Eric Barbour's 6BM8 One-Tube Reverb Imagined...

[dchang0]

Continued thinking-out-loud:

Cathode cap:

Ck = 1/(2*pi*Rk*f-half-boost-k) = 1/(2*pi*0.68k*5k) = 1/21.352 = 0.0468 ~= 47uF

Voltage rating: 25V is fine.

For the strapping resistor, let's choose 1W to be safe.

-----

Now for the coupling section to the reverb send... I'll start with the default values from The Valve Wizard's paralleled ECC82 transformerless reverb example.

Here's the new schematic:



Please let me know if you see any errors in my calculations or shematic--thanks!

[dchang0]

Ra = 6.8K, ra = 1.875K.

Zout = 6.8 * 1.875 / ( 6.8 + 1.875) = 12.75 / 8.675 = 1.47K

[dchang0]

Oops. Turns out that 3W metal oxide or 5W wirewound choices are 5K or 10K for Ra, not 6.8K. 5K has a better load line. At -14V bias, quiescent current is 25mA, just under 28mA for the 8A tank.

-14V/25mA = 560 ohm for cathode bias resistor. We have exactly that in a 1W metal oxide.

Cathode cap:

Ck = 1/(2*pi*Rk*f-half-boost-k) = 1/(2*pi*0.56k*5k) = 1/17.584 = 0.0568 ~= 50uF.

Ra = 5K, ra = 2K.

Zout = 5 * 2 / (5 + 2) = 10/7 = 1.43K

Max power, load should be ra=2K doubled = 4K.

RI should be 20K.

13mA at 12Vp-p at the bias point of -14V on the Ra=5K load line.

[BiBi]

While you are certainly better at math than I am, methinks you aren't getting the impedance issue correct. Driving such a low impedance (8ohm) directly from the plate of the tube ain't gonna work well! Reread what Valve Wizard says about using an output pentode-- you'll want to use the higher impedance tank like the 8F or you would need an appropriate output transformer!

[dchang0]

Yup, you're correct. I spent all last night struggling with the impedance matching. Couldn't figure out the formulas to calculate the proper load. Couldn't even figure out which formulas to use! I'm stuck right now--if you can help walk me through the calculations, that would be awesome!

[BiBi]

Well I would love to walk through calculations but I tend to always get stuck in the mud!
I think the first thing is to find the optimum load for the tube right? So we know that's 8k in SE pentode. Usually I would think up that a bit in triode, say 10k. (Now you can see I'm not very scientific!).
Since the load is resistive not reactive, you'll want to bias it more like a preamp than a power amp connected to a transformer which I think you got already.
So what I do is look at the Philips datasheet, which states 27mA plate current in SE at 270V. From your datasheet, 275V and 27mA puts the bias around 18-20 which gives you a cathode resistor of 680.

[dchang0]

Hey, any help you can give me is very much appreciated. Even if it's as simple as pointing me to the right article or book or online tutorial...

I sorta glossed over the load calculations above, but the optimum load for this triode-strapped pentode 6BM8 as shown is 4K. That's as shown, with all the calculated resistor and capacitor values. I used the formulas shown in The Valve Wizard's online article on reverb drivers, and came up with the first bridging resistor (RI = 20K).

But now, I'm trying to figure out the proper value of the 820K resistor (formerly 1100K) AND the correct impedance of the reverb pan. This requires a bit of RC circuit theory to go back and "reverse-calculate," so I'm having to go back through the formulas and try to figure out which one applies to this topology.

Oh, I forgot to post the latest schematic. This is 2.0a, based on the 5K/5W Ra resistor. Barbour went for the 10K impedance via the transformer, resulting in a lower load line. I went up, towards a hotter load line, which might be too much for any of the Accutronics pans. It is entirely possible that running a triode-strapped pentode in the driver section is a mistake, and I may have to step back down to pentode operation.



P.S. I'm slomping through the circuit theory chapters in the ol' Radiotron Designer's Handbook, 4th. Ed. Sooner or later, I'll figure this out. But man, is it PAINFUL...

[dchang0]

Well, I'm reading the Rod Elliot article referenced by The Valve Wizard's article.

In it, Rod says "the value of R7 (shown in Fig. 5) must be selected based on the coil impedance." Later on, he says "R7 is based on an estimation, where the resistor value is roughly 20 times the coil's 1kHz impedance. Reduce the value of R7 for less treble response and vice versa." Then, he says, " This simple circuit (Fig. 6) has a deliberately limited output impedance.... This is the equivalent of using a resistance in parallel with the coil (R7) as shown in Figure 4-all drive circuits require a high frequency limit."

But all of this theory is explained via solid state circuits such as the one in Rod's Fig. 5. Translating this over to valve circuits is waaaay beyond my skills (or interest).

I think I'll guesstimate my way through this. Start with the 8F tank with its high impedance, cut down the RMS voltage and current of the driver output to where it's close to the safe specs of the tank at full saturation, and call it a day. (Forgetting about optimum impedance matching and tone...)

[dchang0]

Sent an email to Merlin, the Valve Wizard himself, to see if he can help me through the example he gave on the paralleled ECC82s.

