Remote cut-off point?

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Mattsson

New member
Joined
Aug 12, 2004
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4
Location
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Hello!

I've got a few questions about remote cut off tubes ( related to varimu compressors).

1. In the definition of the cut-off point two of my sources give very different stories. The first one says:
"in a valve the cut-off point is the minimum negative grid voltage required to stop the current".

While another source says:
" The same increase in negative grid voltage would have driven a conventional tube into cut-off, before the grid even reached - 35 Volts".

But wasn't the cut-off the grid value closest to 0 (least negative) required to stop the current? Because in the second article it sounds as if the cut-off point is the most negative Voltage that a tube can handle ["the same increase in grid voltage would have..."] Could someone please explain this?=)


2. A second little wonder is whether a louder I/p signal results in a more negative or more positive Grid voltage? In relation to my first question, would the cut-off point be the loudest I/p or the quietest I/p a grid can "handle" [required to stop the current]?

Cheers!

/Robert Mattsson
 
If you look at a data sheet, you see that current decreases as grid voltage becomes more negative. When your first source describes "minimum negative voltage" it is closest to zero as you think. That is not to say it could not be -35 or -10 or -100. It depends on the tube characteristics.

I hope I was not more confusing that necessary.
 
First: real tubes never cut off. That's oversimplification.

Here's a plot of the three types of cut-off actions. All three tubes are picked to pass 4mA at zero grid volts. Plate voltage is held constant, grid voltage is varied, plate current is measured.

cutoff.gif


"A" is an "ideal tube". 1 Volt of grid voltage change causes 1 milliAmp of plate current change, all the way down to zero plate current and cut-off. Since the zero-grid-volt current is 4mA, it takes -4V on the grid to reduce plate current to zero. There are no real tubes like this.

"B" is a typical normal amplifier tube, sometimes noted as "sharp cutoff". Near zero grid volts, 1 Volt of grid voltage change causes 1 milliAmp of plate current change, but as grid voltage "rises" (becomes more negative) 1 Volt of grid voltage change causes less than 1 milliAmp of plate current change. As plate current approaches zero, you have to change grid voltage a lot to keep reducing plate current.

"C" is a remote-cutoff tube. Near zero grid volts, 1 Volt of grid voltage change causes 1 milliAmp of plate current change, but it soon deviates from this path. Around 1mA of plate current, you need to change grid voltage about 10 volts to change plate current 1mA (or 1V grid for 0.1mA plate). Even down at -40V on the grid, the plate is still passing current, but down there a 1V grid voltage change makes only maybe 0.01mA plate current change.

The slope of these curves is the Transconductance or "Gm", which in most amplifiers is a major factor in Gain. Tube "A" has a constant Gm of 1mA/1V or 1,000 microMhos. Tube "B" starts the same but falls to 700, 500, 300 uMhos. Gm gets lower than that, but only at currents so low that the maximum output is too small to be useful. Tube "C" again starts at 1,000 uMho, but is 500uMho at 2mA, 200uMho at 1mA, and maybe 10uMho at 0.1mA. In an AM radio IF amplifier with 100K load impedance, 0.1mA is enough to get 10V output signal, so 0.1mA is not unusably small. But the Gm and Gain have shifted from 1,000uMho to 10uMho, a 100:1 or 40dB change of gain. If you try to push tube "B" to 10uMho, the plate current will probably be around 0.001mA, which with typical load impedances is too small to give good output. Tube "A" of course has the same gain in all conditions, and just clips if you take the grid voltage too far negative.
 
Well that's the clearest explanation I've heard! Thank you.

Good graph too. I love hand-drawn graphs - so much better than excel/poorpoint corporate graphics.
 
Yeah...an exteremly clear explination PRR...Thanks a lot!!!=) And great graphs!

But what I don't get is that a hotter audio signal on the i/p (an increase in grid V) causes a decrease in plate current, starting at 4mV goind down to zero. For example in a normal linear amp, wouldn't an increase in grid V mean an increase in plate curent and O/p as well?


And how is tube "C" operated in a compressor circuit?

Scenario: You have an Audio i/p causing a -5 V grid voltage. Suddenly it jumps up to, let's say, -10 V which happens to be above the threshold. The threshold is around -6V with a very high ratio.

Would this cause the CV to increase the voltage by a huge amount to were the mho is very low (for less amplification) or would the CV decrease the grid V closer to zero?

Or is the limiter action of tube "C" just that the higher i/p the less amplification...so that you can drive it reall hot without really causing it to cut-off? It would just decrease it's own amplification factor.

And in what region is the remote cut off tube usually operated?

Cheers!!!

