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Old 4th January 2017, 01:57 PM   #1
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Default Beginner questions

Hi,

I'm studying AC30 schematics (in particular, this one which has less-interesting signal paths removed) for the first time and have a few questions.
  1. While the chassis is connected to ground, it seems to me that an internal failure (for example, and I don't know if this is realistic, say V1 melts and the grid and plate fuse together) could cause the i/o jacks to become electrically hot. Is this really safe/normal in modern high voltage circuits? My first thought if I were designing the circuit (good thing I'm not) would be to attach a voltage limiter to each input/output contact.
  2. I would like to understand the long-tail pair at V3 a bit better. I believe for small signals the plate currents can be modeled by this system of equations:
    Click the image to open in full size.
    I can see that i2 is in antiphase with i1, which is in phase with v_in. However, I don't see a particular reason why the AC components of i1 and i2 would have similar magnitude or shape to each other. And the ratio of magnitudes between i1 and i2 seems to depend on mu2. Is this supposed to happen? Am I confused?
  3. If the phase splitter output is as asymmetric as it seems, is there any point at all to matching the output tubes?

Any help would be appreciated, thanks!
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Old 4th January 2017, 05:48 PM   #2
Enzo is offline Enzo  United States
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A anything to anything short is possible, but grid to plate short in the input triode is supremely unlikely. I have never seen it nor heard of it happening in my entire career. That was six decades by the time I retired. Doesn't mean it can't happen, but I won't lose sleep over it.

If you really worry about it, simply put a series cap at the input. Something like a 0.1uf would pass any relevant audio.

I would be a LOT more concerned that some unbuffered effect unit at the input would put a few volts DC on the grid. it wouldn't cause harm, but would mess with the sound.

#2 let me try until wiser heads check in. The two triodes should balance DC naturally, as the sides are symmetric. That leaves AC, or signal. here is how I intuit things - right or wrong - The lower triode grid, pin 7, is grounded through C12. The input to pin 2 grid drives current through that triode, which current flows through the cathode circuit. That moves both cathodes together of course, they follow the input signal. So now the lower triode is driven with signal at its cathode while the grid stays at ground. The tube sees only the difference between cathode and grid as input, so the two triodes essentially see the same input signal.

#3 Guitar amps are not hifi. The main plus of matched outputs in my mind is their natural cancellation of supply ripple.
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Old 4th January 2017, 08:08 PM   #3
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On #3, a perfectly balanced PI and power stage will tend to cancel out even order harmonics

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Old 4th January 2017, 08:37 PM   #4
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I think Vbias should be R22 ( i1 + i2 )

Also, I think your first two equations need to allow for AC variation in Vk
which affects Vgk ( = Vg – Vk ) for both triodes. Vg for the first triode caries the input signal and Vg for the second triode is AC grounded.

Last edited by Malcolm Irving; 4th January 2017 at 08:41 PM.
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Old 4th January 2017, 10:30 PM   #5
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although the chassis is connected to ground it should also be connected to earth, so it is protected from becoming live.

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Old 5th January 2017, 12:28 AM   #6
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Quote:
Originally Posted by chaerin View Post
...it seems to me that an internal failure ... the grid and plate fuse together...could cause the i/o jacks to become electrically hot.
The same thing has occurred to me, and I believe you're quite correct.

I do believe the people with decades of tube experience who say that anode/grid shorts are rare events. Still, it would be nice if said rare event didn't electrocute any unfortunate guitarists.

In a typical RCA/Fender 12AX7 triode input stage, even if an anode/grid short occurred in the valve, there is a 100k anode resistor, and a 68k grid stopper between B+ and input jack. So the maximum current out of the input jack would be limited to maybe 2 mA or so, with the typical 350 volts or so of B+.

According to Wikipedia, 2 mA of current falls into the "perceptible but no muscle reaction" category of electric shock. In other words, unpleasant, but unlikely to cause direct injury.

Still, nobody likes being shocked, and if a mild shock causes a guitarist to, say, fall off a stage and break a leg, that is a concern.

Perhaps the logical thing to do would be to solder a couple of reverse-paralleled LEDs (or reverse-series zener diodes) across the input jack. If 3 mA of current did ever appear at the input grid, the diodes would conduct it to ground, and only the small (and safe) LED forward voltage, or zener breakdown voltage would appear on the actual guitar cable.

The only remaining question is whether such diodes would have any audible (or measurable) effect on the guitar signal itself. All diodes conduct tiny leakage currents even when there is very little voltage across them, but think it's unlikely that these leakage currents would be big enough to have any effect on the guitar signal.

Quote:
Originally Posted by chaerin View Post
I would like to understand the long-tail pair at V3 a bit better.
<snip>
However, I don't see a particular reason why the AC components of i1 and i2 would have similar magnitude or shape to each other. And the ratio of magnitudes between i1 and i2 seems to depend on mu2.
Enzo gave you the short and sweet version. The slightly longer version is that IF the two valves are perfectly matched, and IF the tail resistor is infinitely large (or replaced by a perfect constant current source), THEN the two AC signals are perfectly matched to each other.

But "perfect" doesn't actually exist, and, in this case, isn't even particularly important. Some mismatch between the two signals in a guitar amp phase inverter often creates better tone ("better" is subjective, of course), because it lets through some of the even-order harmonics that make SE guitar output stages sound so good.

