Lately, I've been seeing a lot of discussions about slew rate in various media, and frankly, it pisses me off...
A guy says "the old model of this amplifier had a 90v/μs slew rate, the new model has reduced it to 45.. so the amplifier is slower than the old one"...
If you are using your amplifier as an FM transmitter, this can be considered an argument.. Of course, if your transmitter/amp is not transmitting square waves... Bu no, it is still not a logical argument.
Let's do a little math; if an amplifier can handle a rising edge of 45v on the μs scale, it means it can output a signal of at least 500khz at a peak level of 45v. If we reduce the signal to 20khz, the peak value will be 1125v....
Hypothetically, let's say your amplifier is "extremely slow" and only has a slew rate of 5v/μs. + You are listening to heavy metal at extremely high volume, for example 500w rms power (whatever kind of ears you have) and there is such a transition in the song that the signal comes out from 0 to 500w with a 20khz transition.. Can your amplifier respond to this? 500w@8r is ~ 63.5v rms. Even if your amplifier has a weak slew rate of 5v/μs, it reaches this value in about 18μs. This is about 60khz. And if you calculate a rising and a falling edge, we divide this into two and reach a square wave bandwidth of 30khz.
And don't forget that your dear extremely high end speaker will convert this signal to the sine with great pleasure.
So it is meaningless... Even if your amplifier's slew rate is 3v/μs, it is still meaningless.
I would like to draw your attention to the values we used as reference: You are listening to 500w and the dynamics of the music increases with a 20khz, i.e. 50μs, transition from 0 to 500w. I leave it to you to comment on what kind of audiophile (or human) can listen in this way.
A guy says "the old model of this amplifier had a 90v/μs slew rate, the new model has reduced it to 45.. so the amplifier is slower than the old one"...
If you are using your amplifier as an FM transmitter, this can be considered an argument.. Of course, if your transmitter/amp is not transmitting square waves... Bu no, it is still not a logical argument.
Let's do a little math; if an amplifier can handle a rising edge of 45v on the μs scale, it means it can output a signal of at least 500khz at a peak level of 45v. If we reduce the signal to 20khz, the peak value will be 1125v....
Hypothetically, let's say your amplifier is "extremely slow" and only has a slew rate of 5v/μs. + You are listening to heavy metal at extremely high volume, for example 500w rms power (whatever kind of ears you have) and there is such a transition in the song that the signal comes out from 0 to 500w with a 20khz transition.. Can your amplifier respond to this? 500w@8r is ~ 63.5v rms. Even if your amplifier has a weak slew rate of 5v/μs, it reaches this value in about 18μs. This is about 60khz. And if you calculate a rising and a falling edge, we divide this into two and reach a square wave bandwidth of 30khz.
And don't forget that your dear extremely high end speaker will convert this signal to the sine with great pleasure.
So it is meaningless... Even if your amplifier's slew rate is 3v/μs, it is still meaningless.
I would like to draw your attention to the values we used as reference: You are listening to 500w and the dynamics of the music increases with a 20khz, i.e. 50μs, transition from 0 to 500w. I leave it to you to comment on what kind of audiophile (or human) can listen in this way.
Slew rate was a bit of an obsession in '70s commercial gear, especially those Japanese manufacturers desperately trying to out-do each other on the old spec. front, whether it was relevant or not. It seems to be rearing its head again, following the eternal cycle. 😉
Although I certainly agree that slew rate values are widely misused when characterizing audio amplifiers, I think you do not appreciate why slew rate is important, and your math seems to be off.
Let's get the math straight first. A sinewave with the frequency F Hz and amplitude A Volts is the voltage V = A sin ωt = A sin 2𝜋Ft. The rate of change of this voltage if dV/dt = 2𝜋FA cos 2𝜋Ft Volts/second. The maximum rate of change is 2𝜋FA Volts/second. Therefore, a band-limited signal with the maximum frequency F Hz and amplitude A Volts will have the maximum rate of change 2𝜋FA Volts/second. For example, 45 V/μs is the maximum rate of change for a sinewave with the amplitude of 45 V and frequency of about 160kHz.
