Question on parallel stages in 12AX7

[Steve A.]

For a 12AX7 at the input of the amp, if you were to run the two stages in parallel, would you:

- divide the value of the plate and cathode resistor by two

and...

- multiply the value of the cathode resistor bypass cap by two?

I think that sometimes the second half of the equation is ignored, especially if the Ck cap is 25uF, which is plenty big for two stages in parallel. (No sense upping that to 50uF).

I ask because a very famous boutique amp runs the first 12AX7 in parallel with a 68k plate resistor and a 1k2 cathode resistor (which seem fine to me) but a 0.68uf Ck cap, which to my ears makes the sound a bit thin and tinny. I doubled up the 0.68uF cap and I thought it sounded a lot better.

Any thoughts on this?

Thanks

Steve Ahola

[tubeswell]

For a 12AX7 at the input of the amp, if you were to run the two stages in parallel, would you:

- divide the value of the plate and cathode resistor by two

and...

- multiply the value of the cathode resistor bypass cap by two?

I think that sometimes the second half of the equation is ignored, especially if the Ck cap is 25uF, which is plenty big for two stages in parallel. (No sense upping that to 50uF).

I ask because a very famous boutique amp runs the first 12AX7 in parallel with a 68k plate resistor and a 1k2 cathode resistor (which seem fine to me) but a 0.68uf Ck cap, which to my ears makes the sound a bit thin and tinny. I doubled up the 0.68uF cap and I thought it sounded a lot better.

Any thoughts on this?

Thanks

Steve Ahola

Hi Steve

Don't think that the bypass cap value needs to be doubled, as it will still pass .68uF worth of frequency roll-off from the voltage swing at the cathodes (speaking laconically). From what I understand, voltage changes wave amplitude, not frequency, so both cathodes would still be swinging at the same frequencies as if there was only one cathode at that frequency. So maybe you just like the effect of lowering the frequency roll-off slightly?

[Steve A.]

Hi Steve

Don't think that the bypass cap value needs to be doubled, as it will still pass .68uF worth of frequency roll-off from the voltage swing at the cathodes (speaking laconically). From what I understand, voltage changes wave amplitude, not frequency, so both cathodes would still be swinging at the same frequencies as if there was only one cathode at that frequency. So maybe you just like the effect of lowering the frequency roll-off slightly?

So the formula for the RC network would not apply here? E=MC squared... oh, I guess that is something else!

I would think that as the cathode resistance is cut in half, the Ck value would be doubled, to maintain the same frequency rolloff.

Thanks! I guess I won't give the boutique builder hell for allowing such a design to be passed around the internet...

Steve Ahola

[tubeswell]

Actually I guess you must be onto something - I wasn't thinking about that formula.

Speaking of which, what is the correct dang formula?

I have seen it written several ways.

1) On the Valve Wizard site - it is written as; f(half boost) ~ 1/(2Pi*Rk*Ck).

2) The page on the Aiken site has had that part of the formula removed, with a note saying:

"Appendix C: The math behind the lower cutoff frequency due to a partially-bypassed cathode
This section under construction - thanks to Jean-Pierre Trolet for pointing out an error in the original document. I will update it when I get some time to review and rewrite the document."

3) Amp Kat gave me the following formula last year (I really trust Amp Kat) (1/2.pi) x resistance x capacitance = freq.

4) Wakculloch said that was incorrect. He said it should be the reciprocal of 2 x pi x R x C, with R in ohms and C in farads.

5) Tubis had another formula (which I can't paste in here) - something like the square root of (1 + R/Ri)/(2piCR)

Somebody must be an authority on this.

[Steve A.]

Actually I guess you must be onto something - I wasn't thinking about that formula.

Speaking of which, what is the correct dang formula?

I have seen it written several ways.

1) On the Valve Wizard site - it is written as; f(half boost) ~ 1/(2Pi*Rk*Ck).

