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.
Any help would be appreciated, thanks!
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.
- 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.
- 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:
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?
- 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!
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.
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.
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.
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.
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
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
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 AC30‘s 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.
Sent from my LG G2 with CM13 using Tapatalk
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.
Sent from my LG G2 with CM13 using Tapatalk
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
>> 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.
> 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.
And FWIW the AC30 cited has a 220K plate resistor. Half as deadly (not) as a 100K Fender.
________________________________________
I'm sure the designer of the AC30 never looked at this that way. He'd probably take it funny if you showed it to him. If you bought the beer, he might mentor you on understanding tubes "intuitively" before you reach for the algebra. Most questions can be answered by thinking "if this goes up, that goes down, and about so much".
> reason why the AC components of i1 and i2 would have similar magnitude or shape to each other
Input is applied to V3a grid? NO, input is always 2 wires! Input is applied from V3a grid to V3b grid's cap to ground. To a first approximation, V3a and V3b split the input signal across their grid-cathode ends. V3a(gk) = V3b(gk). And their plate circuits are equal. So the output signals are equal.
Wait. V3a(gk) = V3b(gk) is incomplete because R21 and friends leak dynamic current which does flow in V3a but not in V3b. How much? A caveman approximation is to find the cathode impedance, 1/Gm, of V3b. We should know that 12AX7 at Fender-like conditions has Gm of 1uMho and thus Rk about 1K Ohms. So there is a 47K:1K split. V3b gets 98% of V3a dynamic current. There is a few-percent difference in output. As our EL84s are liable to be up to 30% apart, and we are not humping for point-oh-oh THD, this is not critical (and not worth a picture full of math). Although there is a "fix" in many Fenders using a similar stage.
> is there any point at all to matching the output tubes?
I think many new-age g-amp designers polish poop. And I never see them "match" for FULL output, where it *might* affect the sound. They obsess about idle matching, but the amp makes no sound at idle. We do want some equality from the teeniest sounds to the MAX, but it does not have to be a "match". Especially in a self-bias amp which can hardly get out of class A.
________________________________________
I'm sure the designer of the AC30 never looked at this that way. He'd probably take it funny if you showed it to him. If you bought the beer, he might mentor you on understanding tubes "intuitively" before you reach for the algebra. Most questions can be answered by thinking "if this goes up, that goes down, and about so much".
> reason why the AC components of i1 and i2 would have similar magnitude or shape to each other
Input is applied to V3a grid? NO, input is always 2 wires! Input is applied from V3a grid to V3b grid's cap to ground. To a first approximation, V3a and V3b split the input signal across their grid-cathode ends. V3a(gk) = V3b(gk). And their plate circuits are equal. So the output signals are equal.
Wait. V3a(gk) = V3b(gk) is incomplete because R21 and friends leak dynamic current which does flow in V3a but not in V3b. How much? A caveman approximation is to find the cathode impedance, 1/Gm, of V3b. We should know that 12AX7 at Fender-like conditions has Gm of 1uMho and thus Rk about 1K Ohms. So there is a 47K:1K split. V3b gets 98% of V3a dynamic current. There is a few-percent difference in output. As our EL84s are liable to be up to 30% apart, and we are not humping for point-oh-oh THD, this is not critical (and not worth a picture full of math). Although there is a "fix" in many Fenders using a similar stage.
> is there any point at all to matching the output tubes?
I think many new-age g-amp designers polish poop. And I never see them "match" for FULL output, where it *might* affect the sound. They obsess about idle matching, but the amp makes no sound at idle. We do want some equality from the teeniest sounds to the MAX, but it does not have to be a "match". Especially in a self-bias amp which can hardly get out of class A.
Hmm, I see: the grid voltage should be Vin(AC) + R21(i1+i2)(DC) and the cathode voltage should be (R21 + R22)(i1+i2). So the difference is more complicated than I thought. Sorry for being so sloppy.
I will call the differential inputs v+ and v- from now on, even though v- is at AC ground in this particular circuit.
