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Author Topic: A 19OpAmp-2FET approach to the Triode question  (Read 33898 times)

Findeton

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A 19OpAmp-2FET approach to the Triode question
« on: February 16, 2010, 05:07:15 PM »
2 FETand 19 OpAmps:

Since the invention of the transistor, there's always been people wondering if it's possible to emulate the characteristics of a triode with silicon technology in a faithful way. The "tube sound" is still the guitarrists' most wanted sound in the silicon era, and there's been countless attemps to achieve that sound with silicon technology. This, is only another attempt to add to the list.

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. I haven't still tested this "silicon triode" yet, because my free time is sparse. But I'd like to present the first results of my quest:

About a week ago or so I re-read the articule about the Fetzer Valve in runoff groove.com. It's a design that tries to emulate some characteristics of the triode using a JFET and a resistance on the source as feedback in order to obtain a drain to source current that follows the 1.5 power rate that is characteristic in a triode. I thought it was an interesting approach, but, the article said that we should use a certain value of the Rs (source resistance) in order to approximate as much as we could to this 1.5 power/exponent rate. It also mentioned that the value of Rs had been studied on a paper called "On Triode Emulation" written by Dimitri Danyuk.

So, just for fun, I decided to read it and try to emulate the graphs shown on the paper.

Basically, a JFET in Saturation region, follows the equation:

Ids =Idss*(1 - Vgs/Vp)^2

You can see the exponent in that function is 2. The exponent in the current function on a triode is 1.5, and if you add a resistance to the source pin of a JFET, the curve of Ids will fall, so at some point, for a given Rs=k*Vp/Idss, the curve of a JFET will actually be very similar to a power 1.5.

In the paper, Dimitri finds that the best match is Rs= 0.83*Vp/Idss (that is, k=0.83). But that's just the best approximation for the operating point. That is, if the input voltage goes from Vp to Rs*Idss, this Rs is the best approximation when the input is (Vp+Rs*Idss)/2. But the fact is that when we use the "Fezter Valve", we use all the voltage range. So I created a function in matlab that sums the normalized difference from the "Fezter Valve" to the x^1.5 function for each k and this is what I found:



Basically, k=0.64 is a better match when we consider all the input voltage range. The difference is 1.467% when for k=0.84  it's 2.664% (note that for k=0 it's about 16%).

All in all, I think the "Fetzer Valve" is an interesting approach, but, as I had never designed anything with valves before, I didn't fully understand how triodes worked. What the hell did actually meant this 3/2 power law that everyone said triodes followed?

So I installed Pspice Microsim Eval 8 and with this tutorial and the model of a 12ax7a I could finally get to test a triode (or at least the model of a triode). These are the the curves on a triode:



and:



As you can see in both images, the current depends from Vpk (plate to cathode) and from Vgk (grid to cathode)! This is what I didn't understand when I looked to the graphs of a triode, that's why I wanted to check it by myself on the circuit simulator. This fact, tears the Fetzer Valve to shreds, as the current of a triode depends on 2 voltages and the current of the JFET and of the Fetzer Valve only depends (in the first/simplest approximation) on the Vgs (gate to source, the equivalent of the grid to sourcein a triode)!

The next image represents the curve of the current in a JFET. As I said, it follows the equation Ids =Idss*(1 - Vgs/Vp)^2. You can see that it's very similar to the last image I've posted of a triode, being the main differences that it has an exponent 2 instead of 1.5 (the Fetzer valve solves that), and that the curve changes for each Vpk  on a triode:



This is a two dimensional function, a curve. If I wanted to suceed in emulating the charateristics of a triode, I had to know what the current transfer function looks like. So, observing the the model of a 12ax7a for pspice I saw that it followed a not so complex formula. Searching more on the web I also found the formula on this paper  ("SPICE Models for Vacuum-Tube Amplifiers"):

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

Where m and u are constants and p=plate, g=grid, k=cathode. The formula used in pspice introduces a constant. As it's not a difficult modification and it looks like it's a more exact model, I reproduce it here:

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

Knowing the parametters of this function (they are on the pspice model of a 12ax7), and just for fun, and as I didn't find anyone on the web that had ever cared to plot a 3d graph of this function, I went to matlab and here you have the 3d graph:



Look another time to the last image, the current function of a JFET. When Vds changes and Vgs doesn't,  the transfer function doesn't change (in saturation mode, at first approximation). But for a triode, when Vgk=-2, Ip can be 0mA (for example for Vpk=50) or 4mA  (for example for Vpk=200). It makes a hell of a difference!

