Slew Rate

[kyllwtr]

My recently acquired Crown Reference Studio 1, around $4000 in mid 90s
Only
23 V/US

very disappointed, all the other specs are good

see specs

With 23 V/uS slewrate @ 20kHz output can swing up to 183 Vpk before having distortion for slew rate limitation. That means 130 Vrms (for sinousoidal signal) and more than 2kW of power applied to 8 ohm load.
It seems a lot of margin related to the ampli spec of 780 W @ 8 ohm.
Actually the power content of music record in the high frequency is the lowest fraction of the whole power content of the track.

 

[Cbdb2]

Actually the power content of music record in the high frequency is the lowest fraction of the whole power content of the track.

True but you can still get signals with full power, very fast rise times where slew limiting can occur. Look at the freq. spectrum of a square wave, very little power after the first few harmonics,

 

[egellings]

Even if you could get signals with such over the top slew rate, the speakers would not be able to reproduce them and the ear would not be able to hear them in the event that the speaker could reproduce them.

 

[antcollinet]

True but you can still get signals with full power, very fast rise times where slew limiting can occur. Look at the freq. spectrum of a square wave, very little power after the first few harmonics,

Of course - but non of the attenuated harmonics are audible, any more than the vertical edge of the square wave is.

 

[Cbdb2]

Even if you could get signals with such over the top slew rate, the speakers would not be able to reproduce them and the ear would not be able to hear them in the event that the speaker could reproduce them.

What do you mean over the top signals? What I was getting at was, even with 20khz BW limited music with the usual power mostly in the bass you can still get full power 20khz signal rise times.
If an amp slew rate limits theres probably IMD in the audio band.

Fortunately slew rate has been a solved problem for 50 years.

 

[Cbdb2]

Of course - but non of the attenuated harmonics are audible, any more than the vertical edge of the square wave is.

If there below 20khz they are. Take a 100hz square wave and filter out the harmonics one at a time starting with the highest freq. It will progressively sound more and more like a sine as the rise time of the square takes more and more time.

 

[antcollinet]

even with 20khz BW limited music with the usual power mostly in the bass you can still get full power 20khz signal rise times.

Surely if most of the power is in the bass - you are not going to get full power 20kHz. I'm not aware of any music that has full power at 20kHz. It would probably kill the tweeters of any speaker you sent it to.

 

[antcollinet]

If there below 20khz they are. Take a 100hz square wave and filter out the harmonics one at a time starting with the highest freq. It will progressively sound more and more like a sine as the rise time of the square takes more and more time.

Right - but if the slew rate is not slow enough to distort a 20kHz signal, then it won't attenuate any audible harmonics.

The post replied to was complaining about a 23V/uS slew rate - already nearly 5x the slew rate of a 20khz 100W signal into 8 ohm. And I don't think anyone would believe putting 100W at 20kHz into any speaker would do anything but generate smoke. And if it didn't then the result would likely be shredded eardrums.

 

[antcollinet]

you can still get full power 20khz signal rise times.

(My bold)

OK - having thunk, I think I get what you are saying. A 20kHz band limited square wave has a rise (and fall) time waveform that is almost a full amplitude half period of 20Khz. (from peak negative to peak positive - or the inverse).

So yes - to reproduce that you need a slew rate that can do a full power 20Khz sine wave, Even though that power won't exist at 20khz.

 

[DVDdoug]

If an amp slew rate limits theres probably IMD in the audio band.

I don't know about IMD but there will be harmonic distortion. Worst case, a sine wave becomes a triangle wave. If it remains below 1%. let's not worry about it. And since it is a high-frequency phenomenon it should be less audible than other harmonic distortion. With a triangle wave (like with a square wav) you only get odd harmonics so a 6.6kHz or higher triangle (or square) wave doesn't have harmonics in the audible range and it sounds exactly like a sine wave.

...It's really not necessary to "isolate" the cause of distortion - slew induced distortion, crossover distortion. or clipping, etc. If the distortion measurements are good, the amplifier is good! (Assuming it also has flat frequency response and low noise.)

And, I've NEVER heard distortion from ANYTHING that wasn't broken or overdriven.

 

[Cbdb2]

Surely if most of the power is in the bass - you are not going to get full power 20kHz. I'm not aware of any music that has full power at 20kHz. It would probably kill the tweeters of any speaker you sent it to.

Edit:wrote this before your next post.
What I meant is full instantaneous (20khz BW) power, as in an impulse or the leading edge of a square wave, think percussion. You dont need a lot of power in the higher harmonics to get fast rise times, look at the freq. content of a square wave. The point being you can mix different instruments together with most power in the low end and still get very fast rise times in the signal when the freqs. line up. So you can get slew limiting with music even if most of the power is in the low end. If your amp is garbage.

 

[antcollinet]

think percussion.

I get what you are saying


However to correct a common misunderstanding - Percussion doesn't create a square wave. In reality no real music will have full amplitude square waves in it.


Here is an illustration of what a cymbal strike looks like.

