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Old 21st December 2003, 10:42 AM   #11
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Quote:
Originally posted by Jax
One 12AX7 and a pair of 6BQ5? Made me think for a while and came up with an idea using the old paraphase splitter.

Schematic
I like it. Similar to a 6V6 amp I built a while back.

Play with the imbalance and see if you like the sound; 5% seemed pretty nice. But ditch the 12AX7's and put in a decent sounding tube like an ECC99.
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Old 21st December 2003, 10:59 AM   #12
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ECC99 would require some changes to my circuit though.

Change R2 and R3 to 22K 2W, R6 to 5.6K 2W, R16 and R17 to 680 ohm as a suggestion (100V Va at 5 mA).
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Old 21st December 2003, 12:35 PM   #13
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Quote:
Originally posted by Jax
ECC99 would require some changes to my circuit though.
I know.

Quote:
Change R2 and R3 to 22K 2W, R6 to 5.6K 2W, R16 and R17 to 680 ohm as a suggestion (100V Va at 5 mA).
Waaaaaaaaaaaaaay too low an op point; they don't really wake up until 15mA. I used choke and later CCS loading in mine.
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Old 21st December 2003, 01:48 PM   #14
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Quote:
Originally posted by Brett
Waaaaaaaaaaaaaay too low an op point; they don't really wake up until 15mA. I used choke and later CCS loading in mine.

I agree

I just thought it was a bit hot there
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Old 21st December 2003, 02:51 PM   #15
SY is offline SY  United States
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Quote:
No, you don't need an extra cap...If the source is cap coupled, you don't want a second cap.
Frank, you really want DC running through that pot? The required second cap is not to isolate the source but to keep the biasing DC off the pot wiper. Moot, since this is now a design for a CDP, and the "phase splitter" will now need gain.

I still maintain that a simple diff amp is still the best choice here for this application. Gain, low parts count, stability, reasonable balance, and a convenient node for any desired loop feedback.
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Old 21st December 2003, 04:20 PM   #16
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Default Just breadboarding

to learn. Diff amp could be next attempt, Sy.

Tim- I do have about 100k as load to 12ax7 don't I? Makes no difference how many resistors or where they are in the series, no?

Changed R1 to 1k. Changed the input grounding, too. Lost the jumper.

Parts counts on this one is 2.5 tubes (2 6bq5 and 1/2 12ax7).

I should have 50v swing from driver, right? This would be split up to outputs?

Rick
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Old 21st December 2003, 06:06 PM   #17
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Correction: driver tube will swing from about 80 v to about 200v using 100k load resistor to be split between two output tubes. More than enough to drive the 6bq5s.
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Old 21st December 2003, 11:32 PM   #18
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Erm... now you have a floating input. Put a 100k to 1M resistor between grid and the join between R1 and R2, put a capacitor between grid and input, and ground the ground side of input.

Tim
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Old 22nd December 2003, 04:18 AM   #19
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Default Thanks, Tim

Won't the 100k vol pot suffice instead of adding another grid resistor? Capacitor on the input? For CDP input, does the circuit need it? If I ground the ground side of input, won't that effectively put the grid input at -150v bias (below both the load resistor and the cathode bias resistor)?

Appreciate the comments,
Rick
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Old 22nd December 2003, 06:49 AM   #20
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No. The grid has to float on top of the cathode at its circa 75V potential.

Grounding the input will not put the grid at -150V vs. cathode because for one thing, cathode voltage will follow grid voltage naturally (and because of the nature of the circuit, plate voltage will be an inversion of this), so it wouldn't be a full 150V, and two, you use a capacitor so the grid bias is set by R1 instead.

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