Split load Z-out.

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dave slagle
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Split load Z-out.

Post by dave slagle »

There is an age old debate about whether the Z-out of the plate and cathode connections of the cathodyne phase inverter are matched. I'm going ot take the position that as long as the loads are matched (grid of a tube) Not only will the outputs be matched, but the output Z will be that of a cathode follower.

To justify this radical statement I'll let spice do the math for me :-)

I know of two ways to measure output impedance in spice, One way was suggested by Steve Bench to build the circuit two ways, once into a large value resistor to approximate the open circuit (OC) behavior and one into a very small value resistor to approximate the short circuit behavior (SC) then if you divide the V(OC)/I(SC) you get the output impedance. The original discussion that started me down these lines used a 6DJ8 so i'll keep with it.

so According to the steve bench method, the split load has a Z-out of 82.3 ohms in both the plate and cathode.
Image

The other method as suggested to me by Brian Beck was simply to drive the output node with an external source and again apply ohms law. This did give me some interesting results initially but once I realized the output needed to be driven by two inverted sources the results of this approach closely matched the Bench approach. Do note that I added mismatched capacitive loads to this one just to make the traces diverge at high frequencies.
Image

This is essentially where I ended up after a private conversation with some friends (SY, BB, MJ, and a few others.) I'd love to continue the debate if anyone is interested.

The spice file is attached below

dave
Attachments
Screen shot 2009-09-14 at 8.53.29 AM.png
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cathodyne.asc
Tube model is Steve Bench DropDown.
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reVintage
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Post by reVintage »

Hey Dave,

I second you about the equal Zout impedances. I have done my Spice measurements the oldfasioned way where I have gone from almost infinite loadingresistors to the values where signal out was -6dB.

But shouldn´t the version below be more appropriate in this forum 8) ?
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iron.PNG
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Brgds
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dave slagle
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Post by dave slagle »

Assuming equal loads, do you also agree the the Z-out of the plate and cathode is that of a cathode follower?

dave
reVintage
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Post by reVintage »

Yep!
Brgds
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Paul Fawcett
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Re: Split load Z-out.

Post by Paul Fawcett »

Sorry to resurrect an old thread. A slightly more direct way to show this in LTSpice is connect two out-of-phase AC1 current sources to the output nodes, rather than voltage sources as shown. Then you merely click on the node to see voltage (make sure it's linear not dB representation). Each volt corresponds to 1R of output impedance, and you are done.
CPaul
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Post by CPaul »

I don't think anyone disputes that the anode to cathode impedance is approximately 2/gm. The controversy is/was about the single ended, ground referenced anode and cathode impedances.

At the anode, this is a little less than Ra. It is much smaller at the cathode, but still noticeably more than 1/gm. This is contrary to the claim of some that both impedances are about 1/gm.
dave slagle
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Post by dave slagle »

I think this all comes down to application and in its intended use / application the SE output impedance isn't worth discussing unless you plan on using it to drive grids into A2.

dave
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