Discussion in 'Amp Tech Center' started by peteb, Apr 9, 2018.

1. ### petebFriend of Leo's

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Normally a triode works into a plate resistor, this builds up the plate voltage when current flows. It is normally 100K ohms. The cathode may or may not have a cathode resistor. A cathode resistor also builds up a cathode voltage when current flows.

The split load cathodyne phase inverter simply splits the Load in two. Instead of a 100K plate resistor, a 56k or a 47k ohm plate resistor is used along with matching cathode resistor of the same value. The load is splitted between the plate and the cathode. Equal but out of phase signal voltages will appear on the cathode and plate.

But how does it work?

Look at the circuit:

Ground<>cathode resistor<>12ax7<>plate resistor<>ground

Now add the signs on the resistors.

Ground <> (-cathode resistor+)<>12ax7<>(+plate resistor-)<>ground

Note the reversal of the polarity of the resistors and that the cathode resistor will develop its voltage on the right side and the plate resistor will develop its voltage on the left side.

When the AC component of the current flows left to right, a positive voltage develops on the cathode and a negative on the plate. Reverse the current direction and now the plate resistor develops a positive voltage while the cathode develops a negative voltage.

This is meant to show why the signal voltage on the cathode and plate are equal but in opposite phase.

2. ### petebFriend of Leo's

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How do we know that the signal voltage on the plate will be the same as the signal voltage on the cathode?

In a triode all of the cathode current also flows thru the plate. Equal currents, equal resistances above ground reference, means equal voltages.

3. ### LudwigvonBirkTele-Holic

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Do you have a supporting link for this interesting postulation?

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4. ### elpicoTele-Afflicted

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It's a three terminal device so that only leaves two places current from the cathode can go, the plate or the grid. In normal operation the grid is at a lower voltage than the cathode. Electrons flow to a higher voltage, not a lower one, so for the most part they don't want to go into the grid. They want to go around it and hit the plate because that's a good 150v higher than the cathode. There is some tiny current flow from cathode to grid due to unavoidable imperfections in the construction of any real world tube. I think it's in the micro or nano amps for a 12ax7 so typically you ignore it and say 100% of the current leaving the cathode flows to the plate.

Now the key word here was "normal" operation. Us guitar guys don't always do "normal" operation. We like overdrive. When you overdrive a tube you push the grid voltage higher than the cathode and then this all goes out the window. Electrons from the cathode see that attractive, positive voltage on the grid and are happy to dive into it. If I remember right once you make the grid of a 12ax7 higher voltage than the cathode the current path between the two has less than 1k of resistance so now you have something less than 100% of the cathode current reaching the plate.

Last edited: Apr 10, 2018
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5. ### RLee77Friend of Leo's

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If your plate R comes off the tube and connects to ground, you won’t be getting any plate current.

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6. ### BendyhaFriend of Leo'sSilver Supporter

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Simplicity can be so complex.

“Everything should be made as simple as possible, but not simpler.” — Einstein

But...............“Explanations exist; they have existed for all time; there is always a well-known solution to every human problem – neat, plausible, and wrong.” — H. L. Mencken

For................“Nothing is as simple as we hope it will be.” — Jim Horning

Because.........“Simplicity is an exact medium between too little and too much. “ — Joshua Reynold

And...............“The farther you go, the less you know.” — Lao Tsu

....................“Science, my lad, has been built upon many errors; but they are errors which it was good to fall into, for they led to the truth.” ― Jules Verne

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7. ### petebFriend of Leo's

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RLee wrote:

"If your plate R comes off the tube and connects to ground, you won’t be getting any plate current."

Then it must be ok if the plate resistor connects to ground thru the filter capacitors, like they do?

8. ### petebFriend of Leo's

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My first diagram may be overly simplified in that it shows both of the opposite phases in one diagram and lacks the filter caps. Let's break it down to one diagram for each phase.

Ludwig, this is just putting all of the written material into one simple diagram that shows that the plate and the cathode are out of phase.

FIG 1

AC component of signal flows in one direction > the cathode develops a positive voltage relative to ground, the plate develops a negative voltage relative to ground. Plate and cathode AC voltages are developed across their respective resistor by the flow of the AC current component thru the resistors:

Ground>(-cathode resistor+)12ax7(-plate resistor+)>filter cap>ground

FIG 2

Now reverse the direction of the AC component of the signal < the cathode now develops a negative voltage relative to ground, and the plate now develops a positive voltage relative to ground:

Ground < (+cathode resistor-)12ax7(+plate resistor-)<filter cap<ground

Key concept, electrons flowing from ground across a resistor develops a positive voltage relative to ground at the non grounded end of the resistor. Electrons flowing toward ground across a resistor develops a negative voltage relative to ground at the non grounded end of the resistor.

9. ### robrobPoster ExtraordinaireAd Free Member

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It's not the filter caps, it's the power supply pulling electrons through the plate resistor that's missing from your diagram. The caps don't flow any DC.

