cathode circuit corner freq. formula?

[Groover]

What is the formula for determining the 3db down frequency (and slope would be a bonus) for a R and C in parallel as the cathode circuit for a 12AX7? How about for a 6V6, or a cathode biased 2x6V6 push-pull?

Thanks. I hope I don't need to know much trig.

[Wilder Amplification]

What is the formula for determining the 3db down frequency (and slope would be a bonus) for a R and C in parallel as the cathode circuit for a 12AX7? How about for a 6V6, or a cathode biased 2x6V6 push-pull?

Thanks. I hope I don't need to know much trig.

http://www.freewebs.com/valvewizard1...Gain_Stage.pdf

Scroll down to page 26 of that document. It's a rather lengthy and time consuming equation to say the least.

[Groover]

...excuse me, I have some reading to do...

thanks for the link!

[Steve Conner]

Well, it's kind of the wrong question, because there are two break frequencies.

At low frequencies, the gain of the stage is determined by Rk + rk, where rk is the tube's internal cathode resistance, equal to 1/gm.

At high frequencies, the impedance of the cathode bypass cap is negligible, and the gain is determined by rk only.

Going up from low frequencies, the gain starts to rise, and it's 3dB up from its low frequency value at the first break frequency: roughly f1=1/(2*pi*Rk*Ck)

Eventually it stops rising and starts to flatten out. At the second break frequency, it's within 3dB of its high frequency value, and this frequency is approximately f2=1/(2*pi*(Rk || rk)*Ck)

where || means "in parallel with"

With usual guitar amp values, the external Rk is about equal to the tube's own rk, so the two break frequencies are always about an octave apart and the "shelf" is about 6dB.

[jrfrond]

No matter what the calculation tells you, ALWAYS breadboard it to see if it SOUNDS right.

[Wilder Amplification]

Well, it's kind of the wrong question, because there are two break frequencies.

At low frequencies, the gain of the stage is determined by Rk + rk, where rk is the tube's internal cathode resistance, equal to 1/gm.

At high frequencies, the impedance of the cathode bypass cap is negligible, and the gain is determined by rk only.

Going up from low frequencies, the gain starts to rise, and it's 3dB up from its low frequency value at the first break frequency: roughly f1=1/(2*pi*Rk*Ck)

Eventually it stops rising and starts to flatten out. At the second break frequency, it's within 3dB of its high frequency value, and this frequency is approximately f2=1/(2*pi*(Rk || rk)*Ck)

where || means "in parallel with"

With usual guitar amp values, the external Rk is about equal to the tube's own rk, so the two break frequencies are always about an octave apart and the "shelf" is about 6dB.

Correct...you have the "roll up" corner frequency where the transition from un-bpyassed gain to bypassed gain starts to happen. Then you have the "half boost" frequency where the gain is 1/2 way between the two, then the "level off" corner frequency where it transitions to the fully bypassed gain of the stage.

Of course, as jrfrond mentioned and as Joey Voltage and I have previously discusses, none of this shit matters unless it sounds right. The equation is nice to have though if you're trying to reverse engineer a circuit to find out why it sounds a certain way though.

The nice thing about the above posted literature is the graphical presentation of these frequencies with different bypass cap values given a certain fixed Rk value. For higher values of Rk the Ck values must go down for the frequencies shown to match, and vice versa.

[Merlinb]

. At the second break frequency, it's within 3dB of its high frequency value, and this frequency is approximately f2=1/(2*pi*(Rk || rk)*Ck)

Unfortunately that formula doesn't correspond to any particular amount of attenuation (certainly not -3dB), so is pretty meaningless.

For a typical 12AX7, however, the half-boost frequency is closely approximated by:

f = 1 / (2 pi RK Ck)

Which is rather convenient! But it's got nothing to do with -3dB points as Stephen suggested.

[Steve Conner]

Unfortunately that formula doesn't correspond to any particular amount of attenuation (certainly not -3dB), so is pretty meaningless.

Sorry, I stand by it and ask you to prove that it's wrong. :P

It's a step network, so it has to have two time constants. In a typical guitar amp stage, the boost is only 6dB, so the two time constants are about the same, though I argue that one is half the other. But the knees of RC circuits are so soft, that when you have two only an octave apart, it's hard to tell where one starts and the other ends.

