Hopefully, because I find it hard to significantly increase the bandwidth given the capacitive load and current consumption restrictions -and I haven't yet gone to double that for two headphones...
BTW, I just found the following excellent article about recommended headphone amplifier specs that for phase it says: "The error should be under 1 degree from 10hz to 10Khz where spatial information is most critical" which coincides with my current achieved specs but I'd prefer some margin.
For THD+noise it says that due to masking, 0.01% is enough, but 0.005% is definitely inaudible (already achieved), which is why I'm after the lowest numbers: to ensure that any distortions will be way bellow the audible threshold and above any religious/subjective opinions -as the numbers don't lie
NwAvGuy's Heaphone Amp Measurement Recommendations Page 2 | InnerFidelity
Still, I'd like to hear from truly high-end reviewers what specific models worth $$$...$$$ of dolars provide.
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OK, I got it, the graph implies the same delay for the whole signal independent of frequency, so it's useless and meaningless, whether it is 250ns or 100 million ns (=100ms) -it doesn't matter at all for a headphone amp or for loudspeakers unless someone uses different amps to drive multiple speakers simultaneously and wants them aligned to milimeter!
I was confused by the logarithmic plot. I made the frequency axis linear and the graph was a straight diagonal line from 180 degrees and 10hz down to 178.2 degrees and 20Khz (=-1.8deg). That implied a constant delay independent of frequency. Then I set it to show the "group delay" instead, across the same audio range and it showed an arc that only changed the 4th or 5th decimal digits and beyond of 253.8ns =linear.
BTW, before that, I did some changes to the input active stage and now the group delay went down to 60.8ns (180 to 179.56) less than a half degree -but it doesn't matter anyway.
Now the questions is: what does the article above means then by "1 degree at 10khz"?
I was confused by the logarithmic plot. I made the frequency axis linear and the graph was a straight diagonal line from 180 degrees and 10hz down to 178.2 degrees and 20Khz (=-1.8deg). That implied a constant delay independent of frequency. Then I set it to show the "group delay" instead, across the same audio range and it showed an arc that only changed the 4th or 5th decimal digits and beyond of 253.8ns =linear.
BTW, before that, I did some changes to the input active stage and now the group delay went down to 60.8ns (180 to 179.56) less than a half degree -but it doesn't matter anyway.
Now the questions is: what does the article above means then by "1 degree at 10khz"?
OK, I'm glad you got it now!
Either the writer of the article means deviation from linear phase, or he or she hasn't given it enough thought. 1 degree deviation from linear phase means that you, for example, have a phase shift of -1 degrees/kilohertz at low audio frequencies, but -11 degrees instead of -10 degrees at 10 kHz.
By the way, when you want an almost linear phase response over the audio band, be very careful with high-pass filters, such as DC blocking capacitors and DC servos.
Either the writer of the article means deviation from linear phase, or he or she hasn't given it enough thought. 1 degree deviation from linear phase means that you, for example, have a phase shift of -1 degrees/kilohertz at low audio frequencies, but -11 degrees instead of -10 degrees at 10 kHz.
By the way, when you want an almost linear phase response over the audio band, be very careful with high-pass filters, such as DC blocking capacitors and DC servos.
I thought so. I would prefer a more intelligent plot with the deviation alone instead...
I haven't taken care of this yet. I prefer a complete capacitor-less drive and DC offset stabilizer (either with a microcontroller or other circuity). Tiny caps just for feedback compensation are excluded of course. We'll see...
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If that DC offset stabilizer is a feedback loop, you probably get the same kind of phase response as with a normal DC servo or DC blocking cap. You won't get any phase errors at all if you only calibrate out the offset at start-up, but then you don't compensate for temperature drift and things like that. As a compromise, you could first calibrate out the offset and then make the loop very slow to track drift.
Using capacitors, there are tricks to improve the phase response at the expense of getting a bump of a few decibels in the magnitude response in the subsonic region, such as making a DC servo loop with a higher than first-order filter in the feedback path, or using a bootstrap circuit instead of a plain old DC blocking capacitor with bias resistor.
Using capacitors, there are tricks to improve the phase response at the expense of getting a bump of a few decibels in the magnitude response in the subsonic region, such as making a DC servo loop with a higher than first-order filter in the feedback path, or using a bootstrap circuit instead of a plain old DC blocking capacitor with bias resistor.
Why is that a compromise? Of course the less you need a filter in an offset stabilizing circuity, the better, ie it should do the least work possible or be eliminated completely, but even if the amp itself is not drifting, the source might have a dc component, so a clever rather than a dumb circuit would help in this case for faster and more efficient response. It can be done without an mcu, but if I use one for other purposes (safety, health check, etc) why not use it for that important task too?
On my first priorities is also an absolutely flat DC to supersonic response, so I really cannot picture any bumps on that straight line!
