Sometimes a picture is worth a thousand words. I wish to address a common misconception about the output impedance of the cathodyne phase inverter with a couple of screen captures from LTspice.
As can be seen from the above illustration, the cathodyne resembles both a classic gain stage, with a high output impedance output at the anode, but also a cathode follower, with its characteristic low output impedance output at the cathode.
The formula for the anode output impedance is the same as for a normal gain stage:
Zo(anode) = Ra * (ra + Rk*(u + 1))
And indeed, as you can see from the image above, the Zo when considering the anode is approximately 44K. I'm estimating output impedance of nodes by attaching an AC 1A current source to the node of interest, and conducting a small signal AC analysis to see what voltage is produced. By Ohm's law, the voltage at that node in V corresponds to the impedance in ohms. So far, this is the orthodox result.
On the other hand, the formula for the cathode output impedance, when considered individually, is the same as for a cathode follower:
Zo(cathode) = Rk || (Ra + ra)
--------
(u + 1)
Sure enough, we again see a low output impedance of around 1K from the cathode. So what's the problem?
Well, it turns out that the to find out the actual effective output impedance of the cathodyne when acting as a phase inverter, we really need to consider both outputs at once. It turns out that -- so long as the cathodyne has equal anode and cathode loads -- that the effective output impedance is not only low for both outputs, but is also equal - as is the output voltage produced! You can read about the algebra that describes this in this classic paper by Albert Preisman. However, it's way easier just to see this from a simulation:
As you can see, when we consider the output impedance of both outputs simultaneously (in this case using two current sources, 180 degrees out-of-phase, which mimics the phase inverter operation), that both anode and cathode outputs now have very low output impedances, which are moreover nearly equal -- around 700 ohms in this example. The green line in the bode plot represents the anode output, while the blue line represents the cathode output. This so-called differential output impedance is all that really matters as regards behaviour of the circuit. As Merlin Blencowe points out in his fine book on the preamplifier design, it turns out that the formula for this differential output impedance is given by:
where R=Rk=Ra
As can be seen from the above illustration, the cathodyne resembles both a classic gain stage, with a high output impedance output at the anode, but also a cathode follower, with its characteristic low output impedance output at the cathode.
The formula for the anode output impedance is the same as for a normal gain stage:
Zo(anode) = Ra * (ra + Rk*(u + 1))
----------------------
Ra + ra + Rk*(μ + 1)
Zo(anode) ≈ Ra
And indeed, as you can see from the image above, the Zo when considering the anode is approximately 44K. I'm estimating output impedance of nodes by attaching an AC 1A current source to the node of interest, and conducting a small signal AC analysis to see what voltage is produced. By Ohm's law, the voltage at that node in V corresponds to the impedance in ohms. So far, this is the orthodox result.
On the other hand, the formula for the cathode output impedance, when considered individually, is the same as for a cathode follower:
Zo(cathode) = Rk || (Ra + ra)
--------
(u + 1)
Sure enough, we again see a low output impedance of around 1K from the cathode. So what's the problem?
Well, it turns out that the to find out the actual effective output impedance of the cathodyne when acting as a phase inverter, we really need to consider both outputs at once. It turns out that -- so long as the cathodyne has equal anode and cathode loads -- that the effective output impedance is not only low for both outputs, but is also equal - as is the output voltage produced! You can read about the algebra that describes this in this classic paper by Albert Preisman. However, it's way easier just to see this from a simulation:
As you can see, when we consider the output impedance of both outputs simultaneously (in this case using two current sources, 180 degrees out-of-phase, which mimics the phase inverter operation), that both anode and cathode outputs now have very low output impedances, which are moreover nearly equal -- around 700 ohms in this example. The green line in the bode plot represents the anode output, while the blue line represents the cathode output. This so-called differential output impedance is all that really matters as regards behaviour of the circuit. As Merlin Blencowe points out in his fine book on the preamplifier design, it turns out that the formula for this differential output impedance is given by:
Zo(dif) = R * ra
----------------
ra + R * (u + 2)
where R=Rk=Ra
Not quite. With equal plate and cathode loads, the anode to ground impedance is high, about Ra, as given above. The cathode to ground impedance is low, (Ra + ra)/(u + 1), as given above. "Effective impedance" is not a term you will find defined in any electronics text. Differential impedance - the impedance between the plate and cathode - this is what really matters. And it is twice the value given for Zo(dif) above. You can confirm this with LTSpice by connecting a current source between the plate and cathode and dividing the difference between the plate and cathode voltages by the current of the source.
