# Slew Rate

#### MRC01

##### Major Contributor
Brace yourselves for a dumb question: why isn't slew rate simply the slope of the highest frequency in the amp's power bandwidth?
[Of course the slope of a sine wave varies so take it at a designated point, such as 0 crossing.]
I may be over-simplifying this, but it seems to me that:
1. A faster slew rate than this, is not needed.
2. A slower slew rate than this, is impossible - else the power bandwidth would drop off at a lower frequency.
PS: if that is what slew rate is, then it's a redundant measurement - all you need is frequency response.

OP

#### DonH56

##### Master Contributor
Technical Expert
Brace yourselves for a dumb question: why isn't slew rate simply the slope of the highest frequency in the amp's power bandwidth?
[Of course the slope of a sine wave varies so take it at a designated point, such as 0 crossing.]
I may be over-simplifying this, but it seems to me that:
1. A faster slew rate than this, is not needed.
2. A slower slew rate than this, is impossible - else the power bandwidth would drop off at a lower frequency.
PS: if that is what slew rate is, then it's a redundant measurement - all you need is frequency response.

Slew rate is the rate of change, e.g. how many volts change in designated time. It is related to bandwidth as shown in the first post. If you want to achieve flat frequency response to 20 kHz, you need much more bandwidth than that -- a simple first-order roll-off would be -3 dB at 20 kHz if the stated bandwidth were 20 kHz. Low distortion requires feedback, which in turn requires loop gain, which means you have to have some internal gain at the -3 dB point, so again the internal bandwidth (and slew rate) must be higher than the simple datasheet number. Because of circuit limitations, bandwidth changes, usually decreasing as you increase power. Part of that is because there is not enough current to slew fast enough, part is because internal capacitances and such increase with increasing current, and so forth. You also need extra bandwidth to create a stable amplifier, since phase shift gets high at band edges, and it must handle all sorts of nasty loads without oscillating and destroying itself (and/or you speakers, ears, etc.) Certain circuit designs use active devices to enhance slewing and steady-state bandwidth measurements may not show that. All of that leads to small-signal and large-signal (full-power) bandwidth often being very different, with slew rate one of the parameters that helps quantify the difference.

Basically, "it's complicated". For a passive circuit, slew rate and bandwidth are directly related and readily calculated. For an active design, such as a preamp or power amp, the correlation can range from 1 to almost nil.

HTH - Don

#### MRC01

##### Major Contributor
Slew rate is the rate of change, e.g. how many volts change in designated time. It is related to bandwidth as shown in the first post. If you want to achieve flat frequency response to 20 kHz, you need much more bandwidth than that ... Basically, "it's complicated". For a passive circuit, slew rate and bandwidth are directly related and readily calculated. For an active design, such as a preamp or power amp, the correlation can range from 1 to almost nil.
...
I understand that, so I will try rephrasing my question as an example:

Suppose you have an amp having a frequency response of 20 Hz to 20 kHz +/- 0.1 dB, and power bandwidth of 5 Hz to 100 kHz +/- 3 dB, both measured at an output of 40 V into 8 ohms / 5 amps / 200 watts. In this case, you can compute the slew rate from the size & shape of a 100 kHz waveform with 28 V of amplitude. I say 28 V because that's 3 dB less than 40 V. So once you know the power bandwidth, you can compute the slew rate yourself. You don't need them to tell you what it is.

PS: we can try this with a back of the envelope calculation. Derivative of sin is cos, so the slope of a sin wave as it passes through zero is cos(0) = 1. 100 kHz has a period of 1/100,000 second, and it rises in half that time. So the slew rate is 28 V in 1/200,000 second.

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#### Speedskater

##### Major Contributor
Some interesting reading on the subject by Bruno Putzeys.

"Negative feedback in audio amplifiers:"
'Why there is no such thing as too much'

part 1
part 2

This article originally appeared in Linear Audio, a book-format audio magazine published half-yearly by Jan Didden

OP

#### DonH56

##### Master Contributor
Technical Expert
I understand that, so I will try rephrasing my question as an example:

Suppose you have an amp having a frequency response of 20 Hz to 20 kHz +/- 0.1 dB, and power bandwidth of 5 Hz to 100 kHz +/- 3 dB, both measured at an output of 40 V into 8 ohms / 5 amps / 200 watts. In this case, you can compute the slew rate from the size & shape of a 100 kHz waveform with 28 V of amplitude. I say 28 V because that's 3 dB less than 40 V. So once you know the power bandwidth, you can compute the slew rate yourself. You don't need them to tell you what it is.

PS: we can try this with a back of the envelope calculation. Derivative of sin is cos, so the slope of a sin wave as it passes through zero is cos(0) = 1. 100 kHz has a period of 1/100,000 second, and it rises in half that time. So the slew rate is 28 V in 1/200,000 second.
Slew rate is not necessarily directly coupled to bandwidth for active circuits. You can incorporate circuits that will increase or decrease the slew rate compared to what you would calculate from the bandwidth.

#### MRC01

##### Major Contributor
OK, but how can the slew rate be any slower than the bandwidth? If the amp actually produces a 100 kHz wave with 28 V of amplitude, then it MUST be rising at 28 V in 1/200,000 second. If the slew rate were any slower, it couldn't produce that signal!

