Surfing the Glass

[Lectroid]

[Moderator, please move as appropriate]

This is a little long.  I grew up in Southern California and spent much time in and on the water.  Spent many hours watching waves.  For those who don't know, surfing is all-dependent on the shape of the bottom as the waves near the shoreline.  The shape, slope, and rise of the near-shore bottom determines how a wave will shape, which way it will break, etc.  Yah, those days are gone but...

I've built three working amps now and I'm quite pleased with my latest single-channel DR clone (in a PR cab).  I used Hammond organ iron and it turned out pretty hot.  But as much as I've studied tube biasing, and drawn load lines, and wracked my tiny brain, I still have at best a skimpy understanding of how tubes really "work" dynamically, as the amp is under load, and how biasing acts to make a tube distort or overdrive in a musical way. 

I know there are probably rules of thumb I don't know yet, and I'll keeping building and tweaking no matter what.  But I'd sure like to get to where I think some of you are, where plate and screen characteristics actually speak to you.  I can hear them muttering, but it's all gibberish to my untrained ears.

Back home over Christmas I was watching the waves as always and got whacked with an epiphany: that water waves might be analogous to AC signal waves hitting the grid of a tube. 

So I wrote up this analogy and I'm hoping some of you will comment.  Am I totally nuts or is there a teachable moment in there somewhere?  I would love to hear any criticisms or corrections or expansions--whatever you got.  Even hoots of derision.  This may be something people hear in their first year of EE, I don't know.

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In water, the energy of surface waves runs across the surface of a body of water. Half of each waveform (the water wave’s crest) rises above the average surface level of the water, and the other half of the waveform (the trough) drops below the average surface level of the water.  For waves to propagate without any interference, there must be adequate “depth” for the waves to travel without hindrance.  Waves that hit the beach in California have traveled thousands of miles across the Pacific, unimpeded.

When water waves near the shore, the shallower depth is not enough to allow the full peak-to-peak amplitude of the wave.  The lower half of the wave hits the shoreline bottom and this contact deforms the wave shape by causing a drag effect that distorts its perfect shape.

For a vacuum tube the level of the DC bias voltage seems to be analogous to the depth of the water.  In this analogy, the bias voltage is needed to set an appropriate voltage “depth” so that the peak-to-peak amplitude of the AC signal can move up and down freely.  Similarly, when the DC bias voltage on a tube is too small (shallow depth) the AC waves will run out of “depth,” similar to water waves dragging across a shallowing bottom.  This “drag” will change/deform the AC waveforms because the bias voltage is not sufficiently negative.  The wave will not be able to reach its full size and shape, and will be deformed/distorted/clipped.  As long as there is enough headroom—has a large enough DC offset—the AC waves will propagate unimpeded and undistorted.  The DC bias voltage, aka “offset,” provides this depth for the waves to move in.
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Any thoughts?

[PRR]

My immediate objection is: ocean waves have no ceiling. Tubes do. The wave is bent on BOTH top and bottom. (FWIW, inside a drain pipe the "ceiling" has big effect: you get more flow at 70% full than any fuller, because of drag on the ceiling.)

 work" dynamically, as the amp is under load, and how

Does it? Biasing is more for the no-signal case. When you BEAT it, the tube finds a new operating point. In-between you may have a little discretion to set the plate at 50% or 70% of supply voltage, but that won't turn a dullard into a star.

[scstill]

Love your term "Surfing the Glass"

[Lectroid]

PRR,
Yep, the bias-point and a tube's dynamic working point are different, I shouldn't have confused those two.  Your comment about drains is intriguing.  When you say "70%," would 70.7% be more precise, assuming a perfectly circular pipe with zero drag?  As in 2^-1/2 ?

And when water waves flow along the surface, the crest is above, and the trough below, the average water level in the vicinity.  I guess the analogy would be better if I'd positioned the entire wave, trough to crest, as being propagated from a point underwater.  Which happens all the time at the bows of boats; it's the wave that dolphins like to surf on.  (They'd call it "The Endless Wave" if dolphins made surf movies.)

Would the analogy would be closer to reality if we place the bias point voltage, or "depth," halfway between the height of the upper crest just beneath the surface of the water, say, -1.5 foot, or -1.5V, and the lower trough of a 2V p-p wave, at -3.5V?

In that submerged case, is the comparison any better?   To me it still looks like the lower trough of a water wave "dragging" on the bottom aligns pretty closely to the lower half of a cooler-biased AC wave bumping up against the point of saturation and creating distortion.

Anyway, thanks for the critical thinking.  I'm no closer to having an intuitive grasp of operating characteristics, or of predicting distortion or clipping, just from reading the characteristic curves, but maybe that'll come with time.

scstill,
Couldn't resist the pun. 

