Hi Group:
I'm trying to come up with a slew rate measurement (10/90 rise time) of a
tube preamp I'm working on.
I've read all the books I have on the subject but none of them tell what
frequency square wave
I should use or at what output level. My gut tells me just below the
clipping point for the level
and maybe 2 KHZ for the frequency. Using these settings I get a 10/90% of
25V in 3.97uS.
25V/3.97uS gives me 6.29 V/uS rise time. Does this check out with you guys?
If we made the measurement
at a lower output level, It really wouldn't tell us what the amp is capable
of delivering would it?
But the frequency of the square wave I'm not sure what to use.

Any Ideas??

RonL

PS
The square wave generator is a lab grade Tektronix with nS risetimes
so I'm not concerned about the generator error.

[Phil Allison]

Hi Group:
I'm trying to come up with a slew rate measurement (10/90 rise time) of a
tube preamp I'm working on.


** No can do.

Rise time and slew rate are *independent * parameters.

The former is a small signal parameter and the latter a large signal one.

Both can be measured using a fast square wave at the input, but it is two
separate tests with two separate results.



I've read all the books I have on the subject but none of them tell what
frequency square wave
I should use or at what output level. My gut tells me just below the
clipping point for the level
and maybe 2 KHZ for the frequency. Using these settings I get a 10/90% of
25V in 3.97uS.
25V/3.97uS gives me 6.29 V/uS rise time. Does this check out with you
guys?


** NO !!

Rise time is just a TIME - ie so many uS, it is related to the small
signal bandwidth and is a LINEAR parameter.

OTOH "slew rate" is a NON LINEAR parameter related to internal
capacitances in an amplifying stage that must be charged and discharged with
each cycle.

YOU have it *horribly* mixed together.


1. For the "rise time" test, use a low input level and a high enough
frequency to make reading the value off the scope screen easy - ie 10%
to 90 % in so many uS.

2. For the "slew rate" test, use a level that remains below clipping and an
input frequency high enough that the normally gracefully curved rises and
falls turn into STRAIGHT LINES !!!

Then read the up and down slopes in V/uS - they may not be the same.




........ Phil

Hmmm.
So the "rise time" is just the time the signal moves from 10% to 90% and is
not referenced to the voltage?
Where "slew rate" is time referenced to the voltage just below clipping?

I have a commercially produced preamp that has a variable rise time control
on the front panel.
It allows you to adjust the rise time from: 2V/uS to 6V/uS. I'm trying to
verify exactly what it's doing.
It's supposed to emulate the older vintage preamps with the slow "rise time"
to the more modern preamps
with faster "rise times". It seems to me rise time has got to be connected
to frequency response somehow.
If the rise time is too slow to reproduce the high frequencies.



"Phil Allison" wrote in message
...



Hi Group:
I'm trying to come up with a slew rate measurement (10/90 rise time) of a
tube preamp I'm working on.


** No can do.

Rise time and slew rate are *independent * parameters.

The former is a small signal parameter and the latter a large signal one.

Both can be measured using a fast square wave at the input, but it is two
separate tests with two separate results.



I've read all the books I have on the subject but none of them tell what
frequency square wave
I should use or at what output level. My gut tells me just below the
clipping point for the level
and maybe 2 KHZ for the frequency. Using these settings I get a 10/90% of
25V in 3.97uS.
25V/3.97uS gives me 6.29 V/uS rise time. Does this check out with you
guys?


** NO !!

Rise time is just a TIME - ie so many uS, it is related to the small
signal bandwidth and is a LINEAR parameter.

OTOH "slew rate" is a NON LINEAR parameter related to internal
capacitances in an amplifying stage that must be charged and discharged
with each cycle.

YOU have it *horribly* mixed together.


1. For the "rise time" test, use a low input level and a high enough
frequency to make reading the value off the scope screen easy - ie 10%
to 90 % in so many uS.

2. For the "slew rate" test, use a level that remains below clipping and
an input frequency high enough that the normally gracefully curved rises
and falls turn into STRAIGHT LINES !!!

Then read the up and down slopes in V/uS - they may not be the same.




....... Phil

New Info:
According to Valley Wallman, The risetime should be approx the inverse of
..35 X of the bandwidth.

Still fuzzy on this though.




wrote in message
. ..
Hmmm.
So the "rise time" is just the time the signal moves from 10% to 90% and
is not referenced to the voltage?
Where "slew rate" is time referenced to the voltage just below clipping?

I have a commercially produced preamp that has a variable rise time
control on the front panel.
It allows you to adjust the rise time from: 2V/uS to 6V/uS. I'm trying to
verify exactly what it's doing.
It's supposed to emulate the older vintage preamps with the slow "rise
time" to the more modern preamps
with faster "rise times". It seems to me rise time has got to be connected
to frequency response somehow.
If the rise time is too slow to reproduce the high frequencies.



"Phil Allison" wrote in message
...



Hi Group:
I'm trying to come up with a slew rate measurement (10/90 rise time) of
a tube preamp I'm working on.


** No can do.

Rise time and slew rate are *independent * parameters.

The former is a small signal parameter and the latter a large signal one.

Both can be measured using a fast square wave at the input, but it is two
separate tests with two separate results.



I've read all the books I have on the subject but none of them tell what
frequency square wave
I should use or at what output level. My gut tells me just below the
clipping point for the level
and maybe 2 KHZ for the frequency. Using these settings I get a 10/90%
of 25V in 3.97uS.
25V/3.97uS gives me 6.29 V/uS rise time. Does this check out with you
guys?


** NO !!

Rise time is just a TIME - ie so many uS, it is related to the small
signal bandwidth and is a LINEAR parameter.

OTOH "slew rate" is a NON LINEAR parameter related to internal
capacitances in an amplifying stage that must be charged and discharged
with each cycle.

YOU have it *horribly* mixed together.


1. For the "rise time" test, use a low input level and a high enough
frequency to make reading the value off the scope screen easy - ie 10%
to 90 % in so many uS.

2. For the "slew rate" test, use a level that remains below clipping and
an input frequency high enough that the normally gracefully curved rises
and falls turn into STRAIGHT LINES !!!

Then read the up and down slopes in V/uS - they may not be the same.




