Thank you for this contribution! I had spoken of something along these lines in an ancient post, and expressed my laziness about working the thing out explicitly
In particular I have been meaning to get, but still don't have, a nice software package for the matrix manipulations.Brad Wood
I'm familiar with the techniques of 717. Haven't confirmed the math, but I expect there is no need to.
Re: 718, I agree with the driving point imepdance expressions for Plate-ground, Cathode-Ground, which are different.
Also, the impedances applied between ground and P and ground and K(athode) to halve the gains are a bit less than 1/gm. Since they are identical, and the P and K voltages equal and opposite, these impedances are equivalent to twice either one of them connected directly between the P and K, giving the familiar approximation of a bit less than 2/gm as the impedance between the P and K.
Re: 718, I agree with the driving point imepdance expressions for Plate-ground, Cathode-Ground, which are different.
Also, the impedances applied between ground and P and ground and K(athode) to halve the gains are a bit less than 1/gm. Since they are identical, and the P and K voltages equal and opposite, these impedances are equivalent to twice either one of them connected directly between the P and K, giving the familiar approximation of a bit less than 2/gm as the impedance between the P and K.
Chris,
I found in the Dec 2010 issue of AudioXpress a response or yours to a letter from C. R. Morton concerting the Cathodyne. Your response mentioned an error in equation (30) in the RDHB4.
There was a post from rec.audio.tubes in Aug 2008 that described the error like this:
"In another post, John Byrns says that his copy of RDH4 has:
rp' = rp + u * Rk
for equation 30.
My copy of RDH4 has:
rp' = (u - 1)Rk where Rk = RL
which is the same as the downloadable copy of RDH4 at
RDH4 mirror has.
and you (Patrick) say that yours has:
rp' = rp + ( µ + 1 )Rk where Rk = RL
It would seem that there are at least 3 different expressions for equation
30 in various editions/printings of RDH4.
I wonder what other differences there are?"
That same letter from Morton mentions an earlier letter of yours in Oct 2009 AudioXpress titled "More impedance tests". This suggests to me that there may have been an even earlier letter or article. Would you be so kind as to list the various letters, etc., publication locations and dates, concerning the Cathodyne that you have published?
I found in the Dec 2010 issue of AudioXpress a response or yours to a letter from C. R. Morton concerting the Cathodyne. Your response mentioned an error in equation (30) in the RDHB4.
There was a post from rec.audio.tubes in Aug 2008 that described the error like this:
"In another post, John Byrns says that his copy of RDH4 has:
rp' = rp + u * Rk
for equation 30.
My copy of RDH4 has:
rp' = (u - 1)Rk where Rk = RL
which is the same as the downloadable copy of RDH4 at
RDH4 mirror has.
and you (Patrick) say that yours has:
rp' = rp + ( µ + 1 )Rk where Rk = RL
It would seem that there are at least 3 different expressions for equation
30 in various editions/printings of RDH4.
I wonder what other differences there are?"
That same letter from Morton mentions an earlier letter of yours in Oct 2009 AudioXpress titled "More impedance tests". This suggests to me that there may have been an even earlier letter or article. Would you be so kind as to list the various letters, etc., publication locations and dates, concerning the Cathodyne that you have published?
Hi Electrician, my copy of RDH4 has the form of the equation in John Byrns' above.
I don't have ready access to my hard copies of audioXpress, but my Cathodyne-related publications therein are exclusively in the form of letters. They started in response to an article in that same publication by Stuart Yaniger that featured a Cathodyne along with the claim that its plate and cathode impedances were both about 1/gm. I believe that his article was printed in late 2008.
You'll also find two letters to the Editor from me at www.linearaudio.net - home under the Online Resources tab.
Hope that helps. Hey, no unauthorized biographies, OK?
I don't have ready access to my hard copies of audioXpress, but my Cathodyne-related publications therein are exclusively in the form of letters. They started in response to an article in that same publication by Stuart Yaniger that featured a Cathodyne along with the claim that its plate and cathode impedances were both about 1/gm. I believe that his article was printed in late 2008.
You'll also find two letters to the Editor from me at www.linearaudio.net - home under the Online Resources tab.
Hope that helps. Hey, no unauthorized biographies, OK?
Chis, please quote me accurately and completely when you do. My claim (and to date, there has been no experimental refutation nor even a suggestion of an experiment that will refute it) is that the source impedances at the cathode and plate of a cathodyne are equal and low (roughly 1 or 2/gm) when the loads at cathode and plate are equal. That qualifier is VITAL.
My two Thevenin source model is exceedingly simple and gives accurate answers for resistive, reactive, and complex loads. One can get accurate answers as well using more complex models (e.g., TE's matrix methods which seem to require sophisticated software to use), but they are unnecessary for practical purposes with the given constraint- my method uses a hand calculator and 30 seconds of design time.