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i think you are in mood to invent a non existing miracle. i hope you energy will not dissipate soon..
why bother with THD and measurements, if you are listening with ears; not with calibration microphones.
THD to dB - convert percent % to decibels dB percentage voltage % vs per cent converter THD+N total harmonic distortions calculation signal distortion factor attenuation in dB to distortion factor k in percent decibel damping - sengpielaudio Sengpiel according to this, it is about -72dB.
i took a listening test, where nasty sound was mixed into music, i could not hear it at -70 or -75dB.
so, a 2nd (even hd´s) will be absolutely negligible.
even a 2% 2nd is not a big deal.
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Why dissipate? It is fun pushing numbers down.
Since you mentioned it, here is today's numbers: 0.0008% THD = -102db.
Third harmonic is -103.8db down (second and others are lower), but I haven't "treated" those higher-order harmonics yet, as I changed topology -they will be lower.
So I have already surpassed my second goal of 0.001%, (first one was 0.01%) but I'll keep pushing them down a bit more, because that makes me improving the circuit and actually pushing down the real numbers that will be measured when I build it.
And most importantly, those numbers were made with the previous topology after improving it further.
That means cost reduction -less mosfets. So I have left quazi-complementary for now -I don't think I'll come back, although the very low distortion (about the same) proved the rumors false.
Actually, our ears are far better than dumb microphones, not only because they adjust the dynamic range from quiet to bombing, but are also able to decode multiple sound reflections and locate sources in 360 3D space. Try that with a set of "calibrated" microphones. Of course, there are more to learn -we don't know everything yet.
How did you arrive to -72dB?
There can be a huge quality difference between two tests, that doesn't prove anything. Someone else may come with a different test and prove you the opposite.
I'll wait until an extensive, independent and serious scientific research is done.
Note that I'm NOT religious with sound either - if I was, I wouldn't demand a scientific research.
So, until then, having a practically "perfect" amp (at a reasonable cost), guarantees that it will be totally transparent to the music you listen with it.
Cheers.
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Try Marcel's post in really simple words:
Phase shift tells you how much time it takes for a signal to pass through an amplifier.
You might talk about this delay in seconds, but it turns out that low frequencies are hardly affected, but high frequencies are considerably effected.
Let's figure it out. A waveform at 10Khz means 10,000 waveforms in one second, a complete wave starting at 0 degree and moving up to 90 degrees, then to zero, then to -90 degrees, and then back to finish at 0, the crossing point - all on one second. This is a complete cycle of a sine wave counted over one second, from which this concept is based.
A complete cycle will therefore take 1/10,000 of a second since 10,000 waveforms take one second. In microseconds, that is, 1 million uS for one second, this cycle will take [1 000 000/10 000] uS. That is 100uS for one cycle of the waveform moving at 10KHz. You could multiply 10,000 x 100uS (the 'period' of a complete cycle) and get the total uS for 10KHz over a second, and this is of course 1 000 000uS, a million uS, ie 1 s.
We know that a full waveform, a complete cycle, is 0-90-0-minus90-0 = 360 degrees. One cycle of a 10KHz waveform takes 100uS. But this represents 360 degrees of traverse, and if we know that the phase shift is only 0.9 degrees for this waveform, the time this represents is 0.9/360 x 100uS. This works out at 90/360 of a uS, which is 1/4 of one uS. That is better expressed as 0.25uS, or 250nanosecs, written as 250nS.
Kapitz?
Hugh
Phase shift tells you how much time it takes for a signal to pass through an amplifier.
You might talk about this delay in seconds, but it turns out that low frequencies are hardly affected, but high frequencies are considerably effected.
Let's figure it out. A waveform at 10Khz means 10,000 waveforms in one second, a complete wave starting at 0 degree and moving up to 90 degrees, then to zero, then to -90 degrees, and then back to finish at 0, the crossing point - all on one second. This is a complete cycle of a sine wave counted over one second, from which this concept is based.
A complete cycle will therefore take 1/10,000 of a second since 10,000 waveforms take one second. In microseconds, that is, 1 million uS for one second, this cycle will take [1 000 000/10 000] uS. That is 100uS for one cycle of the waveform moving at 10KHz. You could multiply 10,000 x 100uS (the 'period' of a complete cycle) and get the total uS for 10KHz over a second, and this is of course 1 000 000uS, a million uS, ie 1 s.
We know that a full waveform, a complete cycle, is 0-90-0-minus90-0 = 360 degrees. One cycle of a 10KHz waveform takes 100uS. But this represents 360 degrees of traverse, and if we know that the phase shift is only 0.9 degrees for this waveform, the time this represents is 0.9/360 x 100uS. This works out at 90/360 of a uS, which is 1/4 of one uS. That is better expressed as 0.25uS, or 250nanosecs, written as 250nS.
Kapitz?
Hugh
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