....... Phil

[Phil Allison]

Hmmm.
So the "rise time" is just the time the signal moves from 10% to 90% and
is not referenced to the voltage?
Where "slew rate" is time referenced to the voltage just below clipping?


** Read my post again - you still have it all wrong.


I have a commercially produced preamp that has a variable rise time
control on the front panel.
It allows you to adjust the rise time from: 2V/uS to 6V/uS. I'm trying to
verify exactly what it's doing.
It's supposed to emulate the older vintage preamps with the slow "rise
time" to the more modern preamps
with faster "rise times". It seems to me rise time has got to be connected
to frequency response somehow.
If the rise time is too slow to reproduce the high frequencies.


** Read my post - I said exactly that.

To a good aproximation, rise time = 1 / pi x BW


BTW Only assholes top post.



........ Phil



"Phil Allison" wrote in message
...



Hi Group:
I'm trying to come up with a slew rate measurement (10/90 rise time) of
a tube preamp I'm working on.


** No can do.

Rise time and slew rate are *independent * parameters.

The former is a small signal parameter and the latter a large signal one.

Both can be measured using a fast square wave at the input, but it is two
separate tests with two separate results.



I've read all the books I have on the subject but none of them tell what
frequency square wave
I should use or at what output level. My gut tells me just below the
clipping point for the level
and maybe 2 KHZ for the frequency. Using these settings I get a 10/90%
of 25V in 3.97uS.
25V/3.97uS gives me 6.29 V/uS rise time. Does this check out with you
guys?


** NO !!

Rise time is just a TIME - ie so many uS, it is related to the small
signal bandwidth and is a LINEAR parameter.

OTOH "slew rate" is a NON LINEAR parameter related to internal
capacitances in an amplifying stage that must be charged and discharged
with each cycle.

YOU have it *horribly* mixed together.


1. For the "rise time" test, use a low input level and a high enough
frequency to make reading the value off the scope screen easy - ie 10%
to 90 % in so many uS.

2. For the "slew rate" test, use a level that remains below clipping and
an input frequency high enough that the normally gracefully curved rises
and falls turn into STRAIGHT LINES !!!

Then read the up and down slopes in V/uS - they may not be the same.



....... Phil

[Phil Allison]

"François Yves Le Gal"

Yup, slew rate is a large signal parameter and should be measured near
clipping.


** Slew rate can be measured at any output level that exhibits the effect
called "slewing".

Eg, slewing turns a sine wave input into a distinct triangle wave at the
output - the slopes of that triangle do not change with increased input
level or frequency.


Decades ago, Baxandall published in "Wireless World" that 2.2 Khz was the
"corner" frequency for slew rate measurements in audio applications. He
also
showed that a 100 w amp with a 0.5 V/µs SR was sufficient to properly
reproduce music from an LP.


** The two numbers are synonymous.

For a sine wave, the max Slew Rate is given by :

SR = 2 x pi. Vpeak x frequency


Since a 100 watt, 8 ohm amplifier outputs max 40 volts peak:

SR = 2 x pi x 40 x 2200

= 553,000 volts per second.

= 0.55 V/uS


Baxandalls' figure of 2200 Hz was derived from extensive testing with
recorded music programme, via LP.

He found just one LP in hundreds he tested had such high figure.

Certain CDs may be somewhat more demanding.

IME - commercial stereo amps all have at least 800% headroom over
Baxandall's modest criterion.



....... Phil

[Phil Allison]

"François Yves Le Gal"


** Slew rate can be measured at any output level that exhibits the effect
called "slewing".

Eg, slewing turns a sine wave input into a distinct triangle wave at the
output - the slopes of that triangle do not change with increased input
level or frequency.


Such slewing can only take place at high frequencies with very poorly
designed amps entering clipping.


** Dear oh dear oh dear ..........

Another posturing, know nothing, FROG ****WIT with no idea what "slew
rate" refers to.

Shame how every op-amp made and ever power amp has "slew rate " spec.

Go look it up - you PITA prick.






........ Phil

[Ruud Broens]

"François Yves Le Gal" wrote in message
...
: On Wed, 20 Sep 2006 23:42:40 -0500, wrote:
:
: I have a commercially produced preamp that has a variable rise time control
: on the front panel.
:
: Seems like yet another useless gizmo to me...
:
: It seems to me rise time has got to be connected
: to frequency response somehow.
:
: It is. A good approximation is F = 0.35 / Tr with Tr = risetime (s) and F =
: highest frequency (Hz).
:
....which is not difficult to derive:
take a mathematical sine wave graph, vertical axis amplitude
between -1 and 1,
horizontally from 0 to 2 pi degrees phase.
a frequency F means the 2 pi is completed in 1/F seconds.
momentary change in amplitude is largest where the derivative,
or cosine, is max, at the sine's zero crossings.
cos(0) = 1, so for an amplifier stage to perfectly reproduce
up to a highest frequency F, it should be capable of moving
from 0 to 1 in 1/2piF seconds and since cos(pi) = -1,
equally capable of doing that in the opposite direction.

wedding the mathematical and the practical, engineering,
is where it gets somewhat hazy.
as Tr is defined as the interval between 10 % and 90 %
of the amplitude, a perfect reproduction up to frequency F
would work out as a Tr= 0.8*2*1/2piF

-3dB point for a BW of F, or a pi/8 fase shift at F,
leads to Tr=0.8*sqrt(2)/piBW (or about 0.36/BW)

close to phil's Tr=1/piBW

or Tr=0.35/BW in Terman [1], for a signal with less than
5 % overshoot, empirically derived

mathRudy
[1] Terman - Electronic and Radio Engineering, 1955,
p 289

[Patrick Turner]

wrote:

Hi Group:
I'm trying to come up with a slew rate measurement (10/90 rise time) of a
tube preamp I'm working on.
I've read all the books I have on the subject but none of them tell what
frequency square wave
I should use or at what output level. My gut tells me just below the
clipping point for the level
and maybe 2 KHZ for the frequency. Using these settings I get a 10/90% of
25V in 3.97uS.
25V/3.97uS gives me 6.29 V/uS rise time. Does this check out with you guys?
If we made the measurement
at a lower output level, It really wouldn't tell us what the amp is capable
of delivering would it?
But the frequency of the square wave I'm not sure what to use.

Any Ideas??

