Date: Mon, 12 Jun 2000 23:44:55 EDT
Subject: Re: slew rate quandary
X-Mailer: AOL 3.0 16-bit for Windows sub 38
In a message dated 00-06-12 16:15:15 EDT, you write:
<< so steve, i've just about exhausted my mental energies on contemplating slew rate limiting in circuits.. as a result i'm pulling you in on the problem, and seeing if you can shed any light on it for me. >>
Hee hee hee. Been following that thread loosely. You and Ivan both pretty well hit it, but maybe the explanation needs a slightly different twist to make it understandable (or more precisely, differentiable) from interwoven phenomena. See below.
<< as i understand it, it is a simple matter to avoid slew rate limiting--you just insure you have enough current flowing in your driving stage to overcome the current requirements of downstream capacitance, taking into account not only the max frequency of interest, but also the max voltage swing at said frequency. >>
Simple in principle, maybe not so obvious to achieve in practice! I’ll try a slightly different approach below.  (numbers in brackets are referenced in the explanation below and relate to referenced section for that part of the phenomenon. Did that make sense?)
<< i also understand that it is a somewhat different entity than that of bandwidth limiting due to finite driving impedance, using the good old F = 1/2piRC relationship.>>
Actually, there’s some really odd interactions. 
<< thus you could have a high current stage with high driving impedance that was able to sink and source the current, but rolled off the high end due to this TC. conversely, you could have a low current stage, of low driving impedance, that was able to extend quite well into the HF region, but was unable to drive any sort of downstream capacitance without slewing like crazy due to limited drive current. >>
<< am i correct in separating these two effects, or is one contained in the other? i.e. can you also explain the slew rate limiting by calculating the driven impedance @ a particular frequency... i don't think you can, because that neglects the "memory" effect of the capacitance. it's not just a matter of voltage division, right? >>
The effects are different. The capacitance memory is not particularly the issue, it’s the current that’s the issue! 
<< i have also searched my texts, and the net, for some authoritative description of just how slew rate limiting is DIFFERENT than a simple HF rolloff, and found nothing... i started to reason that the HF rolloff type of bandwidth limiting is linear in that if you start with a sine wave, you'll preserve the waveshape (even if it may be phase shifted or altered in amplitude), whereas the slewing circuit produces a wave that isn't anything close to the same shape--the sine wave will be turned into a triangle. >>
Yep. Slew rate issues are “hand-waved” in virtually every text I own; it took solid experience to unravel what was alluded to in the texts.
<< from what i know, you do yourself well to make sure you've got about 5 times the max necessary slew rate current for your maximum HF allowed to pass through your amp. it's a somewhat arbitrary constant, but it works, from what i understand. >>
Reasonable rule of thumb. Not trying to be condescending, but do you understand *why* that rule works?
<< what i really need is an example of the two aforementioned scenarios, but i don't know where to start.. there's a bit of a fray over on RAT (gee, what a surprise) because someone started to associate NFB with slew rate limiting, and i pointed out you absolutely do NOT have to have NFB to have SRL. this met with much resistance, and i have done my best to defend my position with intuition, but i am getting nowhere. what i really need are some numbers to really screw things up!>>
I’ll try. See below.
If you take a voltage source (sine wave) ,and pass it through a R-C network, and vary the frequency of the input waveform, the output level will be a sine wave whose amplitude will decrease as the frequency is increased. The level will be 3dB down at the “pole” frequency f=1/(2*PI*R*C) and decrease at a 6dB/octave (20dB/decade) rate for frequencies above the pole. 
Now if I arbitrarily raise the amplitude of the input signal amplitude, the output at all frequencies will increase by the same proportion as I raised the input, and the waveform will remain sinusoid. THERE IS NO DISTORTION PRODUCING MECHANISM in my ideal case. However, there is an interesting caveat to this scenario. Let’s consider some real world sources. As I arbitrarily raise the input level, the generator is required to supply more and more power, since P=e*e/z. (real part of z actually). This gets into the difference between watts, volt-amps and vars; but that is a different discussion. However, all real world sources have a maximum power they can deliver, as well as maximum voltage. 