Would highly appreciate if you could give as clear explinations on these ones...=)

/Robert
 
Just follow PRR's graph to answer your question. If voltage becomes more negative, current decreases. Remember that there is a feedback loop giving only negative voltage from the output signal back to the input, thus controling the grid voltage.
 
Yes, I get most it...and as PRR;s graph clearly showed plate current goes down with increased grid voltage. And that a remote- cutoff tube can handle more negative voltages before going into cut-off.

But what is hard to understand is how this would be such an advantage in a compressor application? Why do you need a tube that never cuts off and can go that far negative? Well, apart from the fact that a remote-cutoff tube changes its amplification factor the more negative you go. But how is this amplification factor controlled and predicted?

And one question that I would love to get a question to, related to tubes in general, a hotter audio i/p, does that make the grid voltage go more negative or positive? If it is negative, does that mean that the hotter the i/p, the less the amplififaction factor (closer to zero plate current)?

Thank you very much=)=)=)

/Rob :cool:
 
Your second paragraph is correct - the amplification goes down as the slope of the line on PRR's graph goes down. The line gets shallower as the grid voltage drops and that's what does the compression. The control comes from another amplifier and a rectifier contained in the compressor. That section picks the peak voltage and rectifies it into a negative control voltage, which is applied to the grids of the input tubes.

Note that remote-cutoff compressors must be in push-pull (two tubes - look at an Altec 436:
http://recordist.com/ampex/schematics/altec/436atl10.pdf

Let's use that schematic as an example:
You need to look at the whole circuit to get where the signals come from. You now know how the remote cutoff tube changes gain (the slope of the graph) - that's V1 (6BC8). Note it is in push-pull - the input transformer splits the signal so both tubes amplify the same signal but one is opposite in polarity of the other. V2 (6CG7) amplifies the signal and recombines it in the output transformer. The reason for a push-pull configuration will become clear in time.
The output signal is also rectified by V3 (6AL5). Note that the direction of the diodes tend to lower the voltage as the input signal's amplitude goes up. R9 is the release resistance and R12 is the attack resistance. C4 is the control voltage capacitor, and the time constants formed by either R9+R12+C4 or R12+C4 form the release and attack time constants. This control voltage is applied to the grids of the input tubes via the centre tap of the input transformer. This last paragraph is where the control voltage comes from.
The push-pull operation is pretty important here. A change in the control voltage also causes a 'thump' in the plate voltage of V1. In one tube, the control voltage is indistinguishable from signal. So you have to play some trickery to get rid of that 'thump'.
If you can pick a V1 with good matching between the two halves, both plates see the same 'thump' (control voltage change) - but opposing signal voltages. The output amplifier and output transformer see and pass on only differential signals, not common mode signals, so the thump gets cancelled (the thump is a common mode signal) where the signal does not get cancelled - it's differential. Tubes in a simple differential amplifier configuration (resistive cathode current source) do not have the best common-mode rejection in the world so the output transformer does a lot of the work of cancelling the thumping.
 
Well it´s very clear to me that the more "remote" a cut-off point is in a tube, the more signal it can take and compress, without reaching the cut-off point.

Thanks for the great explanation, as always, PRR!

Kubi, think about the signal being amplified, then rectified, then fed at the grid of the first stage again as a negative "bias", and then have another look at the graphs and I´m sure it will be very clear.
 
> remote-cutoff compressors must be in push-pull

Only if you want "best results". There are single-ended vari-gain amplifiers. They distort at much lower input levels than push-pull. They thump. I can't point-out a compressor/limiter example at the moment. But the trick was used for remote volume control in tube PA amps, where levels hardly reach the distortion point of a single-ended vari-gain tube, and volume control action is slow enough that the thump is sub-sonic, at least with the limited bass response of speech PA systems.

And single-ended is common in radio gain control, but they can use tuned circuits.

> question that I would love to get a question to, related to tubes in general, a hotter audio i/p, does that make the grid voltage go more negative or positive?

My question, which I've been trying to be too polite to ask, is: do you have a GOOD grasp of basic tube amplifiers?. It is too confusing to think about variable gain circuits, when you don't have a a strong feel for the nominally fixed-gain circuits.

In a normal tube amp, the grid gets a fixed negative DC voltage (or grid is tied to ground and the cathode is raised positive of ground, same result). The audio signal causes the instantaneous voltage to swing both ways around the fixed DC voltage.

Typical values: grid is 1 volt negative of cathode. Soft audio signals like 1 milliVolt peak will swing the grid from -1.000V to -1.005V to -0.995V and back. Stronn signal like 100mV swings the grid from -1.000V to -1.100V to -0.900V and back. The DC voltage "never" changes. (A little, slowly, with age and power variation, but normally not enough to matter.)