My two Thevenin source model is exceedingly simple and gives accurate answers for resistive, reactive, and complex loads. One can get accurate answers as well using more complex models (e.g., TE's matrix methods which seem to require sophisticated software to use), but they are unnecessary for practical purposes with the given constraint- my method uses a hand calculator and 30 seconds of design time.
Stuart, I will refer everyone to your post in all future references to what you believe. To me, a Cathodyne implies identical plate and cathode loads unless otherwise described.
Your model gives incorrect answers regarding impedance. I have demonstrated this in a number of ways. So has The Electrician. You simply ignore them. Your reasons for doing so are irrelevant. You get the voltages right despite your model for the simple reason that Ik = -Ip in a triode - no output impedances need apply to explain this.
A unity gain current mirror requires no equations to see that it is entirely possible for identical loads to exhibit equal and opposite voltages when driven from different source impedances, the impossibility of which your article takes as a given and uses to "prove" Equal Z. You need to confront this to be taken seriously as a defender of your beliefs against the most simple of criticisms.
If simplicity is a virtue, this refutation is more virtuous than your model. You can't get any simpler than requiring no equations. But you have no answer to it. So once again, tiresomely, you ignore it.
Your model gives incorrect answers regarding impedance. I have demonstrated this in a number of ways. So has The Electrician. You simply ignore them. Your reasons for doing so are irrelevant. You get the voltages right despite your model for the simple reason that Ik = -Ip in a triode - no output impedances need apply to explain this.
A unity gain current mirror requires no equations to see that it is entirely possible for identical loads to exhibit equal and opposite voltages when driven from different source impedances, the impossibility of which your article takes as a given and uses to "prove" Equal Z. You need to confront this to be taken seriously as a defender of your beliefs against the most simple of criticisms.
If simplicity is a virtue, this refutation is more virtuous than your model. You can't get any simpler than requiring no equations. But you have no answer to it. So once again, tiresomely, you ignore it.
A matched pair with a non-linear slope, that behaves different on upswing
than downswing, would give exactly such an incorrect result. Especially so
if that non-linearity might pump any DC onto the coupling caps, to push
even a theoretically perfect Z matched concertina off ideal operating points.
Stuart, once again, you simply ignore what you don't like. Post 673 met your challenge. Let's review how you treated it.
Post 673 proved that all linear matched pair loads result in different Cdyne output impedances. As I progressively challenged and you successively abandoned the claims made in your attempt to refute it, your final word was to reiterate your tired challenge to find loads that confound your model!
Not too impressive: Your circular attempt to refute the proof of a flaw in your model ended in a demand to refute your model !!!!
Kenpeter is correct. Simulate or build a circuit with matched output tubes whose grids are overdriven to accept current. The lack of symmetry in the plate and cathode voltage responses is yet another proof of different Cdyne output impedances.
Post 673 proved that all linear matched pair loads result in different Cdyne output impedances. As I progressively challenged and you successively abandoned the claims made in your attempt to refute it, your final word was to reiterate your tired challenge to find loads that confound your model!
Not too impressive: Your circular attempt to refute the proof of a flaw in your model ended in a demand to refute your model !!!!
Kenpeter is correct. Simulate or build a circuit with matched output tubes whose grids are overdriven to accept current. The lack of symmetry in the plate and cathode voltage responses is yet another proof of different Cdyne output impedances.
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So I’m building this Cathodyne using 1% resistors. I plan to try it out using a triode from my junk box with a measured mu of 30 and an rp of 3K. My cathode resistor measures 10.0K, right on the money, but my plate resistor is a bit high at 10.1K. I figure that’s good enough, right? This Cdyne is pretty well balanced.
I plan to simulate the thing with these values before I fire it up. I use LTspice IV, freeware from Linear Technology. It doesn’t come with models of triodes, but I’ll just use one of their controlled voltage sources with a gain of 30 in series with a 3K resistor (a classic model of a triode) as shown in the first sim.
I drive the “grid” with a 1.00V 1KHz sine wave, and we can see that the plate and cathode voltages are off by the expected 1% because of the resistor mismatch.
I’d like to let the simulator determine the plate to ground impedance. I remember I used to test power amp output impedance by taking the difference between the unloaded and loaded output voltages and dividing by the current going through the load. I figure I’ll do the same thing here. But then I remember that some folks warn that you can’t measure the impedance of a balanced Cathodyne by unbalancing it with a load, because then it’s not a Cathodyne anymore. Hmmm… what to do?
Then it occurs to me that I can choose a plate load that actually brings the Cdyne into perfect balance! I mean, if it was balanced OK before, and now it’s balanced perfectly, where’s the harm, right?
So I load the plate with a 1.01Mohm resistor in the second sim, returning it to B+ (AC ground) for proper DC and AC balance.
