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
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
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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.
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.
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.
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?
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?
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.
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?
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?
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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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That's a good description of feedback. Otherwise, the analogy is inapt; try again with two amplifiers and two loads.
My opinion is that , in Concertina the output impedance of the anode to the ground is nothing else than the output impedance of an common cathode triode amp , and the output impedance of the cathode to the ground is equal to the output impedance of a cathode follower triod buffer . Of course we have equals output voltages in the two circuit if they aren't connected to anything else , or if they connectet to easy loads , but if we connect them to harder loads and start to drive the Concertina harder , we will see that the voltage of the anode start to fall comparing it with the voltage at the cathode , that happens because they haven't equal impedances , the anode have higher one ..
Let us speak in generalities, but in generalities that are compatible with your demand for impedance tests of a balanced Cathodyne that maintain balance during testing.
I have a circuit with nodes A and B between which I want to determine the impedance Zab. To do so, I need to somehow create a loop that puts current Itest into node A, out of node B, past a point outside the circuit, and returns it to point A. I need to measure the change in voltage deltaV between A and B due to that current and that current only. I can then divide. deltaV by Itest to get Zab. If there is no current flowing through A or B, I can’t measure Zab.
In your circuit, node B is ground and there is no current flowing through it. So you can't measure the impedance between nodes A and B.
I have a circuit with nodes A and B between which I want to determine the impedance Zab. To do so, I need to somehow create a loop that puts current Itest into node A, out of node B, past a point outside the circuit, and returns it to point A. I need to measure the change in voltage deltaV between A and B due to that current and that current only. I can then divide. deltaV by Itest to get Zab. If there is no current flowing through A or B, I can’t measure Zab.
In your circuit, node B is ground and there is no current flowing through it. So you can't measure the impedance between nodes A and B.
You can't say that. Because doing measurements changing loads separately you disbalance Concertina, so it is not a Concertina anymore. Try to change both loads at once, equally, to keep it balanced, and calculate both resistances separately.
Edit: What is remarkable, chemist SY understands electronics better than the majority on this forum, including many Electrical Engineers. What is so special in chemical education, so it so superior to education of Engineers?
Edit: What is remarkable, chemist SY understands electronics better than the majority on this forum, including many Electrical Engineers. What is so special in chemical education, so it so superior to education of Engineers?
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What is remarkable, chemist SY understands electronics better than the majority on this forum, including many Electrical Engineers. What is so special in chemical education, so it so superior to education of Engineers?
I think what's throwing so many folks is that the concertina has a floating output but its feedback is not symmetrical. If tested with equal and opposite currents driven into its outputs (IOW, as a concertina works) point impedances will measure equal. But if measured with equal but not opposite currents, or just single currents, driven into its outputs point impedances will not measure equal (because its no longer a concertina).
So we have arguments about angels dancing on the heads of pins.
Thanks,
Chris
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