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Messages - Findeton

Including the changes I said in the last post, and some more to refine the mix, here is the schematic for my solid state 12AX7:

The circles there point out that there are lines in the schematic that are passing one above the other without being actually connected. As I said, RDISTORTION controls the amount of distortion for the load line (which is given by HV and the resistor connected between HV and the plate). RCURRENTGAIN controls the cathode to plate current of the device, and is set at 220 ohms to get the same current than a normal 12AX7 tube would give. Vcc should be about 4 volts higher than the highest grid voltage we want to obtain and 10V will suffice. Vdd must be at least -1V. I'm thinking of designing a voltage multiplier and current pumper that could generate those DC voltages directly using the +/-5VAC heater pins so you have that problem solved :p Ah, and those two new 1n4148 are there for correct biasing. The plate current is:

If you compare it to the graph of a real 12AX7, it is pretty much the same:

The only big difference is that for our SS12AX7, Vgs must be set up 1V higher, because grid current will start to flow at Vgs=0.5V instead of starting at Vgs=-0.5V (and that's why I added those two 1N4148). This is the graph for the grid current for RDISTORTION=9k. As I said before, RDISTORTION should be a 10k log pot:

Again, Vgk for our SS12AX7 must always be between 0.7 and 1V higher than for a normal 12AX7, so you need to rebias cathode voltage 1V lower when you connect the SS12AX7. Taking that into account, this grid current is pretty close to the real deal (if we want a more steep curve and more distortion we just have to lower RDISTORTION):

So that's all for now for the 1:1 solid state 12AX7. On the next post, I'll describe a solid state triode that works with 0 to 10V and works up to about 170mW of power dissipation instead of 1W.  Ideal for stompboxes!
Quote from: mensur on March 24, 2010, 07:49:48 PM
Thanks Findeton,

But can it be done without this many components, or with discrete components, and how can we increase input impedance since there is no 100Meg resistor?

There are no 100meg resistors, you're right! The important thing is that R2 is 100 times R1. Let's say R1=10k and R2=1meg. In order to increase input impedance (which is about 1meg for negative values of Vgk in a real triode), you could disconnect the end of R1 from the grid and connect it to the OpAmp's output, but you may need to connect a negative voltage to Vee's supply of the OpAmp for it to stay lineal.

EDIT: If you just use R1=10k and R2=1meg, it will consume about 300uA of grid current for Vpk=300, and that's a lot more than it should be (for negative values of Vgk). Therefore I reccomend connecting R1 to the output of the OpAmp rather that connecting it to the grid directly. As I said you will need to feed about -1V to the negative power supply (instead of ground) of the OpAmp to maintain it's output lineal. This way the input impedance will be the one of the OpAmp, and that's large enough value :P
Quote from: mensur on March 24, 2010, 11:25:15 AM
here is the spice simulation of Fetron/Triodizer:
Is there any thing I can improve with this design?

You can try this:

It has the same plate current characteristics, and a grid current like this:

It would start to create distortion at about Vgs=0.5V, when it should do that at about Vgs=-0.5V, but modifying that can be a little bit complicated... A way to overcome this is to lower Vk 1 volt, by reducing the resistor connected between ground and the cathode of the stage where the sstriode is used.

Also, the RDISTORTION is, obviously, a resistor that adjusts the level of growth of the distortion. Grid current depends on both Vpk and Vgk, which in the end means that, depending on the plate load line, the grid current curve will be more or less steep. I would use a 10k log pot in series with a 50ohm resistor in the place of RDISTORTION and calculate a good value of that resistor for each load line. I won't show you right now how to calculate this value but it's doable.

As you can see, I am using ground and a Vcc on this ss12AX7. 10V as Vcc would suffice. You could connect ground and Vcc to the ss12AX7 using the heaters pins.

EDIT: added simpler circuit. btw changing C2 to a higher value would remove high frequencies of the distortion.
Quote from: Findeton on March 23, 2010, 11:45:34 AM
I've got a simple question: is Ik=Ig+Ip or Ik=Ip?  Of course I need to know this to properly emulate grid current saturation.