I divide the unloaded minus loaded plate voltage by the load current and get 9783.6 ohms. Pretty good agreement with the theoretical, which is
Now I want to test the cathode. I switch the P and K resistors and run the same sim and calculations on the cathode, getting 402.35 ohms. (I won’t bore you with another schematic and sim - you get the idea.) And sure enough, it checks out really nicely with theory:
Clearly, the Cdyne P and K output impedances are very different.
Next time, we’ll see if we really need to keep that balance when we measure these impedances pesky P and K to ground impedances. ‘Til then….
I plan to simulate the thing with these values before I fire it up. I use LTspice IV, freeware from Linear Technology. It doesn’t come with models of triodes, but I’ll just use one of their controlled voltage sources with a gain of 30 in series with a 3K resistor (a classic model of a triode) as shown in the first sim.
I drive the “grid” with a 1.00V 1KHz sine wave, and we can see that the plate and cathode voltages are off by the expected 1% because of the resistor mismatch.
I’d like to let the simulator determine the plate to ground impedance. I remember I used to test power amp output impedance by taking the difference between the unloaded and loaded output voltages and dividing by the current going through the load. I figure I’ll do the same thing here. But then I remember that some folks warn that you can’t measure the impedance of a balanced Cathodyne by unbalancing it with a load, because then it’s not a Cathodyne anymore. Hmmm… what to do?
Then it occurs to me that I can choose a plate load that actually brings the Cdyne into perfect balance! I mean, if it was balanced OK before, and now it’s balanced perfectly, where’s the harm, right?
So I load the plate with a 1.01Mohm resistor in the second sim, returning it to B+ (AC ground) for proper DC and AC balance.
I divide the unloaded minus loaded plate voltage by the load current and get 9783.6 ohms. Pretty good agreement with the theoretical, which is
Zplate = 10100 || [ 3000 + 10000 · (1 + 30) ] = 9784 ohms
Now I want to test the cathode. I switch the P and K resistors and run the same sim and calculations on the cathode, getting 402.35 ohms. (I won’t bore you with another schematic and sim - you get the idea.) And sure enough, it checks out really nicely with theory:
Zcathode = 10100 || [ (3000 + 10000) / (1 + 30) ] = 402.5 ohms
Clearly, the Cdyne P and K output impedances are very different.
Next time, we’ll see if we really need to keep that balance when we measure these impedances pesky P and K to ground impedances. ‘Til then….
Everyone knows this. Even SY knows this. You've been posting the same thing over and over, and everyone already knew it hundreds of posts ago.
Yes, the plate-to-ground, and cathode-to-ground output impedances are different. Everyone knows this.
But SY's model isn't concerned with that, its purpose is mainly to show how the cut-off frequencies resulting from equal load capacitances are indeed equal, and very high, because the differential output impedance is ~2/gm. It's only meant to be a simple model for back-on-an envelope work.
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Merlin, if "Everyone knows this" can we please hear Stuart say it? As far as I can tell, that's not what he says in post 727 in which he specifically stated what he believes. It's also not what he says in his article. Or do you disagree?
The P - K output impedance is a little less than 2/gm.
The bandwidth is a bit higher than gm/(2 pi C).
Two different parameters. Please sim it if you don't believe me.
The P - K output impedance is a little less than 2/gm.
The bandwidth is a bit higher than gm/(2 pi C).
Two different parameters. Please sim it if you don't believe me.
It further shows that the voltage divider effects resulting from a resistive load is divided down in agreement with a two Thevenin source model (shown in my article). And it clears the common confusion from the "the source impedance is different at the plate" mantra that caused at least two textbook authors to recommend build out resistors at the cathode to "equalize" the source impedances. A tube amp designer frankly doesn't give a rat's patootie about the p-k impedance- that's not what puts poles in his open loop response. My simple two-source model gives correct and useful answers to a tube amp designer who doesn't want to spend hours with Mathematica and doesn't need complex models which (for the purpose for which a cathodyne is intended) give exactly zero more useful information.
Cpaul,
in your spice sims yo do not load the plate and the cathode at the same time. by doing this you no longer have a cathodyne since by definition the loads must be equal which to me means you are measuring the wrong thing.
I have done the sims before and will do then using your method.
Here is the voltage showing matched gain and opposite phase. the green line is 100% hidden by the blue line.
So now that we know that we know we have the proper function, Can we agree that the output impedance can be determined by simply dividing the open circuit V by the short circuit current?
Each of these impedances are measured to ground so how can thy not represent the individual plate and cathode output impedances?
Where is my error?
dave
in your spice sims yo do not load the plate and the cathode at the same time. by doing this you no longer have a cathodyne since by definition the loads must be equal which to me means you are measuring the wrong thing.
I have done the sims before and will do then using your method.
Here is the voltage showing matched gain and opposite phase. the green line is 100% hidden by the blue line.
So now that we know that we know we have the proper function, Can we agree that the output impedance can be determined by simply dividing the open circuit V by the short circuit current?
Each of these impedances are measured to ground so how can thy not represent the individual plate and cathode output impedances?
Where is my error?
dave
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