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The image below shows how a Filterless Class-D amplifier works: enter image description here

The way I understand it: Vout(minus) is just the mirror image of the Vout(plus) but there was a certain phase shift(time delay?) between them that when these two signals are taken as input to a comparator, the Vout(diff) will be the result.

How much of what I said is wrong? Im confused on how this amplifier knows how much degree of time delay there is between Vout+ and Vout- so that the Vout(diff) will SURELY be high when Vin is at the upper half of sine wave and low when Vin is at the lower half of sine wave. I thought that if the input frequency is changed, there will be some kind of overlap between Vout+ and Vout-.

Source:https://www.eetimes.com/filter-free-design-helps-class-d-audio-amplifier-implementations/

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    \$\begingroup\$ The amplifier can output positive voltage, zero voltage, and negative voltage. Vout+ is not just a mirror image of Vout- \$\endgroup\$
    – user253751
    Aug 10 '20 at 14:50
  • \$\begingroup\$ What is the Vout- based on then to ensure difference of Vout+ and Vout- is the correct Vout(diff)? \$\endgroup\$
    – hontou_
    Aug 10 '20 at 14:54
  • \$\begingroup\$ A second comparator fed from a second signal path. Look at the block diagram (Fig 1). \$\endgroup\$ Aug 10 '20 at 14:59
  • \$\begingroup\$ Are you asking why the graphs work, or how the amplifier creates these graphs? \$\endgroup\$
    – user253751
    Aug 10 '20 at 15:02
  • \$\begingroup\$ @Brian Drummond Yes I saw the comparator. Those two Vouts came from separate comparators, right? Then, why is Vout- different from Vout+? (as seen from the image above) \$\endgroup\$
    – hontou_
    Aug 10 '20 at 15:04
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Class-D means using some sort of PWM. What you're showing is just the way the waveforms of a switching bridge looks like, and, in particular, a three-level-PWM (see the 2nd picture in the answer for an example): Vout+ in your picture would be V(x) in the linked example, and Vout- would be V(y). Their difference gives the blue trace in your picture, and V(x,y) as seen in the simulation. What you refer to as "time delay" is the effect of applying such modulation (the two branches are switching differently). In the linked example there is also the source for an LTspice schematic, feel free to use it if you can and are willing.

Since you're also linking an audio class-D amplifier, then you should know that the vast majority of loads for such amplifiers are electrodynamic loudspeakers which, at their bare minimum, have an inductance and a resistance in series (it's more complicated). Driving such a load with a switching signal will bring up noise, cause unnecessary heating (besides increased EMI), but mostly it will cause more distortions which, since it's audio, are not wanted. Even if the switching frequency is greater than the maximum audio frequency (usually 10 times, or more), there are harmonics that are not wanted and, in the case of high-range tweeters, they can be damaging.

This is why, usually, there is a filter at the output of such amplifiers, to attenuate higher frequencies and leave the audio signal, as much as possible. However, these filters are, again, usually, made of inductors and capacitors, and their size and cost is not cheap, so for some cases -- notably woofers and subwoofers -- the filter can be eliminated. The reasoning is that:

  • woofers and subwoofers have large values for the inductance (from a few mH, upwards), so combined with their resistance they can attenuate the switching frequency to a certain degree (though not as much as an LC filter)
  • the audio frequency that these kind of loudspeakers can reproduce is in the low band, so their mechanical properties makes them quite immune to high-frequency noise.
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  • \$\begingroup\$ I'm sorry I don't know anything yet when it comes to digital electronics. But, am I correct to say that the one causing those delays (and the somewhat "mirror-image" of the inverting to noninverting) are the NANDS? Does that mean I can't make these filterless D-amp using using only analogs? (sorry noob here) \$\endgroup\$
    – hontou_
    Aug 10 '20 at 15:30
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    \$\begingroup\$ @IwataniNaofumi As I said in the answer, there is no delay. What you see, or perceive as a delay is nothing more than the effect of three-level modulation. Do you have LTspice? Just for this case, I would recommend installing it, to run the example I linked (uninstall it afterwards if you wish). Ignore the NAND gates, they're there for dead-time, only. If you'll run the schematic, you'll be able to probe the voltages right after the comparators (Schmitt triggers in disguise, A1 and A5). You'll see that the pulses are different, and not just delayed (though they might seem that way). \$\endgroup\$ Aug 10 '20 at 15:35
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If the class D is a full bridge type as shown below, then the voltage applied to the speaker is simply Vch1 - Vch2 because the speaker is connected between these outputs. Note that a filterless class D wouldn't have the LC filter connected across the speaker.

enter image description here

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    \$\begingroup\$ I don't get how Vout+ and Vout- is produced to ensure Vout(diff) will be correct. Those two Vouts are just the result of the signal being differentiated from a triangular wave signal to get the PWM, right? Then why is Vout+ different from Vout- \$\endgroup\$
    – hontou_
    Aug 10 '20 at 14:59
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    \$\begingroup\$ Look at the inputs of the opamps (comparators) the sawtooth is applied to. \$\endgroup\$
    – Colin
    Aug 10 '20 at 15:16

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