I mean, when the grid voltage becomes positive, the grid current starts to rise. Does that current flow to the cathode, like in a BJT or not? Of course there's a reason why we put capacitors in parallel to the cathode resistors (it isn't grid current though) but as I'm trying to emulate this grid current, it makes all the difference because, among other things it would mean being able to simplify the circuit as we wouldn't need this high power OpAmp I'm using.
I've got a simple question: is Ik=Ig+Ip or Ik=Ip?  Of course I need to know this to properly emulate grid current saturation.
Quote from: mensur on March 23, 2010, 10:45:36 AM
We can make high voltage SS tube replacement, by using cascode. high voltage mosfet will be used as a voltage regulator from the HV supply, to a safe level 24V(actually 20V, since mosfet will use 2-4V to conduct),and do the heavy lifting(amplifying the gain):
SS 12AX7.pdf
The gain of this device is very high(in my example,around 570!!!),the gain is around 570 with mathematical approach.(gm is 5.5679 mS(Av= gm x Rl = 5.5679mS x 100K = 567.9 = 55.7dB).
The input capacitance is very low, less than 2pF(due to cascode connection which reduces the input capacitane,and rounds up the clipped signal),identical to 12AX7.With nicely set feedback, we can outperform the real thing,with ss devices, and what is the best, NO MICROPHONICS, NO MORE HEATER TO CATHODE ELECTRON EMISSION, WHICH CAUSES HUM, NO MORE CHANGING TUBES...
Tone Grinder,aka FETRON
Tone Grinder is probably old FETRON, in nowdays production.I doubt it has characteristics of a real 12AX7, more likely version that I posted.

I think those things are called transistors... What we want is not 12AX7 replacements with more gain,but to emulate the distortion that tubes give (and it looks like the Tone Grinder/FETRON) doesn't give that.
Quote from: HENK on March 22, 2010, 08:04:06 AM
Very cool!

Are there Sub´s for the AD815??

Greets Thilo

You could actually use an OpAmp with a lower maximum output current (like TL072), but then the "sstriode" will have a roof for the maximum grid current inferior to the maximum cathode current (and that doesn't happen on a real triode). Therefore you would be unable to reach full levels of distortion. You could also only use low power OpAmps, but that would involve modifying the schematic and adding 2 OpAmps so it's probably not a good idea. AD815AN is just an OpAmp with the ability to give a lot of output current. Any OpAmp with that  characteristic will suffice. These are the characteristics needed:
-support 100mA to 250mA as maximum output current.
-support 10V to 20V as power supply (Vcc-Vee).
-low input offset. Well, this is not actually a big issue because we are using the OpAmps as voltage buffers, but it's always good to use low input offset OpAmps for audio!

So, for example, the Texas Instruments OPA552, OPA551 or OPA567 would suffice, and btw TI sends free samples right to your home ;)

BTW, later today or maybe tomorrow I'll post some rules on how to use this "sstriode" on any schematic involving triodes. I'll also add some final refinements to the sstriode (for example one diode more) because we want the distortion/grid current to start at about Vgk=-0.5 and not at Vgk=3V :P (you can see that on the last picture of my previous message).
Here we go again!

I've been testing the "trioderizer" and it's really interesting. Just using 2 resistors, it actually achieves much of the stuff I've been trying to do.

We want to get a current function of this kind:

Ip = m*(u*Vgk + Vpk+Vconstant)^1.5.

We know that connecting a resistor of certain value to the source of a MOS we get a power 1.5 function.

Now, if we connect 2 resistors as in the picture, we get the "trioderizer":

As the gate current of a MOS is negligible, we obtain that Vg=Vi+Vpi*R1/(R1+R2) . Developing that we easily obtain a current transfer function like the one of a simplified triode.

But the triodizer doesn't achieve everything, I have to point out 2 things:

As we said, the point A marks where the triode's plate current starts to saturate. Yes, it's true, we don't achieve that with the "trioderizer", but that's OK because that thing happens at high power settings (in a point way higher than the maximum rating of dissipation). So we don't have to care about that anymore.

The other thing is that while it emulates the output/plate current of the device/triode, the desired overdrive/distortion of a triode comes from the transfer function of the grid current:

(on the picture above I1 is the current grid)

Basically, when the grid gets positive, it draws incredible amounts of current. And as the grid is not connected to a perfect voltage supply (every stage has a certain impedance, and the guitar is a high impedance for the first stage). That makes the grid to drop its voltage, therefore needing less and less current again. And that's the preamp distortion that many people love (fans of Mesa Boggie for example).

