Semiconductor Devices: Class D Power Amplifiers 

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You're welcome.

Could you clarify what you mean? Are you referring to the action of an inductor?

At the beginning of the video, you explain that the class D amplifier is very efficient beacause when voltage is high, current is low (and P = U.I ...) and vice versa. But I don't understand in practice how the use of FET transistor can realize this phenomenon, because when we have a high voltage at the gate of the transistor, there is a huge current which is going to go through the component, right?

@@ElectronicswithProfessorFiore Would it be possible to contact you privately? I'm a French high school student preparing something about the class D amplifier for the exam I will have at the end of the year and I have a lot of question about this technology.

@@TionebFountain You seem to be thinking in terms of a resistor and Ohm's Law (where a large current produces a large voltage). Although this is not strictly accurate, you can think of the FET as operating like a resistance that is controlled by the gate voltage (high Vgs yields small resistance). Some people compare it to a valve, where applying a higher gate voltage is like turning the valve more "on", increasing current flow. That current flows through the load, and thus, we get a large load voltage. Another way to think of this is like a voltage divider (between the FET and the load). A large Vgs creates a small resistance for the FET, so most of the applied (drain) voltage winds up across the load. Sorry, but I am not available for individual instruction, and anyway, it would probably be best if you worked with your teachers/mentors at your school. That's why they're there.

You mean other than the chapter in my Semiconductor Devices text? (See the video description for details)

We don't really think in terms of normal voltage gain for class D. The size of the signal from the prior (analog) stages is what sets the depth of the PWM, and thus the size of the output signal. If you want a larger output, then increase the input (just as you would with a class B power amp, typically configured as a buffer).

@@ElectronicswithProfessorFiore o... i see. So the gain of class d amp is 1, right?

@@DwiNoviantoUntidar Not really. The very concept of gain is complicated because this is a non-linear system. If we consider the PWM subcircuit to be part of the class D amplifier, then the analogous question would be "Given a certain modulation input signal, what is the size of the load signal?" The ratio of the output signal to the signal into the modulator would be the "gain", or at least the closest thing to what we call "gain" for a linear amplifier.

@@ElectronicswithProfessorFiore so how to set the volume / sound intensity in class D? Should we add adjustable current mirror to control current through the mosfet? Cz in class AB, due to unity gain, if we vary the preamp gain, the output will also vary.

@@DwiNoviantoUntidar The output level is a function of the depth of PWM, which is controlled by the signal feeding the modulator. You do NOT control the intensity of the signal feeding the class D output devices. Those transistors only operate in saturation or cutoff. In other words, the signal coming out of the preamp is what you need to control before it feeds the modulator. A simple pot set up as a voltage divider should work just fine, similar to what you'd do in an all-analog circuit (class A or B) feeding the driver stage. That's your volume control. The size of the signal feeding the PWM circuit depends on the design of the modulator. You could design the modulator so that it gives 80% depth (a typical value for max output) with only 1 volt of input, but you could just as easily design it to achieve 80% with a 5 volt input. Back to your original question, it is important to understand that the audio signal is being encoded as a series of pulses of varying width, not being amplified in an analog manner. Thus, the normal concept of gain does not apply. Does that make sense now?