RonL

PS
The square wave generator is a lab grade Tektronix with nS risetimes
so I'm not concerned about the generator error.

Phil Allison's reply to you about rise time and slew rate was right.

But usually when I test a preamp I aim to have the widest possible bandwidth
at the highest possible output voltage into a rated load.
This output voltage is usually more than 20 dB above the wanted output voltage.

So a typical line stage preamp with gain stage with 1/2 6SN7 driving a volume
pot of 100k
and then cathode follower output buffer with say enough shunt FB around the the
gain stage to reduce the
gain from say 16 to 5, then the bw will be from say 3Hz to over 80kHz into 50k
RL
and be about 50Vrms with slewing of sine waves beginning above 80kHz at about
40Vrms.
at 1Vrms output there is no slewing distortions at any F and the bw is only
very slightly wider.

I may often only ever have to produce 0.1Vrms of output to bring a power amp up

to normal levels of listening and there is no need to measure rise times
because the
unit has excellent dynamic headroom
and bw, and low thd, Rout 600 ohms, and low noise.
Square wave testing with my generator that can make nicely square looking waves

up to 20MHz will give a reasonably square looking wave up to
80kHz at a volt of output at least.

Patrick Turner.

[Phil Allison]

"François Yves Le Gaul Fascist "

Phil Allison

Another posturing, know nothing, FROG ****WIT with no idea what "slew
rate" refers to.

Why don't you **** off and die, ****head?


** Dear oh dear oh dear ..........

Another posturing, know nothing, FROG ****WIT with no idea what "slew
rate" refers to.

Shame how every op-amp made and ever power amp has "slew rate " spec.


Go look it up -


you PATHETIC, ****ING FASCIST PRICK.






........ Phil

[Patrick Turner]

"François Yves Le Gal" wrote:

On Wed, 20 Sep 2006 23:42:40 -0500, wrote:

I have a commercially produced preamp that has a variable rise time control
on the front panel.

Seems like yet another useless gizmo to me...

It seems to me rise time has got to be connected
to frequency response somehow.

It is. A good approximation is F = 0.35 / Tr with Tr = risetime (s) and F =
highest frequency (Hz).

Many amps will manage to make 100kHz of bw at 10Vrms of output.

And this without slewing of sine waves into triangular waves.

The rise time according to Phil A's quoted formula = 1 / ( 3.14 x 100,000 )
seconds
= 3.18uS.

If another amp was able to produce 50Vrms and the same bw without slewing
then the rise time would be the same, but the V/uS would be 5 times better than
the
amp able to make 10Vrms.

Music from either amp should sound identical if it is below the 10Vrms level.

Patrick Turner

Thanks all!
I think I'm clear on it now.
I appreciate your input.


RonL




"Patrick Turner" wrote in message
...


wrote:

Hi Group:
I'm trying to come up with a slew rate measurement (10/90 rise time) of a
tube preamp I'm working on.
I've read all the books I have on the subject but none of them tell what
frequency square wave
I should use or at what output level. My gut tells me just below the
clipping point for the level
and maybe 2 KHZ for the frequency. Using these settings I get a 10/90% of
25V in 3.97uS.
25V/3.97uS gives me 6.29 V/uS rise time. Does this check out with you
guys?
If we made the measurement
at a lower output level, It really wouldn't tell us what the amp is
capable
of delivering would it?
But the frequency of the square wave I'm not sure what to use.

Any Ideas??

RonL

PS
The square wave generator is a lab grade Tektronix with nS risetimes
so I'm not concerned about the generator error.

Phil Allison's reply to you about rise time and slew rate was right.

But usually when I test a preamp I aim to have the widest possible
bandwidth
at the highest possible output voltage into a rated load.
This output voltage is usually more than 20 dB above the wanted output
voltage.

So a typical line stage preamp with gain stage with 1/2 6SN7 driving a
volume
pot of 100k
and then cathode follower output buffer with say enough shunt FB around
the the
gain stage to reduce the
gain from say 16 to 5, then the bw will be from say 3Hz to over 80kHz into
50k
RL
and be about 50Vrms with slewing of sine waves beginning above 80kHz at
about
40Vrms.
at 1Vrms output there is no slewing distortions at any F and the bw is
only
very slightly wider.

I may often only ever have to produce 0.1Vrms of output to bring a power
amp up

to normal levels of listening and there is no need to measure rise times
because the
unit has excellent dynamic headroom
and bw, and low thd, Rout 600 ohms, and low noise.
Square wave testing with my generator that can make nicely square looking
waves

up to 20MHz will give a reasonably square looking wave up to
80kHz at a volt of output at least.

Patrick Turner.

[Ruud Broens]

"Ruud Broens" wrote in message
...
:
: "François Yves Le Gal" wrote in message
: ...
: : On Wed, 20 Sep 2006 23:42:40 -0500, wrote:
: :
: : I have a commercially produced preamp that has a variable rise time control
: : on the front panel.
: :
: : Seems like yet another useless gizmo to me...
: :
: : It seems to me rise time has got to be connected
: : to frequency response somehow.
: :
: : It is. A good approximation is F = 0.35 / Tr with Tr = risetime (s) and F =
: : highest frequency (Hz).
: :: :: :
: ...which is not difficult to derive:
: take a mathematical sine wave graph, vertical axis amplitude
: between -1 and 1,
: horizontally from 0 to 2 pi rad's phase.
: a frequency F means the 2 pi is completed in 1/F seconds.
: momentary change in amplitude is largest where the derivative,
: or cosine, is max, at the sine's zero crossings.
: cos(0) = 1, so for an amplifier stage to perfectly reproduce
: up to a highest frequency F, it should be capable of moving
: from 0 to 1 in 1/2piF seconds and since cos(pi) = -1,
: equally capable of doing that in the opposite direction.
:
: wedding the mathematical and the practical, engineering,
: is where it gets somewhat hazy.
: as Tr is defined as the interval between 10 % and 90 %
: of the amplitude, a perfect reproduction up to frequency F
: would work out as a Tr= 0.8*2*1/2piF
:
: -3dB point for a BW of F, or a 2pi/8 fase shift at F,
: leads to Tr=0.8*sqrt(2)/piBW (or about 0.36/BW)
:
: close to phil's Tr=1/piBW
:
: or Tr=0.35/BW in Terman [1], for a signal with less than
: 5 % overshoot, empirically derived
:
: mathRudy
: [1] Terman - Electronic and Radio Engineering, 1955,
: p 289
:
oops, sloppy errors corrected
that happens when it is great Autumn weather outdoors,
luring [lame exc inc]

anyway, on the errata topic,
i've posted some erroneous bits, on purpose,
to see if anyone is awake out there
lest the Google archive readers in a distant future
are set on the wrong foot (footing, feet?)

