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This assertion sort of makes sense in what perhaps should best be described in an ass-backwards way. But the implied (or explicit) conclusion, that the bigger the filter capacitor the less current there is for the output tube, doesn't follow at all.[Redacted]
First of all, you have to draw a distinction between the source and the load. Power comes from the source and is consumed by the load. We design our circuits (by and large) to operate best from absolutely fixed voltage sources. Therefore, the goal of power supply design is to have as low an output impedance as possible.
To show why the argument in question is absurd, consider an ideal battery of zero internal resistance. This can be modeled as an infinite current source in series with a dead short circuit. By the reasoning presented, no current whatsoever would be available to a load connected across the battery terminals because all of it would flow in the battery's zero internal resistance.
This is just plain wrong. The "capacitor robs the output tube of current" argument follows from the same line of reasoning and is equally wrong.
Here's the real story. Current will always be available to flow to the output tube circuit as long as there is voltage at the output of the power supply. It doesn't matter if the filter capacitor looks like a near-short from the dv/dt point of view. As long as there's, say, 350V across the capacitor terminals, electrons will flow to the output tube. Big, juicy gulps of electrons.
Now, if for some reason the power supply should start to run out electrons, then the voltage across the filter capacitor will drop. As the voltage drops, the number of electrons flowing per second (i.e., the current) out of the power supply will diminish. This is the definition of plate resistance -- for a fixed Vgk, plate current will be a function of Vpk. No mystery here.
So, hand-waving about current is irrelevant. What matters here is the ability of the power supply to maintain its rated output voltage under conditions of varying load. As long as the power supply maintains output voltage, the supply of electrons (hence, current) takes care of itself.
As others have pointed out, the nature of rectification is that the power transformer wants to supply electrons in spurts. We rely on the filter network to do two things: 1) to remove hum and other AC noise from the supply output, and, 2) to provide energy to the load when the transformer/rectifier is in between "spurts."
The bigger the frigging filter capacitor, the more electrons there are in the bank, and the longer it takes for the voltage to draw down, given a surge in demand.
Now, granted, once you have drawn down the voltage across the filter capacitor, all other things being equal, it will take a larger capacitor longer to recharge than a small one. On the other hand, for a given surge in load current, the amount the voltage will draw down will be correspondingly less with a big capacitor.
So, you have a compromise (all engineering is compromise). You can have a power supply that does a relatively poorer job of maintaining its output under changes in load, but that recovers more quickly when the load stabilizes. Or you can have a power supply that does a relatively better job of maintaining its output voltage, but which responds more slowly.
I don't think anyone has really justified why it is comparatively better to have a power supply whose output voltage flaps in the breeze. There are plenty of anecdotal reports from people who prefer the big, fat power supply approach.
The tendency of people to ignore information that is contrary to their beliefs is universal, but that doesn't make it right.
With respect to the slowing down of the power supply due to high reactance filter components, this is a perfectly well-understood principle of engineering. High ripple rejection equals low cutoff frequency equals slow transient response. There is no revelation here.
Let me repeat. The argument that proposes the filter capacitor as a shunt in parallel with and stealing current from the output tube is just plain bunk.
I encourage readers to disregard any arguments based on that premise.
Edits: 03/11/11 03/11/11Follow Ups:
I agree that by and large we design our circuits to operate from a fixed voltage source. Since we do not live in a perfect world and “Ideal” current and voltage sources do not exist, we strive to have as low output impedance as possible. So far, so good.
I take issue with your battery example. Yes an ideal battery would have zero internal impedance, but it would not be modeled as an infinite current source with a short across it. If it had a short across it, it could not develop any voltage potential across its terminals. Therefore it would cease to be a voltage source. Your battery example has no merit.
You are simplistically choosing to overlook the things that make my assertion true.
I’ll restate it here again “Big filter capacitors compete with output tube for current resources”
You wrote “Current will always be available to flow to the output tube circuit as long as there is voltage at the output of the power supply. It doesn't matter if the filter capacitor looks like a near-short from the dv/dt point of view. As long as there's, say, 350V across the capacitor terminals, electrons will flow to the output tube.”
Yes, as long as there is voltage present across the tube, current will flow. However, during the time were the capacitor is recovering charge current; the voltage present will be below the design point of the power supply until the capacitor is full again. A larger capacitor will take longer to recover to the design voltage than a smaller capacitor. If you don’t believe me then model two power supplies in PSUDII with one having half the capacitance for the final C and step the current from 150mA to 100mA then see which one stabilizes to the higher voltage the fastest.
