Negative resistance better than current source for MOSFET follower amp?

[Circlotron]

Often in a source follower amp cct you put in a constant current cct to pull the signal the other half of the cycle. If the amp is driving into an open circuit "load" the output mosfet is very happy because the drain current stays constant for the whole of the cycle and so the source voltage tracks the gate voltage almost perfectly. Now if we attach (for simplicity) a resistive load then the load current varies durning the cycle and so the mosfet introduces some distortion because the now varying drain current means the source voltage doesn't perfectly follow the gate signal voltage any more.

Would it not be better to make the constant current circuit instead a negative resistance circuit (not very much difference) so that for example, as the signal swings positive and the load current is increasing the CC cct is decreasing at the same rate? So make the CC cct negative 8 ohms. That is, for every 8 volts increase across it make it decrease it's current by 1 amp and vice versa. This way the output fet sees a more or less constant current during the cycle and distortion would be very low. As the fet Vds reduces on signal peaks and it's ability to pull current is reducing, the negative resistance cct is easing up as well instead of holding on tight for no good reason. Of course, if all this was applied to a real world loudspeaker load, the wheels might fall off the whole idea. I don't know. What do you think?

GP.

[eLarson]

It sounds a little like one of Nelson Pass's patents. I don't have the US Patent # at hand, though. If you ask on the Pass Labs forum you may get the specifics.

Erik

[janneman]

GP,

Are you sure you are not re-inventing class-A amplification?

Jan Didden

[tschrama]

http://www.audiodesignguide.com/tra/...r_follower.jpg

[eLarson]

See also:
http://www.passdiy.com/projects/zenv2-4.htm
in which Nelson Pass talks about the "Aleph" current source. The patent number I was trying to come up with is mentioned here.

Erik

[Nelson Pass]

The classic Aleph circuit is one where the impedance
is -2 times the load impedance, resistive and reactive.
This makes the load appear to the gain stage as twice
the value is really is.

[Bakmeel]

It would create an ideal situation, if the amp is driving a resistive load. While the current through the resistor increases with the voltage across it rising, the current through the negative resistance drops by an equal factor... (That is, assuming you matched the resistor and the negative resistance.) The two of them together would act as a constant current source, and that in it's place is the first (ideal) situation you described first.

That would make us: "The Perfect Amplifier" but "For this specified load only" If you could make it work for inductive or capacitive or even complex loads, than that would mean gold....

Maybe you should try this: Take one pair of speakers, measure the load the produce at various frequencies, and then try to build the negative of that... (I don't know, using DSP's and some more complex cirquitry not yet conceived) you could at least build a prototype that would be the perfect amp for that speaker only....

But in basics, it would be THE idea.....

Bouke

[Circlotron]

Originally posted by eLarson
See also:
http://www.passdiy.com/projects/zenv2-4.htm
in which Nelson Pass talks about the "Aleph" current source. The patent number I was trying to come up with is mentioned here.

Erik

I see what you mean. I remember looking at that cct a while ago and then I forgot about it. A related thing I am about to try is that with an amplifying stage you can of course replace the drain resistor with a CC source with consequent improvements in linearity and gain. But the improvements are still limited because the further the fet swings downward the less ability it has to go further. This is analogous (I think) to that part of it's characteristic curve where the drain current is *almost* horizontal for a for a large change in drain voltage, but it does slope a little downward as you go to the left. That means as the drain voltage reduces the drain current does too, even if the gate voltage remains the same. So the amount of that slope is the amount the drain current changes with voltage (it's slope(!) resistance). Now, if we had a current source load with a negative resistance *almost* equal to the slope resistance of the fet at the given operating conditions (if it was equal or greater it might oscillate) then we should be able to get almost infinite gain and linearity from that single stage. If the phase response of that single stage is ok we could use local feedback to turn that almost infinite gain into almost zero distortion! That would be nice, wouldn't it? Tell me, has anyone done this before?

GP.

[traderbam]

Backmeel:
I really like the lateral thinking. It has got me thinking about what would happen if the current source was a perfect negative of the speaker load, reactance and all. In this case, the output tranny would see a constant resisitve load under all signal conditions and act like a constant current source. But now, if the output transisitor is just a CCS then I reason that it follows that the special CCS is now the amplifier. So has it just come full circle? The original problem hasn't gone away.

Circlotron:
The ide of making the drain load the -'ve of the drain impdeance is very clever. I have a method in mind for doing this if you would like it. Just a couple of thoughts: each time you talk about increasing the gain of the stage you are doing so by reducing its output impedance. This is ok if the following stage has infinite input impedance, but if not (and this is always the case) then it may be pointless or even degrade the next stage. Something to bear in mind.

[janneman]

Originally posted by Circlotron
[snip]
Would it not be better to make the constant current circuit instead a negative resistance circuit (not very much difference) so that for example, as the signal swings positive and the load current is increasing the CC cct is decreasing at the same rate? [snip]
GP.

GP,

In a class-A amp, the two devices carry a certain current, right? If the signal goes pos, increasing the current in the pos fet or whatever, it decreases by the same amount the current in the bottom device, right? That is what you were looking for, as far as I can see. So, I still think you are re-inventing class-A. What am I missing guys, if anything?

Jan Didden