2) The page on the Aiken site has had that part of the formula removed, with a note saying:

"Appendix C: The math behind the lower cutoff frequency due to a partially-bypassed cathode
This section under construction - thanks to Jean-Pierre Trolet for pointing out an error in the original document. I will update it when I get some time to review and rewrite the document."

3) Amp Kat gave me the following formula last year (I really trust Amp Kat) (1/2.pi) x resistance x capacitance = freq.

4) Wakculloch said that was incorrect. He said it should be the reciprocal of 2 x pi x R x C, with R in ohms and C in farads.

5) Tubis had another formula (which I can't paste in here) - something like the square root of (1 + R/Ri)/(2piCR)

Somebody must be an authority on this.

I think that people holding their breath waiting for Randall to update his site have been passing out due to lack of oxygen... I'm just glad that he has left up what he has there.

As for the RC values, I just figure that as you raise the capacitance you increase the bass response- I look at things proportionally rather than specifically. But I do forget if I should be looking at the ratio or the square of the ratio...

I guess you could say that is like tuning your guitar by ear? I do all of this stuff very intuitively, just waiting for the right inspiration to strike. And sometimes I screw up...
I changed 4 coupling caps going into and out of the PI from 0.01uF to 0.022uF to make the amp a little darker before flashing on the effective 0.34uF value of the Ck cap on V1. D'oh!

But I do try to learn enough about the electronic theory to keep from killing anybody...

I was just Googling my site for RC and came up with this gem of a thread from 2001 (lots of great tips and information in it!):

http://www.blueguitar.org/new/text/t...tra%20Gain.pdf

Those old threads at AMPAGE are like gold!

Steve Ahola

[raiken]

I think that people holding their breath waiting for Randall to update his site have been passing out due to lack of oxygen... I'm just glad that he has left up what he has there.

Heh...I forgot about that article. I do need to get around to writing that up. I devised all the formulas, but never got around to rewriting the paper.

The gist of it is this - the circuit is a shelving filter, so it has a lower and upper frequency breakpoint. There is no 3dB point, only the two frequency breakpoints that form the "shelf" where it transitions from the gain of the unbypassed resistor to the gain of the fully-bypassed resistor. You could characterize it by the center frequency of the transition band and the upper/lower gains, or by the upper/lower gains and the two breakpoints.


Randall Aiken

[Steve A.]

Heh...I forgot about that article. I do need to get around to writing that up. I devised all the formulas, but never got around to rewriting the paper.

The gist of it is this - the circuit is a shelving filter, so it has a lower and upper frequency breakpoint. There is no 3dB point, only the two frequency breakpoints that form the "shelf" where it transitions from the gain of the unbypassed resistor to the gain of the fully-bypassed resistor. You could characterize it by the center frequency of the transition band and the upper/lower gains, or by the upper/lower gains and the two breakpoints.


Randall Aiken

Randall:

Long time no see! I figured that you must be hobnobbing with all of the superstars of the guitar amp world like Mike Soldano and Reinhold Bogner- and we must not forget James Brown, designer of the highly aclaimed Peavey Classic 30!

I guess now would not be a good time to ask if you are putting together any more resistor kits... (I never did buy the one with metal film resistors, just the one with the 2 watt carbon film resistors.) As I recall that ended up being a real fiasco with you saying (with good reason) "never again!"

Steve Ahola

[martin manning]

Actually I guess you must be onto something - I wasn't thinking about that formula.

Speaking of which, what is the correct dang formula?

I have seen it written several ways.

1) On the Valve Wizard site - it is written as; f(half boost) ~ 1/(2Pi*Rk*Ck).

2) The page on the Aiken site has had that part of the formula removed, with a note saying:

"Appendix C: The math behind the lower cutoff frequency due to a partially-bypassed cathode
This section under construction - thanks to Jean-Pierre Trolet for pointing out an error in the original document. I will update it when I get some time to review and rewrite the document."

3) Amp Kat gave me the following formula last year (I really trust Amp Kat) (1/2.pi) x resistance x capacitance = freq.

4) Wakculloch said that was incorrect. He said it should be the reciprocal of 2 x pi x R x C, with R in ohms and C in farads.