Because a larger capacitor holds the output voltage lower for a longer amount of time, this prevents our output tube from operating at the circuit design point. This causes a downward shift in our operating point, thus lowering the current flow through are tube. Plain and simple, this is competition of power supply resources. Our output tube is not operating at the design point because the large capacitor is holding the supply voltage down while it charges.
If you want more available current to charge capacitors faster, then you need to reduce the series resistance of the power transformer and chokes. This would be like using a larger hose to fill a bucket. Once again, if you don’t believe me then model it in PSUDII. Model a low DCR power supply with the same capacitance and inductance values and another with more conventional DCR values for the power transformer and chokes. Step the current from 150mA to 100mA and see which one stabilizes to the higher voltage the fastest.
> I take issue with your battery example. Yes an ideal battery would have
> zero internal impedance, but it would not be modeled as an infinite
> current source with a short across it. If it had a short across it, it
> could not develop any voltage potential across its terminals. Therefore
> it would cease to be a voltage source. Your battery example has no merit.
Sorry, I can't agree with you here, JLH. I'm sure you know that the Thevenin model of an ideal battery is a fixed voltage source in series with zero source resistance. We can transform that into the equivalent Norton model, which gives us infinite current in parallel with zero resistance. It may seem non-intuitive, but in the limit, you certainly do end up with a finite output voltage.
If it makes you more comfortable, think of it as three million Amperes in parallel with one ten-thousandth Ohm. This is Circuits 101, and with all due respect, one of the first things you learn in the EE curriculum after Ohm's Law.
Your current divider argument is problematic. The current divider theorem is usually presented as a constant current being split between two parallel branches. But the upstream power supply components are not a constant current source. The source impedance is low and the available current is high. In a cap-input filter, for example, the input cap charging current is easily ten times the DC load current, but the power supply still functions perfectly.
Even your friend Jeff has posted more than once that power supply performance is improved by the "pseudo shunt regulation" created when you load the supply aggressively with a fixed resistor. This directly contradicts the "it's as simple as two resistors in parallel" argument recently put forth on this forum.
In the end, the bulk of your argument (not the conclusion) is consistent with my posting. The transient response of the power supply is a function of the topology and values of the components. The "current competition" argument, while it might be valid in some narrow way, is really irrelevant. The question is what the output impedance of the power supply is in the complex domain, and how that translates into frequency- and time-domain response. Little else matters.
Obviously, if the output voltage isn't absolutely fixed, the instantaneous operating point of the tube will vary. Since the tube isn't perfectly linear, there will be some intermodulation distortion. The power supply itself isn't perfectly linear, so this is a source of distortion as well. Otherwise, what you observe is just going to manifest as a small (or not-so-small, case depending) linear impedance in series with the amplifier.
Jeff has hinted in the past that you are an electrical engineer. If this is true than you and I should share a common understanding of basic principles and methods of analysis. I don't see much evidence of that in this discussion. An electrical engineer would have recognized the Norton model, for instance. Also, an electrical engineer would realize your argument leads to a logical contradiction. To wit: If a large parallel capacitor starves some current from the output tube, then an infinite capacitor will starve all of the current. Clearly this doesn't happen in the real world.
There are much more sophisticated and revealing approaches to analyzing the situation you are describing. Your presentation suggests that the supply is slewing, i.e., going into current limiting, which is definitely not the case.
You have also failed to justify, having asserted that variations in power supply voltage are bad, why it is preferable to design a power supply that deliberately impairs the voltage regulation in favor of faster transient response. I think this can only be answered in terms of an analysis of the amplifier circuitry itself, which to the best of my knowledge you have never undertaken.
To summarize, I stand on my previous assertion that your "current competition" claim makes very little sense. I don't believe there are too many people on this forum who are willing and qualified to judge our two positions. But, if you're really interested, I'd be happy to take it to the DiyAudio tube forum, or better, the solid-state forum, where the members have the credentials to comment on the merits of our respective cases.
-Henry
Thevenin and Norton equivalents are not very useful with zero output impedance. The correct statement for the voltage across the zero output resistance of the Norton equivalent shorted current source is "indefinite", not finite. I.e., the limit of V (as I goes to infinity and R goes to 0) is any value. Properly set up, the equations will give the load voltage equal to the open-circuit voltage for any finite load resistance, if the source impedance is literally zero.