As you pointed out, 500W into 8ohm is about 63V RMS or 90V peak. At 20kHz, this gives us about 11 V/μs. Note most people don't hear at 20kHz, and music records do not have much energy, if at all, at such a high frequency, so the maximum amplitude at 20kHz may be the worst case for an amplifier, but unlikely to be music.
An amplifier slewing at, say, 15 V/μs, should be able to reproduce that 20kHz 90V peak without slewing. The catch, however, is that without slewing doesn't mean without distortion! In fact, such an amplifier is likely to distort that 20kHz sinewave a lot, because it approaches (even though does not exceed) its slew rate limit. On the other hand, a simlar amplifier slewing at 150 V/μs would be operating far from its slew rate limit and would be able to reproduce the same signal with much less distortion. Constructing an amplifier that remains linear as it approaches its slew rate limit takes special care.
In effect, the marketing message "our amplifier has a 3000V/μs slew rate" means that with ordinary audio signals, the amplifier operates so far from its limit that it distorts less.
Let's get the math straight first. A sinewave with the frequency F Hz and amplitude A Volts is the voltage V = A sin ωt = A sin 2𝜋Ft. The rate of change of this voltage if dV/dt = 2𝜋FA cos 2𝜋Ft Volts/second. The maximum rate of change is 2𝜋FA Volts/second. Therefore, a band-limited signal with the maximum frequency F Hz and amplitude A Volts will have the maximum rate of change 2𝜋FA Volts/second. For example, 45 V/μs is the maximum rate of change for a sinewave with the amplitude of 45 V and frequency of about 160kHz.
Square waves are a little different as they are not band limited - that is, have an infinite spectrum. A practical square wave with a finite rise time has a "knee" in its spectrum, that is, at some frequency its energy starts falling with frequency faster. That frequency can be estimated as F(knee)=0.5/Tr, where Tr is the 10%-90% rise time. That is, to find out how much bandwidth you amplifier needs to transmit an undistorted square wave, you need to specify the 10%-90% rise time.
An amplifier slewing at, say, 15 V/μs, should be able to reproduce that 20kHz 90V peak without slewing. The catch, however, is that without slewing doesn't mean without distortion! In fact, such an amplifier is likely to distort that 20kHz sinewave a lot, because it approaches (even though does not exceed) its slew rate limit. On the other hand, a simlar amplifier slewing at 150 V/μs would be operating far from its slew rate limit and would be able to reproduce the same signal with much less distortion. Constructing an amplifier that remains linear as it approaches its slew rate limit takes special care.
Let's construct an example amplifier (the output stage is an ideal distortionless buffer):
With the 1mA current source in the tail of the long tail pair (LTP) and a 68pF compensation capacitor, the slew rate of this simple amplifier is 1mA/68pF = 14.7 V/μs. It outputs a 90V peak 20 kHz sinewave without slewing. The distortion at that amplitude and frequency is 0.05%, compared to 0.003% at 1kHz.
The major reason for the higher distortion at 20kHz is that charging and discharging the capacitor faster at a higher frequency requires more current from the LTP, which pushes the LTP farther away from its optimal operating point, where the nonlinearities of Q1 and Q2 partialy compensate each other. Here is the capacitor charge/discharge current at 20kHz, 90V peak output as function of the differential input voltage (that between the bases of Q1 and Q2) of the LTP:
You can clearly see that at the extremes, the LTP approaches saturtation. Compare this to the same graph for 1kHz:
No saturation in sight, hence lower distortion.
The remedy for this particular amplier design is well known - increase the slew rate! Here is how you do it:
You can decrease the compensation capacitor, but that extends the bandwidth of the feedback loop and can lead to stability problems. A better way is to increase the LTP tail current - by 10 times is the above example - and add emitter degeneration to the LTP to keep its transconductance, and hence the feedback loop bandwidth, the same. This amplifier has THD @ 20 kHz at 0.006% compared to the 0.05% of its lower-slew-rate version above.
