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Old 29th April 2009, 11:03 PM   #1
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Default Strange bias circuit. WTF?

I was looking at a SCHEMATIC for a Harman Kardon Trio. I dont understand how the bias circuit works.

Is an EL84 PP design with the cathodes of the EL-84 pairs strapped together. Each end of a pot is connected to the cathodes with a 330R resistor and also to the respective coupling cap load side with a 470R resistor. The wiper is grounded. This pot is supposed to balance the cathode bias between the two tubes.

I could see it if the cathodes were not coupled but I'm baffled by this setup. How can this work?

Can someone here explain it? Please.
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Old 29th April 2009, 11:13 PM   #2
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Because it applies the adjustment to the grids, not the cathodes?
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Old 29th April 2009, 11:40 PM   #3
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Quote:
Originally posted by kenpeter
Because it applies the adjustment to the grids, not the cathodes?
Bootstrapping of the 1'st triode is unusual as well.
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Old 30th April 2009, 01:37 AM   #4
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Quote:
Originally posted by kenpeter
Because it applies the adjustment to the grids, not the cathodes?
Well, that makes sense.

On THIS web page, a fellow who seems on top of things informs us that "The output stage is cathode biased and is not adjustable".

So if this is not bias adjustment, I wonder what it is?
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Old 30th April 2009, 01:39 AM   #5
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Quote:
Originally posted by Captn Dave


Well, that makes sense.

On THIS web page, a fellow who seems on top of things informs us that "The output stage is cathode biased and is not adjustable".

So if this is not bias adjustment, I wonder what it is?
It is bias balance adjustable.
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Old 30th April 2009, 01:46 AM   #6
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Quote:
Originally posted by Captn Dave
So if this is not bias adjustment, I wonder what it is?
The schematic says "OUTPUT BAL."

Perhaps it is designed to allow adjustment of the AC balance of the phase splitter upstream of the finals.
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Old 30th April 2009, 02:16 AM   #7
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It's a DC balance so you don't need perfectly matched tubes.

I use it in some of my designs.

Cheers!
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Old 30th April 2009, 02:43 AM   #8
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It allows you to balance the current in the two output tubes. You can't actually change the total current flow but you can adjust it so that each tube gets one half of the total current. Similar circuit was used on the output tubes of the Heath W5. If you can get a copy of the Heath manual it will explain it. The original Williamson not only had a similar circuit but also had an adjustable resistor so that you could vary the total current as well.
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Old 30th April 2009, 02:57 AM   #9
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OK, I get it now. The pot is used to divide DC between the two grids. The pot creates an adjustable voltage divider so that the pair of pentodes can be "balanced" so as to to have equal cathode current. Not a bad way to do it.

A couple of voltmeters to monitor voltage across the cathode resistors and a screwdriver should do it. (That and some new resistors - the vast majority are out of tolerance, especially the power resistors).

Thanks for the face time, professors.
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Old 30th April 2009, 03:11 AM   #10
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No, pair of volt meters won't help you. You need to measure currents. Most probably, some technological plugins between tubes and sockets were used by support technicians who used to replace tubes. I would cut cathode wires from sockets to insert 10 Ohm resistors there, to measure voltage drops across them.
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