pseudo-op-amp FET arrangements

[Eddododo]

This is a design by a talkbass member, Fdeck, or Francis Deck, who also has some diy stuff and a talkbass-popular commercial High-pass filter and piezo preamp. Solid dude, too, super helpful.


this is the schematic in question.
http://personalpages.tds.net/~fdeck/bass/ebpresche.JPG


I get the idea of using the inverting output as a sort-of negative feedback, but what is the purpose/significance of using an active load (i think it is a current source?) .

2) why use two different topologies? (mid vs baxandall)

3) the use of the diode in the baxandall fets?

i have plenty more questions...

[commathe]

That diode appears to be for protection to stop the gate voltage going too high and causing breakdown

[R.G.]

This is a design by a talkbass member, Fdeck, or Francis Deck, who also has some diy stuff and a talkbass-popular commercial High-pass filter and piezo preamp. Solid dude, too, super helpful.
this is the schematic in question.

For what it's worth, *every* inverting amplifier can be used as a feedback amplifier. Even a single NPN gain stage exhibits this behavior, as witness the use in the clipping stages of the Big Muff. In fact, even a single NPN gain stage acts as a differential amplifier, too. The emitter is a "noninverting" input, but with a much lower input impedance than the base. It's used that way in many two- and three-NPN or NPN/PNP gain compounds.

I get the idea of using the inverting output as a sort-of negative feedback, but what is the purpose/significance of using an active load (i think it is a current source?) .

In all the JFET/PNP stages, the PNP is a second gain stage, and a second signal inversion, so that the JFET is a noninverting input and the source of the JFET is an inverting input, so the feedback goes from the output to the source.

The stacked JFETs is a variant of the SRPP stage, using the push-pull arrangement of the JFETs for output, and the top JFET input as a "noninverting" intput. It's an active load setup, but not the standard version.

2) why use two different topologies? (mid vs baxandall)

They're solving different problems. The Baxandall is a bass/treble controll like on most stereos. The mid thing is, as it says, a kind-of parametric eq.

3) the use of the diode in the baxandall fets?

I believe it's part of the biasing arrangement. It's a zener. It looks odd, and I'd have to study it a bit to prove that to myself. If the diode is just for protecting the gate of the top JFET, it doesn't need to be a zener.

[Eddododo]

found in one of his posts, actually, regarding the diode
"I use a 10-V Zener to hold the output at 10 V above the input, i.e., halfway between the power supply rails."

I .. think i get that.

i need to stare at this for awhile..

confused also by some of the caps etc that seem to shunt the drain to source in a couple places
re-read your post and thought more

[Transmogrifox]

I believe it's part of the biasing arrangement.

I agree this is correct.  The JFET looks like an emitter follower in which the emitter is being driven both by the signal coming from the junction at R16 and R36 but also by the current from the drain of J5.  It appears to be a summing node of some sort, but you might have to "study it a bit more" or run the simulation to see exactly how it's interacting.  It is also providing some feedback into the baxandall network.

These circuits really look interesting.  I may have to breadboard them and give them a try.

[amz-fx]

If we break it down into its components, we have an SRPP jfet circuit, with a constant current source atop a basic jfet gain stage as its load (instead of a resistor). The tone circuit is not important at this point.

Remove the capacitors to simplify and look at the dc conditions. if you power the circuit with 9v, then the zener (which is 8.2v rating) does not conduct. In fact, it will not conduct until Vcc is over 16.4v since the dc value at the output is held at 1/2 Vcc by the complementary jfets stages.

Why clamp the output dc conditions?  I'm not sure. If you remove the D1 zener, the circuit works much the same.

[armdnrdy]

Question....what is it! 

The original drawing looks to be an overdrive circuit?

Would this circuit not benefit from making the gain in the first stage adjustable?

[anotherjim]

Question....what is it! 

The original drawing looks to be an overdrive circuit

Would this circuit not benefit from making the gain in the first stage adjustable?

I think it is adjustable -  isn't R6 & R7 inside the dotted ellipses representing a 10k pot position? This is a spice type schematic, I think.

The mid freq control is interesting -  is that a dual gang 100k pot R23/24 to J3 gate?

[PRR]

> If you remove the D1 zener, the circuit works much the same.

If the JFETs are matched. Un-matched, without Vz, it will *tend* to half-supply, but could be quite a bit off.

It is important to note that the bottom of Vz returns to the lower Gate, which is nominally zero voltage. If it gets much off zero, bottom JFET will tend to work to bring it back. (However it can't push Gate negative....)

There's other ways to do this. I don't see a problem with this way.

There is a mind-related "drawing problem". At a glance, it looks like the upper gate is pushing at the bias and Bax networks, driving them. Of course when un-tangled, it is the Source and the 100uFd pushing things around (including upper gate).

> why use two different topologies?

1) He studied (copied) different web-pages, as noted on the drawing.

2) The Bax begs for an Inverter. This Wein-like mid-EQ wants a set-gain non-inverter.

The set-gain non-inverter is also useful for preamp (adjustable gain) and output amp.

[Eddododo]

if you power the circuit with 9v, then the zener (which is 8.2v rating) does not conduct. In fact, it will not conduct until Vcc is over 16.4v since the dc value at the output is held at 1/2 Vcc by the complementary jfets stages.

Why clamp the output dc conditions?  I'm not sure. If you remove the D1 zener, the circuit works much the same.

The circuit is powered by +20 V , for what its worth

[amptramp]

A little additional information about the SRPP circuit:

1. The upper transistor provides balanced swing above and below the quiescent output level is the source resistor is the reciprocal of the transconductance of the upper transistor.  This matches the gain going in the positive and negative direction.  In this case, the 500 ohm resistor is suitable for use with a FET having 2000 micromhos of transconductance.

2. The SRPP circuit works by changing the current through the upper and lower transistor.  If there is no load, the stage simply hits the upper or lower rails with any input.  The SRPP stage requires a finite load so that the difference in current between the upper and lower stage goes to the load.  With insufficient load, it behaves like a fuzz (which may be useful sometimes).

[Eddododo]

Question....what is it! 

The original drawing looks to be an overdrive circuit?

Would this circuit not benefit from making the gain in the first stage adjustable?

it is a preamp- clean and discrete.

in fact...

http://www.talkbass.com/threads/my-new-boo-teek-preamp.600126/

for a picture- nice point-to-point work

[Eddododo]

can anyone help me to understand the significance of the selection of the 2n4393 jfet for the baxandall's inverter?