sleepers thread: 18046 is not equivalent to an E83F
hybrid thread: BC817 hfe ruler flat up to 60 mA (not 6)
hybrid thread: 2H distortion from loadline analyses is
actually 1/2 of what i mentioned ;-)

ok, maybe some others
here and there,
Rudy

[Joseph Meditz]

Ruud Broens wrote:
"François Yves Le Gal" wrote in message
...
: On Wed, 20 Sep 2006 23:42:40 -0500, wrote:
:
: I have a commercially produced preamp that has a variable rise time control
: on the front panel.
:
: Seems like yet another useless gizmo to me...
:
: It seems to me rise time has got to be connected
: to frequency response somehow.
:
: It is. A good approximation is F = 0.35 / Tr with Tr = risetime (s) and F =
: highest frequency (Hz).
:
...which is not difficult to derive:

Hi Ruud,

This proof is incorrect.

take a mathematical sine wave graph, vertical axis amplitude
between -1 and 1,
horizontally from 0 to 2 pi degrees phase.
a frequency F means the 2 pi is completed in 1/F seconds.
momentary change in amplitude is largest where the derivative,
or cosine, is max, at the sine's zero crossings.
cos(0) = 1, so for an amplifier stage to perfectly reproduce
up to a highest frequency F, it should be capable of moving
from 0 to 1 in 1/2piF seconds and since cos(pi) = -1,

The time it takes for a sine wave to go from 0 to 1 is one fourth its
period, T/4,
or 1/(4*F).

equally capable of doing that in the opposite direction.

wedding the mathematical and the practical, engineering,
is where it gets somewhat hazy.
as Tr is defined as the interval between 10 % and 90 %
of the amplitude, a perfect reproduction up to frequency F
would work out as a Tr= 0.8*2*1/2piF


Similarly, the time it takes for a sine wave to go from -1 to 1 is T/2,
or 1/(2*F).

Hence, Phill's approximation of Tr = 1(pi*BW) cannot be referring to
the time for the signal, such as a sine wave, to go from -1 to 1
because that would be shorter than that of the fastest sine wave, i.e.,
faster than 1/(2*F). Instead, the approximation is referring to the
response to a unit step which goes from 0 to 1.

Joe


-3dB point for a BW of F, or a pi/8 fase shift at F,
leads to Tr=0.8*sqrt(2)/piBW (or about 0.36/BW)

close to phil's Tr=1/piBW

or Tr=0.35/BW in Terman [1], for a signal with less than
5 % overshoot, empirically derived

mathRudy
[1] Terman - Electronic and Radio Engineering, 1955,
p 289

[Arny Krueger]

wrote in message
news
Hi Group:
I'm trying to come up with a slew rate measurement (10/90 rise time) of a
tube preamp I'm working on.
I've read all the books I have on the subject but none of them tell what
frequency square wave I should use

Something with a repetition rate that is large enough to give a solid
display on the scope. Pick too low of a frequency, and you'll get a very
faint trace at the sweept settings that you will have to use to see the slew
rate accurately. I'm thinking something in the 1 KHz - 10 KHz range.

or at what output level.

Just below clipping.

My gut tells me just below the clipping point for the level
and maybe 2 KHZ for the frequency.

That could work.

Using these settings I get a 10/90% of 25V in 3.97uS.

The problem is that you can't tell from just this data whether you're seeing
slew limiting, or linear low pass filtering.

If you're seeing just pure slew rate limiting cutting the amplitude to half
will give you the same slew rate.

If you are seeing just low pass filtering, the 10/90% time will be the same
and the slew rate will be cut in half.

25V/3.97uS gives me 6.29 V/uS rise time. Does this check out with you
guys?

The math seems about right.

If we made the measurement
at a lower output level, It really wouldn't tell us what the amp is
capable of delivering would it?

Not really.

But the frequency of the square wave I'm not sure what to use.

Any Ideas??

It would also be valid to measure the frequency response and deduce the
expected rise time from that.

Rise time is a linear property, and slew rate limiting is a nonlinear
property.

The rise time of a perfectly linear amplifier is always the same.

If an amp is slew limiting, the rise time will be doubled if you double the
amplitude.

PS
The square wave generator is a lab grade Tektronix with nS risetimes
so I'm not concerned about the generator error.

This generator should be OK.

[Eeyore]

Arny Krueger wrote:

wrote

Using these settings I get a 10/90% of 25V in 3.97uS.

The problem is that you can't tell from just this data whether you're seeing
slew limiting, or linear low pass filtering.

If you're seeing just pure slew rate limiting cutting the amplitude to half
will give you the same slew rate.

If you are seeing just low pass filtering, the 10/90% time will be the same
and the slew rate will be cut in half.

And that low-pass filtering could even be a passive filter on the input. You
can't tell that by measuring alone. In fact it seems to me that all this fancy
control is doing is simply to adjust the -3dB point downwards.


Graham

[Ruud Broens]

"Joseph Meditz" wrote in message
ups.com...

Ruud Broens wrote:
"François Yves Le Gal" wrote in message
...
: On Wed, 20 Sep 2006 23:42:40 -0500, wrote:
:
: I have a commercially produced preamp that has a variable rise time control
: on the front panel.
:
: Seems like yet another useless gizmo to me...
:
: It seems to me rise time has got to be connected
: to frequency response somehow.
:
: It is. A good approximation is F = 0.35 / Tr with Tr = risetime (s) and F =
: highest frequency (Hz).
:
...which is not difficult to derive:

Hi Ruud,

This proof is incorrect.

take a mathematical sine wave graph, vertical axis amplitude
between -1 and 1,
horizontally from 0 to 2 pi degrees phase.
a frequency F means the 2 pi is completed in 1/F seconds.
momentary change in amplitude is largest where the derivative,
or cosine, is max, at the sine's zero crossings.
cos(0) = 1, so for an amplifier stage to perfectly reproduce
up to a highest frequency F, it should be capable of moving
from 0 to 1 in 1/2piF seconds and since cos(pi) = -1,

The time it takes for a sine wave to go from 0 to 1 is one fourth its
period, T/4,
or 1/(4*F).
........................
....so ?
it does so in a non linear fashion, called a sine function
which changes fastest at the zero crossings - *that* is what is required,
not the average over one quarter period, eh ?

equally capable of doing that in the opposite direction.

wedding the mathematical and the practical, engineering,
is where it gets somewhat hazy.
as Tr is defined as the interval between 10 % and 90 %
of the amplitude, a perfect reproduction up to frequency F
would work out as a Tr= 0.8*2*1/2piF


Similarly, the time it takes for a sine wave to go from -1 to 1 is T/2,
or 1/(2*F).

