I do suppose that 2K$ DAC is the exception. Perhaps it would qualify as
"true high end".
210khz sampling !! , I suppose you would need a fast amp if the media
was a fresh 192K 32bit sampling of a perfect square wave.
Most CD FLAC's (even 24/96kbs ones), would not need 1/10th of that.
Some electronic music transients (square waves) on the CD might
approach a couple volt/uS (before the DAC , measured by the PC).
An emulated 32bit PC synth might take full advantage of that DAC ,
but for typical recordings ....a 210khz - 20V/uS DAC is overkill.
Asus Xonar = 192K ne5532 (8V/us) Asus Essense = 192K Muses 01 (12V/us).
A good match for the wolfson dac's.
These are also typical (high end) CD player spec's , they also use the wolfson wm8718's.
Hmmm ??? would not the buffer amp's slew to be multiplied by the main
amp's gain ? Might still need >200V/us to match some of these new DAC's.
OS
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How does one know if your testing slew rate or rise time. To test rise time the rise time of the input square has to be slow enough not to overload the input but faster than the device rise time.
It is best to test rise time at low levels and slew rate at high levels. Then they separate out. In fact, in some bad cases you can watch the rise time change into slew rate with level.
Mr. Curl is correct. Here are some details:
Start with a square wave of (say) 10mV amplitude and a frequency of about 1/4 the estimated amplifier bandwidth (not critical, the idea is to use a frequency that allows accurate measurements of the rise/fall times). The square wave rise/fall times should be as good as you can get from your generator.
Measure the 10%-90% rise/fall times at the amplifier output (many digital scopes have built in functions for such). The increase the square wave amplitude to 20mV and redo the measurements.
You will note that the output rise/fall times are rather independent of the square wave amplitude. This translate in an apparent SR increasing with the square wave amplitude.
Further increase the square wave amplitude, until you eventually note that the previous rule breaks; the rise/fall times became dependent on the square wave amplitude, hence the apparent SR increase appears to saturate to a constant value. This is usually also accompanied by a significant change in the output waveform: form a symmetrical exponential shaped leading and trailing edges, possibly to an asymmetrical wave form (typical for the Lin topology) with linear leading and trailing edges. You are now in slew rate limitation, and the constant SR value is the "large signal slew rate" that may create TIM nonlinear distortions.
Careful when doing these measurements, your Zobel cell may smoke at large signals. This is because the square wave is rich in harmonics, and all the output upper harmonics power is dissipating in the Zobel resistor. To also improve the measurements accuracy, you may want to disconnect the Zobel, bypass the output inductor and use a pure resistive load for such measurements (to keep the amp unconditionally stable).
Oh God, Save Us!
This thread has taken on all the hallmarks of “personal pet theories” versus real-world double-blind confirmed performance. Measurement VS results. Theory vs reality.
I'm especially amused that the August Mr. Curl has voiced his 10 μs risetime theory without also enveloping the concept that slew rate limited performance only affects an amplifier's highest output. At lower output volumes, the effective slew-rate limited bandwidth increases. Lo, and behold… a solution to most-all problems.
I mean, as how the ears work, its not like you're going to even notice at 100+ dB whether you are getting –0.1 dB or –3 dB or even –6 dB at the nearly painful 10 to 20 kHz band (at those levels). However, when enjoying really, really well mastered music rich with high harmonics, at more enduring levels (70–80 dB average, 100 dB peak), you'll have the extra bandwidth to reproduce the rich harmonics, and you'll even have 'em during those marvelous cymbal crashes and synthesizer square/sawtooth wave transients.
So… if one was to work backward from just this premise, how much slew rate is needed? Say –6 dB at full power, at 20 kHz, with 92 dB/watt speakers, at 3 meters distance, with 100 dB mean SPL, and +15 dB peak SPL.
dB drops about 6 db per doubling of distance. (because of square-law sound pressure diminishment from a point-source). Instead of “20 log₁₀( x )”, its 10 log₁₀( 1/distance² ) in distance-related db loss.
10 log₁₀(⅓²) = –9.5 dB from distance
peak is 85 dB + 15 dB (desired headroom coverage) = 100 dB … in the 'listening seat' for full output.
100 dB + 9.5 dB (lost to distance) is 110 dB more or less. Peak.
110 dB - 92 dB/W = +18 dB of necessary injected power.
Speakers are 4 Ω, nominal.
+18 dB (in power) is 10(1.8) equals … well over 65 peak watts! (tube amps are just fine delivering this)
√( 65 W × 4 Ω ) • √(2) = 22.5 V peak (but not P2P) and 15.9 V RMS.
If we accept 20 kHz as the more-or-less audible limit, AND accept –6 dB performance at peak, then that's
65 W × 10(½) = 32 W of power at 20 kHz.
… repeating math …
15.9 V peak volts.
1,996,000 volts per second slew rate.
