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Old 12th February 2019, 12:50 AM   #21
Printer2 is offline Printer2  Canada
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Quote:
Originally Posted by Gnobuddy View Post
I agree that large grid-stoppers are one of the keys to managing cathodyne distortion. <snip>
I was going to ask you to model it with the grid stopper, good to see it. I could live with that level of distortion.
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Old 12th February 2019, 03:20 AM   #22
Gnobuddy is offline Gnobuddy  Canada
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Quote:
Originally Posted by Printer2 View Post
I could live with that level of distortion.
I was thinking the same thing. With this much drive, the output valves are already distorting heavily. So some distortion in the PI may not be audible on top of that.


-Gnobuddy
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Old 12th February 2019, 03:37 AM   #23
Gnobuddy is offline Gnobuddy  Canada
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Originally Posted by klingo View Post
...will reach -1db at a frequency the guitar speaker & guitarist ear (the golden one, next to the splash cymbal) will ignore.


I used a similar 1/(2 pi RC) calculation to convince myself that 120k grid stoppers wouldn't cause any audible issues in a guitar amp.
Quote:
Originally Posted by klingo View Post
Now we should take care of the total grid load resistance
Exactly! I ran into that same issue, as the grid stopper gets bigger, the grid bias resistor has to get smaller to keep the total within datasheet limits. Which means the source-o-dyne has to work even harder, to drive the smaller grid bias resistor.

Quote:
Originally Posted by klingo View Post
the drawback might be harder grid current clipping from a higher source impedance...should we consider using a 2x115v-2x24v small stepdown toroid load in a LTPI?
Tubelab_com (George) has been using the elegant solution of driving the output valves with a direct-coupled MOSFET source follower. Not only does this eliminate blocking distortion, it can also let you drive the output valve into class AB2 (grid current flow) territory if you want, for more output power.

The downside is that you need an additional negative voltage rail to supply the fixed bias to both the MOSFET and output valve, and you also need an additional positive voltage to supply the MOSFET drain.

But these days switch-mode power supplies are small and cheap, and we can probably put together the MOSFETs and additional power supplies for less money, bulk, and weight than a pair of coupling transformers.
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Originally Posted by klingo View Post
The long tail pair could also suffer from spike effect ...if anode/drain voltage of the inverting device is clamped...
You can certainly end up clipping the signal at either of the LTP outputs because of grid current flow in the output valves. But, unlike the cathodyne / source-o-dyne, if you clip one of the PI output signals, it won't affect the other one. I don't think you can create spikes on one output by clipping the other output.

This is because there is almost no back-coupling* between a MOSFET drain and its gate or source, so doing weird things to the voltage at one FET drain has no way to affect whatever is going on at the other MOSFETs source or gate. I did a quick LTSpice sim, which confirms this (see attachment.)

(* - I mean there is very little mutual conductance from drain to source, i.e., an extremely high "plate resistance". There is a tiny bit of capacitance from drain to gate, which does affect input impedance at the gate at sufficiently high frequencies.)


-Gnobuddy
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File Type: png MOSFET_LTP_w_Grid_Current_Flow_001.png (59.0 KB, 42 views)
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Old 12th February 2019, 09:18 AM   #24
klingo is offline klingo
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Nice, your sim looks good, and nippleless ...a 47k does the trick here. You skip the CCS and (to me) it is a good idea as we (the guitarist) are looking for the perfect imbalance.

loadlines told me that the cathodyne is more "sturdier" under load than LTPI, could be the output to output Zout which is in the cathode follower range....might be the same for MOSFET.

I've been lurking these DC MOSFET follower since the MOSFET folies article, but the +/- additional power supply.......i tried a 3.2va 2x115v-2x18v talema transformer in a dc coupled LTPI today feeding an ECC99, distortion is ending gracefully but it sound a bit too dark & not that gainy (too much steppin' down?), not sure if it is grid current loading or leakage inductance/interwinding capacitance....will add a source follower after the input pentode and a sourceodyne to feed the LTPI, will see.
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Old 12th February 2019, 11:33 PM   #25
Gnobuddy is offline Gnobuddy  Canada
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Quote:
Originally Posted by klingo View Post
...we (the guitarist) are looking for the perfect imbalance.
There is scope for dialing in deliberate imbalance if desired. One could just replace one of the drain resistors with a suitable pot. Alternatively, one of the 1k source resistors could be replaced with a pot.
Quote:
Originally Posted by klingo View Post
loadlines told me that the cathodyne is more "sturdier" under load than LTPI, could be the output to output Zout which is in the cathode follower range...
It seems the cathodyne / source-o-dyne circuit is tremendously misunderstood. Output impedance at the cathode (or source) is indeed very low. But this is not the case at the anode (or drain)!