So, what's the plan? Perhaps there's a hidden reason why I haven't seen this approach to emulate triodes before, but why don't we just use a transistor that emulates this grid current? the gate current of a MOS (or a trioderizer) is negligible, but we can add a transistor to the "grid" of the trioderizer as current source easily!

The transfer function of the grid current is something like this (at least this is my approach):

Ig=k*[ (Vgk)^3 ]*[ (1-Vpk/500)^3 ]

We *could* faithfully emulate that transfer function using a lot of OpAmps and FETs, but it wouldn't be very useful... too many components! A typical stage with one triode would involve a resistor between Vcc and the plate, that means that our output current will move between the points of a line like the blue line here:

An approach can be to consider that 1-Vpk/500  doesn't change that much in comparison with Vgk and therefore deprecate the importance of Vpk. We could take Vpk in consideration but that would imply a voltage multiplier, and that's a big circuit to add in comparison with the simplicity of this approach (one non logarithmic implementation would add about 6 OpAmps to our design). So, if we set Vpk to a certain voltage and take Vk as a parameter, this is the kind of function we'll find:

The following approach is as minimalist as possible. With it we want to emulate the saturation of triodes but  it won't be exact: When Vgk overpasses a certain level (about -0.5V), the grid starts to consume a lot of current, and our solid state triode will do just that, although the law won't be a power 3 function and won't take Vpk into consideration. This is the simplest implementation I can think of it:

Very simple circuit indeed! The TL072 OpAmp is used as low power voltage buffer: we don't want any of the grid current we are generating to go to the "trioderizer" part of the circuit because the trioderizer is supposed to be a circuit that only depends on voltage.

The AD815 OpAmp is a high current device: it will be the responsible for all the grid current we want to generate. We could have used only low power OpAmps, but that would entail adding two OpAmps to the circuit.

Well, that's it! we've finally got a solid state triode that creates distortion in a tubish way, and with only 2 OpAmps! I can hardly wait to test it! In case you don't take my word, these are the graphs I get on PSpice:

On the last graph we can see that when we reach a certain Vgk, Ig starts to raise to never stop. As always, that point can be easily controlled somehow. Here, we could use diodes to control that point (I suggest adding a Zenner diode in series to the 1N4148).

As always, any question, review or comment is welcome. I can hardly wait to test this solid state triode!
I've been thinking more about this subject. Yes, in the last post I described an ideal triode emulated with silicon technology... but as far as I now this ideal model doesn't take into account all the nonlinearities of a real triode. This is a graph found here:

As you can see, I've marked 3 nonlinearities that the simple model of a triode we have implemented doesn't take into consideration at all:

A) Given a certain level of current, the current starts to overdrive/distort, so that the current doesn't exceed a certain limit. It doesn't clip suddenly like an OpAmp would do.
B) The spacing between the curves gets shorter when driving grid more and more negative.
C) Triodes draw current when the grid voltage is weakly negative or even positive. It's not only that, but also that, given a fixated  weakly negative or positive voltage, the transfer function stops being a power three halves function.

Rydel's approach (read this) implements B, but it involves multiplying 2 voltages together, and that's a very difficult thing to do in analog electronics (you could do it with opamps and diodes but it's kind of noisy anyway). I think I read something about another way of implementing this on the teemu.

I haven't tested it yet, but it looks like you can implement C adding a resistor between the gate and drain pins of the MOS, as you can see here (the "trioderizer" uses a JFET but that shouldn't be a problem at all) :

So what about problem A? Should we use the usual clipping techniques, like using diodes on an OpAmp? Well, I've been thinking on something kind of unusual. In fact, this kind of clipping could be applied to other ends apart from emulating a triode, so it might interest you even if you think all the other stuff I've written so far is kind of crazy  :duh

So the idea is:

We've got a function: Iout=f(Vin) and we want to make a soft clipping on that signal. Normally, we would clip the output signal but, as we have about a voltage buffer for every signal, we could do the clipping on the buffered input as well. The input is normally considered to be lineal, so if we don't want the clipped input to exceed a certain value, we have to substract the input a signal that, in the end, is also lineal. This is better explained with a picture:

The input Vin is the red line, which, as we said, is lineal. The clipping signal that we'll substract to the input is the blue curve, which is cero for low voltages (it's logical that the clipping starts at high voltages) and in the end is kind of lineal and parallel to the input... so when we substract the two signals we obtain the green curve, which, just as we wanted, is clipping the signal.