ReplyDeleteThe idea that you can think of this circuit has having two single-ended drivers with identical "effective" impedances of about 1/gm makes things needlessly complex. Worse, it contradicts the acknowledged single ended ACTUAL driver impedances derived above, which are very different from one another and from 1/gm.
ReplyDeleteThe right way to look at this is to start by realizing that identical anode and cathode loads have equal and opposite AC voltages across them. Taken together, this means that the same current flows through each output stage grid resistor. This in turn means that no current flows into or out of ground at the grid resistor junction. So if we simply disconnect the grid resistor junction from ground, the Cdyne can't tell the difference! (Of course, the output stage can!)
Because it makes no difference to the Cdyne whether that ground is there or not, we can think of it as driving only a single load, AC-connected from anode to cathode.
Clearly in such a case it just complicates things to think of two single-ended drivers driving this "floating" load. Instead, we have a single differential driver (made of anode and cathode) driving a single load. And the impedance of that driver is Ra || ( 2 ra / (2 + u) ).
Daddio, thanks for your thoughtful responses, you make some excellent points and it's true I've been a bit sloppy with terminology. However, I suspect that the way that one should properly analyze the situation is dependent on the circuit properties of interest. For instance, I invite you to additionally imagine (or better, simulate) adding equal values of shunt capacitance to the loads being driven -- essentially making a low pass filter by the combination of the output impedance of either the anode or the cathode, and the respective shunt capacitance. If you make these large enough that the result is not confounded by the existing pole, you will find that resulting time constant of these filters is consistent not with the true differential output impedance as you have described, but rather with 1/2 of that value, per the result that is obtained either by the method I have shown above, or, equivalently, by placing the current source between cathode and anode and measuring cathode and anode separately (rather than differentially per your approach). As suggested by the shunt capacitance example, it is this number that will determine the bandwidth of the phase inverter. The formula I gave at the end of the article is not one that I derived -- it is from the wisdom literature from the golden age -- Langford-Smith published it in RDH, and he derived it from even earlier work of Jones. My mistake was indicating that it gives the differential impedance, when in fact it gives half that value.
ReplyDeleteHi Wombat, thanks for your reply.
ReplyDeleteYes, I'm familiar with the RDH analysis and I agree with it.
Let's call the grid resistors Rg. Each is in parallel with a capacitor Cg. Ignoring the high impedances of the Rg's, the load seen by the Cdyne consists of two caps in series, for a total capacitance of Cg/2. The Cdyne differential output impedance is Zca = Ra || (2 ra / (u + 2) ). The time constant is therefore Zca * Cg / 2. It is not that the output impedance suddenly becomes half of its original value because it is driving capacitors predominantly instead of resistors; it is that the load is Cg / 2, not Cg.
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ReplyDeleteAlso, regarding the last simulation you posted with the two ground-referenced current sources 180 degrees apart in phase - there is zero net current flow from these into and out of ground. Replacing the two sources with a single 1A source flowing from Output A to Output K would yield circuit currents and voltages indistinguishable from those with the two sources, and would bring us back to the differential paradigm.
ReplyDeleteHorses for courses... our approaches are *almost* equivalent. Your approach of considering the differential output impedance in conjunction with a lumped load does indeed work fine in the context of perfect cathodyne balance, and uses terminology in a way that is consistent with established theory. No argument. However, it lacks generality in the sense that when the load becomes unbalanced (which will inevitably be the case in the real world) your presumption that there is no net current at the junction of the loads will be violated, and you will no longer derive a correct result for the bandwidth from each of outputs. Whereas my approach of considering each output independently will continue to give meaningful results under simulation regarding the bandwidth from each output. As to your last post about a single vs. two current sources, I actually pointed out the fact of their equivalence in my post above -- "...equivalently, by placing the current source between cathode and anode and measuring cathode and anode separately...", etc. I realize perfectly well that it doesn't matter whether you use one or two. The point is that in either case you can choose to probe the outputs individually, or the difference between them. It's true if you consider them individually, you must then wave your hands a bit and talk about "effective" output impedance or somesuch, but its a perfectly reasonable and practical way to conceptualize the situation.
