Slew Rate Limiting... continued. - Page 3 - diyAudio
Go Back   Home > Forums > Amplifiers > Tubes / Valves

Tubes / Valves All about our sweet vacuum tubes :) Threads about Musical Instrument Amps of all kinds should be in the Instruments & Amps forum

Please consider donating to help us continue to serve you.

Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving
Reply
 
Thread Tools Search this Thread
Old 1st February 2003, 03:43 AM   #21
Joel is offline Joel  United States
diyAudio Member
 
Joel's Avatar
 
Join Date: Oct 2002
Location: Brooklyn, NY
Quote:
Originally posted by choky
...but rule of thumb from ol' good books is-driver stage must have at least 1/10 dissipation of driven stage.
Which books would those be?

The driver current needed has NOTHING whatsoever to do with the dissipation of the following stage.
  Reply With Quote
Old 1st February 2003, 08:39 AM   #22
Zen Mod is offline Zen Mod  Serbia
diyAudio Member
 
Zen Mod's Avatar
 
Join Date: Jan 2003
Location: ancient Batsch , behind Iron Curtain
Quote:
Originally posted by Joel


Which books would those be?

The driver current needed has NOTHING whatsoever to do with the dissipation of the following stage.
you say so;
I say different;
do you believe me or not-it's only your bussiness.
__________________
my Papa is smarter than your Nelson !
clean thread; Cook Book;PSM LS Cook Book;Baby DiyA ;Mighty ZM's Bloggg;Papatreasure;Papa...© by Mighty ZM
  Reply With Quote
Old 1st February 2003, 05:46 PM   #23
gpapag is offline gpapag  Greece
diyAudio Member
 
gpapag's Avatar
 
Join Date: Nov 2002
Location: Athens-Greece
Default A bit delayed reply

Joel
You asked me to shed some light to the "controversy"that you have regarding the problem with slew rate limit on your design. I have not done any SR calculations again, so i had to do my homework first. What i have found though, is, that in theory, both parties may be right. I will explain myself later on, but i think that you may check the validity of your calculations by a scope connected at the anode of the 71A and driving the 6f6 with a sinusoidal variable freq. generator.
You may set the freq. low, say 500Hz, increase the volume up to the point that the output starts to clip, slightly back it up, and then start increasing the freq. and observe the freq. at which you will see clipping at the output.This will be quite a valid method, if your design does not have a global NFB network.
Now to the path and some of the results of my homework which i would like to share with you. I always like to know where formulas derive from. Unfortunately i could not track the origin of the Slew Rate formulas in any book. So starting from the very essentials, i tried toback every step up with a previous equation. All that follows, is a compilation of formulas i found in many different books. There is the possibility that some (or all) of the logical matchings i did, be missleading. I tried to present them here in a trasparent way, so someone to be able to spot the mistake(s). Therefore,any comments are more than welcome.

Basic equations: (Please, be patient. After all, it is only twelve!)

(1) C=Q/V
(2) Q=A*t
(3) t=C*V/A
(4) i=Ie^(-t/T)
(5) v=V[1-e^(-t/T)]
(6) T=R*C
(7) S=Imax/C
(8) Ci=Cgk+[1+((mu*Rl)/(rp+Rl))]*Cgp
(9) e(t)=Vp*sin2pi*f*t
(10) de/dt=2pi*f*Vp*cos2pi*f*t
(11) S=Vp*2pi*f
(12) f=S/2pi*Vp

