Electronic – Direct Coupling and No bias on JFET guitar preamp’s

amplifiercouplingjfet

Why some of JFET preamp's for guitars don't bias the gate to Vgs(off)/2 and just reference it to ground? (Setting Zi)

This also lead to just direct coupling the guitar output to the gate.

As far as I understand with N-JFET's one should apply a negative voltage up to Vgs(off) to control the current the channel. And that on the ohmic region near to Vgs=0 the transfer curve becomes more non-linear.

The direct coupling non-biased design would mean the preamp or voltage buffer would be working on the Vgs=0 non-linear zone and that on the positive semi-cycle of the input sine wave the gate will be forward biased and would not make the channel less resistive (since Ids is already max and the channel saturated) then the output would be quite different from the original.

One example of these designs:

Ruby: MPF102 input buffer for a LM386

Explanations like this one has a DC bias on the gate (that doesn't matters on the AC analysis though).

As far I have figured out if we put our reference on the Source, Rs is biasing both Source and Gate, making the Source more positive than the gate, and the Gate as negative as Vs.

But then on the Ruby circuit, taking the Source as reference, the Gate would be at -9V above the -8V needed to reach cut-off voltage on a MPF102 and -6V with a 2N5457. A positive semicycle above 1V should make a difference but a negative semicycle won't.

My approach is wrong?
I know that DC and AC analysis should be done way different from this, but I don't understand the physics mechanics that are working here.

EDIT:

Setting aside that learn how things work is a good thing, I wanted to understand this to replace that MPF102 with an 2N2302 with smaller specs than the MPF102: Idss=5mA; Vto=-4V. (MPF102 and 2N5457 are not available here)
I was worried if I would have to change values to achieve a proper polarization since my signal in a worst case scenario goes as high as 2Vpp and the MPF102 Vto is -8V. I understood that it was a voltage follower but failed to understand the inner workings so these questions came up.

Best Answer

Maybe this will help: -

enter image description here

For no particular JFET, the characteristic might be something like the above. Importantly, and ignoring when \$V_{DS}\$ is 5V or less, the gate bias voltage sets the operating current for the drain.

So, for JFETs with a simple gate bias connected to 0V, more drain current means more source current which means more source voltage which means a bigger negative bias on the gate relative to the source and this means "regulation" i.e. it finds its own level - there is negative feedback and very elegant it is too. This keeps it in the linear region.

For a guitar amplifier where the input signal is a few hundred millivolt p-p at maximum, the drain current is modulated by the gate ac voltage quite linearly.

If the supply is 10V (easy to see on the curve) and the gate voltage is 3V below the source, the drain/source current will be about 1.4mA and this is self-sustained (regulated) when the source resistance is about 2k2.

Picture stolen from here (a very good website for plenty of stuff).