The following waveforms were captured with MP-7720 single ended Class-D amplifier.
Inductor must follow Ampere’s Law:
di / dt = V / L
There is 6.7% tolerance coming from dead time and component tolerance.
This waveform illustrates how the Class-D modulates the input to output. The voltage gain of this Class-D is -5.
Is this waveform beautiful?
Does Class-D can offer instantly huge output current? Of cause not! Inductor current should be continuously.
di / dt = V / L is a very useful formula.
V means supply voltage minus output voltage (voltage across inductor). In the beginning, (Vs – Vo) is large, so the di / dt is also fast. But di / dt drops when Vo move from GND to Vs.
We have three methods to enlarge the speed and get better response: increase the switching frequency (to reduce the dt), enlarge the supply voltage, reduce the inductance.
However, any of the three modifications will enlarge the ripple current di, it gives the filter capacitor large current stress.
This is a single ended Class-D, so, there is a bulk capacitor between the L-C filter and speaker. It should be charged to 1/2 supply voltage when power on the single ended Class-D amp. In order not to generate pop-up noise, MP-7720 offers a small current to charge the bulk capacitor. Once the bulk capacitor is charged to 1/2 Vs, the small charge current is terminated and enable switching MOSFETs.
Only describe topology here.
Before describe the topology, I have to define the inductor current direction. Right-direction current in the inductor means first quadrant of B-H Loop. Test points are shown in Fig.1 except the latest waveform.
The Duty Cycle is closed to 50% without input signal. The Dead Time occupies a little percentage. I think nobody would have question about Fig.2, ignore the 470uF and 8ohm load.
Dead Time means the interval between Top-Side MOSFET OFF to Bottom-Side MOSFET ON (Also between Bottom-Side MOSFET OFF to Top-Side MOSFET ON). Both MOSFETs will be turned ON simultaneously for a short time and short supply voltage to GND if there is no Dead Time. The operation during Dead Time is very interesting!
At the time of Fig.2, inductor has right direction current to charge 0.68uF. Induct current is at somewhere of first quadrant of B-H Loop. Inductor current should be continuously, it means inductor current can not move from higher current to low current immediately.
At the time of Fig.3, both MOSFETs are turned off. The Top-Side MOSFET no longer supports the inductor current. Inductor current must be returned to Br. Crss of both MOSFETs offer the current path. The crss of Top-Side MOSFET will be charged and the crss of Bottom-Side MOSFET will be discharged very fast because of small capacitance. After the two crss are complete charged / discharged, the body diode of MOSFET take over the inductor current path shown in Fig.4.
There are two interesting things at this moment:
First, Bottom-Side MOSFET is turned on at small Vds (about 0.6V), like Zero-Voltage-Switching. It means less switching loss. On the other hand, when the body diode is turned on, the switching node voltage is 0.6V lower than GND, or 0.6V higher than Supply Voltage. So, there will be a small voltage spike occurred when switching node move from one rail to opposite rail.
At the time of Fig.5, inductor current won’t change direction immediately after Bottom-Side MOSFET is turned on because inductor current needs time to move to Br.
The story is exactly the same between Bottom-Side MOSFET OFF to Top-Side MOSFET ON.