It's pretty much all I need to know to pull this off.

[dchang0]

Taking another approach. I am reverse-engineering the Ampeg Reverberocket II and basically trying to match the performance of the 6BM8 t.s.p to that of the 6U10 triode. They are capable of producing similar output currents. The 6U10 as configured is set for a load line of 34mA at 340V. My 6BM8 is currently set for 55mA at 275V, and that line is on the hot side. Let's try it with a 10K Ra resistor (same as the Ampeg) to bring the load line down to 27.5mA at 275V. The Ampeg appears to be biased at -9V and at quiescent current of 9mA and quiescent voltage of 250V. This is a slightly colder biased tube.

This raises a choice: run the 6BM8 pentode as a pentode or as a triode? The pentode is capable of producing about the same current range as the 6U10, and the t.s.p can product roughly 20mA more. In pentode mode, the curves are far more linear, but the bias voltage range is much wider on the t.s.p (0 to -38V vs. the pentode's 0 to -10V). The 6U10's bias voltage range is 0 to -24. That means the t.s.p 6BM8 is probably a better fit.

Middle of the road bias point of -14V as before / 13.5mA = 1K cathode resistor. That's the same as the Ampeg

Cathode cap:
Ck = 1/(2*pi*Rk*f-half-boost-k) = 1/(2*pi*1k*5k) = 1/31.4 = 0.0318 ~= 33uF. Xicon makes a cheap 450V model. The Ampeg uses a 10uF instead. They must prefer a different roll-off frequency than 5kHz. I'll keep my 33uF.

Ra = 10k, ra = 8V / 3.7mA = 2.162K. Best load = 4.25K.

Power rating is actually 1.8W, so we could use a 2W or 3W for Ra. Use 3W to be safe.

RI = 7.5K/2W. Ampeg does not use an RI or the "constant current" set up as shown in The Valve Wizard's example. I could copy either Ampeg or the Valve Wizard here. Might as well copy Ampeg to save components and also gain this huge gift: I now know that I can use the 8F 1925-ohm high-impedance tank!

Can't seem to find a US supplier of these online yet. Ampeg uses the 1475-ohm 4F tank; I *could* go down to an 8E tank at 800 ohms and just stick a 680 ohm resistor in front. Close enough without getting into the gory details of impedance vs. inductance.


Okay, here's the new schem, Rev. 2.0b. Ampeg-style driver, but using a triode-strapped-pentode 6BM8. Same recovery section as before. (I will double-check the gain produced in the Ampeg to make sure it's about the same.)

[dchang0]

Well, this is interesting. Amplification factor mu = 55 for the 6BM8 recovery triode and mu = 92 for the 6U10 recovery triode (roughly 1/2).

And for the driver sides: mu = 6.25 for the t.s.p. 6BM8 and mu = 17.5 for the 6U10 driver triode.

So the Ampeg circuit has quite a bit more voltage amplification than the 6BM8 setup, with the output currents set up to be similar to one another. I could screw around with the load lines some more to get a better compromise, balancing voltage and current similarities...

There's also the matter of the 8F tank being hard to find and also being different than the 4F tank's impedance.

What to do?

[dchang0]

Awwww, yeah!

It hit me that if I chose the 8E tank instead of the 8F, I'm dropping the impedance from 1475 to 800 ohms. Oversimplifying the driver circuit into V=IR, if V is constant, then I = I0 * 1475/800 = 13.5mA * 1.84 = 24.84mA. The 5K Ra setup I had before was producing about that much quiescent current. AND, because the curves are better in that area, mu = 14, which is much closer to the 17.5 of the 6U10.

So, let's go back to the original Ra=5K design on the driver side.

On the recovery side, we'll keep the Barbour-inspired design (which is middle-of-the-road, most clean headroom, least clipping). It may not have as much gain as the Ampeg recovery, but we can probably control that using the mix resistor. And I don't necessarily want the Ampeg sound, just the Ampeg-style capacitor coupling directly to the reverb tank.

Here is the new schem. 2.0c. This is what I'll build, barring any errors.

[jimboyogi]

I don't think that you'll have a problem with the lower gain on the recovery triode. I'm just completing a single tube reverb with a 12DW7, and for the recovery I've had to reduce the gain down to ~ 30 to keep the reverb from getting out of hand. As you said, you can just tailor the mix resistor to get the balance that you want.

[dchang0]

Thanks, Jimboyogi--I'll leave it as is, then.

Time to order parts!

[BiBi]

Obviously I couldn't help out as much as I would have liked too-- I think you're far beyond me in terms of understanding. One question I had was your calculation of the optimum load for the triode strapped 6BM8. How did you derive that? I've always looked at the internal resistance of the tube-- which is 20k for the pentode but obviously this is much lower in a triode.
As well, I found this schematic yesterday which uses the ECL82/6BM8 as a current source driver. It also uses a second tube the 12AX7/ECC83- which is something you may or may not have to look into. If I remember your schematic correctly you were driving the t.s.p. and a second stage from the plate of the first 12AX7 triode?