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Re: Sweetness VS Load Lines

Posted by Naz (A) on August 11, 2006 at 08:07:30

In Reply to: Re: Sweetness VS Load Lines posted by Tre' on August 10, 2006 at 07:55:16:

I think we are at cross-purposes and perhaps we are both right and wrong. I will preface this with the statement that this is my only interpretation, based on observations and some reasonable (but not wide ranging) measurements.

Lets assume that we fed a 1kHz signal into the first stage, which has a gain of 10 and produces 1% 2ndH distortion. The OP of the stage will be a 10V 1kHz signal and 0.1V 2kHz signal.

If we feed this into an identical next stage we would amplify the 1kHz to produce 100V at the OP, that's a given. But, normally it would have produced 1V of 2kHz in the same phase. What I think, is that in a perfect world this would completely cancel the amplified 2kHz from the 1st tube (0.1V x 10 = 1V) within the tube itself.

Where we differ is that you believe that the 2kHz from the first tube will be amplified and add to the 2kHz produced by the second tube because they are in phase at the OP. I am saying that the 2kHz produced by the second tube cancels the 2Khz at its input BEFORE the input signal ever reaches the OP.

I agree that if it ever did it would have added and your assumption would be correct. But it can't be if you look at the fundamentals of what produces the distortion in the first place, which is the uneven spacing of the transfer curves. To prove it you only have to draw a sine wave as the input to a family of curves and transfer it to the output. It will be more or less egg shaped. Now feed that OP into another identical set of curves with the same load line and the transferred signal will be restored to a sine wave.

The caveat with all of this of course is that it would depend on the transfer curves and the operating points and it can't be achieved in real life. But the principle is what we are talking about and to some extent it does result in some cancellation (not just 2ndH) and it can be measured.

Now, lets come back to the example again but this time let's say the second stage produces 2% 2ndH distortion. I believe that the first 1% would cancel the 2kHz from the first stage and the remainder would be transferred to the OP where it would be as you say, in phase with the 1kHz OP signal.

Now here's the thing, I also think that if the second stage produced less than 1% 2ndH then the 2kHz input signal from the first stage would only be partly cancelled resulting in some 2kHz signal being amplified and appearing at the OP of the second stage, again in phase with the 1kHz. The only difference being that this time the distorted signal would have the opposite appearance from the point of view of which of the lobes (positive going or negative) were the wider.

Does this make any better sense?

Naz

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