> does that mean that the hotter the i/p, the less the amplififaction factor (closer to zero plate current)?

You may be asking a good question for the wrong reason. Go back to the tube biased 1V DC on the grid. Feed a 1V peak audio signal. The instantaneous grid voltage now varies from -1V to 0V to -2V and back. On the ideal tube, output current varies from 3mA to 4mA to 2mA and back, linear. On the typical sharp-cutoff tube, it varies from 3.1mA to 4mA to 2.4mA and back. So as the audio swings positive 1V from idle, plate current increases 0.9mA, as it swings negative 1V from idle plate current decreases 0.7mA. Hmmmm. For symmetrical input, the output is asymmetrical 9/7 or about 28%. I'll skip some math and tell you that this means about 6% or 7% second harmonic distortion. It gets worse as we approach nominal cut-off. That's why we normally bias tubes at fairly high current for their size, grid-cathode voltage as close to zero as possible without any chance that signal swing will push the grid-cathode voltage positive.

The remote cutoff tube has more distortion at high current. You never use it in a fixed-gain amplifier. It is "better" only when you intend to apply a negative DC control voltage to reduce its gain a LOT. A regular sharp-cutoff tube, when treated this way, will get into cut-off before its gain falls very far. It distorts and pretty quick it just flat-lines the signal. The remote cutoff tube lose gain about as fast as its current decreases, so it is possible to bias it to very low gain without touching cut-off.

Understand straight amps, and vari-gain amps, before you try to understand everything that is going on in a Compressor/Limiter with its forward audio path and control feedback path and all the little tricks to minimize flaws.
 
Hey Kubi,

A book you may find useful is John F Rider's "Inside the Vacuum Tube": ISBN 1-882580-42-7

The book has several pages explaining the variable-pitch grid spacing in a vari-mu tube - it is this construction technique that enables it to go into deep GR without being biased beyond its cutoff point, thus keeping plate current above where it would be in a non vari-mu valve.

I've really got to give props to PRR for explaining (in many forms over time now) the principal of the remote cutoff valve to us. I recently spoke to a designer with a master's degree in electronics about the 6386, and even he was unaware of its properties and decreed it was probably overrated - once explained he would comprehend it exponentially better than myself, but this story just illustrates how little documentation the remote-cutoff principal has received over the years. PRR's knowledge on this subject is not only comprehensive, but esoteric and lesser seen even in the broadest sense.

Justin
 
The main problem with PCC189's is that they have a limited max. anode voltage (90-100V) that is much lower than the 6383's. This means that you can't use them like in the 670, who is running them below the uneven "dented" point in the current/gain curve - with many tubes in parallel to acheive a reasonable current.

Rather, you would depend on carefully selecting tubes that are symmetrical enough not to cause problems from the "dent" in the curve.

Jakob E.
 
Kubi!

You mentioned that you had foudn some good online-articles that explain how vari-mu tubes work. Could you please tip me about a couple of good ones.

I've looked for a long time, but haven't got over any clear and good ones.

Thanks!

/Rob
 
I see that Dle used 6BA6 on his fairchild clone. Do you think it´s close enought?

Anyone has cut-off data for this one?
 
Manley use the 6BA6 in their "T-Bar" adaptor, wired as triodes. They claim an identical mu curve to the 6386.

Even if the 6BA6 can mimick the 6386 on paper, a pentode wired as a triode is going to sound different I would have thought. Not suggesting the 6BA6 isn't a great valve to use, but pentodes sound different to triodes on account of their internal construction i.e partition noise etc.

edit: Punch 6BA6 into the search here for data: http://www.mif.pg.gda.pl/homepages/frank/vs.html

Justin
 
I only know the Manley blurb, maybe post a Q for EA at the Geekslutz forum? I asked about it before, try a scan, but I don't think the info there will tell you a lot.

J
 
but which is beter opcion for vari mu compresors ?? 6ba6 in triodo coneccion or pcc189???? :roll: . 6ba6 its direct replacement for 6sk7 ??
 
Tim da P uses multiple PC189's in his EAR limiter, which I have to recomend very highly for many different applications. Probably the nearest to a Fairchild 660 that I can think of out of all the tube compressors i have tried, except Sean Davies' Limiter. (Can't remember what tubes he was using. Such a shame I never bought one whilst he was still making them, but £900 was outside my price bracket . . . . . Then . . . ) . They also have the huge advantage that they are very, very cheep! (the PC189 tubes that is . . . )
 

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