Two electrodes spaced apart, potentialy form a capacitor C.
Eq. (1) C=Q/V says that capacitance C (Farrads) is proportional to
charge Q (Coulomb) of the two electrodes, and inv. prop. to potential difference V(Volts) btn the two electrodes.
Eq. (2) Q=A*t "visualises" charge Q as the product of current flow A (Amperes) times time t (seconds).
If we apply eq (2) into eq. (1), we get eq. (3) t=C*V/A which shows the time that the capacitor needs to be charged. In plane words:
The higher the capacitance, the longer the charging time.
The higher the voltage the cap has to be "charged"up to, the longer the charging time.
The higher the current the charging source can supply, the shorter the charging time.
The value of the charging current i in respect to time t is given by
the exponential (e=2.718) equation (4) i=Ie^(-t/T). "I" shows the
initial current which is the max current the source can supply.
The value of the capacitor voltage v in respect to time t is given by
the eq. (5) v=V[1-e^(-t/T)]. "V" shows the final max voltage the cap will reach at an infinitive long time.
In both eq. (4) and (5), symbol "T" is what we know as the "Time Constant"
of a circuit, the product of the resistance times the capacitance of the circuit, shown by eq.(6) T=R*C
Eq.(7) S=Imax/C is the familiar eq. of the Slew Rate S. This eq. did not fall from the skies.It's origin is eq.(5) v=V[1-e^(-t/T)]. We simply practice the sort of magic Harry Potter has taught us. This is it:
We replace V by it's equivalent V=I*R and T by R*C. We get:
v=I*R[1-e^(-t/RC)]
Then, we differantiate the whole thing in respect to time. We get:
dv/dt=(I/C)*e^(-t/R*C).
Why did we perform the differentiation? Because we had in mind to form an equation that expresses the Rate of Change of the voltage. The dv/dt is exactly this. But this is not enough. We want an equation that expresses the Maximum Rate of Change of the Voltage.
We name this S(Slew Rate)and we have: S=[dv/dt]max=[(I/C)*e^(-t/RC)]max
We turn to some "limits" calculus and the previous eq. becomes:
S=Imax/C, which is indeed eq (7).
For this eq. to be of any practical use, we have to put some meaningful figures into it. To do this , we have to model the circuit into which this formula will be applied. The output resistance R of the driver, the max current Imax it can source, and the capacitance C it will charge.
Eq. (8) Ci=Cgk+[1+((mu*Rl)/(rp+Rl))]*Cgp describes the input capacitance Ci of the triode. From this we see that Cgk is treated as it is. But Cgp is multiplied by what is enclosed in the brackets. This is the "Miller capacitance, which can be many times that of the Cgk, depending on the rp, the Rl, but mostly the mu of the tube.
So pluging in the eq. (8) the data that Joel gave us for his 6F6-71A amp. namely Cgk=3.7pF, Cgp=4.7pF,mu=3 (and i figure rp=1750, Rl=4800) we get Ci=18.73pf.
Unfortunately, 6F6 has to charge not only Ci, but it's own Co, which is 13pF plus some strays dye to wiring of minimum 20pf. Adding all these, makes for a C=52pF
Putting this into eq.(7) S=Imax/C together with Joel's Imax=0.9mA, we get
S=17.4V/uS
How can we see where Joel's 71A will slew limit (suppose we are dealing with sinusoidal signal)?
Eq.(9) e(t)=Vp*sin2pi*f*t describes such a sinusoidal signal.
(Vp=peak voltage, pi=3.14)
Harry Potter again, and the rate of change of this signal is
de/dt=2pi*f*Vp*cos2pi*f*t -eq. (10)-,so the maximum slew rate for a sinusoidal signal, is: S=Vp*2pi*f which is eq. (11). Solving for f, we get eq. (12): f=S/2pi*Vp, which shows the freq. f above which the stage will slew limit.
Joel said that the output swing of the 71A is 80 volts. If this is meant to be Vpeak, then together with S=17.4, eq. (12) gives us f=34.6Khz.
I wish that this is correct. I have my doupts though, and this is why:
Eq. (7) S=Imax/C, is based on eq. (5) v=V[1-e^(-t/T)] (the value of the voltage v in respect to time t while the capacitor is charging). If in this equation we set t=T, we see that at the time t equal to R*C the capacitor will have reached 63.2% of the value that could reach, provided a very long charging time was available. (PS*) Some books commend that, appropriate charging is accomplished at t between 4T and 8T.
I made a small table showing the relation of t=xT, the cap voltage and the respective Sx
t=1T v=63.2%V e^-1 S=S1
t=2T v=86.46%V e^-2 S=S1/2.7
t=3T v=95.02%V e^-3 S=S1/7.4
t=4T v=98.17%V e^-4 S=S1/20
t=5T v=99.33%V e^-5 S=S1/55
t=6T v=99.75%V e^-6 S=S1/147
t=7T v=99.91%V e^-7 S=S1/409
t=8T v=99.97%V e^-8 S=S1/1226
We see that at t=3T (v=95%V) S is 1/7 of what we have assumed. I don't know which "t" knowledgable engineers are utilising on SR calculations. So Joel, i don't know who is correct and who is not.

