The load each tube sees varies as the grid voltage swings in one direction for one tube and the opposite direction for the other. As the grids swings toward zero the impedance of each single tube approaches is 1/4 A-A. The composite tube load remains constant at 1/4 A-A.
So when you are setting one tube to zero, what exactly do you mean? Is anyone here able to construct an equivalent circuit of a cathodyne with a composite tube or illustrate the problem graphically?
John
So when you are setting one tube to zero, what exactly do you mean? Is anyone here able to construct an equivalent circuit of a cathodyne with a composite tube or illustrate the problem graphically?
John
the transition between class A and class B is nonlinear but we are still on the simple plan where you get one or the other so linearity is assured.
once we agree on the black or white.... we can laugh about how everything is grey.
In response to 661: When we talk about shorting a Cdyne output to ground to measure a short circuit current, we aren't really making a DC short. In a kind of electrical engineering shorthand, we're applying an AC short (think really big capacitor) so as not to disturb the Cdyne bias or the small signal parameters of the triode. And so we avoid taking it non-linear.
Class B operation is by definition non-linear; each tube shuts off periodically. There is not much that can be done to linearize such a circuit.
Class B operation is by definition non-linear; each tube shuts off periodically. There is not much that can be done to linearize such a circuit.
Since this discussion has come alive again, it may be of interest to read some newsgroup discussions on the topic from 1999:
Phase Splitter - rec.audio.tubes | Google Groups
and a full out flame war from 2001:
Hood on Negative Feedback - rec.audio.tubes | Google Groups
also, 3 more recent threads:
What's happened to this NG???? - rec.audio.tubes | Google Groups
Phase splitter with capacitance load. - rec.audio.tubes | Google Groups
Audio Cyclopedia - A highly recommended book - rec.audio.tubes | Google Groups
Phase Splitter - rec.audio.tubes | Google Groups
and a full out flame war from 2001:
Hood on Negative Feedback - rec.audio.tubes | Google Groups
also, 3 more recent threads:
What's happened to this NG???? - rec.audio.tubes | Google Groups
Phase splitter with capacitance load. - rec.audio.tubes | Google Groups
Audio Cyclopedia - A highly recommended book - rec.audio.tubes | Google Groups
Electrician, thanks much for the links to discussions on similar matters. It seems that regardless of which is correct, there are significant numbers of people on both sides of the question of C(atho)dyne output impedances. It’s amazing that what seems to be such a simple question, which I hope we can all agree has a single right answer, can generate such controversy.
Let me offer an observation. Regardless of Cdyne output impedances, ignoring parasitic capacitances, and in the absence of grid current in the Cdyne, its P(late) and K(athode) gain magnitudes, time constants, and frequency responses must be equal.Why? Three reasons taken together: identical P & K loads, equal and opposite P & K currents in any triode, and Ohm’s law, which does the rest.
Given such a simple explanation for these equivalences, it is difficult to understand why anyone would claim that they demand any particular Cdyne output impedance model.
Let me offer an observation. Regardless of Cdyne output impedances, ignoring parasitic capacitances, and in the absence of grid current in the Cdyne, its P(late) and K(athode) gain magnitudes, time constants, and frequency responses must be equal.Why? Three reasons taken together: identical P & K loads, equal and opposite P & K currents in any triode, and Ohm’s law, which does the rest.
Given such a simple explanation for these equivalences, it is difficult to understand why anyone would claim that they demand any particular Cdyne output impedance model.
Dave, here’s how. First let’s agree on impedance measurement. If I had a single output circuit with no active AC signals, I could connect an AC signal test source between the output and ground. The ratio of the AC voltage arising on the output to the current flowing through it would be the impedance. Note that there is already a signal on the output that we ignore in our calculations – the DC bias. Similarly, we may ignore any AC voltage that the circuit might have exhibited even before we connected the test source, because that signal did not arise from the test source. So a good definition of the impedance between two nodes is that voltage which arises exclusively due to the current applied between the nodes, divided by that current. (If any of this bothers you, just google superposition theorem.)
