Phaez gives away EL84 design

[GregS]

And FWIW the Marshall 18 watter designs are hit or miss. Even from the factory. Some sound OK and some buzz like bees. There are threads on AX84 that discuss using the right OT and speaker to avoid this. Ok, so now everyone has the same PT, OT and speakers.

Actually, this is the 20W Marshall which originals from the 60's to early 70's (when they were produced) are rare and very expensive when they turn up used. Clones seem to be rare also. Marshall relatively recently reissued it in their hand-wired series as the 2061x. I'm not familiar with the 18W version but perhaps the design is different enough that it doesn't suffer the problems that the 18W does.

Also, I did use clones of the original transformers from Mercury Magnetics and used the same Celestions the 2061x uses and built the cab as closely to spec as I could. Perhaps this has something to do with why it was a hit and not a miss.

One other thing: the brand of tubes made a big difference as well. The new Sovteks I first tried were prone to redplating so I used some new Chinese tubes. They didn't redplate but didn't sound as good as the Sovteks. It now has NOS tubes in it and is a lot better sounding with these.

I didn't want to search for any magic parts so I just designed the flaws out with brute force instead. I would prefere a simpler more eloquent circuit.

But if I build two amps I want them to perform the same. So I designed mine that way

Makes sense. I understand.

Greg

[Bruce / Mission Amps]

Just to clear up any misunderstanding I wasn't arguing with you about what you said. It was the notion presented in some of the other posts that you had to add conjunctive filters, diodes etc in order to make an EL84 power section sound good. That's what I was responding to. My point being that Marshall did it with a simple design without any of that stuff.

I also hate high gain amps.

Greg

Ah... OK.
I have buddy who owns and uses one of the 60's Marshall, small box 20 watt Lead El84 amps... no pedals, no distortion effects... well he does uses a small digital reverb/delay.
It sounds absolutely killer with his 4x10 cab using two reissue C10Qs and two Eminence C10N clones.
Turned down it is still plenty loud enough to cover all the R&B stuff he does too.

[Chuck H]

Yup, those Marshalls, some Traynors and some Besa Moogies run the EL84's over 400Vp and not much lower on the screens. El84's , to my ears anyway, become a whole different tube at that voltage. They get real loud and tight, the bottom end comes up AND they don't seem to fizz and buzz as much. These amps are often biased OVER max too. I had a Subway Blues that was biased at 14 watts per tube with a Vp of 400+. For this extra oomf there are two concessions: Tube life is cut down, a lot IME. And you lose the classic Voxy chime. This last one may not be a concession if your not about that sound and your more interested in getting big and loud. Actually tube life isn't that much an issue either since these tubes are relatively affordable.

My little Subway through it's 10" speaker was a piss ant on stage. But plugged into a 2x12 Celestion cab it could handle a room with no problems.

Chuck

[gui_tarzan]

I fixed a Haze 40w combo a while back and the tone was surprisingly good. The construction sucked, but that's beside the point.

[olddawg]

I guess it depends on your application. All drummers aren’t the same. I would prefer to use a 50 watt Marshall... But MOST of the gigs I play it isn’t a good choice. I use an El84 single 12” combo almost exclusively nowadays. I can turn the master up and it doesn’t kill me to transport it. I play with a guy all of the time that HAS to have his 100 watt Marshall half stack. At this point he has an effects processor plugged directly into the main amp input, so really any power amp would work (but I can never convince him of that). So it just takes up a lot of real estate, weighs a ton, and annoys club owners. And it’s so directional it’s problematic. It’s more coxmanship than reason imho.

[Gingertube]

Yup, those Marshalls, some Traynors and some Besa Moogies run the EL84's over 400Vp and not much lower on the screens. El84's , to my ears anyway, become a whole different tube at that voltage. They get real loud and tight, the bottom end comes up AND they don't seem to fizz and buzz as much. These amps are often biased OVER max too. I had a Subway Blues that was biased at 14 watts per tube with a Vp of 400+. For this extra oomf there are two concessions: Tube life is cut down, a lot IME. And you lose the classic Voxy chime. This last one may not be a concession if your not about that sound and your more interested in getting big and loud. Actually tube life isn't that much an issue either since these tubes are relatively affordable.

My little Subway through it's 10" speaker was a piss ant on stage. But plugged into a 2x12 Celestion cab it could handle a room with no problems.

Chuck

The Subway Blues has another thing happening tone wise, check out the deliberate imbalance in the concertina PI. 68K and 120K for the anode and cathode loads.
I found the sound actually improved dramatically going to 100K and 120K, of-course that was just my preference.


Cheers
Ian

[Chuck H]

The Subway Blues has another thing happening tone wise, check out the deliberate imbalance in the concertina PI. 68K and 120K for the anode and cathode loads.
I found the sound actually improved dramatically going to 100K and 120K, of-course that was just my preference.


Cheers
Ian

Come to think of it I never looked into the balance on the PI for that amp before gifting it to a friend. Now I wish I had. It's not a concertina PI though. More like a LTP with a constant current tail voltage. Then there's an elevation at the grid of the non inverting triode so that it can act as the NFB input. This would cause more local feedback in that stage. In other words... I'm not so sure the design is imbalanced. Especially since Mesa used it on several amps including some that could not benefit from an imbalance in the PI. My old Subway is local and the PI is stock (perhaps the only part of the amp that is ). If I get the chance I'll try to get it and scope it Just the same, if changing to a 100k load for the inverting triode improves the tone I'm all for it.

