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Old 30th November 2011, 10:18 PM   #101
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Dave,
Thanks for "chiming in". It was your spice modelling and that of Brian B. to which I was referring (as having seen) in my post of a few pages back but I refrained from mentioning your name so as to not "drop you in it".

A caveat for those following some of the references - I refer to Morgan Jones's Bevois Valley Amp. In his original Electronics World Article and in the 1st Edition of "Valve Amplifiers" he did state that the output impedances from Anode and Cathode would NOT be equal and that a "build out" resistor would be required in the Cathode to equalise drive impedances. He even calculated the required resistance value. He later changed his mind and the 3rd edition of "Valve Amplifiers" had this section rewritten with his own analysis as to why the impedances would be equal. I built a "Bevois Valley" and used it in my system for probably 2 years before rebuilding it into my Baby Huey prototype.

Now, to be a total fence sitting chicken, when I had the Bevois Valley running I experimented with a build out resistor, using a dual channel oscilloscope with a X10 probe on each of the anode and cathode, I adjusted the build out resistor to balance the outputs into the EL84 grids to 200kHz. To my surprise I found that a build out resistor did help, one of approximately 1/5th the value the Morgan calculated in those original articles before he changed his mind (or learned better?). I subsequently conclude that this result was probably "in the noise" of my experiment and if I swapped the output tubes over I may have got a different result.

I would have to go back through my work books to find the derivations of those formulae I posted but I recall that I followed Morgan's 3rd Edition "Valve Amps" process closely - mostly because I had some trouble wading through the Priesman article's maths.

From a practical stand point I have concluded that the Cathodyne must be used with equal loads on anode and cathode which will require a Class A output stage or a Class A buffer stage between it and the output tubes if they are running Class AB. The "limitations" when running into Class AB Ouput stage (where the loads become unbalanced when one tube cuts off or either tube goes into grid current on positive peaks) are quite euphonic and not seriously objectionable. In fact when those limitations are deliberately emphasized by the use of a low gm, high mu tube, the results are almost perfect for a guitar power amp, but as I said earlier, that ain't HiFi.

Guys - I'm afraid that I'm going to miss the rest of the debate, I'm off on 5 weeks leave at the end of today and will be without internet access for that time. back here early Jan 2012.

A Cool Yule to all.

Cheers,
Ian

Last edited by gingertube; 30th November 2011 at 10:34 PM.
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Old 30th November 2011, 10:34 PM   #102
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Ground is nothing special. But when we say the "plate" impedance, I hope you will agree with me that it is the impedance between plate and ground. Not the impedance between plate and say, cathode. Can we agree on that? To make the measurement, current must flow through both of the two nodes between which the impedance is being measured. If I want to measure the impedance between the ends of a resistor, I have to get current to flow in one end and out the other. This is just common sense.

But we have already established that no current flows through ground in your circuit. Because plate current = - cathode current, all current flowing in the plate flows in the cathode, and everything in the cathode flows through the plate. There is no current available to flow through ground or through the matched P & K loads. Do you agree? So you can't be measuring an impedance in which one of two nodes is ground.

This is a flaw you don't seem to be able to find.

Try this in a sim. Put a 1milliohm resistor between P shorted to K and ground. Measure the current through the resistor. 0, right? Now disconnect the P shorted to K from ground and check the AC voltage at the P shorted to K. Still at AC ground, right? The circuit operates identically, AC voltage and current-wise, whether the ground is connected to P & K or not. Try it if you don't believe me. If you want to do it on the bench, use big caps instead of shorts.

There is absolutely nothing going on with ground in your circuit. You are not measuring the impedances from P or K to ground because there is no current flowing through ground. You can't measure the impedance between points A and B if no current flows through B. Agreed?

Regarding what you call your Thevenin model, look up Thevenin models. They are always defined in terms of two node equivalent circuit. Not a three node or a four node or more. A two node. There is a reason for that. Find me a reference on Thevein equivalents that discusses more than two nodes. Thevenin never did.
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Old 30th November 2011, 10:52 PM   #103
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Quote:
Originally Posted by CPaul View Post
Please see Letters to the Editor under Online Services on the Linear Audio website. You'll see a letter that responds to Stuart Yaniger's article and describes a way to determine Cathodyne plate and cathode impedances.

Because the P and K currents of a triode with no grid current are always equal and opposite, these currents through identical plate and cathode loads will yield equal and opposite voltages at the P and K. It does not follow that the impedances of the plate and cathode of the triode driving them are equal. In fact, you will find all over the place expressions for the impedance Zp looking into the plate = rp + (1 + u)*Rk, and the impedance Zk looking into the cathode = (rp + Rp)/(1 + u). You can see that these are not equal when Rp = Rk. I will leave it to you to reconcile this fact with the assertion that Zk = Zp in a Cathodyne where Rp = Rk.
Your calculations don't take in equation all factors, but use assumptions and approximations. If to take all factors without any omissions and sum them all together you will come back to simple solution based on Ohm's law. Nothing here looks neither into plate, nor into cathode. The tube from plate to cathode is in series with both loads and in series with power supply. Don't complicate it because complicating you loose details and come to wrong conclusions.