\\$
$i_1 = k_1((v_+)_{AC} + R_{21}(i_1+i_2)_{DC} - (R_{21}+R_{22})(i_1+i_2) + \frac{V_C - R_{23}i_1 - (R_{22}+R_{21})(i_1 + i_2)}{\mu_1})^\frac{3}{2}\\$
$i_2 = k_2((v_-)_{AC} + R_{21}(i_1+i_2)_{DC} - (R_{21}+R_{22})(i_1+i_2) + \frac{V_C - R_{24}i_2 - (R_{22}+R_{21})(i_1 + i_2)}{\mu_2})^\frac{3}{2}\\$
[/comment]
So the unreasonably large dependence on mu2 and what seemed like a very small AC magnitude for i2 are both simply because I was missing important terms.
Affine approximation about ${i_n}_0$ (i.e. $i_n$ at $v_+=v_-=0$) followed by eliminating all DC components yields:\\
$i_1 \approx k_1'(v_+ - (R_{21}+R_{22})(i_1+i_2) - \frac{R_{23}i_1 + (R_{22} + R_{21})(i_1 + i_2)}{\mu_1})$\\
$i_2 \approx k_2'(v_- - (R_{21}+R_{22})(i_1+i_2) - \frac{R_{24}i_2 + (R_{22} + R_{21})(i_1 + i_2)}{\mu_2})$\\
where $k_n'=\frac{3}{2}{k_n}^\frac{2}{3}{{i_n}_0}^\frac{1}{3}$.\\
Assuming $\mu_n >> 1$, part of the last term can be absorbed:\\
$i_1 \approx k_1''(v_{+} - (R_{21}+R_{22})(i_1+i_2))$\\
$i_2 \approx k_2''(v_{-} - (R_{21}+R_{22})(i_1+i_2))$\\
where $k_1'' = \frac{k_1'}{1 + \frac{k_1'R_{23}}{\mu_1}}$ and $k_2'' = \frac{k_2'}{1 + \frac{k_2'R_{24}}{\mu_2}}$\\
If ${i_1}_0 = {i_2}_0$, $k_1=k_2$, $\mu_1=\mu_2$, and $R_{23}=R_{24}$:\\
$i_1 - i_2 \approx k''(v_+ - v_-)$\\
$i_1 + i_2 \approx \frac{k''(v_+ + v_-)}{1 + k''(R_{21} + R_{22})} \approx 0.$[/comment]
Okay, so... I think I can see that nothing is seriously broken in the most simplified case. This 12AX7 spec sheet claims nominal transconductance of 1.7mA/V (which I believe I can plug in for k') and nominal amplification factor of 92 (which I believe I can plug in for mu). So I get .6mA/V for k'', or differential gain of around 60, and, umm, "other" gain of 2.
Does this seem correct?
I don't think I really appreciated at all that R21+R22 is big and Rk is small until I understood your post (and without thinking about that, seeing why i1 ~= -i2 is much more convoluted...).
Next, I'm trying to understand how nonlinearities affect the shape of the outputs. It seems that any asymmetrical nonlinearities (in the sense of positive peaks being shaped differently than negative peaks) in the response would be greatly diminished by the symmetry of the circuit. If positive peaks were supposed to be pointy and negative peaks were supposed to be rounded, then that raises questions about what would happen if an inverted version of the signal were applied to the inverting input instead. So I don't think this can happen, other than to the <5% "i1+i2" term?
And then there is the potential for nonlinearity that is symmetric about 0. Is there a useful known way to model how it may affect the signal in the 3/2-law range?
Sorry if this is just a quibble, or if I am misunderstanding something, but I'm not sure why you have R21(i1+i2)(DC) in the above. I think that voltage term has both a DC and an AC component, i.e. (i1 + i2) goes up slightly when Vg1 goes up.
> any asymmetrical nonlinearities ... greatly diminished by the symmetry of the circuit
Yes. If the tail resistor is infinite, distortion is nearly cancelled.
As Malcolm says, "(i1 + i2) goes up slightly when Vg1 goes up", so linearity is somewhat better if input is balanced.
> nonlinearity that is symmetric about 0
To a first order, the rise of Gm on one side is cancelled by the drop of Gm on the other side. Typically Gm goes up slower than it goes down, so there is some 3d order compression, quite small.
Hasn't the long-tail been beaten to death? Extensive analyses have been published. Carry your salt- some are flawed. Here is a bibliography from 1947:
http://www.americanradiohistory.com...ld-1947-07-OCR-Page-0022.pdf#search="schmitt"
Broskie, Merlin and Kuehnel have writ-up the long-tail. It was a standard page in 1950s EE textbooks because most op-amps used it. Parallel studies have been done on the BJT long-tail, the foundation of all chip opamps. There's no need to start from scratch.
