X-101-B, P.I. and "noose"????

larryderouin

I'm VERTICAL and Breathing...most of the time.
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On the X-101-B, I presume the PI is connected to Pin 1 & Pin 3, they being the operative pins.

As the X-101-B is built either right before the 400 or at the same time as the 400, would it have the "noose" that the 400 has?

Right channel scat for P.I. and output.
x-101-B Phase INv..jpg
 
It does (R78), but because of its much higher value in this implementation, has nowhere near the impact on the operation of the phase inverter that it does in the 400's design. Have you had trouble running Russian tubes in that position? That seems to be the recipe that will throw that stage into problems in stock 400 receivers.

Dave
 
Not so far. The only non russian tube is the Rectifier (Mullard). It sounds fine at higher volume settings.

So for the PI adjust, I short the B+ to the stage at the cap, then adjust the pot so pin 1 & 3 are equal. Not gonna worry about the noose if you're not too worried about it. I'll eventually get some Domestic 12ax7's for the PI's but these Tung-Sols seem to be doing ok.

I'm getting ready to bake the 101 for a few hours, and trace out the wiring for the phono vs. tuner and either move them away from each other or shield them like I did in the corner for the phono plugs. One other thing. Mag 2 has a 100K (R1) to ground while Mag 1 has no loading resistor. All of my cartridges require a 47K loading. Can I replace R1 with a 47K(both sides), then add a 47K to Mag 1 (both sides). I'm using Mag 1 right now as Mag2 sounds rolled off in the highs.
Phono section.jpg
 
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With a 100K to ground at the input jack, the effective loading is 50K. R9 is 100k to ground, so anything at the input jack is in parallel with that. If you want it closer to a real 47K load, an 89 or 90k at the input jack would get you there.

Somewhat a sidebar here, but I wonder if the "noose" didn't actually serve a real purpose in the original config with the high value grid resistors. there was a whole discussion about DC blocking distortion and the DC restorer circuit, and I seem to recall the rough conclusion that was arrived at is that the higher value you make the grid return resistors, the more it can actually occur. Maybe Fisher ran into that with the 330k grid resistors they selected, but instead of changing the grid resistors to make it stop, they limited the output capability of the phase inverter to prevent it from driving the output tube grid positive.
 
But MAG 2 is NOT connected to MAG 1. To me there is no way they could be in parallel as they are never connected together, hence my question of the separate 47Kr. Plus MAG1 is Grounded when MAG2/CERAMIC is selected.
 
Larry -- Connect a 91K across the Mag 1 input. When it is selected, it will then represent a 47K load to the cartridge connected to that imput, while its addition will have no effect on the Mag 2 input loading characteristics.

Regarding the noose, as I have studied it over the years, I have found that it addresses a number of areas that Fisher could have been concerned about:

1. Over-driving the output stage is one of them. The noose largely presents that and in so doing, protects the tubes, and could also act to limit blocking distortion as Gadget suggests by limiting the current required to produce the blocking action.

2. Properly bias the phase inverter section -- which I now believe was their primary goal:

Fisher was trying to have their cake and eat it to with the driver stage: Use relatively low value inverter splitting resistors to help emphasize a low output impedance on each side of the push-pull signal, AND maximize the gain from the overall driver stage -- and all at minimum cost. Maximizing the gain meant elevating the plate voltage of the AF Amplifier section, and increasing the value of that section's plate resistor. But the higher plate voltage of the AF Amplifier section in combination with the low value splitter resistance values then underbiased the splitter section: Without the noose, the bias of the inverter section would often either be 0 or even go positive! The noose then lowers the current flow through the inverter section (by shunting some current around it), when then allows it to properly bias itself. And, in some versions, connection the cathode side inverter resistor back to the NFB insertion point also provided a small degree of POSITIVE FB, which then increased gain even more.

It all "works" but:

A. Leaves so little current flow through the inverter section that it leaves its operating point right on the ragged edge of even allowing the tube to operate -- let alone linearly, or with enough distortion free output to fully drive the output stage.

B. Being on the ragged edge of the operating point, it means that some tubes will work OK, while others (most often the Russian tubes, but also some American examples as well) simply will not.

It is easy to see when it doesn't work by checking the power output of each channel with a scope. When the noose is rearing it's ugly head, the clipping gets very lopsided and weird well before full power output is developed. I have seen it happen way too many times on way too many units. At low or normal listening levels, you don't know its there. But push it and it shows up in a hurry.

I still believe that the best cure is to remove the noose, THEN source the B+ for the plate resistor of the AF Amplifier section to the next lower B+ supply point (that used for the line amp/tone control circuits), and then readjust the phase inverter control for balance using the procedure I developed. That approach is what I first used in the improved power amplifier section as published in the improving the 400 thread, and has worked perfectly every time since in any application using the noose.

Dave
 
But MAG 2 is NOT connected to MAG 1. To me there is no way they could be in parallel as they are never connected together, hence my question of the separate 47Kr. Plus MAG1 is Grounded when MAG2/CERAMIC is selected.

Right, what I was getting at is that a 47K wired across mag1 would not present a 47K load to the phono cartridge. The 100K R1 on mag2 is in parallel with the 100K R9, so anything connected to mag2 sees a 50K load. Anything connected to mag1 sees a 100K load in it's originally built configuration. If you want it to see a 47K load, a resistor in the 89-91K range on mag1 and mag 2 (replacing the existing 100K on mag2) would give both inputs a 47K load, or close enough to that value.
 
Close, but not exactly. You also have to consider the effects of R3 and R5 in the Mag 2 input circuit. When they are included in the calculation, the actual input impedance of the Mag 2 input is 44.318K as designed by Fisher (assuming perfect component values). To mimic the actual Mag 2 input impedance at the Mag 1 input, simply connect an 82K resistor across each Mag 1 input jack. This will yield a Mag 1 input impedance of 45.055K, which will then have the two Mag inputs represent a load that is within 2% of each other.

Dave
 
Thanks to both Gadget and Dave for the help. When you add R 9 into the equation it makes sense. 82K it is.

On the P. I. for power I separate the power input between R78 and R80, leaving R78 on the same power section, and run R80 to the next lower Power section, remove R82, then re-adjust for balance????
 
No. Based on the partial schematic you provided, you disconnect R68 from the B+ source R78 & R80 are connected to, and then connect R68 to the next lower B+ voltage tap. Then you remove R78. Then readjust the Phase Inverter control for balance.

Dave
 
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