Eq. 1-7 found in many standard electronics text books
Eq. 8 found in "Vacuum-tube & Semiconductor Electronics", Jacob Millman (1958) McGraw-Hill
Eq. 9-12 and the differentiation of eq.5, found in "Electronics"band of REA's "Problem Solvers" series, 1990, ISBN 0-87891-543-5

Regards
George

(PS*) There is one case in which, capacitor is fully charged in t=T. This happens when capacitor is charged with constant current i=I, that is, the value of the initial charging current.
  Reply With Quote
Old 1st February 2003, 07:35 PM   #24
diyAudio Member
 
Sch3mat1c's Avatar
 
Join Date: Jan 2003
Location: Milwaukee, WI
Send a message via ICQ to Sch3mat1c Send a message via AIM to Sch3mat1c
Yow!

I love algebra

Tim
  Reply With Quote
Old 1st February 2003, 11:28 PM   #25
EC8010 is offline EC8010  United Kingdom
diyAudio Moderator Emeritus
 
EC8010's Avatar
 
Join Date: Jan 2003
Location: Near London. UK
Default Re: A bit delayed reply

Quote:
Originally posted by gpapag

so the maximum slew rate for a sinusoidal signal, is: S=Vp*2pi*f which is eq. (11). Solving for f, we get eq. (12): f=S/2pi*Vp, which shows the freq. f above which the stage will slew limit.
Joel said that the output swing of the 71A is 80 volts. If this is meant to be Vpeak, then together with S=17.4, eq. (12) gives us f=34.6Khz.
I wish that this is correct. I have my doupts though,
You have derived the same equation as in James' earlier post, and it is correct. If you look at my earlier post, suggesting that we simply use Ohm's law and the reactance of the capacitor at the chosen frequency, you will see that they agree. At the instant that you substituted v=Vsin(2pi*f*t) you were constrained to a single frequency, and transient considerations became irrelevant.
__________________
The loudspeaker: The only commercial Hi-Fi item where a disproportionate part of the budget isn't spent on the box. And the one where it would make a difference...
  Reply With Quote
Old 1st February 2003, 11:33 PM   #26
diyAudio Senior Member
 
fdegrove's Avatar
 
Join Date: Aug 2002
Location: Belgium
Default SMART PEOPLE.

Hi,

Quote:
At the instant that you substituted v=Vsin(2pi*f*t) you were constrained to a single frequency, and transient considerations became irrelevant.
Yes,Sir.

Cheers,

\in HH style.
__________________
Frank
  Reply With Quote
Old 2nd February 2003, 07:36 AM   #27
gpapag is offline gpapag  Greece
diyAudio Member
 
gpapag's Avatar
 
Join Date: Nov 2002
Location: Athens-Greece
Originally posted byEC8010:
"At the instant that you substituted v=Vsin(2pi*f*t) you were constrained to a single frequency, and transient considerations became irrelevant".
----------------------------------------------------------------------------------
EC8010 thank you for the positive confirmation of some of the equations i printed. As far as the "constraining" to a single frequency, i think that this is unavoidable, if someone chooses an analytical approach. Except if you are implying that i constrained myself to sinusoidal signals and didn't pay any attention to other signal forms. If this is the case, i am sure that once a reliable and (conservative) way of calculating the SR limit for sinusoidal signals is agreed to be accepted, there will be a way to deal with non sinusoidal waveforms as well.
My main concern is still unanswered though: What "t" is prudent to be utilised in SR calculations, t=1RC, t=2RC, t=3RC......?
My opinion is that t=1RC is not enough. I opt for t=3RC. The price to pay is that S becomes 1/7.4 of the one that is currently accepted (?) as adequate. I would like your comments please
Regards
George
  Reply With Quote
Old 2nd February 2003, 10:55 AM   #28
EC8010 is offline EC8010  United Kingdom
diyAudio Moderator Emeritus
 
EC8010's Avatar
 
Join Date: Jan 2003
Location: Near London. UK
George,

what I meant was that by invoking sine waves, your answer correctly predicted the highest (sinusoidal) frequency that could be handled. In these days of sharply bandwidth-limited (digital) signals, we can be very precise, and say that if it copes with a 20kHz sinewave (CD) then that is perfectly adequate because there are no transients containing frequencies higher than that. If we use LP or SACD etc, then we might wish to allow for a higher frequency.

