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Old 29th September 2009, 07:01 PM   #1
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Default Quick Question on Dc coupling

I am laying out a PP amp and I would like to direct couple the output from the first tube plate (6SN7) to the grid of the split load inverter (1/2 6CG7).

I can work out the voltages alright for bias but:

In calculating the gain of the first stage what do I use as the input impedance of the next stage without a grid load resistor?

Also, suggestion as to grid stopper value or is it even needed?

Plate voltage of the 6SN7 is about 79V and the cathode of the 6CG7 is at about 83V (-4V bias).

see unfinished - unchecked schematic attached
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Old 29th September 2009, 07:20 PM   #2
bigwill is offline bigwill  United Kingdom
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Quote:
Originally Posted by coldcathode View Post
In calculating the gain of the first stage what do I use as the input impedance of the next stage without a grid load resistor?
You can safely take it as "infinity". It's not, but it's not worth worrying about the actual value in the calculation, it will barely load down the previous stage. The effect on the value will be vastly swamped by component imperfection and drift.
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Old 30th September 2009, 03:25 AM   #3
TheGimp is offline TheGimp  United States
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Why not add a cathode resistor between the cathode and lower 62K resistor to set the cathode current, a grid resistor from grid to the bottom of the cathode resistor where it meets the 62K resistor, and a bypass cap around the cathode resistor.

With these changes you can set the cathode current independently of the two phase resistors. This allows you to adjust it so you have 25% of your available voltage on each output resistor at the DC bias point. This maximizes your available voltage swing.
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Old 30th September 2009, 12:51 PM   #4
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You'll need a coupling cap with that circuit - otherwise the operating point of the split-load inverter will still be set by the plate voltage of the voltage amp stage. Good idea if you don't have enough plate voltage... of course it adds another low-frequency roll-off.
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Old 30th September 2009, 04:12 PM   #5
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Your numbers just don't add up. You have 2.6mA of current through the plate resistor of the second stage and 1.33mA through the cathode resistor. That defies the laws of physics and more specifically ohm's law.

Here's another way to approach this: make the current through the first two stages equal. In so doing you solve several problems. The first two stage will be exactly out of phase current wise so they can be fed by the same PS node. To make the two stages have equal current you will need the same tube for each stage (6SN7 and 6CG7 are close enough) and the combined cathode and plate resistors should add up to about the same value. That's why Willliamson used 47k and 22k although he did decouple the two stages. Another reason to use 22k is that the lower value resistors will give you better match plate to cathode. Use about 100V on the plate of the first tube with a 470ohm cathode resistor. Ip will be about 5mA and the stage will have a gain of almost 16. B+ will be about 340V.
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Old 1st October 2009, 01:19 AM   #6
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Need to use a voltage divider of 2 1meg resistors. One from plate to grid and one from grid to ground. Bypass the resistor from plate to grid with a small cap.
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Old 1st October 2009, 02:12 AM   #7
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Old 1st October 2009, 05:14 AM   #8
TheGimp is offline TheGimp  United States
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I figured the"-" signs at the first tube were typo's, and the plate of the second tube was at 190V. That would come close to proper bias (-4V Grid to cathode on the second stage).

Although it would still leave the plate resistor with 65V across a 62K and 85V across the 62K Cathode resistor.

I dono......
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Old 1st October 2009, 05:46 AM   #9
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Sorry I wasn't more clear. No not feedback. Should have said plate of first tube to grid of second tube, then from grid of second tube to ground. Or run the plate of the first tube at 1/3 of B+ and couple plate to grid directly to get the voltages to work out correctly. Running the first tube at 1/3 B+ isnt always a suitable working point as far as symetrical voltage swing and distortion, but it achieves direct couple for dc and ac.If the first tube is run with 1/2 B+ then the second tube would end up biased with a very positive grid and a small voltage across the tube in order to get nearly 1/2 B+ across the lower and upper resistors, since half B+ is forced across the lower resistor because the grid is at that potential. Thus the need for the divider or lowering the plate voltage of the first tube.

Last edited by jerluwoo; 1st October 2009 at 05:58 AM.
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Old 1st October 2009, 06:26 AM   #10
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Examples of correct circuits using 12au7's at ~5mA per tube.
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