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In Reply to: RE: Two Questions posted by FlaCharlie on August 01, 2007 at 10:54:57
"can I omit the plate resistors entirely or put in a much lower nominal value (like 1k) and still have the lower output impedance? Or does this create some other problem?"
No, you cant't do either one. The plate resistor is part of the 'load' seen by the 6SN7s, and shouldn't be set much below a value of 2*Rp (about 15K Ohms).
"Yes and I'd like to eliminate that tube. Is this possible? Does the 120v have to come from a 'pre-driver' tube or can it be supplied from a cap in the PS and have the input signal come from the normal RCA jacks?"
It is possible to eliminate the predriver, but the grid to plate voltage and grid to cathode voltages in the LTP should remain the same. With a 300 volt B+ supply, one alternative would be to connect the plate resistors to +300, tie the grids to ground via a 220K resistor, and move the resistor in the cathodes of the 6SN7s to -115 volts. Yes, that would require an adiitional supply, but that would also allow you to use fixed bias on the 6B4s.
Follow Ups:
So I need at least a 15k load on each plate.
Then you said: "It is possible to eliminate the predriver, but the grid to plate voltage and grid to cathode voltages in the LTP should remain the same. With a 300 volt B+ supply, one alternative would be to connect the plate resistors to +300" . . .
The same voltages or the same ratio? If I have a B+ of 300v and 15k (or more) on each plate my voltage at the plate will obviously be less than 300v.
". . . tie the grids to ground via a 220K resistor" OK, that I get.
Then: ". . . and move the resistor in the cathodes of the 6SN7s to -115 volts."
I'm confused here. I thought the goal was to keep the voltages the same, why would I need to change the cathode voltage from +125v to a -115v? If that's what's needed though, what size resistor would I need?
And: " Yes, that would require an adiitional supply, but that would also allow you to use fixed bias on the 6B4s."
So just another R-C section or does this supply need to be negative voltage? Where would it be connected? to the grids? Also, would this require the use of fixed bias or could I still use cathode bias as planned?
Also :^) . . . regarding Dave Cigna's defense of the paraphase below, if the LTP has only half as much gain, how much of a problem is using a LTP likely to create since I'm already eliminating one gain stage by using only one 6SN7/two 2C22s per channel?
Thanks for all your collective wisdom and (especially!) patience guys! . . . Charlie
Ok I understand all that. Thanks for the clear explanation! But what about the gain issue which was raised? If this does produce half the gain of the paraphase version, will this be an issue considering that I've removed a stage?
. . . . Charlie
Try the PP-1C schematic for a "concept". It is, of course, only one of many possible ways to go.
Russ
Actually, one of the things that I like about the LTP (differential amplifier) is that it is fairly easy to understand. Referring to your original schematic - the 18K resistor that is common to both cathodes has 125 volts across it, so the current through it is 7 mA. This is split equally (in theory) between the two 6SN7 triodes, so the current through each plate resistor is 3.5 mA. So the drop across the 33K is about 115 volts, which explains why the plate voltage is 420 - 115 = 305 volts. But the grids are at potential of +120 volts, so the plate is about 180 volts higher than the grid. If we move the grids to a zero volts and the supply to 300 volts and leave the plate resistors and current the same, the plates will be at 185 volts and the grid to plate voltage will be the same - 185 volts.
In the proposed design, we need to get about 125 volts across the 18K resistor common to the cathodes. The grids will be about 5 volts negative with respect to the cathode, or in other words, the cathodes will be 5 volts positive with respect to the grids (0 volts). So we need to connect the bottom (previously grounded) end of this resistor to -120.
“So just another R-C section or does this supply need to be negative voltage?”
It needs to be a negative supply, but only needs to supply about 7 mA of current.
“Also, would this require the use of fixed bias or could I still use cathode bias as planned?”
You would not have to go to fixed bias, but it would be a possibility.
Ok, following your suggestion, the grids have been dropped to 0v and the difference (180v) between the original grid (120v) and plate (300v) has been roughly maintained by dropping the supply voltage to 300v and keeping the 33k plate resistor. Actually 295v supply would be exact and is the B+ in Pete Millet's schematic.
We now want to have the cathode at 5v (+5v relative to the grid and -175v relative to the plate). Couldn't this be accomplished by simply replacing the 18k resistor with a 720 ohm resistor?
Or am I missing something here?
. . . . Charlie
Your math is correct but, unfotunately, there is an additional requirement. Since the coupling between the stages is via the cathode, the resistor in the cathode is a 'load'. The term 'long tail' reflects the requirement that this resistor (the 'tail' of the differential amplifier) be 'long', i.e., have a high value. In fact, the lower the value of the cathode resistor, the poorer the signal balance between halves. This is because the circuit relies on the cathode resistor being a 'constant current' source. The higher the resistance, the closer to a 'constant current source it becomes.
The article below shows an alternative approach that does not require a negative voltage supply. By the way, the 6SL7 will provide plenty of gain for your application. For an even better tube (the 6SL7 is an excellent sounding tube), try the 6188.
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