So my idea is to use a FET as clipping signal (in fact, the blue curve is just the output current for a FET). It might strike you as a problem that the clipping signal we want to substract to the Vin is a current, but an Opamp can add/substract voltages and currents altogether, and we are already using an OpAmp in the input of our "solid state triode", so it's not a problem at all.

So this would be a general diagram to clip the input signal:

VF controls when does the compresion start and the source resistor of the FET could be a pot that could help us in making the Vin and the current parallel for high voltages (as it should be). In that diagram you see 3 voltage buffers, 2 OpAmps and a FET, but we only have to add a FET and an OpAmp to the design I showed you on my previous post. I've showed you that diagram because maybe someone wants to reuse this clipping method elsewere, and it also might help you understanding it, or we could want to clip the output signal of the solid state triode.

Now we'll compare the output (x^1.5) with the unclipped input, the clipped input, and the output with the clipped input.

The green curve is the clipped input, the red one is the output when the input is not clipped, and the blue curve is the output when the input is clipped. Comparing that blue graph with the kind of saturation of a triode, I think it can be a good approximation:

This is the difference between clipping the input and the output (with this method of clipping):

So I think that, with this method of clipping, you may better clip the input. And that a good thing because that way we can reuse components so we only have to add one OpAmp and a FET to the design I posted on yesterday's post. This would be the schematic:

As I said before, Rpot (the source resistor on the new FET) adjusts the slope of the transistor's current so it's near the "unity" (in cumparison with Vg of the first FET - Vk), and VF2 adjusts at what voltage we want the solid state triode to start compressing the signal.

I would run some tests on this "final" schematic, but I'm out of free time as I've been writing this post for almost 2 hours... With this last schematic, we have implemented the nonlinearity showed in A, and if we added the right resistor between source and drain of the first FET, we would have implemented nonlinearity C as well.
Quote from: J M Fahey on February 27, 2010, 11:51:00 AM
Let us not forget that curve or slope cloning emulates a *clean* triode.
The *clipping* triode emulation is a whole other game, and then we will be able to emulate Preamp distortion only.

I've been searching more info on the saturation of a triode. On one of the links of this this webpage, I found the complete graph of a triode.

This is the normal, incomplete graph of a triode:

This is the plot of a triode's output current, including the saturation effect:

Quote from: J M Fahey on February 27, 2010, 11:51:00 AM
Let us not forget that curve or slope cloning emulates a *clean* triode.
The *clipping* triode emulation is a whole other game, and then we will be able to emulate Preamp distortion only.
We still need to emulate power pentodes, both clean and overdriven, transformer nonlinearities, the works.
Just as a side note: it can and *has* been done.
A very intelligent guy I know from a Brazilian Forum ("Amplificadores Valvulados" in Orkut) did an amazing work as his final Project to become Electronics Engineer there.
AFAIK what he did was to create this *huge* Spice model of a Mesa Boogie, a DC-5 or something like that, including transformer nonlinearities, speaker response, power supply sag, everything. :duh
I told him he was crazy, a Champ would do to prove his point.
End of the story: he reached a point (after 1 year work, involving almost all of his class and a couple Professors) where he could play 10 seconds of Blues with his trusty Strat into his soundboard, and hear them some 30 seconds later, undistinguishable from his miked Mesa whatever.
Almost real time. :o
Mind you, it was not DSP work, but fully a Software solution.
Yea, I know, you have Guitar Rig and others there who *claim* to emulate 500 classic amps. Do they? :trouble
This one did emulate just *one* .... faithfully.
Unfortunately he became Engineer and now he's working full time in a completely unrelated field, say Electrical utility company or something like that.

I'd like to read more about his project, perhaps I can learn something. Could you tell me more about it?
Quote from: dimitri on March 02, 2010, 10:55:21 AM
QuoteBut it looks like most of the desired tube distortion comes from the output power tubes and the rectifier tubes, am I right now?

This is correct for the amplifiers with single/dual triode preamp - blues
For the triple/quad triode high gain preamps - heavy metal this will have less pronounced effect.
You forgot the loudspeaker specific tone, when loudspeaker is driven that hard that the spider loose it linearity and cone self-resonance modes are excited.  