Well, if you define
I(x) = 300 * x
Y(x) = x
then
V(x) = I(x) / Y(x)
where
I(x) is the Norton equivalent generator current
Y(x) is the Norton equivalent source conductance
V(x) is the Norton equivalent open-circuit load voltage
So, by L'Hôpital's rule
lim(x-> inf) V(x) = lim(x-> inf) I'(x) / Y'(x) = 300 / 1 = 300
Which is not to say that you need L'Hôpital's rule to solve this one, but the general point holds.
Maybe that's too simplistic (I suck at math), but I think anyone well enough versed in math to care about this distinction should also recognize that envisioning what happens in the limit is a common engineering analytical technique, and it doesn't mean we are actually proposing real-world solutions that approach those limits...
-Henry
As I said, this calculation is correct in the limit.
However, I stand by my remark that a Norton equivalent with zero impedance is not a useful model, nor is a Thevenin equivalent with infinite impedance.
Well, I'm afraid we'll have to agree to disagree. It seems to me that if the conditions of L'Hôpital's rule are met, the open-circuit voltage is defined exactly. And the concept itself does seem useful (to me, anyway) in helping to show why the "two parallel resistors" argument fails in this case.
Infinity is not a real number, so an infinite capacitor is undefined, too, as is an infinite resistor. That doesn't mean it isn't useful to imagine what happens to a circuit if a component value approaches this limit.
If I had the time and resources, I would build it and see for myself, ha, ha.
As I said, you can always set 'x' to a million, give or take a couple orders of magnitude, if you're uncomfortable with limits.
-Henry
----To wit: If a large parallel capacitor starves some current from the output tube, then an infinite capacitor will starve all of the current. Clearly this doesn't happen in the real world.
Now I think it would; an infinite capacitor *WOULD* starve the tube of all current( and deliver no voltage ). Such a capacitor would behave like a full short circuit in parallel with the load.
Getting such a capacitor would not be possible ov course, not to mention the designer would fail to spec such a marginal power trans that could not charge what ever capacitor got selected. In practice, the supply will deliver voltage( and thus current) in the fashion you described...but your example is not so water-tight as you might have originaly thought...:)
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
> Now I think it would; an infinite capacitor *WOULD* starve the tube of
> all current( and deliver no voltage ). Such a capacitor would behave like
> a full short circuit in parallel with the load.
I see your point, but again there is a satisfactory engineering answer.
It's enough to shoot down the example to say (as you have) that an infinite capacitor is impossible. But if we imagine such a capacitor, we can also imagine giving it an infinite amount of time to acquire the infinite charge required to build up the needed B+ voltage across its terminals. Once the steady-state has been reached, we can talk about the AC behavior. And in that case, the capacitor would behave like a perfect battery.
-Henry
Of course there is a satisfying answer!!! that is afterall the whole thesis/driving reason behind this whole discussion.
Now do be careful with terms like 'engineering' and 'scientific'...many marketing departments have worked tirelessly to hijack their meaning.
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
> > I’ll restate it here again “Big filter capacitors compete with output tube for current resources” < <
> > Yes, as long as there is voltage present across the tube, current will flow. However, during the time were the capacitor is recovering charge current; the voltage present will be below the design point of the power supply until the capacitor is full again. A larger capacitor will take longer to recover to the design voltage than a smaller capacitor < <
This is a case of comparing apples with oranges. A "large capacitor" or a "small capacitor" ... compared to what?
Assuming that a small capacitor means one that discharges appreciably with load changes compared to a large one that doesn't, this becomes a case of swings and roundabouts. What is best, for the cap voltage to drop appreciably with load transients but recover quickly or for the voltage across the cap to remain stable so it doesn't need to recover?
Yes, the load competes with the capacitor for current but the point is lost that the load should present significantly higher impedance than both the supply impedance feeding the cap and the cap itself. Thus, the "competition" is insignificant and has little effect on discharging the cap. In this case the supply current feeding the cap is substantially constant and not taxed to a point that its ability to supply the cap is diminished.
How all of this affects sound is another matter altogether.
Naz
Henry didn't actually say: "It would not be modeled as an infinite current source with a short across it."
He said: "Consider an ideal battery of zero internal resistance. This can be modeled as an infinite current source in series with a dead short circuit." (Shouldn't that be a fixed voltage source?)
