Can you *please* stop calling "RMS" the Volts we are talking about?
We are trying to describe instantaneous values varying very fast, not interested on eventual heating of some element.
My head hurts reading your posts, as much as listening to mistuned instruments playing, no kidding.
As of chipamps, *all* of them are more than capable to slew fast enough (and then some) to play any kind of music (including Orchestral 😀 ) at full rated power.
Well if you limit your audio bandwidth to 20kHz nothing in that band can be faster than a 20kHz sine wave full level.
Some people think that if you add several signals the 'speed' adds too but that is not the case.
Example:: add a 5V 20kHz and another 5V 20kHz; the result is 10V 20kHz that has faster 0-crossings than the original two. But at the end you can never get 'faster' than the 20kHz full level in your system.
If you have a signal that has faster than max 20kHz rise parts, that means that signal has freq components above 20kHz.
Aagain, some poeple think that within a 20kHz band you can have transients faster than the 20kHz full level but that is physically impossible.
Jan
Quoted by maty tinman
20kHz includes more or less everything we can hear. Some people in their late teens and early twenties may be able to hear a little higher; most of us can't even hear to 20kHz.
This thread is rapidly turning into a mire of misinformation!
I can't be bothered to do the maths right now, but I suspect that the highest slew rate which a band-limited audio system can possible experience is that arising from a band-limited low frequency square wave, not a high frequency sine wave. I seem to recall that this will have a slew rate twice that of the sine wave. However, this has nothing to do with the OP question as we don't listen to full amplitude low frequency square waves - well, I don't anyway!
No it doesn't! The 20kHz limit was originally established by getting people to listen to music, not sine waves.
No they are not. Fundamentals only go up to a few kHz. Above that are harmonics, overtones and percussive noises.
20kHz includes more or less everything we can hear. Some people in their late teens and early twenties may be able to hear a little higher; most of us can't even hear to 20kHz.
That would depend on what you mean by "fast".maty tinman said:
The so-called audio bandwidth was established by changing the amplifier bandwidth until people could hear no further improvement. So 20kHz is the amplifier bandwidth required to hear music in all its glory.
This thread is rapidly turning into a mire of misinformation!
I can't be bothered to do the maths right now, but I suspect that the highest slew rate which a band-limited audio system can possible experience is that arising from a band-limited low frequency square wave, not a high frequency sine wave. I seem to recall that this will have a slew rate twice that of the sine wave. However, this has nothing to do with the OP question as we don't listen to full amplitude low frequency square waves - well, I don't anyway!
Vp = 1.41 Vrms
28.28 x 0.2 = 5,6 Vrms/uS (20 Khz) = 7.9 V/uS
28.28 x 0.2 x 5 = 28 Vrms/uS (100 Khz) = 39.5 V/uS
28.28 x 0.2 = 5,6 Vrms/uS (20 Khz) = 7.9 V/uS
28.28 x 0.2 x 5 = 28 Vrms/uS (100 Khz) = 39.5 V/uS
Curious coincidence
High voltage amplifier - Bandwidth, high speed and slew rate
100 Khz -> 40 V/µs
High voltage amplifier - Bandwidth, high speed and slew rate
5000 Khz -> 2000 V/µs
100 Khz -> 40 V/µs
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Amplifiers don't have memory therefore can't know what occurred a wavelength or two ago, nor can they predict what is going to occur. They only deal with a voltage level and its rate of change over a given time period. So this complex waveform speak is nonsense.
The interesting thing with music is that, although harmonics can go beyond 20KHz, the actual energy is already very low and pretty easy for even a slow amplifier to deal with.
The interesting thing with music is that, although harmonics can go beyond 20KHz, the actual energy is already very low and pretty easy for even a slow amplifier to deal with.
100 W / 8 Ohms -> 7.9 V/uS
Ignoring the bandwidth, do you agree with the minimum (100 Wrms at 8 Ohms, with 20Khz)?
28.28 x 0.2 = 5,6 Vrms/uS (20 Khz) = 7.9 V/uS
Ignoring the bandwidth, do you agree with the minimum (100 Wrms at 8 Ohms, with 20Khz)?
28.28 x 0.2 = 5,6 Vrms/uS (20 Khz) = 7.9 V/uS
I have had a quick look at the maths and I now believe the square wave slew rate factor is not 2 but 4/pi. So a band-limited square wave at a sufficiently low frequency (which doesn't have to be that low, but down into the bottom decade of audio frequencies) will have a slew rate characterised by 8 x f_max (where f_max is the HF cutoff frequency), instead of the 2 pi x f_sine for a sine wave of the same peak amplitude.DF96 said:
We are beyond a nice request.Could you pleas stop that nonsense with Vrms/us?
We should demand he stops this nonsense on this Forum.
Sorry, I don't understand your calculation. No such thing as Wrms - I think you mean W average, which is what you get when you use Vrms. When you say 'S' (Siemens - the measure of conductance) I assume you mean 's' (seconds - the measure of time).maty tinman said:
100W at 8 ohms needs a peak voltage of 40V. A 20kHz sine wave with peak value 40V requires 5.03V/us. A 100Hz square wave with peak value 40V requires 6.4V/us.
I also believe so, and it can be checked by derivating the fourier transform of a square wave, the derivative being the slope we derivate it as well to find the trivial solution t = 0 for the second derivative to be equal to zero.
The third derivative being non zero confirm that this is an extremum, its negative sign indicate that this point is a global maximum.
From here reporting the value t = 0 in the first derivative eliminate all the cosinus terms of this expression and what remain is 8.f.k, with k being the number of harmonics.
I am having second thoughts!
The peak amplitude of the fundamental of a square wave is 4/pi times the peak amplitude of the square wave. If the square wave is at frequency f_max/n (where f_max is the HF cutoff frequency, and n is a biggish number) then the fundamental contributes slew rate as (4/pi)x(2pi)x(f_max/n)=8f_max/n. Each harmonic at frequencies kf_max/n, where k is a set of odd integers up to n, contributes the same slew rate as the frequency is multiplied by k but the amplitude is divided by k. My mistake was to think that we have n such contributions, but we actually have about n/2 of them as k only uses odd integers.
So the slew rate of a low frequency square wave has factor 4 x f_max, not 8 x f_max as I claimed earlier. Have I got it right now?
The peak amplitude of the fundamental of a square wave is 4/pi times the peak amplitude of the square wave. If the square wave is at frequency f_max/n (where f_max is the HF cutoff frequency, and n is a biggish number) then the fundamental contributes slew rate as (4/pi)x(2pi)x(f_max/n)=8f_max/n. Each harmonic at frequencies kf_max/n, where k is a set of odd integers up to n, contributes the same slew rate as the frequency is multiplied by k but the amplitude is divided by k. My mistake was to think that we have n such contributions, but we actually have about n/2 of them as k only uses odd integers.
So the slew rate of a low frequency square wave has factor 4 x f_max, not 8 x f_max as I claimed earlier. Have I got it right now?
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