I assume you are Dimitri Danyuk. It's an honor to read you, thank you for coming by this forum! If I've understood you well, high gain preamp triode amplifiers actually use those triodes to generate tube distortion. Then, maybe this can still be used for something useful :P

It's true that loudspeakers change their output characteristics when driven hard, tough this also happen in solid state amplifiers. If you want to emulate the whole "tube sound" at low levels of volume,you have to emulate this too.

Quote from: dimitri on March 02, 2010, 10:55:21 AM
Hello Findeton,

Let me thank you to push this further. What was explained in my article is soft 3/2 nonlinearity and published circuit  is suited for "aural exciter"-like applications. To emulate hard triode nonlinearity in guitar amp you should take grid current into account. The decoupling cap from anode to the next stage grid is charging with grid current, and the average grid voltage becomes more negative, thus the output waveform becomes similar to half-wave rectifier. After overload decoupling cap discharges through the grid resistor.  You don't need to have 19 opamps in the circuit, you need a couple of extra components per stage.

I don't need 19 Opamps in the circuit. That's true, I've come to the conclusion that my idea was kind of crazy. I haven't  understood very well your "decoupling cap" explanation... but I have thought about it and this is my new approximation to the triode, taking grid current into account (I use one OpAmp and 4 voltage buffers):

I simplified the schematics a little, deleting some connection pins. In the schematics, this component depicts a positive unity gain voltage buffer:

Real examples of this component are BUF634, EL2002A, EL2072... or you could build one with two FETS (VF must be a fixed value in order to use one of the FETs as current source) or one OpAmp.

And this is how I depicted a negative unity gain voltage buffer:

Again, real examples of this component are, for example, INA134, INA2134 and you could build one with an OpAmp or 2 FETS.

The model of a triode showed on the first image of this post will accurately follow an output current function like the one of a triode:

Ip = m*(u*Vgk + Vpk +Vconstant)^1.5.

Indeed, let the current of the chosen Fetzer Valve (FET+Rs) be:


Then the current output of the "solid state triode"  would be something like:

Ipk =K3*K1(K2*Vgk + Vpk + Voffset - Vth)^1.5

Which is similar to the triode's transfer function. So that's it, we've got a triode made of one opamp and 4 voltage buffers. Of course, real components are everything but perfect so this solid state triode won't be that exact, but it will be a very good approximation. Also, I don't know how much noise this circuit will generate (noise figure) but it shouldn't create much noise, as we are using a FET (which is a low noise device), some voltage buffers (also low noise devices) and an OpAmp as inverting amplifier (this is the active device that will add most of the noise, I think). We should also use low noise resistors (film resistors) with low tolerance in order to reduce noise and deviations from the desired triode emulation.

As we are using solid state components, and given that triodes are low wattage (1.2 watts) and high voltage devices, if we want this "solid state triode" give the same amount of power we should escalate tensions/currents. Let's say, for example that we use Vpp-Vss=24V, then, with the total power being 1.2Watts the maximum output current should be 50mA. That's a reasonable output current, I think (we don't need to manage that much current on the OpAmp as we are working just with voltages there).

I'd do some simulations or build a prototype, but I don't have much free time. This solid state approximation of a triode should work well when trying to emulate  preamps.

As always, any feedback, questions and critics are welcome.
Quote from: phatt on February 18, 2010, 04:38:58 AM
Hello Findeton,

I quote; "I have a jtm45 hand-made by myself, but i'd love to have a pedal that is able to get the classic tube distortion without having to mod my amplifier nor having to pump up the volume. There are many stompboxes out there, even some with tubes in them, but I like to do things by myself."

So I'll take it That this is the AIM of Your exercise?

My aim at the moment is just to enhace the design of the "Fetzer Valve" (which tries to emulate a triode). If that helps anyone in the design of a stompbox, it would be great though. But it looks like most of the desired tube distortion comes from the output power tubes and the rectifier tubes, am I right now?

Quote from: phatt on February 18, 2010, 04:38:58 AM
Take the classic Bassman/ Marshall swap as an example;
Without changing anything (other that a rebias) change the 6L6/5881 to a EL34.

This turns a mildly hot bassman into a monster compressed OD machine.

