Slew rate effects on audio
Premise
In electronics, slew rate is defined as the change of voltage or current, or any other electrical quantity, per unit of time. Expressed in SI units, the unit of measurement is volts/second or amperes/second or the unit being discussed, usually written as units of V/μs in electronic datasheets.
It’s no secret that slew rate is a really relevant parameter in electronics. In audio power amplifiers it is certainly one of the three most important parameters, together with the measurement of distortion and the damping factor (output impedance).
However, it seems that this is largely forgotten by everyone, probably because first historically and then for marketing issues, the THD parameter has become the one and only concern of those who buy an amplifier.
This also ends up dividing the audio world: some believe they have perfect and impeccable amplifiers only because they have infinitesimal distortions, others have stopped trusting the measurements, since although two amplifiers may have similar levels of distortion, in reality they produce different signals and manage the load differently, resulting in different sounds.
From here, the distortion parameter, for those who trust only their hearing, begins to lose importance, until they stop believing that measurements really matter and end up believing in the magic of amazing cables and ground boxes filled with sand.
The main problem lies in education regarding the subject, fortunately the world is slowly becoming more and more scientific and increasingly oriented towards measurements, but the process is really long.
The final point is that a single parameter is not enough to define how an amplifier will behave towards a certain input signal or a certain load at its output. So now let’s talk about a very important one, the slew rate.
How slew rate changes the audio signal
Slew rate comes into play when we talk about the passage of a signal in any electronic device, whether active (like an amplifier) or passive (like a passive crossover) since all of them will inevitably modify the final speed of the signal, which will ALWAYS be lower than the input speed. Having a higher output speed than input speed is not possible in physics, it is like saying that the efficiency of the device would be less than zero.
In audio, the slew rate is measured by introducing a square wave signal into the device, and then dividing the amplitude (expressed in volts) by the time which represents the rise time of the square wave, or its fall time.
The slew rate can also change greatly depending on the output load, if the load impedance lowers the slew rate will tend to lower due to the greater current demand.
What can immediately be noticed is that the very fact of having a lower output speed represents a sort of distortion of the shape of the signal itself. This distortion is always present, just as there will ALWAYS be a certain level of distortion introduced by the device itself, which can never be truly neutral.
The point, however, is to understand how much distortion is acceptable and what is not.
For example, how much THD+N a device can produce before it’s perceived by the ear o changes in a significant way the signal?
It depends on which device it is measured and how much of this can be perceived by the human ear. If the device, for example, were an audio source, this should have the lowest possible distortion: it is not an exaggeration to say, for example, that a THD+N below 0.005% is necessary.
In the example above, the signal is amplified and passed through a resistive load. The final distortion of the operational amplifier is very low anyway. In the second circuit, a source of 100uV of a second signal is introduced to simulate the distortion of 0.01% from the source. The final distortion is amplified as much the signal is amplified and then the device distortion is added.
The problem is not only the THD, but the noise too, as the input distortion is amplified the noise is amplified in the same way. So, the noise of the audio source will be added to the total noise, plus the noise generated by the amplifier.
Problem is: until is not amplified, the distortion is not really perceived, but at high levels it can really degenerate the sound, generate artifacts in the speaker excursion and a lot of background noise. It’s really important to have a very low distortion equipment, the source more than all.
There is no specific and declared standard of acceptable quantity of distortion, nor there is a standard threshold of distortion hearing (that’s because the perception really varies a lot from person to person).
However, it is common in the audio community to take CD as the standard: 16bit depth, that means a dynamic interval of 96dB, with a theorical THD of 0.0015%. With those performance, distortion or noise is practically inaudible by anyone.
But what is the level of acceptable decrease in speed? In simpler words, how much slew do I need to not have a distorted signal? For sinusoids it is simple to calculate:
Depending on the frequency, you need a certain quantity of output speed to reproduce that frequency perfectly. In electronics the slew rate is nothing other than the maximum speed with which the signal can vary. For this reason, bandwidth and slew rate are two things that go hand in hand. If a device can reproduce a certain frequency, it means that it has sufficient speed to do so: it means also that the higher the frequency the higher the slew rate required.
In the example above, is calculated the slew rate required to reproduce a 20kHz sine wave or rather the maximum audible frequency. It means, that for each output Volt is required a slew rate of 0.125V/us to not distort the signal.