[HotBluePlates]

PRR ...  When you say "70%," would 70.7% be more precise, assuming a perfectly circular pipe with zero drag?  As in 2^-1/2 ? ...

Sure, but 70% = 70.7% near enough, or 1/√2 if we're trying to be exact.

The goal was to point to the relationship between RMS and Peak values of a sine wave, where RMS is 1/√2 times the peak value, and the peaks are where the signal runs into the upper or lower limit imposed by the operating point.

Would the analogy would be closer to reality if we place the bias point voltage, or "depth," halfway between the height of the upper crest just beneath the surface of the water, say, -1.5 foot, or -1.5V, and the lower trough of a 2V p-p wave, at -3.5V?

PRR's point is the "air above the water's surface" is never a hard ceiling, but the grid current we get when we drive the grid positive of the cathode (during the positive-going parts of the signal cycle) very much are a ceiling.  Absent a specialized driver circuit, they will cause our wave-tops to be lopped off flat.


Ignore the "Good" & "Bad" below.  Instead see that these should be symmetrical sine waves, but their tops are lopped off.  The scope-shot is of EL84 grids being driven positive of their cathodes, which altered their input signal.  Here we are not monitoring the EL84 plate current or output signal to know if these tubes are also going into cutoff on the negative-going part of the signal cycle (though it's possible they are).  So "floor & ceiling" apply to tubes, or water flowing in a pipe that is constrained on the top as well as the bottom.

[Lectroid]

HBP,

Thanks for helping me understand this. I adjusted the analogy so the wave was propagated completely below the surface of the water.  I think it's useful after that.  But let's ditch the the analogy and stick to tubes.

From my understanding from reading and your comments, I think you're saying that:

• Biasing a tube too cool puts the bias point too far to the right on the load line, where the area of saturation  lessens the available headroom for the lower half of the wave.  This flattens out the bottom half of the AC signal wave and yields a softer, 'overdrive' sounding distortion.
• Biasing a tube too warmly, past where there is sufficient headroom for the upper half of the signal to fully form can push a tube into grid current.  This causes a hard clipping during the period that grid current flows, and yields a harsher, clipped form of distortion.

Is that an accurate picture?

I prefer the softer, (Marshall-like?) overdrive type of distortion.  But I don't understand how to bias a tube and choose a plate voltage that will: either produce an amplifier circuit with lots of headroom and clear amplification, or to deliberately drive the tube into overdrive. I've read a lot of explanation but they all seem to stop short of my understanding, just explaining the facts but leaving the "therefore..." to the read.  So far, I haven't got it.

Is it as easy as pushing the bias point cooler to cause that softer overdrive distortion?

Can the bias voltage and plate voltage be adjusted together in some formula/relationship to produce either effect or anything in between?  Something like a rule of thumb? 

If this is more than you want to get into, is there a concise explanation online you can point me to?  I also have some of the better-known books on tube amps

Thanks

[HotBluePlates]

... From my understanding from reading and your comments, I think you're saying that:

• Biasing a tube too ...

Let's not focus on "biasing" because your original analogy was about "limits."  That is, the ocean wave propagated fine until the "floor moved up," running into the bottom of the wave, and caused the wave to fall apart ("distort"?).

I swiped a graphic from Merlin's explanation of a basic gain stage to show "limits."

     Merlin put a vertical line at 350v on the plate, because data sheets give a plate voltage limit of approx. this amount.

     Merlin also drew a red curve corresponding to 1w of plate dissipation, again because data sheets have roughly this limiting value.

     I added the Blue cross-hatched area denoting where grid current is certain to happen if input signal ventures to this extreme along a load line.  This is certain-distortion on one side of the signal cycle, where the signal will be abruptly clipped. 

     I also added a light gray box down at the low plate current region, where the grid lines get very-bent.  If the input signal drives a single-ended tube all the way along the loadline to zero plate current, the tube will clip.  But the light-gray area represents increasing distortion before actual "clipping."



Now if we design a gain stage, we can bias anywhere inside that "Safe Operating Area" and our loadline can be literally any angle.

Afterwards, the 2 ends of our loadline will run into one of the four Limits that bound our operating area:
     Grid Current
     Low (or Zero) Plate Current
     Plate Dissipation Limit
     Plate Voltage Limit

One end of the loadline is then our "Floor Limit" and the other end is the "Ceiling Limit" (though they're generally positioned "upper left" and "bottom right").

Along the way, the angle of our loadline (which is the amount of plate load Resistance) will determine how much Output Voltage change we get for an Input Voltage change (our "Gain").  And that is going to depend a lot on the relative slope of the grid curves (which is really the Internal Plate Resistance; we're balancing external load resistance against internal resistance to garner "Gain").

Where we Bias just amounts to where we start along the loadline, and whether we've optimized the stage to accept a large input signal.


