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Old 31st December 2004, 10:51 PM   #1
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Default Phase Splitter as input stage

I have a Dyna 70 clone, (it is a point to point wired version of Ned Carlson's circuit with the EF86 and the El34's triode wired) and I would like to convert it to a PP6B4g Amp.

I have read Lynn Olson's Amity article and would like to drive the output tubes with an ECC99, but rather than an input tranny I was wondering about 1/2 of another ECC99 wired as a phase splitter.

In his tubecad notes, J Broskie indicates that the Split Load Phase Splitter sets the standard for balance and suggests overcoming its limited voltage swing by placing it in the very first stage of the power amp.

The topography of the Dynacclone would work well since you would be bringing the input directly to the center dual triode and then on to the driver stage.

I would appreciate any thoughts on this idea since I have never seen it used before.

Pete
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Old 1st January 2005, 01:59 AM   #2
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I don't see any reason why this wouldn't work provided you had enough input voltage available. This phase splitter operates at less than unity gain. You would be required to capacitively couple the input however to allow the grid of the splitter triode to reach proper mid rail bias.
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Old 1st January 2005, 08:37 AM   #3
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If you're using a split-load phase splitter as the first stage, and since this has less than unity gain, you'll basically need another stage to have some sort of reasonable input sensitivity. Since there are now 2 phases, that means an additional two valves (or sections thereof), bringing the total to three. That means you can't use a single ECC99 for the entire input/driver section.

I think this was what rcavictim was eluding to, but I'm not sure so I'll say it anyway. You could get away with an amp with a very low input sensitivity if you had a preamp capable of large-ish voltage swings at its outputs (i.e. larger than the normal 2V or so).... but then that just becomes a three-stage arrangement anyway.
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Old 1st January 2005, 01:20 PM   #4
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Quote:
" You would be required to capacitively couple the input however to allow the grid of the splitter triode to reach proper mid-rail bias."
Thanks for the responses. I understand the issue of sensitivity however, I am not certain what you mean by the above statement. Could you elaborate please?

Pete
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Old 1st January 2005, 02:01 PM   #5
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A cathodyne splitter does just that, it splits the phase. It does nothing else in terms of gain. (It also has limitations due to the high plate Zo vs. low cathode Zo, particularly against capacitance and thus frequency response, but I digress.) Thus putting it before or after the required gain stages makes little difference, except in the amplitude you expect it to produce.

So. Let's put it just before the output tubes. Gain comes first, as what amounts to a preamp stage. The output tubes get little drive, which may or may not be an issue - triodes have large miller capacitance, while pentodes escape that, with higher gain to boot. (Certainly, voltage output is not an issue, if you bias and supply it properly. It requires a lot of voltage, since it produces the peak to peak drive voltages simultaneously!)

Now let's put it at the front instead. It produces NO distortion ("none" is impossible, but seriously, I mean, almost total NFB here...). It splits phase and now a balanced amplifier stage is required, taking an extra tube (or two). Balance can be improved using a common cathode resistor, particularly a large one (or a CCS, i.e. LTP). This is important if HF balance is required, especially if a weak (low current) cathodyne is used (assuming triodes (high miller C) for the gain stage, which is pretty much certain). The output tubes are driven well, depending on the driver tubes. But you use two tubes instead of one.

For high gain situations, a cross between the two is used - a preamp front end, usually direct coupled to the phase splitter, which then drives a balanced amp which drives the output. The basic Williamson topology. Unfortunately... it *really* sucks. You get, at best, three phase shift elements in your loop (two pairs of coupling caps and the OPT inductance), making for motorboat city if you aren't careful.

So where does that leave us? Well, a pentode has a lot of gain, so let's use one of those. LTPs are great because they have gain and produce an unconditionally balanced output, even when driven with an unbalanced signal. And we can stand the LTP on top of a tube CCS, allowing us to direct couple the pentode to the LTP and eliminate a coupling cap from the loop. Wow, it suddenly works! And with only *TWO* tubes, at that.

(Well okay, the attached doesn't show *quite* that. You could use the 6U8 triode for a CCS instead, if you do it right. It might cost some extra voltage though.)

Tim
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Old 1st January 2005, 03:02 PM   #6
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Quote:
(It also has limitations due to the high plate Zo vs. low cathode Zo, particularly against capacitance and thus frequency response, but I digress.)
I thought that this myth had been exposed enough in this and other forums but maybe not.

The output impedances at anode and cathode of a split load phase inverter are equal given that cathode and anode are equally loaded. This has been mathematically described by several earlier and it is easy to convince one self about this by either measuring output impedance on a real circuit or by doing spice simulations.

Regards Hans
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Old 2nd January 2005, 02:51 AM   #7
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No, not really...

Tim
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Old 2nd January 2005, 03:29 AM   #8
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Hi,

Quote:
The output impedances at anode and cathode of a split load phase inverter are equal given that cathode and anode are equally loaded.
How can this possibly be?

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Old 2nd January 2005, 04:43 AM   #9
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You have to start with the assumption that the plate load and cathode load are symmetrical...
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Old 2nd January 2005, 04:57 AM   #10
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Hi,

Quote:
You have to start with the assumption that the plate load and cathode load are symmetrical...
Sure but while that would give a (more or less) equal output from the cathodyne I fail to see how that would make for an equal Zout at both ends however.

No electronics engineering book I have seems to mention this. I recall we went through this some time ago but I still fail to see to logic....
Surely resistively loading plate and cathode equally doesn't make for equal Zout at both ends as they never are. Hence "assumption", perhaps ...?

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