Hence, Phill's approximation of Tr = 1(pi*BW) cannot be referring to
the time for the signal, such as a sine wave, to go from -1 to 1
because that would be shorter than that of the fastest sine wave, i.e.,
faster than 1/(2*F). Instead, the approximation is referring to the
response to a unit step which goes from 0 to 1.

Joe

........................
you also seem to miss out on things like the difference between bandwidth and
determining Tr required for a given F, Tr being defined as between 10 & 90
perc...

read again
R.

[Joseph Meditz]

Ruud Broens wrote:
"Joseph Meditz" wrote in message
ups.com...

Ruud Broens wrote:
"François Yves Le Gal" wrote in message
...
: On Wed, 20 Sep 2006 23:42:40 -0500, wrote:
:
: I have a commercially produced preamp that has a variable rise time control
: on the front panel.
:
: Seems like yet another useless gizmo to me...
:
: It seems to me rise time has got to be connected
: to frequency response somehow.
:
: It is. A good approximation is F = 0.35 / Tr with Tr = risetime (s) and F =
: highest frequency (Hz).
:
...which is not difficult to derive:

Hi Ruud,

This proof is incorrect.

take a mathematical sine wave graph, vertical axis amplitude
between -1 and 1,
horizontally from 0 to 2 pi degrees phase.
a frequency F means the 2 pi is completed in 1/F seconds.
momentary change in amplitude is largest where the derivative,
or cosine, is max, at the sine's zero crossings.
cos(0) = 1, so for an amplifier stage to perfectly reproduce
up to a highest frequency F, it should be capable of moving
from 0 to 1 in 1/2piF seconds and since cos(pi) = -1,

The time it takes for a sine wave to go from 0 to 1 is one fourth its
period, T/4,
or 1/(4*F).
.......................
...so ?

So, take another look at what you wrote which I'll quote. I've added
parentheses.

" ...so for an amplifier stage to perfectly reproduce up to a highest
frequency F, it should be capable of moving from 0 to 1 in 1/(2piF)
seconds "

The units here are _seconds_. The time it takes for the sine wave to go
from 0 to 1 is not 1/(2*pi*F). It can't be because that would be
faster than the highest frequency. The correct answer is 1/(4*F).
Later you took your _erroneous_ value, doubled it, took 80% of it and
showed it agreed with Phil's approximation. Good grief!

it does so in a non linear fashion, called a sine function
which changes fastest at the zero crossings - *that* is what is required,
not the average over one quarter period, eh ?

We are talking about rise _time_, Tr, which has units of seconds and
not slew _rate_ which has units of V/s. What is required here is the
time it takes for a step function to go between two points. And when
put in a band limited system it will follow a trajectory. And its rate
of change V/s over that trajectory will not be constant. But that is
not of concern for rise _time_.

Apparently you are as confused as the OP.

Joe


equally capable of doing that in the opposite direction.

wedding the mathematical and the practical, engineering,
is where it gets somewhat hazy.
as Tr is defined as the interval between 10 % and 90 %
of the amplitude, a perfect reproduction up to frequency F
would work out as a Tr= 0.8*2*1/2piF


Similarly, the time it takes for a sine wave to go from -1 to 1 is T/2,
or 1/(2*F).

Hence, Phill's approximation of Tr = 1(pi*BW) cannot be referring to
the time for the signal, such as a sine wave, to go from -1 to 1
because that would be shorter than that of the fastest sine wave, i.e.,
faster than 1/(2*F). Instead, the approximation is referring to the
response to a unit step which goes from 0 to 1.

Joe

.......................
you also seem to miss out on things like the difference between bandwidth and
determining Tr required for a given F, Tr being defined as between 10 & 90
perc...

read again
R.

[[email protected]]

On Thu, 21 Sep 2006 12:47:52 +0200, François Yves Le Gal
wrote:

On Thu, 21 Sep 2006 18:27:09 +1000, "Phil Allison"
wrote:

** Slew rate can be measured at any output level that exhibits the effect
called "slewing".

Eg, slewing turns a sine wave input into a distinct triangle wave at the
output - the slopes of that triangle do not change with increased input
level or frequency.

Such slewing can only take place at high frequencies with very poorly
designed amps entering clipping.


Not really... slewing is the effect of the amp not being able to increase its
output voltage as fast as the input V x Gain. Therefore, a slewing 'problem' in
an amp can be caused by too high a frequency at a low level, or too high a
signal output requirement at a lower frequency.

IE your amp might output 10khz for an output of 1 watt no problem, but when
trying to output 1000 watts, the amp can't "slew" fast enough to output that
much voltage.

A 'step input' test can determine maximum slewing, note that the amp has no true
feedback during a step slew test, it is basically uncontrolled at that time, the
feedback signal having it's own slewing problem. Generally, the feedback path
should slew faster then the amp.

[Eeyore]

wrote:

On Thu, 21 Sep 2006 12:47:52 +0200, François Yves Le Gal
wrote:

Such slewing can only take place at high frequencies with very poorly
designed amps entering clipping.

Not really... slewing is the effect of the amp not being able to increase its
output voltage as fast as the input V x Gain. Therefore, a slewing 'problem' in
an amp can be caused by too high a frequency at a low level, or too high a
signal output requirement at a lower frequency.

The source of the limit on slew rate is normally an internal capacitor which forms
part of the overall gain / phase setting characteristics.

It's commonly driven from a stage that operates in constant current mode so the slew
rate is determined by......

SR = I/C.