Speaker "Speed"

I know on ASR, they are so big on measurements and basically saying if you can’t measure it it’s not real, but this seems like a pretty fundamental example for how not everything can be measured. Not agreeing on what to measure != not able to measure it.

audiosciencereview.com

 

[danadam]

A 20kHz band limited square wave has a rise (and fall) time waveform that is almost a full amplitude half period of 20Khz.

AFAICT it's more like a full period:

(that's around 1 second mark of the attached signals)

 

[Sokel]

If you want to test difficult but common in music signals, try recordings with loud applause.
It's a good listening test too, it's hard to sound realistic.

 

[kyllwtr]

AFAICT it's more like a full period:

View attachment 463723
(that's around 1 second mark of the attached signals)

Let's make some calculation.
For sinusoidal signal slew rate is:

Thus, with Vpk = 1 and 20 kHz, SR is 125600 V/s or, better, 0.126 V/us.
For square wave, SR calculation is usually made measuring ΔV over Δt between 10% and 90% (rise time).
Again, in your waveform, ΔV is 2V and Δt is (110-72)us, thus 0,053 V/us.
Thus, it seems, that a 20kHz bandwith limited square wave of 100Hz has a slew rate of less than half of the 20kHz sinusoidal signal (actually 0,42).

 

[DonH56]

The first post of this thread notes the simple slew rate equation is for a single-frequency sine wave. That is the maximum slew rate which occurs around the center (zero-crossing) of the signal, and slew rate decreases as you move away from that point. There is a chart in the first post from which you can interpolate or extrapolate if you want a quick estimate without running the numbers. For other waveforms, e.g. square or complex waveforms, minimum rise or fall time is typically used with the end points (10-90, 20-80, etc. percent) defined to determine the amplitude (A in the equation -- note this is peak, not peak-to-peak, amplitude). One must be careful interpreting the slew rate when comparing dissimilar waveforms; a 100 Hz square wave bandlimited to 20 kHz with slew rate calculated over say 10% to 90% is usually much more stressful than a sine wave of the same amplitude since the sine wave's maximum slew rate is for a smaller fraction of the waveform. See the "building a square wave" article.

As @kyllwtr and others have said, 23 V/us is more than adequate for the Crown amplifier in question. Higher slew rate can help reduce distortion in some cases, but also requires greater bandwidth which in turns means greater noise and more power to support the higher bandwidth, and potentially less stability (more sensitivity to the load). Like many things, too much of a good thing can be a bad thing.

 

[NTK]

The slew rate of a square wave band limited with a brick-wall LPF can be analytically determined:

A sine wave at the highest component frequency (strictly speaking, for N ≥ 2) with the same V_pk as the square wave will have a higher slew rate. Therefore, if an amplifier can reproduce a sine wave with the same amplitude at the highest component frequency of the band limited square wave, it's slew rate is more than sufficient to reproduce the square wave.

 

[DonH56]

The slew rate of a square wave band limited with a brick-wall LPF can be analytically determined:
View attachment 464044
A sine wave at the highest component frequency (strictly speaking, for N ≥ 2) with the same V_pk as the square wave will have a higher slew rate. Therefore, if an amplifier can reproduce a sine wave with the same amplitude at the highest component frequency of the band limited square wave, it's slew rate is more than sufficient to reproduce the square wave.

View attachment 464048

Nice job! I decided to not walk through the expansion; just as well since your presentation is cleaner than the one I have written down in my notes from the primordial past (that include the impact of band limiting and Gibbs effect but are much messier equations). Two comments:

1. For those who might not recognize it, angular frequency w = 2*pi*f where w (omega) is in radians/sec and f is Hz (cycles/second).

2. Note my previous comment about a square wave having higher slew is at the fundamental frequency. That is, a 100 Hz square wave will in general have higher slew rate than a 100 Hz sine wave of equal amplitude. This does not conflict with @NTK's post; if the square wave is bandlimited to say 10 kHz, then a 9 kHz sine wave of the same peak amplitude as the 100 Hz square wave will have higher slew rate as he said. However, the 9 kHz component of a 100 Hz square wave is much smaller in amplitude -- note the terms go down as 1/N in the first y(t) equation @NTK presented -- and thus exhibits much lower slew rate (because the amplitude, A, of the 9 kHz component in the square wave is much smaller - 1/9 - than the fundamental amplitude - 1 - of the complete square wave).

I have also neglected Gibbs in this and the square wave posts; in my (former) day job, it was a royal pain as Tx outputs often exceeded DSO bandwidth thus making true amplitude difficult to measure, but for my simple analyses decided Gibbs introduces complications I did not want to explain to a lay audience.

Aside: I am rarely on ASR these days but check in now and then. I accidentally noticed the additions to this thread; in general, if you would like me to respond, please tag me (@DonH56 or drop me a PM) otherwise I am unlikely to see your response.

 

[Holmz]

Even if you could get signals with such over the top slew rate, the speakers would not be able to reproduce them and the ear would not be able to hear them in the event that the speaker could reproduce them.

I would bet that the person could hear it for a split second, before they fell down in pain.

 

[egellings]

I would bet that the person could hear it for a split second, before they fell down in pain.

They'd get knocked over by the hyper-velocity sound wave impacting them. Then the pain would hit.