≈ 2.0 V/μs
And that's that. At lower sound pressure levels, you're going to get performance all the way up to John Curl's 100 kHz level from the same amplifier, so long as it is also not bandwidth limited, but only slew-rate limited.
Tada. One solution to rule them all.
Nice thing is, you can take that to the bank, too… If you decide that your amplifier is going to chug along at a higher peak and average power (because say … your speakers are less efficient, or because the listening chamber is bigger, and your listing point is further away), all the math continues to work.
Even John should be happy.
2 V/μs. Higher for more output (√( of power ratio )) more, (or less) to be exact.
GoatGuy
This thread has taken on all the hallmarks of “personal pet theories” versus real-world double-blind confirmed performance. Measurement VS results. Theory vs reality.
I'm especially amused that the August Mr. Curl has voiced his 10 μs risetime theory without also enveloping the concept that slew rate limited performance only affects an amplifier's highest output. At lower output volumes, the effective slew-rate limited bandwidth increases. Lo, and behold… a solution to most-all problems.
I mean, as how the ears work, its not like you're going to even notice at 100+ dB whether you are getting –0.1 dB or –3 dB or even –6 dB at the nearly painful 10 to 20 kHz band (at those levels). However, when enjoying really, really well mastered music rich with high harmonics, at more enduring levels (70–80 dB average, 100 dB peak), you'll have the extra bandwidth to reproduce the rich harmonics, and you'll even have 'em during those marvelous cymbal crashes and synthesizer square/sawtooth wave transients.
So… if one was to work backward from just this premise, how much slew rate is needed? Say –6 dB at full power, at 20 kHz, with 92 dB/watt speakers, at 3 meters distance, with 100 dB mean SPL, and +15 dB peak SPL.
dB drops about 6 db per doubling of distance. (because of square-law sound pressure diminishment from a point-source). Instead of “20 log₁₀( x )”, its 10 log₁₀( 1/distance² ) in distance-related db loss.
10 log₁₀(⅓²) = –9.5 dB from distance
peak is 85 dB + 15 dB (desired headroom coverage) = 100 dB … in the 'listening seat' for full output.
100 dB + 9.5 dB (lost to distance) is 110 dB more or less. Peak.
110 dB - 92 dB/W = +18 dB of necessary injected power.
Speakers are 4 Ω, nominal.
+18 dB (in power) is 10(1.8) equals … well over 65 peak watts! (tube amps are just fine delivering this)
√( 65 W × 4 Ω ) • √(2) = 22.5 V peak (but not P2P) and 15.9 V RMS.
If we accept 20 kHz as the more-or-less audible limit, AND accept –6 dB performance at peak, then that's
65 W × 10(½) = 32 W of power at 20 kHz.
… repeating math …
15.9 V peak volts.
1,996,000 volts per second slew rate.
≈ 2.0 V/μs
And that's that. At lower sound pressure levels, you're going to get performance all the way up to John Curl's 100 kHz level from the same amplifier, so long as it is also not bandwidth limited, but only slew-rate limited.
Tada. One solution to rule them all.
Nice thing is, you can take that to the bank, too… If you decide that your amplifier is going to chug along at a higher peak and average power (because say … your speakers are less efficient, or because the listening chamber is bigger, and your listing point is further away), all the math continues to work.
Even John should be happy.
2 V/μs. Higher for more output (√( of power ratio )) more, (or less) to be exact.
GoatGuy
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Many of you have it wrong
It does not matter what the ears can hear.
What matters is the worst case signal that the electronics might get under worst case conditions!
What if there is a 19KHz signal mixed with a 20KHz signal?
What is the sum?
What if there are other frequencies included?
These IM frequencies might occur after the last band limiting filter but still be present in the electronics due to nonlinearities.
So, design for worst case rather than typical so that you can prevent overloads which are clearly audible even though the signal may not be.
It does not matter what the ears can hear.
What matters is the worst case signal that the electronics might get under worst case conditions!
What if there is a 19KHz signal mixed with a 20KHz signal?
What is the sum?
What if there are other frequencies included?
These IM frequencies might occur after the last band limiting filter but still be present in the electronics due to nonlinearities.
So, design for worst case rather than typical so that you can prevent overloads which are clearly audible even though the signal may not be.
39kHz, but it will be at a low level unless the electronics is significantly nonlinear. The biggest contributions to signal slew rate will be the original 19kHz and 20kHz, which together will have a slew rate slightly smaller than 20kHz.
Are you arguing that 'high-end' electronics, with its poor linearity, may need better slew rate performance than well-engineered stuff? An interesting thought.
There s no confusion other than your interpetation of my sayings, what i said is that if an amp require 10us to reach a given value then a 20KHz sine of this amplitude value will not be reproduced accurately as the amp slew rate is too limited.