ReplyDeleteWombat, I believe you’ve misread me. I said that there was no net current into ground in the simulation above. I should have been clearer and added that what was being tested in that sim was still the K-A impedance, not two single ended impedances. Thevenin allows us to test the impedance between one pair of nodes at a time. I am unaware of any theorem allowing us to simultaneously test the impedances associated with more than two nodes in the same circuit.
ReplyDeleteThere are several problems with the effective impedance approach: it cannot be measured, it contradicts actual impedances, and in fact it must be abandoned in favor of actual (single ended) impedances in the very practical instance of the unbalance that occurs when the output stage draws grid current.
To illustrate, consider a simulation where the output stage inputs are approximated by the parallel combination of a diode with a grounded cathode and a negative (voltage) biased grid resistor. When the cap-coupled Cdyne outputs respond to a single full cycle of a sine wave input of adequate amplitude, the current driven by the cathode into its diode is much larger than that driven by the plate into its. Actual single end impedances can explain this – the equal effective impedances cannot.
I do not believe that we should invent a new term with the problems mentioned above when conventional differential and actual single ended impedances are perfectly adequate in the balanced and unbalanced instances above, respectively.
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ReplyDeleteThevenin’s theorem allows us to arrive at the expressions given as Zo(anode) and Zo(cathode) by Wombat at the beginning of this blog. I believe that no worthwhile model of a circuit should contradict this theorem. Can we conceptualize a generalized Cathodyne without this contradiction?
ReplyDeleteIt turns out that we can think of a triode as consisting of two ground-referenced sources driving arbitrary plate and cathode loads Zp and Zk. The following two equations should generate no controversy:
The grid to plate gain Agp = –u*Zp / [ Zp + rp + (1+u)*Zk]
The grid to cathode gain Agk = u*Zk / [ Zp + rp + (1+u)*Zk]
Of course, the gains are equal and opposite when Zk = Zp. Let’s factor those equations:
Agp = < -u > * [ Zp / ( Zp + {rp + (1+u)*Zk} )]
Agk = < u / ( u + 1) > * [ Zk / ( Zk + { ( rp +Zp ) / ( 1 + u) } )]
< > are the gains of sources. Terms in [ ] show a voltage divider with the sources’ impedances { } driving loads Zp and Zk. { } in parallel with their respective loads give the Cdyne impedances listed by Wombat. And so we have our two single-ended sources.
Note that { } are the classical impedances looking into the plate and cathode of a triode. Even when Zk = Zp (the balanced cathodyne case), they are anything but equal.
Some may find this model disturbing because of a belief or intuition that the two { } must be equal. But we don’t need this to be true to explain the Cdyne. As long as no grid current flows, the plate and cathode currents (which therefore must be equal and opposite) produce equal and opposite voltages across identical loads. And the equations above bear out the expected time constant of about C/gm, where C is the capacitance of each of the identical loads Zp and Zk.
Classical expressions for Cdyne impedances, voltage symmetry and bandwidth with identical loads, asymmetric behavior with non-symmetrical loads (output stage grid currents), and Thevenin’s theorem: this conceptualization is consistent with all of them.
Hello Wombat,
ReplyDeletethanks for the cathodyne simulationes and formulas for calculating the output impedance. It helped me.
I developed a cathodyne that has equal output impedances:
Click here
Hi oldeurope, your circuit does indeed have nearly equal output impedances and nearly equal and opposite output voltages. What do you see as its practical value?
ReplyDeleteDaddio, the circuit proposed by oldeurope could see practical application I guess as a balanced line driver (the normal cathodyne is terrible for this, because in this case it really is the individual output impedances of cathode and anode that matter). The drawback is that the output impedances from oldeurope's proposal are still a bit on the high side - but maybe within a useful range for some applications.
DeleteGood point, Wombat, thanks for reminding me that matched Zouts are important in addition to equal and opposite outputs for line drivers. I would think that a long tailed pair made up of matched triodes whose cathodes are biased from a current source or at least a very impedance would also suffice.
ReplyDeletevery high impedance, that is.
ReplyDelete