A 6L6 is a Beam Power Tetrode and an EL34 is a pentode, and pentodes are able to achieve higher current outputs and wider output voltage swings. So yes, it makes sense that, for a given input and a maximum voltage, a EL34 will more easily overdrive the signal.

Quote from: phatt on February 18, 2010, 04:38:58 AM
"""""Pentodes where invented to overcome the limitations of triodes."""

The sound you hear from you JTM is **Output compression** and has very little to do with triodes. (though the PI may hold some secrets).
Triodes can't do it With the aid of a transformer so what hope is there with only a triode?? :duh

Tube power stages are just big oversized, overweight, expensive *Compressors* 8|
Darn problem is that by the time they compress well they are to darn loud,
as you have noted.

The curves of a triode look (way) simpler to emulate faithfully than the characteristics of a pentode. This is a good reason to start modeling triodes rather than pentodes. I guess that, ater this, I could try emulating rectifier tubes and, after that, pentodes (I said try, not succeed!). I've given a glance at the models of a pentode for pspice and yes, there are some characteristics that are quite difficult to emulate.
Quote from: J M Fahey on February 17, 2010, 07:44:45 PM
Hi findleton.
Sorry to say so, but I think you are over engineering this too much.
Personal opinion, everybody can disagree, is that we want to get a certain *sound* here.
Triodes are one way to go, yes.
Emulating their transfer curves is *part* of the problem, there are a zillion other factors to consider, specially when they are driven *beyond* the linear, "polite" part of their curves.
I'm sure what you are doing here will grant you a scolarship at any great University you wish, but it's a little over our heads.
Maybe you should post it at some IEEE's convention, it will be met with much acclaim.
Maybe I'm mistaken here, it just happens I am a Minimalist , I prefer to get 90% of a triode with 1 resistor rather than 95% with 19 Op Amps, but of course, I insist, I may be a backwards kind of guy.
Good luck and excellent work.

Don't worry if you think it's a too big/complex circuit for emulating just a triode. I'm over engineering because I want to get as near as possible to the behaviour of a triode (so no one can say again "you just can't get a triode out of silicon technology"  :P). But, actually, this design I'm doing would only make sense if it's implemented on an Integrated Circuit (a chip). Otherwise it's of almost no use.

But, if you get the ideas of this design and simplify just a little, you can obtain a very good triode with just 3 OpAmps and 1 FET. Would you fancy that a little more? Let's say a Fetzer valve approach gets 50% of the triode sound and this complex design gets about 95% of the triode's sound: well a 3 OpAmp design can get you about 80% or more of the triode sound. I'll present the simplified approach when I finnish to explain the complex design though.
This is the roadmap:

- First we'll see that we can solve the first problem that a "Fetzer valve" approach has: the simulated cathode is connected to ground.

- Once we solve this, we'll see the limitations of the design and some ways to overcome them.

- Then we'll have a nice device with a current function resembling Ids=k*(Vgs)^1.5 . But we'll still have to implement something like Ipk = m*(u*Vgk + Vpk)^1.5 . Using the same building blocks we used before, we'll easily achieve our dreamed silicon triode. I'll leave this part for the weekend, because I still have to test it, though I already have the design (nothing out of the ordinary, just a bunch of OpAmps).

- Conclusions, and some more ideas.

The basic building blocks we're gonna use are OpAmps, resistors and FETs (it doesn't matter if they are JFETs or MOSFETs). If you want to understand the circuit you should just review 3 OpAmp configurations: as summing amplifier, as voltage follower/Buffer and as inverting amplifier. We'll actually use very easy basic stuff.

As I already said, the fetzer valve approximates the current function as Ids=k*(Vg-Vp)^1.5 :

The first and main problem with this device is that we want a device that has one input pint (G) and two output pins (D and S), so we want Ids to be Ids= k*(Vgs)^1.5 . The next configuration solves the problem. Let G, D and S the input pins of our device and G1, D1 and S1 the pins of the first JFET/MOSFET we use in out device.  First, as the Fetzer Valve has an output current resembling(note that Vp is negative in a JFET):

Ids1 = k*(Vg1-Vp)^1.5

We use some Opamps to adapt this Vgs1 to Vgs1=Vg-Vs+Vp, so we get Ids1=k*(Vg-Vs)^1.5 :

Note that I only used 1k resistors. The important thing is to use resistors of the same value, the value doesn't matter if we only talk about the current function, only matters if we want to talk about noise or power consumption.