With the 1mA current source in the tail of the long tail pair (LTP) and a 68pF compensation capacitor, the slew rate of this simple amplifier is 1mA/68pF = 14.7 V/μs. It outputs a 90V peak 20 kHz sinewave without slewing. The distortion at that amplitude and frequency is 0.05%, compared to 0.003% at 1kHz.
The major reason for the higher distortion at 20kHz is that charging and discharging the capacitor faster at a higher frequency requires more current from the LTP, which pushes the LTP farther away from its optimal operating point, where the nonlinearities of Q1 and Q2 partialy compensate each other. Here is the capacitor charge/discharge current at 20kHz, 90V peak output as function of the differential input voltage (that between the bases of Q1 and Q2) of the LTP:
You can clearly see that at the extremes, the LTP approaches saturtation. Compare this to the same graph for 1kHz:
No saturation in sight, hence lower distortion.
The remedy for this particular amplier design is well known - increase the slew rate! Here is how you do it:
You can decrease the compensation capacitor, but that extends the bandwidth of the feedback loop and can lead to stability problems. A better way is to increase the LTP tail current - by 10 times is the above example - and add emitter degeneration to the LTP to keep its transconductance, and hence the feedback loop bandwidth, the same. This amplifier has THD @ 20 kHz at 0.006% compared to the 0.05% of its lower-slew-rate version above.
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what is the "correct" way to measure slew rate?
with the input LP filter in place, or with the input filter disabled?
with the input LP filter in place, or with the input filter disabled?
Slew-rate "per se" means practically nothing, at least for semiconductors amplifiers.
A problem is that slew rate is generally viewed like a clipping limit: hard and well-defined. When some amps approach their SR limit, they begin to degrade in some way.
Thus, a sensible approach is to give yourself some margin: with 20/µs, you are safe in all practical domestic uses, much less for "reasonable" power levels.
Anyway, most reproduction media have a much lower inherent SR
A problem is that slew rate is generally viewed like a clipping limit: hard and well-defined. When some amps approach their SR limit, they begin to degrade in some way.
Thus, a sensible approach is to give yourself some margin: with 20/µs, you are safe in all practical domestic uses, much less for "reasonable" power levels.
Anyway, most reproduction media have a much lower inherent SR
You should remove the input LPF when measuring the SR. The LPF is there to stop possible HF present at the input from driving the input stage away from its linear region, as shown in my previous post. With a well designed LPF connected, you'd be measuring that LPF's impulse response, not the SR.
Douglas Self's "Blameless" amplifier (which I think of as a baseline these days) has 60 V/µs (6mA/100pF). But the question is not in this number or that number, It is in the HF linearity as measured by e.g. the IMD 18+19kHz test.
Inadequate slew rate causes triangular distortion, mostly due to being unable to charge the Miller capacitor rather than the speakers.
You might be right about the distortion issue. However you're still working on 0 to 500w rms transition.. And I am asking which psycopath listens this kind of transititon? While only audiophiles interested in with distortion.. And listen generally in first watt.
Very true. It is caused by a current source limit charging a capacitor somewhere within the amplifier. Negative feedback can not correct it. If it is within the global feedback loop, it is difficult to find the spot where it happens. It is seldom the output stage, although the SR is usually specified as referred to the output slew limited voltage per rise time. It is best practice to apply a LP filter right at the input stage.
It is my opinion that if your power amplifier can produce a 'nice clean' 10Khz square wave at *full power [ *same power as sine clip ]
then your amp is 'good to go'. As someone said > "We're not trying to transmit at radio frequencies".
However, if you have 'ringing' or significant 'rounding', then things are not all good.
PS.
Speed/rise time/slew rate DOES play an important role regarding many 'complex waveforms in music'.
then your amp is 'good to go'. As someone said > "We're not trying to transmit at radio frequencies".
However, if you have 'ringing' or significant 'rounding', then things are not all good.
PS.
Speed/rise time/slew rate DOES play an important role regarding many 'complex waveforms in music'.
A typical voltage feedback amplifier has three stages namely the input differential stage, the integrator / VAS and the output / drive stages.