Yes. If the tail resistor is infinite, distortion is nearly cancelled.
As Malcolm says, "(i1 + i2) goes up slightly when Vg1 goes up", so linearity is somewhat better if input is balanced.
> nonlinearity that is symmetric about 0
To a first order, the rise of Gm on one side is cancelled by the drop of Gm on the other side. Typically Gm goes up slower than it goes down, so there is some 3d order compression, quite small.
Hasn't the long-tail been beaten to death? Extensive analyses have been published. Carry your salt- some are flawed. Here is a bibliography from 1947:
http://www.americanradiohistory.com...ld-1947-07-OCR-Page-0022.pdf#search="schmitt"
Broskie, Merlin and Kuehnel have writ-up the long-tail. It was a standard page in 1950s EE textbooks because most op-amps used it. Parallel studies have been done on the BJT long-tail, the foundation of all chip opamps. There's no need to start from scratch.
The BJT analysis is considerably simpler, actually; the differential in / differential out voltage transfer function is simply the tanh() function, the hyperbolic tangent of the input signal voltage. (Assuming perfectly matched devices, and a constant tail current.)
Because of the cancellation between the exponential transfer functions of the two individual BJTs, this transfer function is linear to first order for small voltage excursions - much more so than one individual BJT.
And, as PRR says, the first distortion term is cubic, so in principle, there is no 2nd harmonic distortion. In fact, all even harmonics disappear from the output. Only odd harmonics are present. Much like the hypothetical perfectly balanced push-pull output stage.
Do a Taylor Series expansion on the tanh() function, and this result drops right out.
-Gnobuddy
My thinking was that R19/R20 are too big to pass AC signals to the grid because C11a/C12 act as a short circuit. So any dependence on (i1+i2)(AC) in the grid-cathode voltage would be due to variation in the cathode-ground voltage, making the grid-cathode voltage (Vin(AC) + R21(i1+i2)(DC)) - ((R21 + R22)(i1+i2)) in total.
Quibbles are super welcome. I didn't study EE in school, so if I write something that is wrong on a technicality, it's likely it's because I actually don't understand it.
I checked out Broskie's web site, and it seems quite interesting. I am already looking into getting the Merlin and Kuehnel books.
As for tube-era literature, is there a good source of it online anywhere? I imagine there are a lot of expired copyrights, but I don't find much when I search for it.
This is the library:
Technical books online
You'll love them, there is a lot of math in there. I think in pictures so I don't, much. Pages and pages on LTP, Pierce oscillators, astable multivibrators and such, devices you can buy now four for a dollar. Or find them on the curb left out for the trash pickup.
Technical books online
You'll love them, there is a lot of math in there. I think in pictures so I don't, much. Pages and pages on LTP, Pierce oscillators, astable multivibrators and such, devices you can buy now four for a dollar. Or find them on the curb left out for the trash pickup.
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Here is a 1939 audio paper reviewing the common splitter/driver schemes, and describing the long-tail as "new". 1.8MB PDF
Here is a scan of Cathode Phase Inversion, O H Schmitt, Journal of Scientific Instruments, Volume 15, Number 3, March 1938. 250KB PDF The scan is not great. You can buy the article from the current copyright owner, but 20 GB Pounds! ($33)
Otto did a bunch of stuff.
https://en.wikipedia.org/wiki/Otto_Schmitt
Here is a scan of Cathode Phase Inversion, O H Schmitt, Journal of Scientific Instruments, Volume 15, Number 3, March 1938. 250KB PDF The scan is not great. You can buy the article from the current copyright owner, but 20 GB Pounds! ($33)
Otto did a bunch of stuff.
https://en.wikipedia.org/wiki/Otto_Schmitt
> Merlin and Kuehnel books.
I encourage supporting authors; but both have extensive "free tease" websites. Have you found them?
If you really love math, ask Kuehnel for the first edition of 5F6a. He solved the whole thing with a matrix method I had never seen, known but impractical in the 1950s when a "computer" was a grad student.
I encourage supporting authors; but both have extensive "free tease" websites. Have you found them?
If you really love math, ask Kuehnel for the first edition of 5F6a. He solved the whole thing with a matrix method I had never seen, known but impractical in the 1950s when a "computer" was a grad student.
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