But you could easily construct a 12AX7 stage with, say, a 10k cathode resistor, and then (assuming 1/gm = 1k) the two break frequencies would have to be a factor of 10 (11?) apart.

If you bypassed that 10k with 0.1uF, then the response would be flat at low frequencies, then start rising 3dB up from its LF value at 159Hz, then it would keep rising at 6dB/octave until 1750Hz where it would be 17dB above its LF value, then start to flatten out at nearly 20dB gain.

[Merlinb]

Sorry, I stand by it and ask you to prove that it's wrong. :P

Consider it done:
http://www.freewebs.com/valvewizard2...BypassCaps.pdf

It's a step network, so it has to have two time constants. In a typical guitar amp stage, the boost is only 6dB, so the two time constants are about the same, though I argue that one is half the other. But the knees of RC circuits are so soft, that when you have two only an octave apart, it's hard to tell where one starts and the other ends.

Yes, strictly it does have two TCs, but the formulae you quote do not correspond to +-3dB gain because the loop gain simply isn't high enough (with triodes) to make that kind of assumption. For example, in fig. 4 of the article above you can see that the true -3dB gain freq' is actually below the +3dB freq! The concept of conventional 3dB break frequencies is no longer useful under such circumstances, and the half-boost freq' becomes more useful.

Or perhaps I should say that although your formulae do correspond to meaningful frequencies, it is misleading to call them +-3dB points as it implies the gain at the anode really will be down by that much.

[Steve Conner]

Yes, strictly it does have two TCs, but the formulae you quote do not correspond to +-3dB gain because the loop gain simply isn't high enough (with triodes) to make that kind of assumption.

I thought I took the finite gain into account by adding the tube's internal cathode resistance into the formula?

Or are you saying that mu, plate resistance and so on, reflect back onto the cathode and decrease the gain even more? In which case, yes, you're probably right...

So, the circuit I described, with a 10k cathode resistor, would have a gain of 10 at low frequencies? Or are you saying it shouldn't be 10 because I need to reckon the tube's internal plate resistance in parallel with the load? I agree that it can't go up 20dB from 10, because then it would be 100, but a 12AX7 stage never has more gain than about 50.

[Merlinb]

I thought I took the finite gain into account by adding the tube's internal cathode resistance into the formula?

Yes, but the pole and zero are so close that ultimately they don't result in an actual gain/loss of 3dB at the anode, unless you use a really massive cathode resistor resulting in a high feedback factor.

So, the circuit I described, with a 10k cathode resistor, would have a gain of 10 at low frequencies?

It will be nearly 10 yes, as the feedback factor is pretty large in this case so gain will approach Ra/Rk. (I make the gain roughly 8.5). In this case your pole/zero frequencies will also start to approach +-3dB at the anode. But you don't normally use such large Rk value is practice.

[Dave H]

Attached are plots for 1k5 100n and 10k 100n both with 100k anode resistor.

Dave H.

[Steve Conner]

OK, so Merlin is right and I am wrong, the upper break frequency is a good octave below what my second formula predicts.

Sorry for the disinformation, folks

[Merlinb]

OK, so Merlin is right and I am wrong, the upper break frequency is a good octave below what my second formula predicts.

Sorry for the disinformation, folks

Don't be too harsh on yourself! Your calc's and correct and work perfectly if you draw the idealized frequency response, they just don't usually correspond in an actual 3dB gain or loss at the anode. And nobody cares about that kind of error!

[Steve Conner]

And nobody cares about that kind of error!

Well, except you!

You made me think about it though, thanks!

I have actually used this very circuit in one of my homebuilt amps. It started with a cathode follower first stage, but I modified it for gain by bypassing the 50k cathode resistor with a pot and series cap, and taking the output from a 33k resistor in the anode instead. I undersized the cathode capacitor too, so it would boost mids and highs more than bass, and more so the more the gain pot was turned up.

That circuit worked fine for me, but as you pointed out, it doesn't work quite the way I thought it did...