If I assume a very power limited generator, what will I notice happening at the output of my filter? At low signal levels, my “frequency response” will be just that predicted by the frequency, and R and C. However, at higher signal levels, the frequency response will be different (lower apparent cutoff frequency). In fact, you will note that at some high frequency, no matter how much I try to raise the level, the output level WILL NOT CHANGE. Yet if I lower the frequency, I will still see a change in output level as I change the source level. This is one manifestation of slew limiting. 
L------------------------------------\ .....................................\ E--------------------------------------\ ........................................\ V-----------------------------------------\ ...........................................\ E-----------------------------------------\..\ ...........................................\..\ L.............................................\..\ f r e q u e n c y
In this picture, the lower “E” and “V” levels are not slew limited, but the L and upper E levels are slew limited. Note that the “pole” of the V and lower E is the same frequency, whereas it gets progressively lower for the L and upper E curves. “V” is at the onslaught of slew limiting. [2, 4]
What’s happening is CURRENT related. To understand that, you have to recall the voltage-current relationship of a signal across a capacitor. This is:
i = C * dV/dt
In words, the current flowing in a capacitor is proportional to the time rate of change of voltage across the capacitor; the constant of proportionality is called the capacitance value. [There's another great side issue associated with this formula... if the capacitance varies with voltage... but that's again an entirely different topic].
If we re-arange this formula, we get
i/C = dV/dt
dV/dt is sometimes called slew rate.
If you like integral calculus rather than differential calculus, the relevent formula is
Vc = (1/c) * Integral I dt + K
which in words is Voltage across the capacitor is 1/c * integral from -infinity to infinity of the current flowing through the capacitor with respect to time, plus a constant (which is the DC component across the capacitor). Both formula say the same thing.
So let’s summarize to this point…
In a circuit containing a R-C (or just C for that matter), with an ideal voltage source (sinusoidal), and supplying unlimited current, the output will always be a sine wave. The bandwidth will be defined by the RC network as described above. There are no additional effects.
In a circuit with a limited current capability, the maximum available current controls the maximum time rate of change of voltage across the capacitor. 
How does this manifest itself?
Perhaps as a first example, lets apply a square wave to the input of our network. If we have no current limitations, the square wave will have exponentially rounded sides to it. We’ve all seen pictures like that. Why? Because the source acts as a true voltage source. If you were to measure the *current* produced by the generator, you would see a maximum at the “edge” of the square wave, diminishing “away” from the edges.
If however, our generator was current limited, we would see an altogether different waveform. Instead of an exponentially rounded waveform, we would see a constant slope edge (triangular) until (1) no more voltage was “called for” or (2) the required current dropped enough to be provided by the source. This can be clearly seen in the integral form of the equation above… if I is constant (because it’s limited), then the integral is a linear ramp. [calc 101]
Now, Instead of the square wave, lets place a sine wave onto the network.
First, for high frequency signals (near, at, or above the “pole”), where is the maximum current flowing in that circuit? If you guessed at the amplitude peak, you guessed wrong. Why? Look at the formula. I = C dV/dt. For a sine wave, dV/dt [which is the SLOPE] is maximum at the zero crossings. Note that d(sin(x))/dt = cos(x). cos(x) is maximum at the SIN’s zero crossing.
[Aside... in a real world circuit, you also have to consider the effects of the resistances, capacitances and inductances to determine where the maximum current demand is... which, of course is frequency dependent].
Sooo, driving our network can produce a rather odd effect. Recall the square wave “turned into” a triangular wave as a “worst case” slew limit condition. Well, the sine wave ALSO turns into a triangular wave as a worst case slew limit condition.