[Dave H]

I'm not so sure the design is imbalanced. Especially since Mesa used it on several amps including some that could not benefit from an imbalance in the PI. My old Subway is local and the PI is stock (perhaps the only part of the amp that is ). If I get the chance I'll try to get it and scope it

When you measure it you'll probably find that it's balanced. The negative feedback only goes to the non inverting triode grid (and not also to the tail as the usual Fender LTP circuit) which makes the balance worse. I suspect they had to make the plate load resistors 68k/120k to correct this.

[Gingertube]

You are right of-course - not a concertina but a LTP.
https://schematicheaven.net/boogieam...ubwayblues.pdf
The cathodes of V3 will sit at around, or just less than +2V giving a total of say 17V across that 15K tail resistor. Will tend to force V3a and V3b total current to approximately 1.1 mA but otherwise no real current source (hi Z) behaviour.

Assumming that V3a and V3b share this current equally then each triode has Ia of 0.65ma and that infers rp of 70K.

Doing a quick and dirty calc of Gain V3a (A1) vs Gain V3b (A2) from
A1/A2 = 1 + [Ra +rp/Rk(u+1)]
gives
A1/A2 = 1 + [68K + 70K/ 15K (100 +1)]
= 1.09

That is, the gain of V3a is 9% higher than V3b.

From that I would assume that for good balance that 120K should be 68K x 1.09 = 74K

OR if leaving the 120K then the 68K should be 120K/1.09 = 110K.

That would give best balance before FB is applied which should be your aim.

Does this make sense to you?

Cheers,
Ian

[Chuck H]

That would give best balance before FB is applied which should be your aim.

Does it make a difference that the NFB is applied at the cathode coupled, non inverting triode rather than at the tail? It seems to me that it would.

[Helmholtz]

You are right of-course - not a concertina but a LTP.
https://schematicheaven.net/boogieam...ubwayblues.pdf
The cathodes of V3 will sit at around, or just less than +2V giving a total of say 17V across that 15K tail resistor. Will tend to force V3a and V3b total current to approximately 1.1 mA but otherwise no real current source (hi Z) behaviour.

Assumming that V3a and V3b share this current equally then each triode has Ia of 0.65ma and that infers rp of 70K.

Doing a quick and dirty calc of Gain V3a (A1) vs Gain V3b (A2) from
A1/A2 = 1 + [Ra +rp/Rk(u+1)]
gives
A1/A2 = 1 + [68K + 70K/ 15K (100 +1)]
= 1.09

That is, the gain of V3a is 9% higher than V3b.

From that I would assume that for good balance that 120K should be 68K x 1.09 = 74K

OR if leaving the 120K then the 68K should be 120K/1.09 = 110K.

That would give best balance before FB is applied which should be your aim.

Does this make sense to you?

Cheers,
Ian

I don't quite agree with your calculations. The gain formula used seems to be incomplete. Using the same numbers and equal triode currents of 0.65ma (resulting in a total current of 1.3mA !) with the exact formula from the attachment below, I get a gain ratio of 1.7 (quite close to the ratio of the plate resistors as expected).

But what makes things complicated is that the triodes must not be viewed independently. Rather they are coupled by the common cathode resistor. When the current of the first triode increases, the increased cathode voltage causes a decrease of the current of the second triode. This leads to mutual gain degeneration. Also the second triode is forced to follow the first one (being driven by the common cathode voltage), which causes the gain of the second one to be lower with equal plate resistors. For low value cathode resistors, the gain of the second triode approaches 50% of the first one for equal plate loads. So the different plate resistors are needed to compensate for the difference.

As I don't see an easy way to do a complete calculation I hope that someone would offer to simulate the circuit.

[nickb]

As I don't see an easy way to do a complete calculation I hope that someone would offer to simulate the circuit.



[ATTACH=CONFIG]52482[/ATTACH]

Sim results:


Using equal 68k plate resistors
Gain Gaa input A to output A = 24.9 (i.e. same triode)
Gain Gab input A to output B = 21.8 (i.e. other triode)
i.e. Gaa/Gab = 1.14

To get Gaa=Gab you need to make the B plate resistor 78.5K.

I would have thought that you would need it be be balanced for both inputs. If you get it right for the signal input then the NFB will be way off and so fail to operate properly.

I do like the simplicity that using the negative supply brought. But given they did that it would have been easy to replace the 10K tail resistor with a simple current source.

[Gingertube]

Thanks nickb,

I thought I was close.

I used this as a guide.
http://www.valvewizard.co.uk/acltp.html

As Helmholtz says, the circuit has to be treated as a differential amplifier where V3a grid is the +ve input and V3b grid is the -ve input.

What I understand (or possibly miss-understand) is:
With any differential amplifier (SS or tube or whatever)
When balance is perfect:
- 2nd harmonic distortion is cancelled
- what distortion remains is mostly odd harmonics and is a result of non-linearities in the devices (triodes in this case) themselves.

Deliberatly unbalacing a differential amp is a way to allow more of the naturally resulting 2nd harmonic distortion (from the diff amp) to remain (not be cancelled).

The Push Pull output stage is a differential amp too, if it is perfectly balanced then the same 2nd harmonic distortion (from the output stage) cancellation occurs.

Feeding the output stage unbalanced inputs causes increased 2nd and other even harmonic distortions to occur.

AND (a point which is often missed)
The mechanism which produces Harmonic Distortion also produced Intermodulation Distortion, you simply can not have one without the other. It is often the Intermodulation Distortion products which can sound nasty.

Cheers,
Ian

[Helmholtz]

Thanks!

Maybe they chose this asymmetric design to introduce some even harmonics in the powerstage spectrum.

(simul-posting)

[Helmholtz]

I thought I was close.

OK, now I see. The formula from the valvewizard page A1/A2 = 1 + [(Ra+ra)/(Rk(mu+1))] is valid only for equal plate resistors. But you omitted a pair of brackets. So I couldn't verify your results.