6U8A reverb driver pentode input impedance question

[rob_h]

Hello again you wizards of the worlds of space and electron beams!                                                                         
I inherited a couple of reverb pans; an Accutronics 4FB2A1C, and a Hammond.  The Accutronics Zin=1475R and the Zout=2250R.  After looking around I found an article by Merlin on reverb driver and recovery amps.  In it he describes the standard Fender type with a 22.8k/8R driver transformer type, and others.  Then he describes, lucky for me, a driver example for the Accutronics reverb unit like I have.  He uses 2 12AU7s in parallel.  This translated, I believe, almost exactly into the single 2.5w triode of a 6U8A, leaving the 3w pentode for the recovery amp.  I used the RCA "Resistance Coupled Amps" table #12 for the 7199 pentode for the 6U8A pentode, biasing for Vo/Vi=82 with 180vdc to try and keep the noise and microphonics low.
     Then I read a "Note" in the RCA material that left me wondering what to do.  It reads "Note: The listed values for Eo are the peak output voltages available when the grid is driven from a low impedance source. The values listed for the cathode resistors are optimum for any signal source.  With a high impedance source, protection against severe distortion and loss of gain due to input loading may be obtained by the use of a coupling capacitor connected directly to the input grid and a high value resistor connected between the grid and ground."
     Can anyone define low vs high resistance source, and a "high value" resistor; 1M, 470k?        Schematic & RCA tables attached.  Thanks! Rob H

[terminalgs]

    Can anyone define low vs high resistance source, and a "high value" resistor; 1M, 470k?        Schematic & RCA tables attached.  Thanks! Rob H

I suspect by low Z sources, they are thinking <=500ohm.  with 2250K Zout of the tank, I'd try a 22K grid leak resistor and connect the tank directly to it.,  Like an Ampeg Gemini I, or a Magnatone M10A.  

on the driver side, impedance matching is more important because you are delivering current.  on recovery, I think you are just picking up a weak little single and applying gain.
 

[PRR]

To answer the immediate problem:

> pentode, biasing for Vo/Vi=82
> peak output voltages available when the grid is driven from a low impedance source.

Your selected condition can output 32V peak.

Your reverb tank is unlikely to deliver much over 10mV (0.010V) rms, say 14mV peak.

14mV * 82 is 1.15V peak output.

Even if there's some mis-guessing here, you seem to be NOwhere near the 32V output limit.

So it won't distort.

(Also you'd *never* want 32V in the middle of an amplifier. Certainly not before your dry/wet mixer.)

[terminalgs]

Accutronic's page has good stuff under "Application"

http://www.accutronicsreverb.com/

including:

One type of equalization that should be considered in all applications is to add a low cut filter to the recovery amplifier. A filter with a 50 Hz to 100 Hz cutoff will lessen the effects of rumble when the unit is mechanically shocked.

[rob_h]

Thanks for your reply!  I have to work after reading your responses to comprehend and process the information that rolls out.  I enjoy it greatly!
Please correct or confirm or add to anything I have written here.

"Your selected condition can output 32V peak." _Vo/Vi=82, so at 32v out the input would be 33/82=390mv, correct?
"Your reverb tank is unlikely to deliver much over 10mV (0.010V) rms, say 14mV peak."  _Very good to learn.
"14mV * 82 is 1.15V peak output."  _That is very good, I think, compared to the "65 Super Reverb" at 100mvac, and the "65 Deluxe Reverb" at 80mvac.
If these numbers are correct, the reverb wet signal should be 1.15vpeak max available into the Reverb Pot.  Does this, being more than the reverb amps mentioned, mean that this bias selection would be acceptable if the output of the tank remains below the maximum of 14mv?

Thanks!

[HotBluePlates]

    Can anyone define low vs high resistance source, and a "high value" resistor; 1M, 470k?        Schematic & RCA tables attached.  Thanks! Rob H

I suspect by low Z sources, they are thinking <=500ohm.  ...

It's much simpler than that.