REAL WORLD SLEW EFFECTS:
Let’s assume the source current and sink current capability are the same. [Which, strictly speaking is definitely *not* real world, but keeps the analysis simple]. Then at high frequencies, and low enough levels, the system is linear, and a sine wave in produces a sine wave output, neglecting distortion produced in the tubes, transistors or whatever active devices you care to consider. The frequency response will be limited by the effective R and C of the system. At higher levels, slew rate limiting can occur, if the current required to meet the I = C * dV/dt is not available. This will first be noticed at or near the zero crossings, since this is the highest slope condition. If even more level is asked for, this will distort more and more of the waveform. As the slew limiting distortion approaches the maximum amplitude of the sine wave, further increase in input level (demand) will not cause the output to increase; and the waveform will look sort of triangular, but with a somewhat rounded top.
Note that this differs from a simple clip condition where the top or bottom or both of the output waveform will be “clipped” or flattened. However, as we will see below, this can also occur.
Note that I did not need to invoke any feedback mechanism in order to cause slew limiting to occur. Well, does feedback affect slew limiting? You bet it does. Why? The feedback “error” signal will indicate to the system that “more” signal is needed, and the driver will merrily try to pump more signal into the slew limited node. This will cause the output to look even more triangular, and probably clipped as well, since the error signal will never “correct” the errant condition.
This is *very* similar to what happens in non-slew limiting condition when a feedback amplifier “clips”. This drives a high level signal into the clipping stage. In fact, this is THE one example where clipping AND feedback can actually induce slew limiting that might not otherwise occur in a non-feedback amplifier. [Since the amplifier may try to pump, say, 20dB more signal than normal into the node, 10x the slew rate is also being called for. This may slew limit the driving stage while the driven stage is clipping].
[Another interesting aside is the very high voltages that *might* be generated across the output transformer in a clip condition with feedback, but that's yet another topic].
Can we draw any conclusions?
1. Slew limiting is a different effect from R-C frequency response limitations.
2. Slew limiting is a different effect from clipping.
3. Slew limiting *can* occur whether or not feedback is present.
4. Feedback can worsen the effect of clipping.
5. Feedback can worsen the effect of slew limiting.
Hope that helps explain what’s going on.
Date: Tue, 13 Jun 2000 21:55:34 EDT
Subject: Re: slew rate quandary
X-Mailer: AOL 3.0 16-bit for Windows sub 38
In a message dated 00-06-13 19:30:16 EDT, you write:
<< thanks. am digesting. >>
Let me know what doesn’t make sense or is unclear. That’s one of those topics that is easy to oversimplify and overcomplicate at the same time.
Incidentally, note that by current starving a capacitive circuit, you end up with a “design issue”. The electrical “dual” to this issue is “voltage starving” an inductive circuit … but no one brings that one up! [ Remember, E = L di/dt and it's integral twin ]
<<>Reasonable rule of thumb. Not trying to be condesending, but do you
>understand *why* that rule works?
none taken… well, no, i don’t know WHY that rule works. you’ll end up with 5 times as much headroom–or the ability to pass a signal 5 times louder, or 5 times higher in pitch. beats me why 5 though. why not 7, for good measure? you’ll be able to charge more >>
That’s almost lost to urban legend status. You exactly hit the nail on the head. It IS headroom. As Ivan pointed out, up to *exactly* the slew rate limiting point, there is *no* slew limiting; that is, the phenomenon either exists or it doesn’t. There’s no such thing as “partial” degradation due to “almost” slew limiting. Well, with *normal* program material in Hi-Fi kinds of applications, the peak to average ratio turns out to be more-or-less 14 dB (5x). Thus, to avoid slew limiting *peaks*, a 5x headroom fudge factor is usually sufficient. ‘Tis a “soft” number, though, so it is interpreted differently by lots of people. E.g., does that mean 100kHz ought to be slew limited point (5x 20kHz) -or- 5x voltage at 5x 20kHz (250kHz) ought to be the limit point etc? Or conversely, since you don’t want to -clip-, the average signal presented to the amp ought to to be 14dB below the amp rating, and thus it’s OK if you avoid slew limit conditions only as high a frequency as you expect to pass (20 or 30kHz). I’ve seen *reasonable* people interpret it all three ways (and a few additional variants).
7x or 10x are even better, of course, but usually overkill, and thus, in some sense, less “optimum” of a solution.
Again, let me know what seems obscure, and I’ll try to clarify it a bit more.