5) Tubis had another formula (which I can't paste in here) - something like the square root of (1 + R/Ri)/(2piCR)

Somebody must be an authority on this.

Tubes, go back and look at the Valve Wizard's Common Cathode Gain Stage article again; P. 25 Eq. VI has the details included.

Note that a paralleled stage is passing twice the current at the same voltage drop, and accordingly Ra and Rk are halved, as is ra (anode resistance) since the two triodes are in parallel. The numerator of the equation will therefore have the same value for single or paralleled triodes (voltage gain mu is constant), leaving you with 2(pi)RkCk in the denominator... so, you must double Ck to make up for halving Rk if you want the same frequency response.

MPM

[mooreamps]

Not that I do mine this way, but here is what others have done.....



http://www.schematicheaven.com/newam...s_chieftan.pdf

http://www.schematicheaven.com/newam...ss_clubman.pdf

-g

[HTH]

Steve,

The last time I used parallel triodes on the first gain stage I had it setup with a 220k plate, 1k5 cathode and a 22uF bypass cap. I tried a 100k plate but it was too weak for my needs - the 220k plate gave just the right amount of gain without any compression or fizzies.

The preamp was essentially an AC30 top boost with some tweaks to give you perspective.

[MattT]

kg has had a small thing on it on his site for years...here.

I did this in my Rocket and did what you said Steve...except I didn't double the 25uF because I thought 25uF was plenty (just like you said in the op). Specifically it's 100k/1k/22uF right now.

There's also a link on kg's site to a little test (with graphs!) that Steve Bench did on this but it was more about what happens when they distort...which isn't really applicable for the input stage (I wouldn't think?).

HTH says when he used 100k Rp on his AC30-inspired preamp (the Rocket is the same thing) that he didn't get enough gain. I'm getting about the same gain as I would if I had used a single triode with a 220K/1k5/22uF set up. I put 100mV in and get 6.2v out (true RMS)....edit: and that's with a supply voltage of only ~215vdc (Vp is like ~115vdc). Besides...the unbypassed Rk on the second stage can be tweeked (or bypassed) to get any lost gain back if needed.

[BK-Amps]

Paralleling the tube essentially creates a 'hybrid' 12AX7 with half the internal plate resistance of 1 section. Keeping the values the same, i.e. Rp=100k and Rp=1.5k will yield a little more gain than one section alone and have lower noise. No need to double, halve or whatnot.

Use Randle Aikens equations with 1/2 Rp for a 12AX7.


BK

[martin manning]

Paralleling the tube essentially creates a 'hybrid' 12AX7 with half the internal plate resistance of 1 section. Keeping the values the same, i.e. Rp=100k and Rp=1.5k will yield a little more gain than one section alone and have lower noise. No need to double, halve or whatnot.

Use Randle Aikens equations with 1/2 Rp for a 12AX7.BK

Think of it this way:

Suppose you have two identical stages hanging off of the same B+ node. They have identical plate and cathode resistors, and each is using one side of a twin triode. Same value of cathode bypass cap on each too.

Ideally all of the node voltages will be identical, and the current will be the same through each stage.

Now connect all of the like nodes together to parallel each component with its mate in the other stage. Everything should remain as it was as far as the operation of each stage is concerned.

But, now what do you have? The effective plate load resistance is halved, as is the cathode resistor value, and the cathode bypass capacitance is now doubled. Internal plate resistance is halved, too, as you said. Voltages are the same from top to bottom, and the current is twice that for a single stage.

So to preserve the operating point and response for paralleled triodes you halve the resistors and double the cathode bypass cap used for a single triode. The output impedance is halved, though, and so the paralleled stage has more current driving capability.

MPM

[Merlinb]

Actually I guess you must be onto something - I wasn't thinking about that formula.

Speaking of which, what is the correct dang formula?

Somebody must be an authority on this.

The correct formula is derived in this short article:
http://www.freewebs.com/valvewizard2...BypassCaps.pdf