Again, show me how current variation through 2 equal impedances connected in series can cause unequal variations of voltage drops on them. It is simple. Does not matter what the tube consists of. Like, on one pilot's exam it did not matter what students had to do when if lost form the plane Her Majesty The Queen, or a bag of a sand, no difference; they have to perform the same tasks they have to perform loosing part of plane load. The same way, no difference where is cathode, where is anode, if both loads are equal and connected in series.

Or, I can show you similar trick with wrong approximations demonstrating how can be proved that sum of lengths of 2 sides of triangles are equal to the length of the 3'rd side. It is exactly the same trick as you use, when wrongly used approximation as if prove basic facts wrong.
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Old 30th November 2011, 10:53 PM   #104
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Hi Dave, I trust you can speak for yourself.

The jist of my point is this. If I have an arbitrary circuit and I want to measure the impedance between an arbitrary pair of nodes in it, I would cause a current to flow around a loop including those nodes and some point outside the circuit under test. I would measure the change in voltage between the nodes due only to this current flow. I would divide by the current to get the impedance. Sound reasonable?

Now I take it that you feel it is important to keep the Cath balanced: identical loads, equal and opposite ground-referenced AC currents driving the P and K, so equal and opposite P and K voltages. So be it. Let’s do it.

Now the voltages appearing on Vp and Vk are affected by both ip and ik, correct? That is, Vp = Vp(ip) + Vp(ik) and Vk = Vk(ik) + Vk(ip), correct? But from the definition of impedance, we want to calculate Vp(ip)/ip, not [Vp(ip) + Vp(ik)]/ip. If this is not clear, consider for a moment a “Cathodyne” where the loads are unequal, and you don’t feel the need to keep the circuits balanced. To find Zp, you’d set ik to 0 and calculate Vp(ip)/ip. If ik were anything other than 0, it would affect Vp and give you a false result for Zp. The same is true in a balanced Cathodyne.

So how do we measure impedance and keep your Cathodyne balanced at the same time? Well, I don’t see how to do it with a sim, but if you did and algebraic calculation, you could. Comments up to this point?
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Old 30th November 2011, 11:00 PM   #105
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Quote:
Your calculations don't take in equation all factors, but use assumptions and approximations. If to take all factors without any omissions and sum them all together you will come back to simple solution based on Ohm's law. Nothing here looks neither into plate, nor into cathode. The tube from plate to cathode is in series with both loads and in series with power supply. Don't complicate it because complicating you loose details and come to wrong conclusions.
Far too vague to even comment. Be specific.

Quote:
Again, show me how current variation through 2 equal impedances connected in series can cause unequal variations of voltage drops on them.
they can't. I already told you.

Quote:
The same way, no difference where is cathode, where is anode, if both loads are equal and connected in series.
Don't understand this comment.

Quote:
Or, I can show you similar trick with wrong approximations demonstrating how can be proved that sum of lengths of 2 sides of triangles are equal to the length of the 3'rd side. It is exactly the same trick as you use, when wrongly used approximation as if prove basic facts wrong.
You are being too vague in all of your comments. You have to be more specific with your arguments. I really do doubt that arguments about geometry are "exactly the same" as those about electronics.
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Old 30th November 2011, 11:06 PM   #106
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Right. And what would happen to accuracy of your impedance measurement at the output of the amp if you simultaneously varied the load at the output and an additional load between some other pair of nodes inside the amp? You'd get a wrong result at the output due to interference from the second load inside the amp.

Even if you feel you must keep the Cath balanced during all impedance measurements, you must find a way to keep the load at the plate from messing up the result at the cathode with the cathode load, and vice-versa.

I know how to do both: keep the Cath balanced at all times, and keep the other load or current source from mucking up the results at the output of interest. Interested?

Last edited by CPaul; 30th November 2011 at 11:09 PM. Reason: Added a paragraph at the end
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Old 30th November 2011, 11:27 PM   #107
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Quote:
Originally Posted by CPaul View Post
Far too vague to even comment. Be specific.

they can't. I already told you.

Don't understand this comment.

You are being too vague in all of your comments. You have to be more specific with your arguments. I really do doubt that arguments about geometry are "exactly the same" as those about electronics.
I can draw you a very vague schematic later. Now I am at work and have no appropriate software here.

Are you familiar with equivalent schematics, or they would be too vague for your understanding?

Arguments about geometry use the same trick you use with electronics, trying to use approximations to prove that strict axioms are wrong.
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Last edited by Wavebourn; 30th November 2011 at 11:29 PM.
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Old 30th November 2011, 11:33 PM   #108
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Quote:
Originally Posted by CPaul View Post
Right. And what would happen to accuracy of your impedance measurement at the output of the amp if you simultaneously varied the load at the output and an additional load between some other pair of nodes inside the amp? You'd get a wrong result at the output due to interference from the second load inside the amp.
That's a good description of feedback. Otherwise, the analogy is inapt; try again with two amplifiers and two loads.
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Old 30th November 2011, 11:36 PM   #109
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As an Electrical Engineer of 30 years, I'm familiar with equivlent circuits.
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Old 30th November 2011, 11:38 PM   #110
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If the amplifier employed zero feedback, your remark would be irelevent. Try again.
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