While you are correct that Class D amplifiers are switching devices, you are incorrect in stating that they are "completely analog from input to output". There is a digital element. Class-D amplifiers convert an input signal to a power 1-bit digital output signal, that is the output signal is either a 1 (say +70V) or a 0 (say -70V). The rate at which the output switches is much higher than audio frequencies (say 500kHz). This rate is controlled by a reference clock signal. The ratio between the average time spent at +70V and -70V varies in proportion to the input signal. That is if the input signal is say more than 0V, then the output correspondingly spends more time on average at +70V than -70V and so on. This ratio is also controlled by the clock signal. The 500kHz switching signal is removed by analogue filters so that only the average signal passes to the loudspeaker, and as stated, this is controlled to be in proportion to the corresponding input signal.
We will clarify this when the results are published and explain what Class D is all about.
Hey All, again, thanks for doing this, and for all of your hard work.
James - Keep in mind that due to the nature of the comparators in all of these devices that the timing ratios of the on/off state of each transistor is infinite.
It could be considered digital in that there are only two states that the output can exist at, and digital typically means that a signal could be replicated by 'ones' and 'zeros'; however.... these ones and zeros have to have an associated sampling frequency in order to be meaningful. However, because of the nature of an analog input and a triangle wave comparator as used in most of these devices, the duration that the output will be either a one or a zero is neither linked to some higher sampling frequency, nor are the possibilities of how long the output can remain in that state less than infinite. The one thing that is certain, is that the output will switch states at a frequency greater than double that of the frequency of the triangular wave used in the comparator, but we know nothing more than that, and that alone does not make the system digital.
Now, you might thing that this sounds like PWM (Pulse Width Modulation), and it is, because the width of the pulse in infinitely variable. What you are referring to is PCM, where there is a time duration associated with the width of the pules, i.e. there is a relationship to some sampling frequency... this is like sony DSD etc, the time spent in one state will ALWAYS be related to the sampling rate in one way or another.
Here is a good article on Class D amplifiers:
http://en.wikipedia.org/wiki/Switching_amplifierAnd notice how this article on PWM:
http://en.wikipedia.org/wiki/Pulse-width_modulationdiffers from this one on PCM:
http://en.wikipedia.org/wiki/Pulse-code_modulationBecause these amplifiers are comparing a triangular wave (purely analog of course) to an analog input, and switching the output once per intersection, there is no possible way to quantify how long the output will remain in one state before it will switch to the other state, other than we know it will be in less than one half of one cycle of the triangular wave.
At least this is the way all of the amplifiers I build work, and I am about 95% sure that the ICE and Tripath chips work the same way.
Correct:
...the output signal is either a 1 (say +70V) or a 0 (say -70V). The rate at which the output switches is much higher than audio frequencies (say 500kHz). This rate is controlled by a reference clock signal. The ratio between the average time spent at +70V and -70V varies in proportion to the input signal. That is if the input signal is say more than 0V, then the output correspondingly spends more time on average at +70V than -70V and so on
Incorrect:
This ratio is also controlled by the clock signal
It is controlled by the intersection of the clock signal and the input signal, and the only way for this to work at all is if the clock signal is a triangle wave (common), or a sawtooth wave (uncommon). Clock sometimes implies square wave, so I just wanted to make that clear.
Hope this helps,
Paul Hilgeman
