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