Considering voltage slew rates is an awkward way of looking at the problem, and is equivalent to replacing spark plugs via the big end. The problem is one of current. All we need to do is to calculate the reactance of our capacitor at the highest frequency of interest. We then find the highest voltage that could be applied without overloading the stage, and combine the two with Ohm's law to find the maximum current required. Because of the phase shift caused by a capacitor, this current is required when the applied voltage is zero. We then reach for the anode characteristics of the previous stage and their loadline, and plot the current swing above and below the operating point. If we combine those two points with the maximum voltage swing points, we have an elliptical loadline. We can now make an informed choice as to whether we need more quiescent current because we can look at the linearity of the stage as it delivers this current. I usually find that I want a lot more quiescent current in order to swing the (small) signal current with any degree of linearity.

The results using this approach tend to agree with the various "rules of thumb" that have been suggested - with the difference that our choice is far more informed.

I hope this clarifies matters.
__________________
The loudspeaker: The only commercial Hi-Fi item where a disproportionate part of the budget isn't spent on the box. And the one where it would make a difference...
  Reply With Quote
Old 4th February 2003, 02:29 PM   #29
diyAudio Member
 
Join Date: Nov 2002
Location: North Herts, UK
George,

Many thanks for your post. I have found that my amplifiers sound best with 5 times the calculated current drive available in the driver stage. If we assume in Class A that only half of the bias current level is available for drive in a driver stage - then I think this is inline with your three times RC suggestion...

For those who disagree then I recommend you try it and see...if it works for you great - if it doesn't then you have not lost anything by trying it... It is perfectly possible in any given system, listened too by an individual for the low current driver to be preferred...

ciao

James
  Reply With Quote
Old 4th February 2003, 06:22 PM   #30
Joel is offline Joel  United States
diyAudio Member
 
Joel's Avatar
 
Join Date: Oct 2002
Location: Brooklyn, NY
Quote:
Originally posted by James D.
...It is perfectly possible in any given system, listened too by an individual for the low current driver to be preferred...
Well, there really isn't any "low" or "high" current value - there is a certain amount needed to neutralize the capacitance in the tube, and that's it - no more, no less.

If you choose to use a driver with 5 times that value, be my guest - but any difference, or "improvement" in sound is due to the lower mu, and differing plate resistances the high current tubes enjoy, as opposed to low current + high Rp + high mu tubes.
I can't think of any situation where you could do a 1 to 1 switch and have a fair comparison. So, maybe we can't generalize like that.

But don't get me wrong - I like the idea of a triode strapped 6V6, or 6BQ5 as a driver stage - but just because it's different - we've proven mathmatically that one doesn't need the current.
  Reply With Quote

Reply


Hide this!Advertise here!
Thread Tools Search this Thread
Search this Thread:

Advanced Search

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off


Similar Threads
Thread Thread Starter Forum Replies Last Post
Slew rate calculations - how much do I need? peranders Solid State 232 20th November 2011 07:02 PM
IcePower slew rate Rikard Nilsson Class D 24 29th April 2009 05:39 AM
slew rate adjustment... chipmonster Parts 7 25th January 2008 11:01 AM
Slew rate value Luke123 Chip Amps 5 7th April 2005 01:30 AM
slew rate and power output. JBL Everything Else 2 16th September 2001 11:51 PM


New To Site? Need Help?

All times are GMT. The time now is 07:10 AM.


vBulletin Optimisation provided by vB Optimise (Pro) - vBulletin Mods & Addons Copyright © 2014 DragonByte Technologies Ltd.
Copyright ©1999-2014 diyAudio

Content Relevant URLs by vBSEO 3.3.2