This is one area where even Blencowe's book got it wrong. The whole section in the book about output-to-output impedance is tragically wrong, a complete red herring. Have you noticed that in actual use, a cathodyne does not drive a floating load suspended between cathode and anode? Instead, both anode and cathode drive a load (grid bias resistors) with the other end grounded. There is no floating load anywhere!

The confusion stems from the fact that you get equal voltage outputs at both cathode and anode when they are both loaded equally. This has been wrongly interpreted as proof that the output impedance is the same at both cathode and anode. But in fact you cannot draw that conclusion from that data at all! (See attached figure - equal output voltages, from two voltage sources with extremely different Zout.)

What you have to do - independently, at both cathode and anode - is measure the voltage, change the load resistance, and measure the voltage again. Then you can use those two separate voltage measurements, along with the corresponding change in load resistance, to calculate Zout.

The easiest way to do this - independently, at both cathode and anode - is to set the load resistance to infinity first (no load at all!), measure the output voltage, then connect an actual load, and measure the output voltage again.

If you do this, you will come to the correct conclusion: Zout at the cathode is very low, like a cathode follower. Zout at the anode is very high, like a gain stage with heavy local negative feedback due to a large unbypassed cathode resistor.

The key fact that has been overlooked is that when you hook up a load to the cathode output, it changes the voltage gain to the anode, increasing the output voltage at the anode. Then when you connect the same load to the anode, the voltage falls dramatically, and ends up at the same voltage as the cathode signal. But that large voltage drop is what tells you the Zout is much higher there (at the anode.)


-Gnobuddy
Attached Images
File Type: png Unequal_Zout_Demo_001.png (52.8 KB, 26 views)
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Old 13th February 2019, 07:50 AM   #26
klingo is offline klingo
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Hi, Floating load...yes you can... see attachement.

Not sure MerlinB got it wrong, he describes the issue we are talking about, i mean grid current draw from the output tube which is like paralleling a resistor to the cathode/source resistor and by reducing feedback it increase the anode/drain gain.

I will not criticize MerlinB, because his books help me a lot to understand what i'm doing to these poor valves amps, and to improve my english as well

By "sturdier" i got it wrong, it's just that paralleling a 220k load to a 56k rotate the loadline less than paralleling a 220k load to a 100k. But if the cathodyne has 100k anode/cathode resistor it will not be sturdier than the LTPI with 100k anode resistor.

Thanks for your time and explanation.
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File Type: png Parafeedodyne.png (5.3 KB, 22 views)
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Old 13th February 2019, 03:28 PM   #27
Gnobuddy is offline Gnobuddy  Canada
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Quote:
Originally Posted by klingo View Post
Hi, Floating load...yes you can... see attachement.
If you have a coupling transformer, why use a phase splitter at all? A centre-tapped transformer is an excellent phase splitter, and there are no capacitors to produce blocking distortion!

In the schematic you posted, I notice there are no values on the resistors. So we have no reason to believe that the cathode resistor equals the anode resistor. I think the schematic is just showing a gain stage with a (small) unbypassed cathode resistor for local negative feedback, and transformer coupling to the next stage.

Also note that in the schematic you posted, there is only one phase from the output of the transformer - you can't use this to drive a push-pull pair!

Look at any tube/valve guitar amp schematic with a cathodyne PI and you'll see what I mean. Each output drives a grid bias resistor with its other end grounded. There is no load wired directly between anode and cathode.
Quote:
Originally Posted by klingo View Post
Not sure MerlinB got it wrong
Please don't misunderstand, I love Merlin's guitar preamp book. It is the best book on valve guitar amps I have ever read, and I learned a lot from it. I still re-read it every now and then, and still learn new things from it.

But that doesn't mean there are no mistakes in it. The section on calculating cathodyne output impedance - using a floating load and coming to the conclusion that cathode and anode output impedances are the same - that section is wrong.
Quote:
Originally Posted by klingo View Post
But if the cathodyne has 100k anode/cathode resistor it will not be sturdier than the LTPI with 100k anode resistor.
Actually, it will - but only at the cathode! The output impedance at the cathode side will be less than 1k, even with a 100k cathode resistor. Very "sturdy"!

But the output impedance at the anode side will be fifty to a hundred times higher (depending on internal anode resistance ra of the triode.) It will be in the neighbourhood of 50k - 100k, instead of 1k! The cathode output is not "sturdy" at all.

This is exactly the mistake I'm talking about in Merlin's book. He comes to the conclusion that cathode and anode output impedances are equal - and that is the wrong conclusion.


-Gnobuddy
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