Suppose that capacitor is 470pf and the current is 5mA.

SR = 5.10^-3/470.10^-12 = 1.06.10^7 Volts/second = 10.6 Volts/ microsecond.

Graham

[Phil Allison]

"François Yves Le Gaul Bloody Fascist "


But please read again what I've written:


** Why ??

It was absolute ********.


Such slewing can only take place at high frequencies with very poorly
designed amps entering clipping.


** Still is.




......... Phil

[Ruud Broens]

"Joseph Meditz" wrote in message
oups.com...

Ruud Broens wrote:
"Joseph Meditz" wrote in message
ups.com...

Ruud Broens wrote:
"François Yves Le Gal" wrote in message
...
: On Wed, 20 Sep 2006 23:42:40 -0500, wrote:
:
: I have a commercially produced preamp that has a variable rise time
control
: on the front panel.
:
: Seems like yet another useless gizmo to me...
:
: It seems to me rise time has got to be connected
: to frequency response somehow.
:
: It is. A good approximation is F = 0.35 / Tr with Tr = risetime (s) and F
=
: highest frequency (Hz).
:
...which is not difficult to derive:

Hi Ruud,

This proof is incorrect.

take a mathematical sine wave graph, vertical axis amplitude
between -1 and 1,
horizontally from 0 to 2 pi degrees phase.
a frequency F means the 2 pi is completed in 1/F seconds.
momentary change in amplitude is largest where the derivative,
or cosine, is max, at the sine's zero crossings.
cos(0) = 1, so for an amplifier stage to perfectly reproduce
up to a highest frequency F, it should be capable of moving
from 0 to 1 in 1/2piF seconds and since cos(pi) = -1,

The time it takes for a sine wave to go from 0 to 1 is one fourth its
period, T/4,
or 1/(4*F).
.......................
...so ?

So, take another look at what you wrote which I'll quote. I've added
parentheses.

" ...so for an amplifier stage to perfectly reproduce up to a highest
frequency F, it should be capable of moving from 0 to 1 in 1/(2piF)
seconds "

The units here are _seconds_. The time it takes for the sine wave to go
from 0 to 1 is not 1/(2*pi*F). It can't be because that would be
faster than the highest frequency. The correct answer is 1/(4*F).
Later you took your _erroneous_ value, doubled it, took 80% of it and
showed it agreed with Phil's approximation. Good grief!


................
good grief, indeed,
i thought it was clear, apparently not to all. doubling comes from taking
full swing instead of upwards half, 80 % comes from the *definition*
of Tr, 1/2piF instead of 1/4F because it has to comply with the greatest
rate of change, hence i used the derivative. sqrt(2) comes in because
instead of perfect reproduction, a 3 dB drop at F is allowed for (BW def)
what is unclear ?
not to mention, the result is within 3 % of the empirically derived law
known since the late fourties...
................

it does so in a non linear fashion, called a sine function
which changes fastest at the zero crossings - *that* is what is required,
not the average over one quarter period, eh ?

We are talking about rise _time_, Tr, which has units of seconds and
not slew _rate_ which has units of V/s. What is required here is the
time it takes for a step function to go between two points. And when
put in a band limited system it will follow a trajectory. And its rate
of change V/s over that trajectory will not be constant. But that is
not of concern for rise _time_.

Apparently you are as confused as the OP.

Joe

.......................
you also seem to miss out on things like the difference between bandwidth and
determining Tr required for a given F, Tr being defined as between 10 & 90
perc..., etc.

read again
R.

[Joseph Meditz]

Ruud Broens wrote:
"Joseph Meditz" wrote in message
oups.com...

Ruud Broens wrote:
"Joseph Meditz" wrote in message
ups.com...

Ruud Broens wrote:
"François Yves Le Gal" wrote in message
...
: On Wed, 20 Sep 2006 23:42:40 -0500, wrote:
:
: I have a commercially produced preamp that has a variable rise time
control
: on the front panel.
:
: Seems like yet another useless gizmo to me...
:
: It seems to me rise time has got to be connected
: to frequency response somehow.
:
: It is. A good approximation is F = 0.35 / Tr with Tr = risetime (s) and F
=
: highest frequency (Hz).
:
...which is not difficult to derive:

Hi Ruud,

This proof is incorrect.

take a mathematical sine wave graph, vertical axis amplitude
between -1 and 1,
horizontally from 0 to 2 pi degrees phase.
a frequency F means the 2 pi is completed in 1/F seconds.
momentary change in amplitude is largest where the derivative,
or cosine, is max, at the sine's zero crossings.
cos(0) = 1, so for an amplifier stage to perfectly reproduce
up to a highest frequency F, it should be capable of moving
from 0 to 1 in 1/2piF seconds and since cos(pi) = -1,

The time it takes for a sine wave to go from 0 to 1 is one fourth its
period, T/4,
or 1/(4*F).
.......................
...so ?

So, take another look at what you wrote which I'll quote. I've added
parentheses.

" ...so for an amplifier stage to perfectly reproduce up to a highest
frequency F, it should be capable of moving from 0 to 1 in 1/(2piF)
seconds "

The units here are _seconds_. The time it takes for the sine wave to go
from 0 to 1 is not 1/(2*pi*F). It can't be because that would be
faster than the highest frequency. The correct answer is 1/(4*F).
Later you took your _erroneous_ value, doubled it, took 80% of it and
showed it agreed with Phil's approximation. Good grief!


...............
" good grief, indeed,
i thought it was clear, apparently not to all. doubling comes from
taking
full swing instead of upwards "

Yes, I understand. But, Phil's approximation is for the case of going
from 0 to 1. For a fair comparison to your value, you would have to
double Phil's as well.

"half, 80 % comes from the *definition*"
No argument there.

" of Tr, 1/2piF instead of 1/4F because it has to comply with the
greatest
" rate of change, hence i used the derivative. "

So, you took the derivative of a sine wave of unity amplitude and
frequency F. Yes, the derivative of that is 2piF. But, that is the
derivative for one particular amplitude, namely, unity. Any other
amplitude will give a different derivative for that same F. So, this
is far from being a general case.

Then you took that derivative whose units are rad/sec and flipped it
over and called it the rise time Tr giving rise time units of sec/rad.
This is incorrect.

