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Old 8th April 2008, 03:55 PM   #3841
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Quote:
Originally posted by PMA
Let's go on a bit. See effect of degeneration at single JFET, lambda = 10m, source resistor = 50R
Could you describe the circuit in more detail - Vds and the value
of any Drain resistance?
 
Old 8th April 2008, 04:08 PM   #3842
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Quote:
Originally posted by Nelson Pass


Could you describe the circuit in more detail - Vds and the value
of any Drain resistance?
Sure.
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Old 8th April 2008, 04:12 PM   #3843
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For completely dc coupled, result is the same.
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Old 8th April 2008, 04:40 PM   #3844
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Right on, PMA. It is OK now to go forward and your simulation shows what we would expect. The only minor question is why you did not set Lambda to zero? Just to give the equations a base line, and then add increased Lambda?
 
Old 8th April 2008, 04:48 PM   #3845
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This is a very good point. Even for lamda=0, there is a 3rd and 5th order distortion with degeneration, only a little bit smaller, then for lamda = 10m. Without degeneration and Rd near to 0, there is only pure 2nd.
 
Old 8th April 2008, 04:54 PM   #3846
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The answer is, I would guess, that bias, therefore Vgs is affected by degeneration resistor voltage drop, Vgs is one of the variables in equation showed yesterday.
 
Old 8th April 2008, 04:57 PM   #3847
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Syn08, my assocate did come back with a slightly different set of equations. The problem is to interpret them into real world (even if idealized) devices.


So far I have:


single device: H2 = b2/b1 = 1/4 Via/(Vgg-Vt) (I was advised to make Vt = 1 for convenience).

diff, devices H3 = b3/b1 = 1/32 [Via/(Vgg-Vt)] squared.


You have most probably said just as much in your equations, thus far.

Unfortunately it is still difficult to compare the two distortions at this point as one is controlled by square law.
However, to just equate them, and find a reference point might be interesting, as it would show where on the operating point that the 2'd harmonic of a single fet would equal the 3rd harmonic of an ideal diff pair.

Then we might put in realistic lambda for the fets and note the difference.

After, we might try resistive degeneration, much like PMA has already started.

Finally, we might find what complementary differential fet input does and how the two fet pairs interact, and if there is an ideal operation point or condition for best overall distortion, especially maintaining lower order and minimizing higher order, such as 5th and especially 7th.

I do have the class notes that my associate sent me, if you want me to forward them. They are only a few pages, and some of what I got was from Don Pederson's book was well.
 
Old 8th April 2008, 06:46 PM   #3848
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I think that I should Vt to -1 from +1.
 
Old 8th April 2008, 07:14 PM   #3849
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Quote:
Originally posted by john curl
Finally, we might find what complementary differential fet input does and how the two fet pairs interact, and if there is an ideal operation point or condition for best overall distortion, especially maintaining lower order and minimizing higher order, such as 5th and especially 7th.
John,

probably you will hate me now, because I again only fill in unbacked information.... see it (this goes to the other contributors as well) as food for thought:

Assume we define a tolerance band within gm might change for a given maximum drive. If the band is virtually zero width, only D2S qualifies.

If we allow a small change my math intuition tells me that we then need a gm curvature with the smoothest rate of change to get the lowest amount of higher order harmonics, most probably at the cost of increased lower order amounts and increased (non-optimum) overall distortion in the small signal area (say, 1/10th or lower max drive).

That's the hypothesis: the more rapidly gm changes while the signal cycles through the whole range the more higher harmonics are produced. Assume a step change in gm, it will produce a step change in the output waveform also, obviously creating lots af nasty high order stuff. The rate of change has a "infinity" spot in it, but any sudden gm slope (step change in rate of change of gm) will likewise produce higher harmonics.

The smoothest rate of change is a constant rate of change, which then means the gm vs. drive curve should approximate a parabola.

- Klaus
EDIT: spelling & removing too much speculative things...
 
Old 8th April 2008, 07:36 PM   #3850
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Klaus, please take this is the nicest possible way. My first answer to you is:"Well, DUH!"
 

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