The worst case signals an amp can encounter are probably not audio signals.
So, design to avoid the intrusion of these undesirable signals the audio path.
By the way, do not forget that quite a lot of british amps in 1970-1980is had their bandwidth limited to around 30 kHz and had the reputation of sounding very smooth.
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In the audio path, there are two transducers, one at the input, converting sound pressure to voltage, and one at the output, converting voltage to sound pressure.
Which kind of sound has such a level of high frequency content that, once converted to voltage by a microphone, causes the following amplifying stages to suffer from slew-rate limiting ?
I agree with John Curl when ha said "Real life high end audio demands a 10uS risetime square wave at any audio frequency".
It seems not to be too difficult to design electronics having full voltage output perfectly able to handle signals shorter than this rise time.
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Sleep over it. If any EE in good standing uses the ultra well known formula I posted here will find that a rise time of 10us is exactly equivalent to a closed loop bandwidth of 35KHz. Enough to pass a 20KHz signal without any attenuation (and of course no non linear distortions, since I am talking about the small signal "rise time" not the large signal "slewing effects").
Your statement:
doesn't make any sense, and here is, step by step, why:
a) "to reach a given value" - that is? this makes a wealth of a difference, since it may involve small vs. large signal.
b) "a 20KHz sine of this amplitude" - you are confusing the sine signal max dv/dt (which is around the zero crossing) with the rise time. A passive low pass filter set at 35KHz has a rise time of 10us, and will pass a 20KHz signal just fine, with no attenuation to speak of, or any "slewing distortions".
c) "will not be reproduced accurately" - what exactly means this? Some may think that a signal attenuation @20KHz of 0.2dB (ref: 0dB @1KHz) is "not accurate", other may disagree. Linear vs. non linear distortions dilemma.
A correct statement would be:
"if an amp has 10V/us slew rate, then a 20KHz sine will not be reproduced accurately as soon as the output exceeds 80Vpk, as the amp slew rate is too limited".
Of course, the above doesn't consider any common x10 safety margin usually considered in audio engineering. An amp with 80Vpk output (that is, 400W/8ohm or 800W/4ohm) is usually designed for 10x the slew rate above, so at 100V/uS. Which is an overkill by any engineering metric, but why not making the Golden Ear Brigade happy? 100V/us is a pretty easy design target, anyway.
I hope this explanation was not a waste of time...
Thank goodness for the 5-10X safety factor. You guys want to live in slew city, it would seem. '-) We did all this testing 35-40 years ago. Analog was worst case at first, but digital has caught up and can now exceed analog. Keep up the slew rates, guys. That is why we design power amps with 50V/us or greater almost all the time, if they have 100W output (my smallest amp) or greater.
Yep, that's pretty good.
Perhaps a “key value to remember”: 0.125 V/μs for limiting slew rate on 20 kHz signal at exactly 1 VOLT peak. Easy to remember. ⅛ V/μs for 1 volt.
And it scales linearly, just as you say.
Now, as to whether audio amplifiers at their top rated output even need to offer 10× better slew rate … is a matter mostly of marketing materials. Tell you this: if I was being bombarded by 50+ watts of output (per channel) AND the system was capable and tolerant enough to be outputting occasional peaks of 800 watts (which is only +12 dB more than average), I sure wouldn't be very qualitatively critical about whether the highest frequencies were perfectly accurately reproduced. Really.
Indeed… if anyone cares to remember, this is why on now-quite-popular tri-amped venue amplification systems, the woofer-subwoofer has a power of 10 units, the low-mid to upper mid has 3 units, and the high frequency tweeter has 1 unit of amplification power, with intentional limiters to keep from burning out the poor tweeter arrays.
Y'all remember that, yes?
Why? Because our ears are really quite insensitive to low frequencies the lower they go. And because our ears are adequately sensitive to higher frequencies up to about 5 kHz, then it all starts to fall off gradually. Yes, younger people CAN hear more of those higher sounds. But generally, there's ⅓ the real energy demand from 5 kHz to 20 kHz, than there is from 100 Hz to 5 kHz. And to get the 'thump 'n' grind' to happen below 100 Hz takes 3× more power (or for some venues even more…)
Just saying.
GoatGuy
Why, "weary" of course 😀
Which in a way is related to my being tired/sick of endless Baroque arguments .... sorry .... "threads" 🙄 ....... where hundreds of posts are added to a thread where everything useful which can be said was so done in the first 10/15 posts and the rest is endless repetition of the same 5/6 ideas ... if that many.
You wouldn't believe it, but there is a thread about a certain preamp, whose name doesn't come to my mind at this very moment, which was so long that Forum System Memory run out of digits to record posts and a new one had to be started 😱 (or so it seems 😛 )
As a side note, weary is an English Language word and has its place in proper Dictionaries ... while wery is at most an idiom or slang term.
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