OP

#### DonH56

##### Master Contributor
Technical Expert
I was speaking in general. Small and large signal bandwidths are different. Slew enhancement can increase slew rate, slew limiting decreases it, not necessarily tightly coupled to bandwidth.

#### RayDunzl

##### Grand Contributor
Central Scrutinizer
SR = 2*pi*f*A where
pi = 3.14159…f = frequencyA = amplitude

Does that little morsel take into account that the required rate of slew is 0 at 90 and 180 degrees of a sine wave and maximum at the zero crossings?

Seems like there should be a "where am I in the cycle" factor - degrees or radians.

The formula looks like the minimum requirement to produce a triangle.

Which would imply the amp could also produce higher frequency harmonics, so maybe it's good.

I dunno.

OP

#### DonH56

##### Master Contributor
Technical Expert
Does that little morsel take into account that the required rate of slew is 0 at 90 and 180 degrees of a sine wave and maximum at the zero crossings?

Seems like there should be a "where am I in the cycle" factor - degrees or radians.

The formula looks like the minimum requirement to produce a triangle.

Which would imply the amp could also produce higher frequency harmonics, so maybe it's good.

I dunno.
The formula is for the maximum slew rate of a sine wave.

#### pma

##### Major Contributor
Slew rate of my PA2 amplifier (2009) -

#### kyllwtr

##### Member
OK, but how can the slew rate be any slower than the bandwidth? If the amp actually produces a 100 kHz wave with 28 V of amplitude, then it MUST be rising at 28 V in 1/200,000 second. If the slew rate were any slower, it couldn't produce that signal!
There are two different aspects to keep in mind.

-Bandwidth: the bandwidth limitation has effect on all signal levels depending only on its frequency. Thus even a 'small' signal is affected by this limitation. If, for example, you have -3 dB at, say, 10kHz, it means that all output signals at that frequency will be 1.41 time lower regardless its input level (even a low volume one).

-Slew rate: this take in account two different factor: frequency and voltage. If the voltage output is half, the frequency limit (due to slew rate) is double. That means that an amplifier with not sufficient slew rate at full power, can still reproduce with fidelity at lower power level (if it has sufficient bandwidth and in contrast with bandwidth limitation case).

Now, since the power content of recorded music decrease a lot with frequency, an amplifier could be actually works fine even with not sufficient slew rate for 'full power-full frequency operation' (but not completely true for what said by DonH56 about bandwidth limitations).

A spectrum envelop of a commercial song, taken as sample. In this case, at 11kHz the max. signal level is quite at -33dB while at more central frequencies it rises at about -10dB or more.

#### MRC01

##### Major Contributor
Ah, so in summary, if the slew rate is limited (slower than the actual rise time of a full power wave near the upper end of the amp's bandwidth), then the amp may have flat full spectrum response at low volumes, but will attenuate high frequencies when you crank it up.

#### kyllwtr

##### Member
Ah, so in summary, if the slew rate is limited (slower than the actual rise time of a full power wave near the upper end of the amp's bandwidth), then the amp may have flat full spectrum response at low volumes, but will attenuate high frequencies when you crank it up.
Actually distort, not attenuate.
Bandwidth limitation attenuate (but in complex signal reproduction this could be considered distortion, too).

#### MRC01

##### Major Contributor
Ouch. Goodbye tweeters.

#### MRC01

##### Major Contributor
Does that little morsel take into account that the required rate of slew is 0 at 90 and 180 degrees of a sine wave and maximum at the zero crossings?
Seems like there should be a "where am I in the cycle" factor - degrees or radians.
The formula looks like the minimum requirement to produce a triangle.
Which would imply the amp could also produce higher frequency harmonics, so maybe it's good.
...
It's worth noting that the maximum slope of a sine wave is 1:1 or 1.0, and it achieves this max slope as it crosses zero.
Put differently, the derivative of sine is cosine, cos(t) <= 1.0 for all t, and cos(0) is 1.0.
So if the slew rate can handle that, it can handle the rest of the waveform.

#### NikJi

##### Member
The slew rate on my 30w per channel stereo amplifier is 500. ‍

#### egellings

##### Major Contributor
That is vastly more than any audio signal will ever need for its reproduction. Kudos for the engineering involved in achieving that, though.

#### NikJi

##### Member
That is vastly more than any audio signal will ever need for its reproduction. Kudos for the engineering involved in achieving that, though.
Yea. It was originally designed for the Kawai Digital Piano CS X1.

#### Cbdb2

##### Major Contributor
Slewing is a recovery from "overloading" the front end. The signal is too fast for the feedback to follow, usually caused by the capacitance of the second stage. In the test, the +ve in goes to 5v, the -in fed by the feedback loop is still at zero (no instant response) so the input + to - , which is Vout/open loop gain (millivolts) in its operating mode is now 5v. Not in its linear region. From the bible: Analysis and Design of Analog Integrated Circuits by Grey, Hurst, Lewis, and Meyer

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OP

#### DonH56

##### Master Contributor
Technical Expert
Overload recovery is not slewing though recovery from excess slew or amplitude is a critical part of any design.

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