It was his following sentence that made no sense: "By the reasoning presented, no current whatsoever would be available to a load connected across the battery terminals because all of it would flow in the battery's zero internal resistance." This sentence implies that the zero internal resistance would be in parallel with the current (voltage?)source.
Sorry to interrupt, just being pedantic!
Sorry, Ray, but what I said was correct. See my reply to JLH.
-Henry
I just did that, thanks. Read it a couple of times. Now it's my turn to say sorry.
As usual your logic is spot on Henry and IME although using a good quality cap with sufficient storage to cater for the lowest frequencies (at least 10Hz) is all you need, going higher on the last cap does not degrade sound quality. Perhaps the imperfection of capacitors explains in part why some prefer lower values, the worse the cap the greater the degradation with higher values. More likely, the slightly leaner sound has better synergy in some systems ... excessive mid bass can muddy sound.
What I don't understand is the outcome of some of my recent experiments. I had always wondered what it would sound like running a tube Phono from a string of SLA batteries, including heaters. The result was absolute CRUD until bypassed with a significant high quality cap, which made things pretty darn good. But even this could not compare with my stock (tiny C)LC, tube regulated PSU, but only when using the best tubes.
The real surprise was just how audible the rectifier tubes are, despite all the circuitry that follows. Changing a single one is as audible as changing any tube on the input stage. Even replacing a rectifier with a diode strapped triode took on the EXACT characteristics of the same triode used in a gain stage!
Furthermore, mixing the tubes anywhere in the PSU chain could be used for fine tuning the overall balance but again, only with “good” sounding tubes to begin with. Oh, and it didn’t much matter whether CCS was used on the Phono gain stages, everything was still just as audible! I’d have expected a reasonable amount of shielding from the (sound of the) PSU to take place with CCS.
Technically I thought I could explain why bypassing the SLAs made such a huge sonic difference and to some extent I think I did. For anyone who thinks that HF response well beyond 20KHz is unimportant think about this. I could not detect any voltage drop whatsoever, even on transients but measuring the response of the Phono did reveal a slight reduction in OP at very high frequencies (200KHz) when running from unbypassed SLAs!
Naz
"But even this could not compare with my stock (tiny C)LC, tube regulated PSU, but only when using the best tubes.
The real surprise was just how audible the rectifier tubes are, despite all the circuitry that follows."
When it comes to preamps and other low current circuits, I believe that audible rectifier differences point to inadequate isolation through the filter. While it's true that the resistance of the rectifier can influence filter performance, I would expect the output Z of the supply to be so low as to overwhelm that factor.
--------------------------
Buy Chinese. Bury freedom.
exactly what I used to think, I'm not convinced any more. If this were true the CCS on top of each tube in this particular Phono should almost completely isolate the entire PSU but it certainly doesn't. Every tube change right back to the rectifier is immediately and convincingly heard!
There is something more going on than impedance would suggest or more likely, it's more to do with the incredible sensitivity of our ear brain network. It seems that from any given reference point, whether it's good or bad we can discern even the most minute changes; ones barely measurable. A perfect example is the fact that despite my aging ears which can barely hear beyond 13KHz these days, I can hear an appreciable difference when switching from standard RIAA to Enhanced, and here we are talking about a 0.6db difference at 20KHz and 3db at 50KHz.
Interestingly, the closer to natural sound the easier it is to detect seriously small changes.
Naz
A filter that does not ring with one rectifier might ring with a different type.
I've seen the happen.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
True, but I guarantee you that this is not happening in the case I describe. I've spent countless hours measuring and listening for any clue that might answer the magnitude of the difference. If anything I have concluded that the better the overall sound the more sensitive we are to minute changes for the better or worse.
And it's not just me hearing the difference, three other audio friends who share "music nights" with me (and who had not heard my system for some months) immediately described EXACTLY what I was hearing and they had no clue what, if anything I had changed! What's more they have all made similar changes to their respective systems since, for the better ... yes I've got them all running Tube Phonos at least, even one di-hard!
Naz
multi-section cap cans versus individual caps, in parallel?
Just curious. Thanks!
----I don't think anyone has really justified why it is comparatively better to have a power supply whose output voltage flaps in the breeze.
I did put exactly that question up, and got no reasonable answer...much to my dissapointment.
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
In a way, the method used to pose a question is as important as the question itself. Likewise for statements.