This is the output we get on the simulation:

For this particular JFET, Vp is -1.42356 and a Idss of about 19mA. This gives us a Rs=k*Vp/Idss=48 ohm (k=0.64). The input range of a Fetzer Valve would be [-|Vp|, k*|Vp|], but as this is not a Fetzer Valve, the input now is [0, |Vp|*(1+k)], though in fact as you can see the device behaves well out of that range. As you can see, at Vgs=|Vp|*(1+k)=2.33, Ids=19.356mA. It is an acceptable number because this shift from 19mA to 19.356mA is mainly caused because of channel lenght modulation: in a second order approximation Ids depends also on Vds and I measured Idss for a different Vd than 20V. Anyway, this is not something we should care at this moment because:
- Rs will play a more important role in determining the operating point, and we can modify it if we make Rs a pot.
- We'll talk about channel lenght modulation later on.

Now we have the current function we wanted, but, we don't have a device with G as input and D and S as outputs. My idea is a simple current mirror, we use another JFET, we copy Vgs1 to this new JFET: Let it be Vs2 the source voltage of the new JFET, we set Vgs2 to Vgs1. On first approximation, the current of a JFET only depends on Vgs, so we'll basically be copying the current to this JFET. The source and drain of this new JFET will actually be D and S of our device. This is the design:

Remember that the value of the resistors is the same for all of them (1kohm) so don't worry if the image is not big enough to see it clearly. In order to copy Vgs1 and G I've added some voltage buffers. Apart from that, I just used the second JFET as current mirror. I set the value of Vd to 20V and the value of S to 1V (we didn't have a S terminal with a Fetzer Valve, ¡now we have one!). This is what we obtain (we sweep the voltage of Vg from Vs-1 to Vs+4):

The green line is the current of the first JFET and the red line is the current of the new JFET (which pins D and S are the output pins of our device). Objective accomplished! Both lines are basically the same! Now we have a transfer function that resembles:


where G is an input terminal and D and S are the output terminals.

We have now completed the first point of our roadmap: to improve the Fetzer Valve in order to get a 3 pins device (the Fetzer valve had only two pins, as the source of the JFET was connected to Rs and Rs to ground).

I said it before and now you've confirmed it's true: we've used the simplest configuration of Opamps. In fact, they are simple, lineal operations that use unity gain, this will mean low noise, low distortion and low power consumption. The only things that will give us any problem at all are resistor mismatching and nonlinearities of the transistors.

So lets talk about the limitations of this design. The first one being the channel length modulation: in order to get the last graph, I set the value of Vd to 20V, just like I set the value of the drain of the first JFET, and that's cheating indeed! It's cheating because every FET has channel length modulation, Ids also depends on Vds:


lambda normally is a small value, but, as you can see in the next graph, it's quite noticeable if you change, for example, Vd from 20V to 7V:

The range considered for Vgs in the graph  is from -1 to 4V, but our calcs only took into account from 0 to |Vp|*1.64=2.33V so the difference between the current of the first JFET and the secon one is not that great. Anyway there actually are some ways to overcome these inconvenient nonlinearities. This means that this channel length modulation is affecting us on the first JFET, but also on this current mirror to the second JFET. Actually, as in the first JFET Vd1 is fixated (in this example, it's fixated to 20V) and Vs1 doesn't change that much, the channel length modulation hardly affects the first JFET, it's in the second JFET where we have to point our eyes.

The most practical is to use another FET that has a much lower lambda. Other way to overcome this distortion could be to add yet another JFET and yet another OpAmp (just one more please! :P), but I said I would use 19 OpAmps and I won't use no more! Actually, I'll continue this weekend and explain this and other limitations (or maybe this friday) because I've already run out of time and I have no time tomorrow. And there are many things that still remain unsaid.

Until I come back, think about this: we have a device with a current transfer function Ids= (Vgs)^1.5 and we want a transfer function  Ipk = m*(u*Vgk + Vpk +Vconstant)^1.5. We just need some more OpAmps: using the same buffers, summing amplifier and inverting amplifier OpAmp configurations, we'll finally obtain the desired transfer function!

Any input, observation and of course critics are welcome!