... "Note: The listed values for Eo are the peak output voltages available when the grid is driven from a low impedance source. ...  With a high impedance source, protection against severe distortion and loss of gain due to input loading may be obtained by the use of a coupling capacitor connected directly to the input grid and a high value resistor connected between the grid and ground."

Can anyone define low vs high resistance source, and a "high value" resistor; 1M, 470k?

In your RC chart info, look at "Diagram No. 1" in the column next to the note you quoted. Notice Rg (grid resistor of the following stage) in particular.

For a.c., Rg is connected by C so that it is in parallel with Rp. Look at the values of Rp and Rg on the chart; they are either equal, Rg is 2x Rp, or Rg is 5x Rp. Because the resistors are in parallel due to C, if Rg of the following stage is not several times bigger than Rp, it reduces the effective value of the plate load for the tube, which reduces gain.

Ex. Rp = 100kΩ, Rg = 100kΩ, 220kΩ, 470kΩ or 1MΩ.
Total load resistance (assuming "large" coupling cap):
Rp = 100kΩ, Rg = 100kΩ:  R load = 50kΩ
Rp = 100kΩ, Rg = 220kΩ:  R load = 67kΩ
Rp = 100kΩ, Rg = 470kΩ:  R load = 82.5kΩ
Rp = 100kΩ, Rg = 1MΩ:    R load = 91kΩ

In all cases, if you calculate gain using a particular plate load but the following stage's grid resistor is only 2x the value of the previous stage's plate resistor (or lower), gain may be noticeably reduced.

... Can anyone define low vs high resistance source, and a "high value" resistor; 1M, 470k?

If your amp's output tubes have 100kΩ resistors from their grids to the bias supply, then you can't have a phase inverter with 470kΩ plate load resistors. The "source impedance" (the 470kΩ plate resistors) is too high and the smallish grid resistors of the following stage cause the actual output voltage and gain to collapse compared to the values you might calculate for the phase inverter without considering the following stage.

Conversely, if you have a preamp stage with a 100kΩ plate resistor, and the "grid resistor" for the following stage is a 1MΩ pot, then the 100kΩ looks like a "low source impedance" by comparison. Gain and output voltage are what you expect.

All of the above (and the R-C Chart itself) mostly applies to a voltage amplifier stage, like most preamp stages. The reverb driver is a power amplifier stage because output current is important in shaking the reverb pan input transducer (as Merlin points out in his info). So just follow Merlin's example in how to calculate a desirable driver stage.

[rob_h]

Response to Terminalgs:
Thanks for your input.  I have read the applications a number of times and still have some questions.  The RCA paper I used the tables from for the pentode design says that the values are for 100hz, which I assume is the cutoff frequency.  They also show how to increase or decrease the frequency if wanted.  I didn't think about the "rumble" when the springs are jarred, good thinking.  Would lowering the cutoff allow 60hz hum in, or it not a concern in a reverb driver application?
Thanks

[PRR]

> I found an article by Merlin on reverb driver and recovery amps.

Maybe this?

http://www.valvewizard.co.uk/reverbdriver.html

Did you want to post your curves and plots to compare to Merlin's? While we shouldn't post huge chunks of someone's work here, links to Merlin's site and short quotes are reasonable use.

If we are getting into reverb design we should maybe also find info about Fender's two prime designs (the 6F6/6V6 standalone and the 12AT7 used in many heads).

