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Old 28th September 2003, 01:35 AM   #1
SHiFTY is offline SHiFTY  New Zealand
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Unhappy Direct coupling question...

I am building a PP 807 amp that uses a direct coupled pentode in the front end. Only problem is my voltages are all out of whack, possibly because I am using EF37A instead of 6J7 which I suspect this circuit was designed for.

In the schematic of the input stage below, how crucial are the voltages? If I adjust the power supply resistors to get the plates of the 6SN7 to 290V, and the grid 110V, is it all OK?

How does direct coupling really work anyways?
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Old 28th September 2003, 02:03 AM   #2
Colt45 is offline Colt45  Serbia
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if its over 110v, the grid of the 6SN7 will start pulling current.. if its too much over it will melt

if its too far under, the operating point will shift (the other direction from melting)
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Old 28th September 2003, 02:57 AM   #3
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Nope, the 6SN7 cathodes are resting on resistors, not a voltage source. A change in grid voltage will change current level, but it's pretty soft. Come on Colt, you know this... remember? Cathode follower?

I'd feel better with a CCS in place of that 20k for the 6SN7 cathode load, but whatever.

Lesse... to get 6SN7 plates in range... BTW I'd feel more comfortable with them around 220-250V... and 6J7's plate a bit lower too... anyway... change the 6J7's 3.3k cathode bias resistor to a 5k pot and go adjustin'.
Frankly I can't believe a direct-coupled design going a few stages wouldn't have a bias control. It leaves no adjustment for different tube characteristics.

Tim
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Old 28th September 2003, 03:06 AM   #4
Colt45 is offline Colt45  Serbia
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yeah yeah, ive been awake 24 hours.. didnt put too much thought into it ;-P
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Old 28th September 2003, 04:29 AM   #5
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Default direct coupling

I'm a newbie but direct coupling simply means that the output tube cathode biasing is set above the quiescent point for the driver tube, I think.

Perhaps some hypothetical numbers will help.Say your operating point voltage for the driver is 120 and you swing maybe 10 volts both ways. You would need to set your output tube cathode bias at above 120 plus the room for the AC signal. With capacitor coupling, the 120 DC is knocked down essentially leaving the output tube at ground. Then the output tube would only "see" the AC signal. The output tube would be biased so that the AC signal would not be clipped.

Probably not the best explanation but I'm sure others will correct or revise this.
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Old 28th September 2003, 05:56 PM   #6
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Default Re: direct coupling

Quote:
Originally posted by fragman56
I'm a newbie but direct coupling simply means that the output tube cathode biasing is set above the quiescent point for the driver tube, I think.
It means that any two or more stages are DC-coupled together.

It's possible to have the whole thing done, but to do it well requires adjustments to counter day-to-day settling and differences in tubes, and lots of DC NFB to keep the operating point stable.

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Old 28th September 2003, 11:26 PM   #7
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Hi,

Quote:
It means that any two or more stages are DC-coupled together.
Great, now we all understand how DC coupling works....

I'll 'splain the principle of operation later.

Cheers,
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Old 29th September 2003, 01:09 AM   #8
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Yeah well whatever, I needed a nap...(hi Colt)

It is what it is. If a circuit uses coupling caps or transformers, it's AC coupled. If there is a DC path from one stage to another, such that the following stage responds to changes in the proceeding stage's DC voltages, then.. ya.

But a big definition like that just seems to muddify things.

Coupling can be with constant-voltage devices such as zeners or gas tubes, if you need a large voltage difference. Usually it's just done with several power supplies.

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Old 3rd October 2003, 09:49 PM   #9
PRR is offline PRR  United States
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Yank the 6SN7 and get the pentode stage working sweet. The Plate voltage should probably be about 2/3rd of the supply voltage (point C). Adjust the Screen resistor to get in this range. If you need a large adjustment of the Screen resistor, like more than 2:1 away from the schematic value, adjust the 3K Cathode resistor too. Does not matter what small pentode you use, you can get the Plate voltage about 2/3rd of the supply voltage.

The 6SN7 stage is all ratios. It is a little odd with plate resistors about the same as the common cathode resistor; I would use larger plate resistors. But stick with the plan shown. Again the Cathode-Plate voltage should be about 2/3rd of the Cathode-to-B+ voltage. The Cathode-Ground voltage will be about twice the Plate-to-B+ voltage. (20K/(25K/2))=1.6. The Plate-to-B+ voltage should be about half the Cathode-Plate voltage. So pretend the Cathode-Plate voltage is 100 (to make the math easy). The Plate-to-B+ voltage should be 50. The Ground-to-Cathode voltage should be 1.6*50= 80. The total supply voltage is then 100+50+80= 230V. But we have a 375V supply, so everything multiplies by 375/230: Ground-Cathode is 130V, Cathode-Plate is 163, Plate-B+ is 81V.

To get the Cathode to sit at 130V, we set the Grid a little negative of 130V. As a rough-guess, 0.6*Vkp/Mu. If the Cathode-Plate voltage is about 163, and the Mu is 20, then 0.6*163/20= 5 volts. Set the grid at 130-5= 125V. The plan says 110 volts. Since some voltages are black and some are red, this plan may not agree with itself. Anyway tubes are not precision DC amps. Get the Grids around 110-130V and it will work.

To adjust the Pentode Plate voltage to the voltage that the grids want: not shown on your plan, but very critical to this amp, is a resistor in the power supply from point B to point C.(*) The DC condition in the Pentode depend critically on this resistor (if you ignore the AC bypass caps you will see that the B-C resistor is a DC load/bias resistor for the Pentode). Adjust this resistor to get 125V at the Pentode Plate. Or if the "375V" varies as you trim the B-C resistor, get the Pentode Plate to sit at about 1/3rd of the point B voltage.

(*) Point C really should come from Point B. Everything is ratios, so they should all ratio from the same supply voltage. In this amp, getting Point C from a different supply than Point B is likely to give severe stability trouble. The two supplies will drift in different ways, and the diff-drift all lands on the 6SN7 grid, socking it around for no musical reason. Point C should be dropped from point B with a resistor.
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