If tubes were perfect devices, there would be Zero Distortion until the signal hit the ends of the loadline & clipped.  But they're not perfect, and there will be some distortion before "clipping."  York's Amplifiers does a good job in Chapter 3 of explaining how to perform graphical analysis, and how to recognize the "distortion before clipping" (which is unequal amplification on each side of the signal cycle, resulting in an output waveform that is not the exact same shape as the input waveform).

[HotBluePlates]

... I don't understand how to bias a tube and choose a plate voltage that will: either produce an amplifier circuit with lots of headroom and clear amplification, or to deliberately drive the tube into overdrive. ...

It will greatly benefit you to have a death-grip on Ohm's Law, the Equation for Power, and how to convert between Peak and RMS values of a sine wave.

When designing from a blank sheet of paper, we start with the speaker & a desired power output, and work back towards the input jack.

Assume we're making a 40w Super Reverb.  The voltage across the speaker terminals is
     Volts = √(Power x Impedance) = √(40w x 2Ω) = 8.944v RMS

The output transformer is 4kΩ to 2Ω.  The Voltage (or Turns) Ratio is the square-root of the Impedance Ratio:
     4kΩ / 2Ω = 2000 : 1 ----> √2000 = 44.72 : 1 volts ratio

Let's use the transformer's Volts Ratio to figure the RMS volts on the primary side for our 40w output:
     8.944v RMS x 44.72 = 399.98 volts RMS

The figure above is the voltage swing across the entire primary, or from "plate to plate."  But we commonly calculate the required supply voltage & power output with respect to one side of the push-pull output stage.  So let's divide our RMS Volts in 2 parts, to get the part contributed by one side:
     399.98v RMS / 2 = 199.99v RMS

But we were talking earlier about "limits" and that the signal is only 1/√2 (0.7071 or 70.71%) times as big as its limit.  So let's convert from RMS sine volts to a peak value:
     199.99v RMS x √2 = 282.83v Peak

We now know that to make our 40w of output, the 6L6 on one side will reach a peak plate voltage swing of nearly 283 volts.  But the supply voltage needs to be "Tube Output Volts" + "Tube Volts."  That is, there needs to be some voltage left across the tube for it to continue working & amplifying.  There might even need to be some extra "safety margin."

The upper graph of Page 6 of the 6L6GC Data Sheet shows the limiting 0v gridline for a number of screen voltages.  The "Knee" of the curve is where it goes from being mostly-horizontal to mostly-vertical.  Our loadline should stay just above the knee on the horizontal part, and this implies a lower-limit of plate voltage swing.

     The knee of the curve for E_c2 = 400v lands pretty close to 100v on the Plate Voltage Axis.  If there were an E_c2 = 450v curve, we might guess we would need closer to 120v at the 6L6GC plate.

Now we add our required minimum plate voltage of 120v to our peak signal output:
     120v + 282.83v Peak = 402.83v required

Now let's look at the Super Reverb schematic: 460v plate & screen.  About 50v of extra margin allowed for supply voltage sag when the amp pushes maximum output power.

[HotBluePlates]

I know people mostly bias by plate dissipation today, and that generally works.  But what did Fender proabbly do to determine the required bias for the Super Reverb?

Grid Input Volts result in a Plate Current Output of the 6L6GC.  This plate current is pulled through the output transformer's primary impedance to create the "plate volt output" we calculated earlier.  How much plate current?  We calculate by looking at one side of the push-pull power section:
     Transformer Impedance at Max Power is 1/4 the plate-to-plate impedance:  4kΩ / 4 = 1kΩ
     Peak Plate Current = Peak Plate Volts / Transformer Impedance = 282.83v Peak / 1kΩ = 282.83 mA Peak

How much plate current a tube pulls for a fixed amount of grid-voltage change is the tube's "transconductance" (or "Gm").  Gm is not a constant value, but changes with plate current.  But we usually see 6L6 Gm figures cited between 5-6 "micromhos."
     Mho is the inverse of "Ohm" so it is "Conductance = Amperes / Volts" (where Ohm's Law is Resistance = Volts / Amperes)
     Europeans say "5 micromhos" as "5 milliamps per volt" or "5mA/volt"
     This means 1v of change at the 6L6 grid yields a 5mA change of plate current

How much Grid Voltage Input do we need to get ~283mA of Plate Current Change?  Let's guess a middle-Gm of 5.5mA/volt or 5500 micromhos:
     283mA Peak / 5.5mA/volt = 51.45 Volts Peak

Look again at the Super Reverb schematic:
     If the peak of the Signal Input exceeds the bias voltage, we push the grid positive of the cathode & draw grid current.
     That grid-current case is a "limit" we mentioned earlier.
     We see Fender chose a bias voltage of -52v, just a bit over the needed signal input calculated above.

The above should show why Fender landed on 450-460v of supply voltage for their Super Reverb, and why they selected a particular bias voltage. 