But there's even more! Even if, for a moment, we didn't consider this "little thing" called Vpk in our model of a triode, the Fetzer Valve does a good job approximating the 1.5 exponent and therefore the behaviour of a triode... Well, of a triode... with the cathode connected to ground! (and most designs do not connect the 12ax7a cathode to ground) Because, of course, if we added another resistance/capacitor/whatever to the path from the Rs of a Fetzer Valve, the current transfer function will vary.. in a veery different way than the transfer function of  a triode would do. In order to imagine up to what point it differs, take into account again that an ideal JFET/Fetzer Valve has a curve as transfer function and that a triode has a whole surface as transfer function!

But, not everything is lost, we can overcome ALL those problems and obtain a silicon circuit that implements the function that so many circuit simulation programs use to modellate a triode, and, that's done modifying the Fetzer valve approach. Of course, the pspice model of a triode is not infinitely accurate, and the implementation of that model that I've found is not perfect and although it solves all the aforementioned problems, it also adds it's own kind of problems (19 opamps, resistors mismatching, power consumption etc). But, hey, it will be a better approach than the one of the Fetzer valve (though a little bit more complex). I think we could even create a "silicon triode" that could be connected to the socket of a triode (the 3 pins of the heaters should be replaced with Vcc, gnd and Vss) and work just like a triode. It is not my goal, but theoretically possible, and it would need a bigger power supply (or, if you want to implement a guitar amp that has an-all valves power supply, you would need a separated all-silicon power supply to feed the "silicon valves").

Having about 19 Opamps, the design shouldn't be used for that kind of crazy stuff but for a pedal, or a preamp, and of course only in the case you prefer to implement a valve through 19 Opamps instead of using the real stuff. Although if this design actualy works well (and not only on our simulations), it could be possible to create "silicon valves" so anyone that wanted to use them could simply buy one instead of soldering 19 Opamps.

I'm sorry no to present the work yet, but here in Europe it's 11pm, I've been writting this post for hours and I'm early tomorrow so I'll continue another day and present the actual circuit (with pspice simulations)!

EDIT:  I had the second of the triode's graphs missing, I've now added it.
« Last Edit: February 17, 2010, 08:32:40 AM by Findeton »

J M Fahey

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Re: A 19OpAmp-2FET approach to the Triode question
« Reply #1 on: February 16, 2010, 06:25:12 PM »
Nice beginning, please post some more, sounds very interesting.

Findeton

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Re: A 19OpAmp-2FET approach to the Triode question
« Reply #2 on: February 17, 2010, 05:42:54 PM »
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:

Ids=(Vgs)^1.5

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:

Ids=k*(1+lambda*Vds)*(1-Vgs/Vp)^2

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!
« Last Edit: February 17, 2010, 05:53:54 PM by Findeton »

J M Fahey

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Re: A 19OpAmp-2FET approach to the Triode question
« Reply #3 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.

Findeton

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Re: A 19OpAmp-2FET approach to the Triode question
« Reply #4 on: February 18, 2010, 04:21:52 AM »
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.
« Last Edit: February 18, 2010, 04:22:55 AM by Findeton »

phatt

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Re: A 19OpAmp-2FET approach to the Triode question
« Reply #5 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?

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.

"""""Pentodes where invented to overcome the limitations of triodes."""

Triode output stage has a distinct rounded top when driven hard only Tetodes and better still Pentodes can do the marshall *Square Wave* compression trick.

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

Go the whole hog,, swap your JTM45 to 6V6 output and you will be lucky if it rattles on 10 as they have very low TC.

Or just triode strap you output tubes in the JTM45 to make them work like triodes?  I doubt you will like the thinner weaker sound. :'(

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.