When the input error signal into the amplifier is large, almost all the input stage bias current flows into one of the legs of the input differential pair, but this current being limited, can only charge the compensation capacitor at a limited rate, called the slew rate of the amplifier given as:
SR = Cp * Ib / 2; where Ib/2 is the max current switch-able by the input differential pair.
Thus, for the perfect reproduction of any arbitrary waveform, an infinite slew rate becomes necessary. Slew rate is best measured by applying a step signal into the amplifier, as it ensures that the max charging rate (dV/dt) is easily visible on the output,
The slew rate in the above case is easily seen as being slightly less than 1V / us.
When the input error signal into the amplifier is large, almost all the input stage bias current flows into one of the legs of the input differential pair, but this current being limited, can only charge the compensation capacitor at a limited rate, called the slew rate of the amplifier given as:
SR = Cp * Ib / 2; where Ib/2 is the max current switch-able by the input differential pair.
Thus, for the perfect reproduction of any arbitrary waveform, an infinite slew rate becomes necessary. Slew rate is best measured by applying a step signal into the amplifier, as it ensures that the max charging rate (dV/dt) is easily visible on the output,
The slew rate in the above case is easily seen as being slightly less than 1V / us.
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it depends on the input filter how much rounding is seen, no indication of an issue.
Its only the loud high frequency components that generate fast slewing, white noise is a good example of a demanding signal, or something ultrasonic like a 30kHz full-scale sine wave (easier to detect slew rate limiting than with noise). Such signals are demanding on the ear too, typical music signals are more benign as high frequency energy is usually much lower than low-frequency.
Complex doesn't have to imply fast...
The search term you need is "power bandwidth" or "full-power bandwidth". Here's a decent overview: https://en.wikipedia.org/wiki/Power_bandwidth
A 100 W (8 Ω) amplifier would, thus, need a minimum slew rate of 5 V/µs to satisfy the 80 V peak-to-peak swing of a sine wave at 20 kHz at 100 W into 8 Ω.
Another way to approach this would be to assume that you're using a digital source at, say, 192 kHz sampling rate and that you go from the most negative voltage to the most positive from one sample to the next. In that case you'll need:
SR ≥ (VCC + |VEE|)/Ts -> SR ≥ (VCC + |VEE|)*Fs
So if your amp runs on, say, ±65 V rails (VCC + |VEE| = 130) you need 130*192000 = 24960000 V/s = 24.96 V/µs. For the 100 W amp (80 Vpp swing) you'll need 15.4 V/µs minimum slew rate.
You can analyze your music collection to see how many such transitions exist in it. I suspect you won't find many, if any at all.
Tom
A 100 W (8 Ω) amplifier would, thus, need a minimum slew rate of 5 V/µs to satisfy the 80 V peak-to-peak swing of a sine wave at 20 kHz at 100 W into 8 Ω.
Another way to approach this would be to assume that you're using a digital source at, say, 192 kHz sampling rate and that you go from the most negative voltage to the most positive from one sample to the next. In that case you'll need:
SR ≥ (VCC + |VEE|)/Ts -> SR ≥ (VCC + |VEE|)*Fs
So if your amp runs on, say, ±65 V rails (VCC + |VEE| = 130) you need 130*192000 = 24960000 V/s = 24.96 V/µs. For the 100 W amp (80 Vpp swing) you'll need 15.4 V/µs minimum slew rate.
You can analyze your music collection to see how many such transitions exist in it. I suspect you won't find many, if any at all.
Tom
If you're running 20 kHz at 90 V then your tweeters are either robust beyond belief or a small pile of ash.
Also consider any realistic source is band-limited so demanded slew rate will be constrained. A 15 kHz square wave through the CD process is a sine wave, and even at 192 kHz sampling rate you'll get only the first two harmonics above the fundamental: 45 kHz and 75 kHz. A little bit of math tells us the slew rate will be 6π * 15 kHz * Vpk [1], where Vpk is the maximum voltage of the wave, or about 0.3 V/µs * Vpk, so for an amp with 60 volt rails you're looking at about 18 V / µs -- if you're insane enough to try on 15 kHz square waves at 200 watts!
