveforms from Figure 1 illustrate the effects of power-supply pumping. The time from t1 to t2 represents the negative half cycle of the output waveform (bottom trace), which causes VDD to increase (top trace) and VSS to decrease (middle trace)
(Click on images to enlarge)
The time interval from t1 to t2 represents the negative half cycle of the output sinewave, which is similar to the current direction in Figure 1. You can see that VDD (Channel L) rises to a more positive voltage between t1 and t2 . As the sinewave output becomes positive at t2 , the voltage increase on VDD discharges from C1. At the same time, VSS (Channel R) becomes more negative as an increasingly positive output voltage dumps charge into C2. As the speaker voltage goes negative, VSS discharges and the cycle repeats.
The problem of pumping is relatively larger during low-frequency operation, because low frequencies provide longer periods for pumping charge into an output capacitor, before the opposite-phase waveform discharges that extra voltage. The resulting increase in supply voltage can cause component damage by exceeding the absolute-maximum ratings for C1 or C2 (or both), and M1 or M2 (or both) within the Class D amplifier.
Solutions for power-supply pumping
Three simple methods can reduce this problem:
- The IC in Figure 1 has two single-ended outputs (of which one is shown), which can be bridged to provide a single differential channel as shown in Figure 3 . Output signals in this bridged-output configuration are out of phase with respect to each other. Thus, any power-supply pumping in one channel cancels pumping in the other.
- Low-frequency energy in the music of most stereo audio tracks is monaural (i.e., left and right channels are the same). One can therefore reverse the polarity of Channel R with respect to Channel L (or vice-versa). As shown in Figure 4 , the audio input on Channel L is inverted by the input stage, so the Channel L and Channel R outputs are out of phase with each other. Any power-supply pumping that occurs in one channel now cancels that in the other channel. You must then ensure same-phase audio outputs by swapping the speaker connections on one channel (Channel R in this case).
- Larger power-supply capacitors (>1000 μF) also help to reduce the magnitude of power-supply pumping. The larger capacitors accept more current before the power supply gets “pumped up,” and while receiving current they allow more time for the opposite-phase signal to discharge the pumped supply voltage.
Figure 3: Configured as shown, the input op amps of this IC let you configure a bridge-tied load output.
(Click on images to enlarge)
Figure 4: To cancel power-supply pumping while preserving the two channels of the IC, invert the L-channel signal by configuring that input amplifier as an inverter, then swap the R-channel speaker connections to restore similar signal phase from the speakers.
(Click on images to enlarge)
Equation 1 below models a good approximation for power-supply pumping. For simplification, we assume
VDD = VSS , and RDS(ON) = 0 for M1 and M2.
ΔVDD = 1/C × A/RLOAD × [1/(2 × π) – A/8VDD ] × Tsine
- C is the bypass filter capacitor on the VDD supply (C1),
- A is the amplitude of the output sinewave,
- RLOAD is the speaker impedance,
- VDD is the nominal value of positive supply voltage, and
- Tsine is the period of the output sinewave.
Equation 2 describes a parabola whose peak occurs when the amplitude A = (2/π) × VDD . Substituting this expression into Equation 1 gives the maximum power-supply pumping voltage:
ΔVDD (max) = [VDD /(2 × π × π)] × [Tsine/(RLOAD × C)]. (Equation 2)
For a system with ±15 V power supply, 8 Ω speaker loads, 20 Hz sinewave output, and 220 μF power-supply capacitors, VDD can pump up to 21.6 V. Increasing the size of the power-supply capacitors reduces ΔVDD (max), however. Installing 1000 μF capacitors, for instance, reduces the maximum supply-pumping voltage to 4.7 V.
Discussion of methods
Method #1 prevents power-supply pumping by adding an inversion stage. You add only an op amp, so the cost is relatively low. And, if your stereo-amplifier IC is a MAX9742, you don't need the additional op amp, because its input-stage op amps can be configured as required (see Figures 3 and 4). But this approach is not always practical because it turns two amplifiers into one. If you need stereo outputs, you would have to add another MAX9742.
The second method for preventing power-supply pumping, which inverts the pumping in one channel and uses it to cancel pumping in the other, is the most elegant. The circuit (Figure 4) includes an op amp at the input of channel R and requires room on the PC board for that op amp, unless your Class D amplifier is a MAX9742, which lets you configure the inputs as required to avoid power-supply pumping. Most other implementations will require revision to the PC board, unless you account for power-supply pumping as part of the power-amplifier design.
The third approach is the most common but also the most costly. Power-amplifier designers are usually in a hurry to meet production deadlines, and since most power amplifiers already include bulk power-supply capacitors of 100 F to 330 F, it's convenient to simply increase this capacitance to 1000 F or so.
Adding an op amp and the associated passive components costs $0.10 to $0.15, but a 330 μF capacitor only costs about $0.10. Adding two 330 μF capacitors per supply voltage (VDD and VSS ) will therefore increase the bill of materials by $0.40. Or, removing the 330 μF capacitors and adding two 1000 μF capacitors increases the cost by $0.25 to $0.30. Thus, increasing the capacitance costs more but offers lower risk from the designer's standpoint, because it's easier to increase a capacitor value than to modify the PC layout for adding an op amp.
Although the problem of power-supply pumping and its solution is well understood, IC designers have not yet integrated all the necessary circuitry into a chip. They've so far produced some creative solutions such as the MAX9742, whose differential inputs include op amps that can be configured as required (Method #2). If the IC isn't properly configured, its supply voltages will be pumped to levels greater than their absolute maximum ratings. (Every IC has an absolute maximum voltage rating, which if exceeded can permanently damage the part.) The solution of adding power supply capacitance (Method #3) will probably never be integrated into an IC because capacitors in silicon are expensive, and are measured in picofarads rather than microfarads.
Again, power-supply pumping is a problem that occurs when playing low-frequency audio signals through a single-ended-output, Class-D amplifier driving a low-impedance load. This problem can be troublesome, but you can reduce it to safe levels (for the amplifier and external components) by understanding the phenomenon and applying one of the solutions presented above.
About the author
Marshall Chiu is a Strategic Applications Engineer in the Multimedia Business Unit of Maxim Integrated Products Inc., Sunnyvale, CA. Before joining Maxim in 2006, he was a Staff Applications Engineer at Tripath Technology, and a Senior Applications Engineer at National Semiconductor. His 13+ years of system design covers high-fidelity home audio, car audio, and portable audio for consumer electronics. He holds one U.S. patent, and has written numerous articles relating to audio semiconductors. He is a regular contributor to the EE Times China website, writing an audio column and answering questions on an online audio forum. Marshall received a BSEE from San Jose State University in California. He can be reached at Marshall.email@example.com.