[terminalgs]

Response to Terminalgs:
I didn't think about the "rumble" when the springs are jarred, good thinking.  Would lowering the cutoff allow 60hz hum in, or it not a concern in a reverb driver application?
Thanks

In a guitar amp, I don't think you'll care much about anything lower than 100Hz coming from the reverb tank. However, you can leave The 'low-rumble cap' off completely, If you have 60hz (or 120hz) power supply noise  at that point, you probably have a systemic filtering issue that probably appear elsewhere as well.   I wouldn't worry about it at this point.

looking at your schematic: I would build it initially without the cg1 cap,  make rg1=22K,  I'd add a grid stopper of 10K.  At that point the recovery grid is identical to an Ampeg's recovery for a similar tank (one exception, they also added a .005uf cap from G to K, which you can play with once you have it working).   look at an Ampeg reverbrocket or Gemini, mid-60s)

In fact, thats pretty much exactly how I'd tie it back into your primary signal, Ampeg style:  loose C20 and P20, and that 330p, (what is C20 for? you already have the previous stage's coupling cap  C5 to keep DC off  the grid).  If the signal is too hot you can add a resistor where C20 and P20 and let that resistor and the 470K grid leak be a voltage divider.

Instead of having R20 100K and R25 330K to mix,  Ampeg and Fender made R20 be 1M to 2.2M and dropped R25.  Sometimes the 1M R20 was paralleled with a 10pf or something like that.  I don't see why your mix wouldn't work tho...

[HotBluePlates]

... In a guitar amp, I don't think you'll care much about anything lower than 100Hz coming from the reverb tank. ...

Lows going into the reverb tank can turn the effect into a muddy mess pretty quickly. But I agree that he can find this out the hard way and adjust to shave bass going into the tank as-needed.

[rob_h]

Quote: "Your selected condition can output 32V peak.
Your reverb tank is unlikely to deliver much over 10mV (0.010V) rms, say 14mV peak.
14mV * 82 is 1.15V peak output.
Even if there's some mis-guessing here, you seem to be NOwhere near the 32V output limit. "

I chose the gain of 82 because I knew the 12ax7 gain is 100max.  I guessed that a gain of 82 would be ok because most 12AX7s aren't used with a 100% gain.  Is there a problem with the max voltage even though the input, as you stated, is to be 14mv which yields 1.15v peak output? 

[HotBluePlates]

Is there a problem with the max voltage even though the input, as you stated, is to be 14mv which yields 1.15v peak output? 

PRR's just saying you won't get anywhere near the maximum voltage swing the stage is capable of producing, so there's no need to sweat it.

... the 12ax7 gain is 100max.  I guessed that a gain of 82 would be ok because most 12AX7s aren't used with a 100% gain.  ...

You'll find over time that triodes rarely have a stage very close to their mu (amplification factor). Maximum stage gain of a 12AX7 in typical guitar-amp use is around 60. [There are circuits that can help a 12AX7 get closer to a gain of 100, but they either waste another triode, use solid-state, or make the stage more susceptible to hum/noise.]

[rob_h]

Thanks, I'm sweating it again, but not for long.
I have a 380Vct transformer for this amp, which @ x.707=268.66v minus tube rectifier drop.  It is a low power amp, <4w, so minimal drop across the rectifier, but the voltage “seems” to be giving me concern about the reverb driver.  I can use a FWBridge and drop down to 300v, anyway, question:
     
If I use 300v in the 6U8A triode graph and get the needed 12ma p-p at Zac=11k, the Vg1q=-4v with Ipq=6ma. Ip=(1ma to 14ma), so, 13ma p-p with an 8v p-p swing on Vg1q.  The 8.25ma+ and 4.5ma- is a (1.83:1) ratio.  [6U8A TRIODE 300V22k240v4v]PDF
     
If I use 250v, the best I can get with Zac=11k is 12.3ma p-p with Vg1q=-5v and Ipq=5.75ma.  That is a ratio of (3.23:1), 9.7ma+ and 3ma- around the Vg1q of 3ma. [6U8A TRIODE 250v22k215v5v] PDF
     
My concern is that the current being so high above Vg1q compared to below it will be a problem.  If not, there is not a problem.  I know this is a property of real triodes, and that only a perfect triode current will be 50% + and – from Vg1q, but is there a standard limit to the ratio?  I just don't know the sound of the waveform I know I'll see.  If it's nothing, no sweat!

Thanks

[HotBluePlates]

... If I use 300v ... get the needed 12ma p-p ...