When the slew rate is not enough for a sinusoid the signal is distorted until the amplitude and frequency are reduced (the period is larger in time domain) and the wave is transformed in a triangle wave, thus reducing the total bandwidth of the device.
If a power amplifier with a standard load of 8 Ω has to deliver 100W of power it will need about 28V of output voltage, but this has to be doubled, 56V, because we are talking about alternating current, so the power passing through the load goes both in positive and negative direction and the complete wave reaches both positive and negative peak, slew rate works on the entire wave function, thus we have to consider the Vpp.
At 20kHz of frequency, with a signal of 56V the slew needed is 7V/us for undistorted output: still talking about a sinusoid.
The slew requirements are therefore dependent by the output voltage and the maximum frequency of the device: this determines the “power bandwidth”, or rather the bandwidth that a device can output at certain voltage.
This topic stumps many people when it comes to understanding what the effective bandwidth of an audio amplifier is. The bandwidth is not always the same at any output voltage.
To simulate a slew reduction, an inductor is introduced
As shown, the resulting bandwidth is directly affected by the lack of slew rate, and vice versa.
In this example, there are two identical circuit, but in the second one the input voltage is 10 times larger. As shown by the graph, the output of the first circuit has the correct amplitude and the wave is undistorted. Meanwhile, the second voltage (same frequency) has reduced amplitude and distorted to a triangle wave. This is clearly a demonstration on how the output voltage interfaces with the maximum speed of the amplifier: the higher the voltage, the lower the reproducible bandwidth.
For the maximum bandwidth an amplifier can have at a certain output voltage, is possible to use the inverse equation for the slew rate:
A good example could be produced using the bandwidth of a class D amplifier. To filter the output square wave and obtain a sine wave, switching amplifiers specifically reduce the speed and therefore the output bandwidth. The reduction of the latter is sometimes really excessive in poor designed switching amplifiers.
The best class D amplifiers normally have a fairly linear band from 20Hz to 20kHz, with just -0.5dB of attenuation over 20kHz: one could say “perfect”, because the amplifier reproduces all the frequencies of the audio band equally.
But is this really true? Is a flat band up to 20kHz sufficient for an amplifier? The answer is no.
By ignoring the frequency shift created by a cut too close to the useful band, the output filter heavily reduces the maximum speed on the load, resulting in a slew no more of 12-14V/us even for the best class D amplifiers.
But why? All class D amplifiers have a low slew rate? The answer is: no, only the one which have the feedback point after their LC filter network (pretty much the majority of them).
This is a simplified schematic of an amplifier with its feedback point after the LC filter.
As long as the feedback factor is greater than one (i.e. for frequencies below the unity-gain point of the feedback loop) an amplifier shows a frequency response which is approximately the inverse of the feedback network’s frequency response.
With the feedback point placed in this way, the LC filter will force the overall amp to behave like a first order lowpass. If the slew demand increases (increases the input frequency and amplitude) the NFB will force the modulator/output stage into transient saturation while trying to emulate the lowpass response of the overall amplifier, thus lowering the output slew.
Even using an ideal amplifier with infinite slew and output current, the maximum rise/fall time of the wave and the amplitude is strongly decreased.
Talking about measured real data, with a slew between 10 and 14V/us, and picking the maximum audible frequency, 20kHz, the maximum output voltage with undistorted signal is about 111V, meaning (divided by positive and negative swing) 55.5V, or 385W on 8 Ω.
Surely a pretty high power, but this means that even the best class D amplifiers cannot reproduce 20kHz at their maximum power: both because maximum power can be over 2000W and because there are no more (or equally) precise class Ds with wider bands than these.
However, this ONLY:
- IF the slew is actually this high, because many of the times the measured slew at that bandwidth is no more than 10-11V/us. In low and mid quality class D amplifier the slew is no more than 2-3V/us, meaning a power bandwidth of only 8.5kHz at 100W output, and standard linear amplifiers no more than 5-7V/uS.
- IF we are talking only about sinusoids, because the required slew to reproduce a square wave is much greater.
- IF we are calculating a single sine wave.
The calculations were only for a single sinusoid, but in audio, we will not have a single signal. Any audio signal can be represented as a sum of sinusoids, in a given delta time there could even be over 10 different signals, at different amplitudes.
For example, a square wave, that is a very simple signal, is a sum of the fundamental frequency plus all his odd harmonics with a slope of 8dB/oct.