[1] 6π because the contribution from each harmonic ends up being 2π since the increased dV/dt is exactly matched by the lower amplitude of the harmonic.
Now charging the loop capacitor will create slight amounts of distortion below the slewing limit as shown by an earlier post with graphs, so higher slew is not completely a fool's errand even though music won't exercise an amplifier nearly so hard.
Also consider any realistic source is band-limited so demanded slew rate will be constrained. A 15 kHz square wave through the CD process is a sine wave, and even at 192 kHz sampling rate you'll get only the first two harmonics above the fundamental: 45 kHz and 75 kHz. A little bit of math tells us the slew rate will be 6π * 15 kHz * Vpk [1], where Vpk is the maximum voltage of the wave, or about 0.3 V/µs * Vpk, so for an amp with 60 volt rails you're looking at about 18 V / µs -- if you're insane enough to try on 15 kHz square waves at 200 watts!
[1] 6π because the contribution from each harmonic ends up being 2π since the increased dV/dt is exactly matched by the lower amplitude of the harmonic.
Now charging the loop capacitor will create slight amounts of distortion below the slewing limit as shown by an earlier post with graphs, so higher slew is not completely a fool's errand even though music won't exercise an amplifier nearly so hard.
Yes, exactly.
In plain language, the input - output relationship in an amplifier is not immediate / instantaneous. The input voltage drives an error current which (over a small time) fills charge into a capacitor whose voltage is in turn buffered to give the output. This makes the rate of filling the capacitor (SR) an important factor, especially at high frequencies.
At high frequency depending on the power transistors.
The error signal that the feedback must correct is a much higher slew.
Your assuming a distortion free signal.
Everyone assumes the pretty signal at the end.
Nobody looks at error signal being corrected.
With low bias in that "first watt" at high frequency.
Error correction is almost a straight line.
Getting .001% distortion at 1k is easy.
Not living in a fantasy world.
Most 500 watt amplifiers at 12k have horrible distortion.
At those power levels.
Actually many magical 7 watt amplifiers are garbage at high frequency.
The slew rate of the devices and the error correction/feedback
Has to be high, doesn't matter if 5 or 50 or 500 watts.
Then again, exactly why pay 5 grand for a amplifier. If a 5 dollar chip amp does the same.
The other favorite is, speakers just make 1% to 3% distortion anyways, what is the point of doing better.
Nobody would cry or whine listening to 80 year old wire recording then.
Distortion is eliminated every step of the way, and was done because people ignore excuses to never be better.
" Calibration" is actually just a comparison to the highest known standard. Not a adjustment.
A plastic ruler at the dollar store makes a straight line right? Ask the Calibration team at Nasa
if a plastic ruler is acceaptable as a calibration standard. LOL
The error signal that the feedback must correct is a much higher slew.
Your assuming a distortion free signal.
Everyone assumes the pretty signal at the end.
Nobody looks at error signal being corrected.
With low bias in that "first watt" at high frequency.
Error correction is almost a straight line.
Getting .001% distortion at 1k is easy.
Not living in a fantasy world.
Most 500 watt amplifiers at 12k have horrible distortion.
At those power levels.
Actually many magical 7 watt amplifiers are garbage at high frequency.
The slew rate of the devices and the error correction/feedback
Has to be high, doesn't matter if 5 or 50 or 500 watts.
Then again, exactly why pay 5 grand for a amplifier. If a 5 dollar chip amp does the same.
The other favorite is, speakers just make 1% to 3% distortion anyways, what is the point of doing better.
Nobody would cry or whine listening to 80 year old wire recording then.
Distortion is eliminated every step of the way, and was done because people ignore excuses to never be better.
" Calibration" is actually just a comparison to the highest known standard. Not a adjustment.
A plastic ruler at the dollar store makes a straight line right? Ask the Calibration team at Nasa
if a plastic ruler is acceaptable as a calibration standard. LOL
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