Another clue can be found in some of the *early studio compressors* (names escape me here but) some of them actually used *phase splitters* and *Transformers*. (hint)
Though they worked well for audio they still could not pull off the same trickery as guitar amp outputs. (maybe because they only used triodes)

You obviously have the mind for vastly complex stuff (You leave me for dead) but I fear your base data is way off.

As I've come to understand with most things scientific,,,If you miss the basic concept all the maths in the world won't make it right.
Have fun, Phil.

J M Fahey

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Re: A 19OpAmp-2FET approach to the Triode question
« Reply #6 on: February 18, 2010, 09:26:41 AM »
Hi Findleton
Quote
(so no one can say again "you just can't get a triode out of silicon technology"  Tongue).

Quote
this design I'm doing would only make sense if it's implemented on an Integrated Circuit (a chip).
Fully agree on both points.
1)*Any curve* can be simulated with Op Amps, that's what they were invented for, that's what Analog Computers do.
2) Yes, if somebody puts it in a chip where you don't worry about the internals, and only have cathode, plate and grid pins, plus aome ground and +/-15V pins to feed it, and can sell it for less than, say, $2, it'll sell like hotcakes.
Impossible?: just look at PICs, full CPUs with processor, storage and interface pins for less than $2 !!!! In a DIP8 !!!!

Findeton

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Re: A 19OpAmp-2FET approach to the Triode question
« Reply #7 on: February 18, 2010, 03:21:34 PM »
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?

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.

"""""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.

phatt

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Re: A 19OpAmp-2FET approach to the Triode question
« Reply #8 on: February 19, 2010, 08:06:34 AM »
Hello Findeton,
Explanation excepted, 8|

Be aware that (to the novice),, comments like that need clarification otherwise it
implies that you just need to emulate a triode to get a full blown tube powerstage
sound and *That is Obviously not the case*.

As one of the lesser motals who do not have the gift of being able to grasp
complex things it took me many years of wading through the endless claims from the
experts who often don't even play guitar.

I've now at least learnt enough to know when something does not sit right.

I respect the fact that every electronics teck has their own idea on what is the *Best* approach to a given subject and I think the wise amongst them respect that there are often many ways to effect the same thing.
At the end it may not make any audible improvement,, just different.

ie; Find 2 technicians who will agree on how to correctly bias a power tube? :lmao:

That's not a criticism of teck's, just an amateurs observation of what I
personally encountered while trying to track down some valid useable information
when I was starting out playing with circuits.

Now I've interupted your work for long enough so Carry on,,  :tu:
Best of luck with it all.
Phil.
« Last Edit: February 19, 2010, 10:46:39 AM by phatt »

dimitri

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Re: A 19OpAmp-2FET approach to the Triode question
« Reply #9 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.

Quote
But 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. 





 

Findeton

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Re: A 19OpAmp-2FET approach to the Triode question
« Reply #10 on: March 16, 2010, 04:43:53 AM »
Quote
But 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.


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:

I=K1*(Vgs-Vth)^1.5

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.
« Last Edit: March 16, 2010, 08:24:30 AM by Findeton »

Findeton

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Re: A 19OpAmp-2FET approach to the Triode question
« Reply #11 on: March 17, 2010, 07:26:52 PM »
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.
« Last Edit: March 17, 2010, 07:40:47 PM by Findeton »

Findeton

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Re: A 19OpAmp-2FET approach to the Triode question
« Reply #12 on: March 21, 2010, 12:12:52 AM »
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!
« Last Edit: March 21, 2010, 05:56:49 AM by Findeton »

HENK

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Re: A 19OpAmp-2FET approach to the Triode question
« Reply #13 on: March 22, 2010, 08:04:06 AM »
Hej
Very cool!

Are there Sub´s for the AD815??

Greets Thilo

« Last Edit: March 22, 2010, 08:05:07 AM by HENK »

Findeton

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Re: A 19OpAmp-2FET approach to the Triode question
« Reply #14 on: March 22, 2010, 10:37:15 AM »
Hej
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).
« Last Edit: March 22, 2010, 04:05:36 PM by Findeton »