Using a central frequency of 1kHz we calculate the slew rate required for each volt of a square wave:
| Frequency | 1 kHz | 3 kHz | 5 kHz | 7 kHz | 9 kHz | 11 kHz | 13 kHz | 15 kHz | 17 kHz | 19 kHz |
|---|---|---|---|---|---|---|---|---|---|---|
| Level | 0 | -9.5 dB | -14 dB | -16.9 dB | -19.1 dB | -20.8 dB | -22.3 dB | -23.5 dB | -24.6 dB | -25.6 dB |
| Voltage | 1 | 0.335 | 0.2 | 0.143 | 0.111 | 0.091 | 0.077 | 0.067 | 0.059 | 0.052 |
| Required Slew (V/uS) | 0.0063 | 0.0063 | 0.0063 | 0.0063 | 0.0063 | 0.0063 | 0.0063 | 0.0063 | 0.0063 | 0.0063 |
A simple sine wave of 1 volt at 1kHz of frequency require only 0.0063V/us of slew rate. A square wave, is a sum of the odd harmonics to a fundamental frequency. In the table we consider only harmonics below 20kHz (the ones that will reach the ear), the required slew is calculated by multiplying each harmonic base required slew with actual voltage.
As we can see, the required slew is constant for each harmonic, so we can summarize in:
Where “h” is the number of odd harmonics to consider (all before 20kHz). If we sum all the required slew values, we obtain that a square wave of 1kHz frequency and with amplitude of 1V require 0.063V/us. Which seems like little, but it’s actually 10 times the value needed for a sine wave.
This means that for a 100W output on a 8 Ω load the slew required will be then 3.5V/us, in the case of a single sinusoid it would barely be 0.35V/us.
Now, this is what happens with a simple square wave, but the slew required for a real music signal is even higher.
In a real audio signal, the number of intermodulated signals and their amplitude is unpredictable. There could even be over 20 different signals intermodulated at the same time.
Now arrives a difficult step to understand: as for test, if we added the slew required for each single frequency from 1kHz to 20kHz, jumping 1kHz at a time, all at the same amplitude we would obtain 1.323V/us of required slew (required slew for voltage of 1kHz + 2kHz + … + 20kHz).
The output voltage for each signal is exactly 1V, intermodulating those signals we obtain a complex wave of about 1V.
If we need 2.646V/us (the double of slew for each direction of each volt of output, this means that slew rate must be really high to have an undistorted signal for high power outputs.
If we were to take the example of the best quality class D from before, whose maximum possible slew is only 14V/us, then the maximum undistorted voltage output with such a signal will be 5.29V: only 3.5W on a 8 Ω load before the amplifier produces harmonics and thus slew induced distortion.
For a 100W output on 8 Ω the slew required by the amplifier for the same type of test, will then be about 74V/us, definitely a fairly high value for over 90% of amplifiers on the market.
Without using 20 different waves, let’s use only 8 intermodulated waves.
In the simulation is used a universal ideal amplifier, without any load and at unitary gain, capable of amplifying with zero distortion: because we need to isolate the effects of the lack of slew.
Applying about 28V of voltage at the input (corresponding at 100W on 8 Ω load), 8 fundamentals (1, 2.5, 6.5, 8.5, 13, 14, 17, 20kHz): the resulting slew required to amplify those sinusoids at that specific amplitude is about 26V/us, the amplifier maximum slew is 60V/us (as the slew of KRIEG Nemesis V1 and V2), the resulting distortion is negligible (-110dB).
Using same parameters, we now downgrade the maximum amplifier slew to 14V/us (as the slew of one of the lowest distortion class D amplifier), obtaining a resulting distortion of -65dB (0.056%): 180 times higher than the amplifier with 60V/us of slew rate.
This test excludes any possible distortion introduced by the sources and the amplifier, even excluding the load, demonstrating how the slew actually has a heavy and serious impact on the final distortion performance. This case is definitely a real case, the quantity of simultaneous signals is very low, and despite this the performances are strongly degraded.
Another test, made just for fun, with a slew of 800V/us, but demonstrates that if the slew is indefinitely high, then the distortion would be zero.
Real measurement demonstration of distortion induced by low slew of an amplifier:
Conclusions
This is the mathematical and experimental demonstration that, even if an amplifier could be perfect from the point of view of distortion measurement, in reality it produces a lot of distortion when the signal is no longer a simple sinusoid but something complex like music.
It also demonstrates that the slew rate is a value that should absolutely not be underestimated and that it is as important as the THD measurement, and as such, the higher it is, the better.