Often, one's (perceived) reputation precedes them.
Cheers
RCD
> > We design our circuits to operate best
> > from absolutely fixed voltage sources.
That is the crux of the matter.
The other camp "tunes" the PS so
that it complements the output.
They regard the PS response to be
part of the output tube's plate load.
DanL
My Leak St 20's - written up here in 'reviews' - use quite large 470uf 400V big brown Nichicons, bypassed with MKP and MKS Wimas. I went down this path because good sensibly priced GZ34's were unobtainium, more detail on all that and NoS caps in the article. I'm kinda betting they still aren't for sensible money.Off a snubbed anyway, over-sized, soft recovery, 4-diode bridge.
They are used to biamp a pair of largish 2-way 8" spheres, that have no low-pass. The mid and bass duty one has 4 of the big nichis and runs in pentode mode, and the Treble duty one has just three and runs in triode mode.
I put one of them in the GZ34 hole just to have a laugh at the 'don't touch a thing' originalist types. Mostly investors who don't want to listen to the gear and want to tell me what to do with my amps.
A few other myths in audio piss me off:-
i) $$$$$$$ fashion / lemming caps and R's!!!???
ii) ALL cone drivers esp. mid/bass ones will have break up modes - > 0db - above their pass-band, well Foster/x FW202's don't! Just rolls off with a tiny -3db blip at about 3.3k.
iii) An active crossover will always be better than the OEM passive xover. Really!? I kinda think it might depend on how good an engineer and listener you are.
Anyone here want to volunteer to activate a Wilson, eh? :-)! And sell the box on the open market?
iv) Baffle diffraction is 'a minor issue'. Depends on whether you're used to having smooth diffraction, for years, like I am. The recent article in Stereophile by Keith Howard might change your mind.
I could go on you know!
Warmest
Timothy Bailey
The Skyptical Mensurer and Audio Scrounger
And gladly would he learn and gladly teach - Chaucer. ;-)!
'Still not saluting.'
http://www.theanalogdept.com/tim_bailey.htm
Edits: 03/11/11
Would using two smaller say 50uF caps instead of a single final 100uF cap have benefits in this regard, or does just the total capacitance matter?
In the context of this discussion, it would make no difference at all.
-Henry
It depends on the type of cap.
I've not measured any "significant" difference from paralleling motor runs.
Lytic caps, absolutely. Panny TSxx series.
But you do state you don't like to use any more than 50 uF in your post earlier this week.Jeff Medwin
Edits: 03/11/11
"High ripple rejection equals low cutoff frequency equals slow transient response." People often use this statement to imply that the audio output response to the speakers will always be "slow" as well, by definition.I always find that operating a power supply at significantly less than it's maximum output current capability solves a lot of problems. Engineering compromises are more acute when cost plays a dominant role.
___
Long Live Dr.Gizmo
Edits: 03/11/11
> With respect to the slowing down of the power supply due to high
> reactance filter components, this is a perfectly well-understood
> principle of engineering. High ripple rejection equals low cutoff
> frequency equals slow transient response. There is no revelation here.
Right. Slow transient response refers to the change of output voltage with respect to time. The voltage may be varying slowly (as it should), but the change with respect to time of current is still lightning fast, assuming a good quality output capacitor.
-Henry
To an extent. Chokes prefer to be run "fairly" close to their rated current, and sized according to their load.:-)
Although my experience has been, Hammond's will saturate close to, or at their rated load.
Yes, large reserve capacity equal reasonably low DCR & WELL filtered. No mini chokes for me.
How high in value(H) chokes?
Choke values in H, are determined (Hanna chart) by the current passing through them.
my estimate is just critical L times 2, IMHO
I email Jack Elliano with the specs. The chokes arrive in a couple of weeks.
If you turn the phrase around as in the header, zero = zero. I say build a well filtered low DCR PS without any ringing and be done with it.
When you say low DCR, just how low? Also, over the entire power supply, how low?
Antek transformers typically have 10-20 ohm secondaries. Some damper tubes are capable of over 300ms CCS each. Chokes? 20-35 ohm each. I use mostly LCLC with last cap to the power tubes & likely a LC downstream for preamp/drivers.
Typical PS for my projects are 4HY to 30uF to 4HY to 70-100uF @ 120-150ma current in the 250 to 400 volt range.
Sometimes I will shunt regulate the entire amp PS, but makes a lot of extra heat.
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