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10. ### petebFriend of Leo's

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Rob, Thanks.

DC:

The power supply provides high voltage DC for the plates. The power supply generates the high voltage and puts it on the plates. This high voltage powers the tube and creates DC current flow on the tube. The AC signal on the grid then causes the DC current to become pulsing DC, or overall DC current with a smaller AC component.

AC:

The AC on the plate has a totally different source than the DC on the plate. The overall flow of the DC on the schematic originates at the rectifier and goes up and to the left to the plates. The AC signal originates in the guitar then travels thru the amp inputs to the right, thru the tubes, OT and speaker. The AC on the plate, and the AC on the cathode, result from current flow, AC current flow, or at least the AC component of the current flow, that flows thru the plate and cathode resistors at all times during AC signal amplification.

Rob, the power supply puts the high voltage DC on the plates. Does the power supply need to be shown on my AC diagram?

Again, the diagrams are AC. In my diagrams the filter caps are passing AC not DC. And now I think I understand why the plate resistors are connected to ground thru the filter caps and not direct. I think this is Rlee's point. If the plate was connected to ground thru the plate resistor, you would lose the high voltage DC on the plate, no current flow, the show is over.

I find it interesting that caps often serve a DC purpose as well as a different AC purpose.

11. ### BendyhaFriend of Leo'sSilver Supporter

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A.C.= Alternating Current....AC current flow = Alternating current current flow.

There is current flowing from the ground to the ground, it is a circuit, break it anywhere and it will cease to function.

This current is present on the cathode, and plate in equal amounts...because it is all part of the same current.
With a good power supply, in quiescence, this will be a constant unidirectional flow of non-alternating direct current.
A small fluctuating voltage presented to the grid will modulate the flow of current through the tube, imparting upon the direct current a proportion of alternating current - BUT... it is the same current, from the same source, heading for the same destination. Get rid of the large direct current supply, and the input current generated by your guitar pickups will not be amplified...this source is, and remains tiny.

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12. ### petebFriend of Leo's

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Ludwig,

I think the diagrams and explanation in post 10 are good.

i think it is broken down into small elements and self justifying, that's the reason for the diagram (justification).

is there any part that you feel would benefit from real documentation from a real source?

Or any part that sounds spurious?

thanks

13. ### petebFriend of Leo's

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Bendyha,

thanks for post 11 and other comments.

is post 11 in agreement or disagreement with post 10?

thanks

14. ### BendyhaFriend of Leo'sSilver Supporter

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When you postulate;

Then, my attemps at the dissambiguation of post 10, in post 11, should be taken as unsupporting.

Last edited: Apr 11, 2018
15. ### petebFriend of Leo's

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If you accept the fact that the signal voltage on the cathode develops across the cathode resistor, and the signal voltage on the plate develops across the plate resistor, then the rest is an obvious and direct result.

Post 10 stands as an easy way to understand how the cathodyne phase inverter provides equal and out of phase signals at its cathode and plate.

16. ### BendyhaFriend of Leo'sSilver Supporter

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You don't mention signal voltage in post 10 like you do now in post 15, just DC high voltage, any reference there to signal is altenating current....you are not making this cathodyne easier to understand.

I think I prefer my attemp here.... The cathodyne, a simplified explanation of how it works.

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17. ### petebFriend of Leo's

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I did, in post ten, but it is my mistake to say post ten is the important post.

The explanation was given in post one, fixed up in post 8, and then clarification was added in post ten.

The best part is the two diagrams in post eight. In two lines of text it clearly shows how the plate and cathode have to be 180 degrees out of phase, based on the orientation of the plate and cathode resistors with the tube and ground.

18. ### RLee77Friend of Leo's

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Almost, but not quite. The plate signal is not taken from across the plate R; it’s ground-to-plate. The plate voltage is b+ minus the voltage across the plate R, so the same current that causes the cathode voltage to rise causes the plate voltage to fall.

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19. ### BendyhaFriend of Leo'sSilver Supporter

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No....where....it doesn't come up on my screen

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20. ### petebFriend of Leo's

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Thanks RLee,

You know the tubes well.

I think we are more in agreement than you are saying.

The plate signal is read at the plate, referenced to ground, it could in fact be the voltage read across plate R as the other end is direct referenced to ground thru the filter cap. The main thing I have been saying about plate R is that the signal voltage develops across plate R as a result of the signal current flowing thru plate R. I don't see that we are in disagreement here.

The plate voltage is B+..........

We do see things differently here now, but I think that you are going to agree with me.

The plate voltage in my diagrams is AC:

The difference between our views is that I have separated AC and DC completely.

The signal AC that appears across plate R is due to the AC current component flowing thru plate R.

And it is worth noting that the overall current flow thru plate R and the tube and the the cathode resitance, is overall DC, electrons moving in one direction only, but there is an AC component to the current, and the AC current component truly is bidirectional.

"So the same current that causes the cathode voltage to rise causes the plate voltage to fall."

Well said, we can agree on that

Last edited: Apr 11, 2018
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