Lastly, the derivative has nothing to do with rise time. For a band
limited signal, the rise time for a 1 volt step is the same as the rise
time for a .01 volt step even though the signals do not have the same
maximum slope.

I'm sorry to say that your derivation is a mess.

Joe

sqrt(2) comes in because
instead of perfect reproduction, a 3 dB drop at F is allowed for (BW def)
what is unclear ?
not to mention, the result is within 3 % of the empirically derived law
known since the late fourties...
...............

it does so in a non linear fashion, called a sine function
which changes fastest at the zero crossings - *that* is what is required,
not the average over one quarter period, eh ?

We are talking about rise _time_, Tr, which has units of seconds and
not slew _rate_ which has units of V/s. What is required here is the
time it takes for a step function to go between two points. And when
put in a band limited system it will follow a trajectory. And its rate
of change V/s over that trajectory will not be constant. But that is
not of concern for rise _time_.

Apparently you are as confused as the OP.

Joe

.......................
you also seem to miss out on things like the difference between bandwidth and
determining Tr required for a given F, Tr being defined as between 10 & 90
perc..., etc.

read again
R.

[[email protected]]

On Fri, 22 Sep 2006 09:15:52 +0200, François Yves Le Gal
wrote:

On Thu, 21 Sep 2006 21:07:51 -0400, wrote:

Therefore, a slewing 'problem' in
an amp can be caused by too high a frequency at a low level, or too high a
signal output requirement at a lower frequency.

Agreed.

But please read again what I've written:
Such slewing can only take place at high frequencies with very poorly
designed amps entering clipping.

um... Phil said that slewing in an amp can be shown to affect a sine wave by
turning it into a triangle... actually the triangle is a slope representing the
rise time of the amp, and for large signals will be a fixed large ramp... you
don't need a sine wave input of course, since the amp will turn EVERYTHING above
the slew rate into that slope! Also, the fall time slope may be of a different
integration, if that's the right word...

So your statement that it affects only high frequencies is wrong, any and every
design has a max slew rate, and clipping doesn't have a thing to do with it!

Perhaps it's a language problem? Si vous demandez en Francais dans un newsgroup
Francais, c'est possible que vous le comprend plus bien!

[[email protected]]

On Fri, 22 Sep 2006 02:18:21 +0100, Eeyore
wrote:



wrote:

On Thu, 21 Sep 2006 12:47:52 +0200, François Yves Le Gal
wrote:

Such slewing can only take place at high frequencies with very poorly
designed amps entering clipping.

Not really... slewing is the effect of the amp not being able to increase its
output voltage as fast as the input V x Gain. Therefore, a slewing 'problem' in
an amp can be caused by too high a frequency at a low level, or too high a
signal output requirement at a lower frequency.

The source of the limit on slew rate is normally an internal capacitor which forms
part of the overall gain / phase setting characteristics.

It's commonly driven from a stage that operates in constant current mode so the slew
rate is determined by......

SR = I/C.

Suppose that capacitor is 470pf and the current is 5mA.

SR = 5.10^-3/470.10^-12 = 1.06.10^7 Volts/second = 10.6 Volts/ microsecond.

Graham

Interesting... could you say the amp has become an integrator? Do you need to
calculate each stage or is there a way to do the whole amp?

[Eeyore]

wrote:

On Fri, 22 Sep 2006 09:15:52 +0200, François Yves Le Gal
wrote:

On Thu, 21 Sep 2006 21:07:51 -0400, wrote:

Therefore, a slewing 'problem' in
an amp can be caused by too high a frequency at a low level, or too high a
signal output requirement at a lower frequency.

Agreed.

But please read again what I've written:
Such slewing can only take place at high frequencies with very poorly
designed amps entering clipping.

um... Phil said that slewing in an amp can be shown to affect a sine wave by
turning it into a triangle... actually the triangle is a slope representing the
rise time of the amp

Not AIUI. The slew rate is determined by the charge and discharge of typically one
critical internal node under large-signal conditions.

Rise time is a 'small-signal' parameter.

Graham

[Eeyore]

wrote:

On Fri, 22 Sep 2006 02:18:21 +0100, Eeyore wrote:
wrote:

On Thu, 21 Sep 2006 12:47:52 +0200, François Yves Le Gal
wrote:

Such slewing can only take place at high frequencies with very poorly
designed amps entering clipping.

Not really... slewing is the effect of the amp not being able to increase its
output voltage as fast as the input V x Gain. Therefore, a slewing 'problem' in
an amp can be caused by too high a frequency at a low level, or too high a
signal output requirement at a lower frequency.

The source of the limit on slew rate is normally an internal capacitor which forms
part of the overall gain / phase setting characteristics.

It's commonly driven from a stage that operates in constant current mode so the slew
rate is determined by......

SR = I/C.

Suppose that capacitor is 470pf and the current is 5mA.

SR = 5.10^-3/470.10^-12 = 1.06.10^7 Volts/second = 10.6 Volts/ microsecond.

Graham

Interesting... could you say the amp has become an integrator?

No.

Do you need to
calculate each stage or is there a way to do the whole amp?

It's typically one section that limits before anywhere else and that sets the figure.

Graham

[Phil Allison]

um... Phil said that slewing in an amp can be shown to affect a sine wave
by
turning it into a triangle... actually the triangle is a slope
representing the
rise time of the amp,


** No way !

Rise time is a TIME interval !!

Slew rate is a RATE of voltage change.



......... Phil

[Chris Hornbeck]

On Sat, 23 Sep 2006 12:12:49 +1000, "Phil Allison"
wrote:

Rise time is a TIME interval !!

Slew rate is a RATE of voltage change.

Right; or alternatively a time rate of change of
current into a node (for single dominant
pole amplifiers).

Your point about keeping discussions of rise time
and slew rate separated can't be over-emphasized.

Much thanks, as always,

Chris Hornbeck

[Phil Allison]

"François Yves Le Gal"

So your statement that it affects only high frequencies is wrong, any and
every
design has a max slew rate, and clipping doesn't have a thing to do with
it!

What part of

Such slewing can only take place at high frequencies with very poorly
designed amps entering clipping.

is unclear?


** All of it, apparently.

Since what it says is that only defective amp designs ever undergo slew rate
limiting.

Which is totally false.



