Slew Rate Limiting... continued.
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Joel
diyAudio Member

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.

Zen Mod
Official Court Jester
diyAudio Member

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.
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 1st February 2003, 06:35 PM #24 Sch3mat1c   diyAudio Member     Join Date: Jan 2003 Location: Milwaukee, WI Yow! I love algebra Tim
EC8010
diyAudio Moderator Emeritus

Join Date: Jan 2003
Location: Near London. UK

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.
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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...

fdegrove
diyAudio Senior Member

Join Date: Aug 2002
Location: Belgium
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.
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Frank

 2nd February 2003, 06:36 AM #27 gpapag   diyAudio Moderator     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
 2nd February 2003, 09:55 AM #28 EC8010   diyAudio Moderator Emeritus     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...
 4th February 2003, 01:29 PM #29 James D.   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
Joel
diyAudio Member

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.

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