You graphs show ~12mA and ~14mA peak. If you need 12mA peak, you have to idle around 6mA or more. In other words, current has to swing as far down as up.

That's for class A anyway; you might be confused because you don't have to do that with push-pull class AB, but in your case there is no "pull" side and no other tube to be swinging current when your 6U8A goes into cutoff.

You'll need a different operating point with less bias voltage and more idle current.

[rob_h]

Sorry, I didn’t explain what I graphed very well.  On the “250v” graph, Vplate=250v.  The DC impedance, resistance, is from a 22k plate resistor “Rp";   DC load line in red. The AC load line, in violet, is from Rp||Rp1, Rp1=22k also.  So the AC load, being = Rp||Rp1=22k||22k=11k.  The AC and Dc load lines intersect at the quiescent point of Vg1=-5vdc.  The input signal swing of 8v peak-to-peak will cause an output voltage of 78v to 209v from Vq=181v, and an output current of 0 to 12.3ma Vp.  This will be split by Rp and Rp1 giving each 0 to 6.15ma.  

I found what I was questioning, in a book of all things, to be the ubiquitous harmonic distortion.  The equation for it is:

Where A=minimum peak voltage, B=quiescent voltage, and C=max peak voltage, %Harmonic distortion = H%=(100(AB-AC)) / (2(AB+BC)).

In the 275v graph, the H% would be equal to 10.61%.  If this is correct, the distortion I was concerned about is ok, is it not?

Values for 250v triode on .sch attached.

[HotBluePlates]

Sorry, I didn’t explain what I graphed very well. ...

No, I followed your graph.

But you're not following what I'm saying. Look back at the Valve Wizard page, and where the green dot is for his idle condition. It's halfway between the upper and lower current extremes, and the lower current extreme doesn't reach cutoff.

Your idle point is 3mA; you expect to swing down to 0mA but up to 12mA. This won't work with a single tube without severe distortion (and tell me when you heard a distorted reverb rather than a clean reverb of a distorted guitar sound).

Look at the attached graph; the green is where you have a roughly symmetrical current swing. This is only about half the current swing you want, and is already distorted because it takes more voltage input on the low current portion than the higher current portion for the same current excursion. From the -5v midpoint of the green line, a positive swing of 2v reaches the upper current peak, while a negative swing of 5v is required for a similar current change in the other direction.

What I'm saying is your idle point needs to be shifted up to about 6mA, or the halfway point of the swing you want. Squinting hard, that might be a bias of ~2.25v to 2.5v. You leave your d.c. loadline where it is, and shift your a.c. loadline to intersect at ~6mA of current.

A byproduct is that distortion will be reduced somewhat. Unfortunately, your formula for distortion may not apply because if the input signal was as big as you assumed by using voltages only, you've already partially half-wave rectified the output of the tube (again, your bias is pushed towards class AB, but this isn't a push-pull stage so it doesn't work). To bias the tube towards class A, you have to idle at a current that is at least half of your planned peak current.

[rob_h]

Ok, I think I have it.  One more question:
If I increase Vp to 300 and lower Zac to 10k from 11k, the graph is much better, from Vg1=(-4v to 0v) Ip=(6ma to 12ma), however if I bias at -4v, when Vg1 drops to -8v, Ip is still a little over 1ma.  So, does this mean that the realistic current swing on the attached graph is 10ma, Blue, because I need to take 1ma off the top too, leaving 5ma+ and 5ma-?

[PRR]

> take 1ma off the top too

1mA is small in this case. Smaller than tube-to-tube variability. I would not sweat it.

What I see is 7mA-8mA idle, 3V bias, giving a larger more symmetric swing. That does mean 1.4W Pdiss; can this tube handle it? And a few more mA for the power supply to make and filter, though you were already promising 6mA so 8mA should not break the deal.

[davew]

For anyone still interested in this here is another variation

http://www.prowessamplifiers.com/schematics/misc/SJB_Ant.html

This use a 1475ohm in and 600 ohm out tank.