[Lectroid]

Mr. Plates,

First thank you so much for making that grand effort in laying all that out in such detail.  I appreciate so much your willingness to help me learn.  It must have taken you hours and I am in your debt. 

Second, WOW!  That was the most complete, concise answer I've ever gotten on any forum.  Ever.

Third, my main question was really about preamp tubes, but you expanded it to answer many other basic questions on tube operation that I've been curious about for a couple of years at least.  I'm saving your posts out into a doc so I can refer back to  it.  My head is kind of spinning (in a good way) and I know I'll keep reading and assimilating this for a long time. I still have questions about how to drive tubes into overdrive or distortion by adjustment of the bias but I'll save them until I really have a handle of what you've already written.  I know I asked for rules of thumb, but ... damn! 

I know that tubes are devices that exhibit variable resistance, but no other author has explained how operating tubes behave with their interlocking parts so clearly, start to finish, and tying all the pieces together using Ohm's law.  Some books have used your starting-at-speaker-and-work-backward approach, but always left me missing something, like they were expecting me to draw some 'obvious' conclusion myself.  They never managed to connect all the dots for me, but you did it in nine or ten paragraphs.

Fourth, have you ever thought about writing a book?  Seriously.   

[HotBluePlates]

...  That was the most complete, concise answer I've ever gotten on any forum.  Ever.

... have you ever thought about writing a book?  Seriously.   

Thanks for the compliments!

But I'll give the "Concise Award" to PRR.  He says a lot with few words, and it sometimes requires outside study to grasp all the implications.

I've also learned a lot from PRR over a lot of years (I steal from the best!).  You might not know, but he's done peer review & editing for some folks who do write the books on guitar amp tube electronics.

I'm glad you found value in the write-up!  It should show that the basics of tube guitar amps are less about Magic and are quite straightforward.  "The Magic" (when there is some) comes about by understanding how all the straightforward bits interact with each other, and making use of those interactions to arrive at a desired result.

[Lectroid]

Thanks for the compliments!
But I'll give the "Concise Award" to PRR.  He says a lot with few words, and it sometimes requires outside study to grasp all the implications.
I've also learned a lot from PRR over a lot of years (I steal from the best!)

I agree that he is, uh...'concise.'  That's why your completeness was so helpful, connecting everything together down at my level.  I've never had classes or systematic training, so my learning journey has been pretty haphazard, even with outside study.

I too steal from PRR any chance I get.  Regretfully not as often as I'd like because his explanations often sail right over my head.     Just being a noob, I've often wished he was a little more wordy, but he's always worth reading.

[scstill]

Magic of course comes thru the player as well :-)

Speaking of Magic, I find that a lot of good sounds come out of vintage tubes (30s, 40s, 50s), and if fact I seek them out.

Then I read this article that recommends replacing tubes quite often, Is there an opinion on this? are they trying to sell more tubes?? or just giving non tech types a simple solution?? My experience is that old tubes sound better and I rarely find bad tubes. Just be careful for microphonic types.
https://www.fuelrocks.com/how-to-tell-if-your-guitar-amplifiers-tubes-need-to-be-replaced/

[Lectroid]

I like the old tubes.  I have a lot of them from old Hammond organs, GE & Magnavox tube stereos, and other places.  They all date from no later than the 60s and all seem to work well enough.  They're all close enough to spec, just like with modern tubes.

[HotBluePlates]

... I read this article that recommends replacing tubes quite often, Is there an opinion on this? ...
https://www.fuelrocks.com/how-to-tell-if-your-guitar-amplifiers-tubes-need-to-be-replaced/

I strongly disagree with numerous points made in that article, and do not consider its advice authoritative.  Rather than re-write, here is a recent TGP thread on this exact issue.

The arbitrary "replace every 6 months" is 150% bogus, and is simply restating the advice of someone who sold tubes (and gave bad advice that amounted to "change your tubes more often" and "change tubes other people don't say to change").

I have amps that haven't "needed" a tube replacement for 20 years.

[scstill]

Thank you for confirming my suspicion.

I have a Silvertone 1484 that has all its original tubes from 1965-ish, except for one that needed replacing.
It is one of the best sounding amps I have (great advice from this forum to get it running). Tubenit even donated the footswitch. Even its original reverb (that pretty much no one likes) sounds amazing after resto.
Still looking for an original Silvertone 12ax7 to replace the replacement :-) .

https://stillampd.com/silvertone-1484-restoration

[Lectroid]

Man, that takes me back.  My very first band's lead guitarist had a 1484 while our rich-kid rhythm guy had a BF Bandmaster.    I couldn't tell any difference between them.  The Silvertone's reverb wasn't badly thought-of back then; everyone liked rackety, splashier reverb.  First Saddleback and now this.  Thanks for the memory jogs.