Care to demonstrate that a properly designed amp running way below
clipping
exhibits slewing at low frequencies ?


** No need to show any such thing as that is NOT what you claimed.



SR = Max (DVout(t) / Dt)
Sr 0.5 V/µs for crappy amplifiers or opamps (such as the 741)

Now punch in values of DVout and Dt in a typical audio context.


** Been done by Mr Baxandall and posted in this thread.

Shame you have no interest in facts.



Perhaps it's a language problem?

It's not. At least not on my side.


** Le Gaul Stone has the same comprehension problem no matter what lingo he
posts in.




........ Phil

[[email protected]]

On Sat, 23 Sep 2006 12:12:49 +1000, "Phil Allison"
wrote:




um... Phil said that slewing in an amp can be shown to affect a sine wave
by
turning it into a triangle... actually the triangle is a slope
representing the
rise time of the amp,


** No way !

Rise time is a TIME interval !!

Slew rate is a RATE of voltage change.



........ Phil


Yes Phil,

RATE of change over a certain TIME period... as in 1 volt per microsecond...

picky picky

Show me a rate of change not involving any time...

[[email protected]]

On Sat, 23 Sep 2006 03:06:57 +0100, Eeyore
wrote:



wrote:

On Fri, 22 Sep 2006 09:15:52 +0200, François Yves Le Gal
wrote:

On Thu, 21 Sep 2006 21:07:51 -0400, wrote:

Therefore, a slewing 'problem' in
an amp can be caused by too high a frequency at a low level, or too high a
signal output requirement at a lower frequency.

Agreed.

But please read again what I've written:
Such slewing can only take place at high frequencies with very poorly
designed amps entering clipping.

um... Phil said that slewing in an amp can be shown to affect a sine wave by
turning it into a triangle... actually the triangle is a slope representing the
rise time of the amp

Not AIUI. The slew rate is determined by the charge and discharge of typically one
critical internal node under large-signal conditions.

Rise time is a 'small-signal' parameter.

Graham

yes yes I know, sheesh You guys really nit pick... your balls must be really
clean!

[Eeyore]

wrote:

Show me a rate of change not involving any time...

Bandwidth.

Graham

[Eeyore]

wrote:

On Sat, 23 Sep 2006 03:06:57 +0100, Eeyore wrote:
wrote:

um... Phil said that slewing in an amp can be shown to affect a sine wave by
turning it into a triangle... actually the triangle is a slope representing the
rise time of the amp

Not AIUI. The slew rate is determined by the charge and discharge of typically one
critical internal node under large-signal conditions.

Rise time is a 'small-signal' parameter.

Graham

yes yes I know, sheesh You guys really nit pick... your balls must be really
clean!

It's simply a case of knowing what you're talking about instead of bull****ting it !

Try taking a 3 yr course in pro-audio !

Graham

[Phil Allison]

um... Phil said that slewing in an amp can be shown to affect a sine
wave
by
turning it into a triangle... actually the triangle is a slope
representing the rise time of the amp,


** No way !

Rise time is a TIME interval !!

Slew rate is a RATE of voltage change.



RATE of change over a certain TIME period... as in 1 volt per
microsecond...


** Read what YOU wrote !

" actually the triangle is a slope representing the rise time of the amp,"


is completely WRONG.



.......... Phil

[Chris Hornbeck]

On Sun, 24 Sep 2006 03:19:01 +0100, Eeyore
wrote:

Show me a rate of change not involving any time...

Bandwidth.

Arf!

Maybe, to bring this discussion back nearer Earth,
we should start to draw distinctions between small
and large signal behavior.

Small signal is (assumed to be) linear; large signal
is non-linear. The distinction is critical to this
discussion because they're being conflated.

Linear, small signal, response means that the output
of a [black box] follows the input with only
minimum-phase differences. Non-linear, large signal,
response, includes amplitude distortions, new
frequency components not in the original signal.

So, in this framework, let's add in the difference
between rise-time and slewing. The former is a small
signal property and the latter is a large signal one.

Rise time is a property of linear devices; slewing
is a property of non-linear devices. Could, of course,
even be the same device at different signal levels.


For complete, definitive, historical, and exhausting
discussion, readers should find these:

_Audio_ magazine June, July, August 1979, Jung,
Stephens, Todd "An Overview of SID and TIM"

_Audio_ magazine February, March 1980, Cordell
"Another View of TIM"

and, for gluttons for punishment, Otala's earliest
papers:

IEEE Trans. 1970 no.3 p.234-239 "Transient Distortion
in Transistorized Audio Power Amplifiers"

JAES 1972 p.396-399 "Circuit Design Modifications for
Minimizing Transient Intermodulation Distortion in
Audio Amplifiers.

My copies of the old Otala papers are now effectively
lost (any kind souls out there?), but if no legitimate
source of the Jung or Cordell isn't available, serious
folks should contact me.


Much thanks, as always,

Chris Hornbeck

[Chris Hornbeck]

On Sun, 24 Sep 2006 12:48:22 +1000, "Phil Allison"
wrote:

** Read what YOU wrote !
" actually the triangle is a slope representing the rise time of the amp,"
is completely WRONG.

Could we begin by agreeing that there's no such critter
as "the rise time of the amp"?

There is a small signal time rate of change of output
in response to a step change input, and a large signal
time rate of change of output in response to a step
change of input.

Like all religious arguments, lotsa smoke, little fire.

Much thanks, as always,

Chris Hornbeck

[Eeyore]

Chris Hornbeck wrote:

Could we begin by agreeing that there's no such critter
as "the rise time of the amp"?

No.

Graham

[Phil Allison]

"Chris Hornbeck"

** Read what YOU wrote !
" actually the triangle is a slope representing the rise time of the
amp,"
is completely WRONG.

Could we begin by agreeing that there's no such critter
as "the rise time of the amp"?


** NO !!

Cos there IS - you ****ing half wit .




There is a small signal time rate of change of output
in response to a step change input,


** WRONG !!

Rise time is simply a TIME interval characteristic of a given amplifier.



....... Phil

[Chris Hornbeck]

On Sun, 24 Sep 2006 04:49:36 +0100, Eeyore
wrote:

Could we begin by agreeing that there's no such critter
as "the rise time of the amp"?

No.

Please reread the entire post, with emphasis on content.

Chris Hornbeck