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Old 10th April 2006, 05:20 AM   #1
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Question Help with inverter CCS+NFB

I am working on a P-P 7189 amp, class A (probably around 275v/10k load... haven't decided yet) and I want to rig up a simple-as-possible CCS phase inverter. Since the drive requirements for the 7189's are so low, a 12ax7 LTP should be fine to drive them from a line level input, so I don't have a driver stage to return feedback to. I drew up this inverter and wanted someone to check it out. I returned the feedback to the non-inverting grid. Simulations looked OK, but I want some opinions/ other options options before I breadboard.

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Old 10th April 2006, 05:25 AM   #2
hacknet is offline hacknet  Singapore
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i always find 12ax7s failling to sound dynamic when forced to swing more than a couple volts. why not try some tube like the 5687 that is more fit for this role?
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Old 10th April 2006, 05:39 AM   #3
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Quote:
Originally posted by hacknet
i always find 12ax7s failling to sound dynamic when forced to swing more than a couple volts. why not try some tube like the 5687 that is more fit for this role?

Only because I have a pile of 12ax7's around. If the design ends up being sound, I'll probably redesign around a better tube.

Does the circuit itself look good? I could always readjust the plate resistors and CCS for a different tube later.
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Old 10th April 2006, 07:24 AM   #4
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Quote:
Does the circuit itself look good?
Yes, apart from one thing - the CCS needs a bit of headroom. You need either to povide a negative rail (say -35v) for the CCS or to bias the LTP above ground.

The second method can be achieved by connecting the grounded grid to a potential divider from B+ to ground, with a cap of ~0.33uF from that grid to ground, in series with your feedback resistor (which can be smaller than you have shown here - say 1k). You need a 1Meg resistor between the grids, instead of from grid of the left half of the LTP to ground.

It can also be achieved by dierect coupling from the previous stage (if there is one). You will still need the bypass cap and feedback resistor on the grounded grid and the 1Meg resistor between the grids.
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Old 10th April 2006, 03:45 PM   #5
Nafty is offline Nafty  Israel
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Default Phase split

Hello.
See if that idea meay be usefull to you.
You nay reduce considerably NFB.
Best regards
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Old 10th April 2006, 04:27 PM   #6
Nafty is offline Nafty  Israel
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Default sorry

See if that one is usefull to you:
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Old 10th April 2006, 04:46 PM   #7
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Quote:
Originally posted by ray_moth


Yes, apart from one thing - the CCS needs a bit of headroom. You need either to povide a negative rail (say -35v) for the CCS or to bias the LTP above ground.
So the -12v that I have it referenced it to is low? I can up that to -30v or more if I have to. The fet I had is only rated at 25v, but that's no biggie. How do I calculate for that required headroom there?

Quote:
The second method can be achieved by connecting the grounded grid to a potential divider from B+ to ground, with a cap of ~0.33uF from that grid to ground, in series with your feedback resistor (which can be smaller than you have shown here - say 1k). You need a 1Meg resistor between the grids, instead of from grid of the left half of the LTP to ground.
I have used that method too... I am just trying to keep this PI as simple as possible, going for a minimalistic thing.

Quote:
It can also be achieved by dierect coupling from the previous stage (if there is one). You will still need the bypass cap and feedback resistor on the grounded grid and the 1Meg resistor between the grids.
No previous stage, so that doesn't work out.
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Old 10th April 2006, 04:51 PM   #8
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Default Re: sorry

Quote:
Originally posted by Nafty
See if that one is usefull to you:

That looks pretty simple. What is the theory behind biasing the transistor to the junction of the 1M resistors? Do I need to generate a negative rail there as well?
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Old 10th April 2006, 05:32 PM   #9
Nafty is offline Nafty  Israel
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Default simple.

This diff pair can work with only one input.
The 1M resistors are error detector and at same time they pull the transistor on wich provide current to the diff pair. See that if any ac in the midle of the two resistors, is considered error and compensated on the cathode current.
You may use NFb or just ground the second grid.
NO -V requiered.
Regards.
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Old 10th April 2006, 05:58 PM   #10
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Default Re: simple.

Quote:
Originally posted by Nafty
This diff pair can work with only one input.
The 1M resistors are error detector and at same time they pull the transistor on wich provide current to the diff pair. See that if any ac in the midle of the two resistors, is considered error and compensated on the cathode current.
You may use NFb or just ground the second grid.
NO -V requiered.
Regards.
Nafty
OBS: 3.6, are Kilos.
Thank you. I'll give it a try. Any suggestions on transistors?
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