Quote:
Originally Posted by chaerin View Post
If the phase splitter output is as asymmetric as it seems, is there any point at all to matching the output tubes?
Yes, the point is to make more profit for those who sell matched valves, and to lighten your wallet more effectively.

I have seen a few guitar amp designs where there is actually a pot that lets you dial down the signal to one output grid. This lets you dial the amp from normal push-pull class AB at one end to single-ended at the other. In effect, you go from a normal level of "matching", to a complete and utter absence of any matching whatsoever.

Tubelab (George) on this forum is a smart engineer with decades of valve experience. I've read posts by him where he described using two completely different output valves in a guitar amp push-pull output stage, for example, one 6V6 (beam tetrode) and one EL84 (true pentode).

This guarantees different (and quite mis-matched) distortion spectra from the two output devices, and that, according to him, translates to rich and interesting guitar tone. The concept makes total sense to me, and some day I will build an amp to try it out.

On a simpler level, I deliberately unbalanced (by about 1 dB) the cathodyne phase inverter in my current valve guitar amp project. This was easily done by making the anode load resistor a different value than the cathode resistor. The idea, once again, was to prevent complete cancellation of the sweet-sounding second harmonic distortion in the output stage.

I didn't have access to a way to measure the resulting distortion spectrum, but I think I heard an improvement in both clean and slightly overdriven tone from the amp. It's an easy and easily reversible tweak to try, in any case.

-Gnobuddy
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Old 5th January 2017, 12:44 AM   #7
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Quote:
Originally Posted by ArcticBreaze View Post
although the chassis is connected to ground it should also be connected to earth, so it is protected from becoming live.
The OP's concern was that the "hot" wire in the guitar lead could become "hot" in the wrong way, as a result of a short between input stage anode and grid. In the scenario he outlined, this will happen even if the chassis is connected to earth.

Your comment did make me realize, though, that (with a properly grounded chassis) any current leaking out of the input jack will be grounded through the electric guitar pickups. Those have typically no more than roughly 12 kilo ohms DC resistance, and, at 60 Hz, not a lot of additional inductive reactance.

That means a 3 mA, 60 Hz, AC current leaking out of the guitar amp input jack, can only cause a maximum of maybe 40 volts AC anywhere inside the electric guitar. That is probably safe for most people most of the time.

-Gnobuddy
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Old 5th January 2017, 04:02 AM   #8
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This all goes back to the requirement for guitar amps to colour the guitar's sound in a pleasing way. Even order harmonics are generally considered to be musical and warm, and hence desirable for this purpose. The AC30's lack of NFB also helps these harmonics to come through unsuppressed, which along with the chime of the EL84s, is a key part of the AC30s distinct tone.

As for the tube matching thing, it's generally believed that the classic guitar amps of the past weren't normally fitted with matched tubes. At the same time they shouldn't be too far out of whack, or the amp would hum badly or suffer premature burnout of tubes, due to one side of the pair running too hot relative to the other. In my own builds, I therefore prefer to use approximately matched tubes, so that hum is minimized, but you still get good warm tones.




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Old 5th January 2017, 04:29 AM   #9
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Quote:
Originally Posted by AquaTarkus View Post
...they shouldn't be too far out of whack, or the amp would hum badly or suffer premature burnout of tubes, due to one side of the pair running too hot relative to the other.
This is a very good point which I forgot to mention earlier. If using very different output tubes, each individual output tube's bias should be adjusted so that both run at about the same quiescent (bias) current. This takes care of any unwanted saturation effects in the output transformer due to mis-matched idle currents in the two output devices, and also makes sure that hum cancellation still occurs.

Incidentally, the hum issue largely goes away if you use solid-state rectification, and larger filter capacitance values, both of which were unavailable (or too expensive to use) when these vintage amps were designed.

-Gnobuddy
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Old 5th January 2017, 05:50 AM   #10
PRR is offline PRR  United States
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>> i/o jacks to become electrically hot

> Fender 12AX7 ...100k anode resistor, and a 68k grid stopper


Insignificant quib: usually we have two 68K in parallel (depending which jack). So 134K from 300+V to user. OK, this is exactly your "maybe 2 mA or so", which I agree is not too lethal, and below a GFI's 5mA trip (so some experts thought 5mA was pretty safe.)

Except: assuming (as Gnobuddy does) a guitar is plugged in, there's 5K-50K of bleed there. So the pre-person voltage is lower, and the shock current has an alternate path through the guitar. (And no, it won't burn-up, do math.)

0.1uFd at 60Hz is 26K, so not a large change in total impedance. 0.01u would be ample going into a 1Meg grid, so you can add 260K of 60Hz impedance without any change in normal tone. Some OVER-drive units work different if the input is naked or cap-coupled, a minor consideration.

Also we rarely touch the input wire. Doing that makes the amp BUZZ!! So the habit we develop of not touching tip incidentally saves us if all heck breaks loose in the tube one day.

And I too have never seen a G1-P short.

And if such things were common, and serious, we'd all be long dead.

> a guitarist to, say, fall off a stage and break a leg

I think beer (in belly or on slick floor) is greater risk to legs. Also the long drive home late at night just as the bars are letting out.
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