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Old 30th January 2003, 06:20 PM   #21
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I could argue your point of:

Quote:
I'll happily repeat again, and I would welcome somebody trying to disprove it, but: ALL THE ELECTRODES IN ANY TUBE MUST HAVE A DC PATH TO THE CATHODE.
If you cap-coupled an AC signal to the grid, then when the signal goes positive, it will draw current from the G-K diode. Where does this current go? Why, clearly, it constitutes a negative bias voltage due to the rectification of the applied signal.

And don't get pissy with me about leakages...we are ignoring those in this scenario.

So where's the DC g-k path, hmm?

Tim
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Old 30th January 2003, 06:30 PM   #22
Joel is offline Joel  United States
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Quote:
Originally posted by Sch3mat1c
it will draw current from the G-K diode.
The grid and cathode do not form a diode. The grid is negative with respect to the cathode, so how can any diode be formed?

Quote:
And don't get pissy with me
Excuse me?
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Old 30th January 2003, 06:42 PM   #23
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Ummm....how can they not form a diode....

Besides, a diode doesn't have to be forward-biased to be a diode.
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Old 30th January 2003, 06:53 PM   #24
Joel is offline Joel  United States
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Tim, a diode is an anode and a cathode. It functions because we apply a positive potential to the anode. The electrons are attracted to it.

What happens if I apply a negative potential to the anode? Think about a rectifier.

You see?
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Old 30th January 2003, 07:47 PM   #25
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I repeat...

It doesn't have to be forward-biased to be a diode.

Or are you saying that, in a rectifier, the tube alternately becomes a diode and a - nothing? - every 60Hz?

Well anyway, still - what about the bias voltage? There's no DC path out of the grid, only in...

Tim
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Old 30th January 2003, 07:57 PM   #26
Joel is offline Joel  United States
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Tim, the statement I made about all electrodes needing a dc path to the cathode can be found in the first few pages of Radiotron. It is stated very clearly. I did not make it up, and I'm pretty sure RCA and Mr Langford-Smith knew what they were talking about.

You're going to have to retract later on, but please continue if you feel it's necessary.
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Old 30th January 2003, 07:58 PM   #27
Joel is offline Joel  United States
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Quote:
Originally posted by Sch3mat1c
Or are you saying that, in a rectifier, the tube alternately becomes a diode and a - nothing? - every 60Hz?
In fact, that's exactly what it does.

Here is some reading online:
http://www.tpub.com/neets/book6/20f.htm

And this describes diodes when presented with an AC voltage:
http://www.tpub.com/neets/book6/20b.htm


cheers
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Old 30th January 2003, 09:56 PM   #28
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Default GRID LEAKS.

Hi,

Let me get this straight from the start:I don't want to stirr things up any further but:

Quote:
And no, this is not "internal" in any tube I've ever seen.
Neither have I, however when you use a cathode bias and don't connect anything to the grid you will still notice a negative voltage developping at the grid.
This take time and by adding the gridleak resistor the biasing process is sped up.

As Gabe pointed out the tube will function.
Sure enough it will function as a diode until a load is added to the grid.

While I toyed around with omitting bleeder resistors from the preamp end I noticed an opening up of the sound,with more air and splashier highs on rimshots.
As if a small veil was lifted.

Quite likely there must have been an impedance mismatch somewhere along the line.

Later on I increased the value of the bleeder resistor from 100K to 1M and I came close to the same result as without the bleeder.

Hope this clarifies some,
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Old 31st January 2003, 02:29 AM   #29
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Joel wrote... somewhat erroneously:

Quote:
The grid and cathode do not form a diode. The grid is negative with respect to the cathode, so how can any diode be formed?
In a triode biased in class B in a detector circuit, the grid and cathode do indeed form a diode detector, with the grid as the "anode". The same function of grid as "anode" or "plate" is true in a pentagrid converter. The second grid is actually the "plate" of the first "triode" grid.

The definition of a diode, functionally (the key word), is to "rectify", or allow current to flow in one direction but not the other. When the grid is biased negative enough to just cut off the tube, or class "B", then anything that brings the grid toward positive turns on the tube. Anything below that voltage level, does not turn the tube on. So... the grid/cathode circuit becomes a rectifier... a diode. Plate current follows the changes and with a load resistor, "amplifies" them.

Don't they have that in the RDH? It is in the seven or eight books, and countless electronics magazines, I have read about it since I was 10.

As for the path... when the tube conducts, there is a path already made between the cathode and the grid... simply because the grid is in the way of current flow. The grid can impede the flow of current with AC simply because the negative peaks will cause it.

Re the grid as a plate, again to bring up an icon in the business, Steve Bench has made amplifiers where the output is the grid. He says they sound very good.

Cheers!
Gabe
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Old 31st January 2003, 02:45 AM   #30
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Joel,

Please refer to the example of DeForest's experiments here:

http://www.tpub.com/neets/book6/20e.htm

It shows that the triode he worked with flowed 5 milliamps with nothing connected to the grid, while it flowed 10 milliamps with a positive voltage to the grid, and 2.5 milliamps with the same amount of negative voltage.

But note the non connected one. Grid not connected to anything... tube works. Tube theory from the one who invented the thing!

A pure AC on the grid will make the tube fluctuate. But, as those pages you posted brings out, one would not want the AC to bring the tube into saturation or cut off, so applying a DC bias voltage allows one to have an AC voltage such that plate current will vary within the cut-off/saturation points.

The RDH was likely written to discourage engineers from designing less stable tube circuits. Same with solid state. I have seen many SS circuits where only a resistor from base to B+ was used, the most unstable circuit imaginable, but the circuits work. In all the textbooks I have read for solid state, this practice is strongly discouraged.

I rest my case.

Gabe
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