So let’s determine the P(late) to G(round) impedance Zpp and the K(athode) to G impedance Zkk in a C(atho)dyne whose P and K AC voltages will remain equal and opposite throughout our test.
In this Cdyne, no AC signal is applied to the grid. Instead, equal and opposite ground-referenced AC current sources Ip and Ik are connected to the P and K. We use infinite impedance current sources so as not to disturb Cdyne impedances. It can be easily demonstrated that the P and K AC voltages in this circuit are equal and opposite.
Now, we derive expressions for the P and K voltages Vp and Vk. Their general forms are
because each current source contributes to each voltage. We don’t care about Zkp or Zpk – they are irrelevant to the definition of impedance. Only Zpp and Zkk matter. I’m not going to do the derivation here; you’ll either have to do it yourself or trust me (Oh, alright, if you twist my arm you could get me to supply it in a subsequent post):
And that’s how you measure P-G and K-G impedances while maintaining a fully balanced Cdyne. As you can see, these impedances are quite different, even when Rp = Rk as in any proper Cdyne.
But there’s one more thing worth noting. Zpp and Zkk are composed solely of combinations of circuit parameters – they cannot include any AC signals, because these would produce non-linear results when multiplied by Ik and Ip, and the Cdyne is a linear circuit. A complete list of circuit parameters is as follows: Rk, Rp, gm, rp and μ. Note that not a single one of these parameters is affected by the signal levels of Ip or Ik, and therefore neither is Zpp nor Zkk. So I could set Ik to zero, unbalancing the Cdyne, and Zpp would not change. That means I don’t need to go through all the trouble of balancing the Cdyne in the first place to measure its impedances – all I had to do to measure Zpp is apply Ip. Similarly, all I had to do to measure Zkk is apply Ik.
Clearly, there is no need to maintain a balanced Cdyne while determining its P to G and K to G impedances.
So let’s determine the P(late) to G(round) impedance Zpp and the K(athode) to G impedance Zkk in a C(atho)dyne whose P and K AC voltages will remain equal and opposite throughout our test.
In this Cdyne, no AC signal is applied to the grid. Instead, equal and opposite ground-referenced AC current sources Ip and Ik are connected to the P and K. We use infinite impedance current sources so as not to disturb Cdyne impedances. It can be easily demonstrated that the P and K AC voltages in this circuit are equal and opposite.
Now, we derive expressions for the P and K voltages Vp and Vk. Their general forms are
Vp = Zpp · Ip + Zkp · Ik
Vk = Zkk · Ik + Zpk · Ip
because each current source contributes to each voltage. We don’t care about Zkp or Zpk – they are irrelevant to the definition of impedance. Only Zpp and Zkk matter. I’m not going to do the derivation here; you’ll either have to do it yourself or trust me (Oh, alright, if you twist my arm you could get me to supply it in a subsequent post):
Zpp = Rp || [rp + (1 + μ)·Rk)]
Zkk = Rk || [(rp + Rp) / (1 + μ)]
And that’s how you measure P-G and K-G impedances while maintaining a fully balanced Cdyne. As you can see, these impedances are quite different, even when Rp = Rk as in any proper Cdyne.
But there’s one more thing worth noting. Zpp and Zkk are composed solely of combinations of circuit parameters – they cannot include any AC signals, because these would produce non-linear results when multiplied by Ik and Ip, and the Cdyne is a linear circuit. A complete list of circuit parameters is as follows: Rk, Rp, gm, rp and μ. Note that not a single one of these parameters is affected by the signal levels of Ip or Ik, and therefore neither is Zpp nor Zkk. So I could set Ik to zero, unbalancing the Cdyne, and Zpp would not change. That means I don’t need to go through all the trouble of balancing the Cdyne in the first place to measure its impedances – all I had to do to measure Zpp is apply Ip. Similarly, all I had to do to measure Zkk is apply Ik.
Clearly, there is no need to maintain a balanced Cdyne while determining its P to G and K to G impedances.
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