Triode Emulator

[tca]

Hi,

I've been trying to do the calculations in Dimitri's paper http://www.scribd.com/doc/65330169/Triode-Emulator-by-Dimitri-Danyuk.

But I'm always getting

Rs=1.66*Vp/Idss  instead of

Rs=0.83*Vp/Idss.

Did anyone tried to repeat the calculations in that paper? Any comments?

Thanks

[tca]

I forgot that Vp was negative. 

[Davelectro]

Did anyone tried to repeat the calculations in that paper? Any comments?

Not really. I trust the fetzer valve calculator. 

[mac]

I never read that paper, but i guess that the fet I-V eq can be expanded in a Taylor series, compared to the expanded tube eq, and then one can see under what conditions the fet eq can be "forced" to take the shape of the tube eq.
Under clipping, is this condition satisfied?

Instead of doing this, go high voltage.

mac

[tca]

I've just managed to reproduce the results (figures included) with a simple numerical script in Octave (http://www.gnu.org/software/octave/). The value o k=0.83 just pops out easily and the precision is enough for all practical implementations.

Thanks.

[tca]

Actually the exact result is:

Rs=2^(-1/4)|Vp|/Idss= 0.840896415253715...|Vp|/Idss

Who says that math doesn't pay off?

[PRR]

> "forced" to take the shape of the tube eq.
> Under clipping, is this condition satisfied?

As I read it: no.

In fact it does the wrong thing for any large swing.

His math (confirmed by TCA) shows a 3/2 law but only AT the half-way operating point. Fig 5 plots this for the entire range. The exponent actually shifts from 2.0 toward 1.0, and passes-through 3/2 (1.5) at one point.

OTOH, an ideal tube in space-charge is reliably 1.5. Actual tubes are 1.6 to 1.3 depending on geometry, but fairly consistent over a useful range of current.

And I *think* the tube tends to slant the other way from this degnerate FET.



So small signals may be tube-like, but I doubt "fuzz-tone" is tube-like.

And the big objection: "One of a musicians reported that there is certain quality a tube amplifier must have and this design has not. It is the tubes. The reviewer could not peer inside the ventilation windows in the case and watch the tubes glow."

[stm]

Actually the exact result is:

Rs=2^(-1/4)|Vp|/Idss= 0.840896415253715...|Vp|/Idss

Who says that math doesn't pay off?

Congratulations, I looked hard at the explanation in the paper about getting the 0.83 value but couldn't actually get to the exact math behind it.

[stm]

To add some perspective, Danyuk's triode emulator is intended for music reproduction, i.e. Beethoven's 5th Symphony.  In this context, triodes in a hi-fi amplifier do not get close to clipping limits, and so this circuit doesn't address clipping characteristics.  The idea is to add a subtle to moderate amount of harmonic distortion, let's say between 1% to 5% depending on the listener's tastes.  There is a mix knob that blends the pure signal with the same signal processed by a JFET circuit that approximates to the 1.5 exponent.  If you think about this, a mix of these two signals will no longer follow the 1.5 exponent rule anyway!

The relevant part here is that adding a controlled amount of low-order harmonics (mostly 2nd a bit of 3rd) is pleasant to the ear.  Whether these harmonics follow a 1.3, 1.5 or 1.7 exponent rule is not as relevant as having the *right* amount in the mix, which by the way will be different from person to person.

In addition, a practical 12AX7 triode stage with bypassed cathode won't follow a pure 1.5 exponent law either, even if popular knowledge says the triode plate current follows a 1.5 exponent law with regards to the grid voltage. Why?

Let's take Child's law for the plate current:  Ip = (Vg + Vp/mu)^1.5

There you have the gate voltage (Vg) and the 1.5 exponent, but you also have another term (Vp/mu) which to make things worse depends on the instantaneous plate voltage as well.

The plate voltage will be something like Vp = Vcc - Rp*Ip
where Vcc is the supply voltage and Rp is the plate resistance (here we assume there is no load on the stage)

Replacing into the first equation you get:

Ip = [ Vg + (Vcc - Rp * Ip) / mu ] ^ 1.5

Notice that Ip is on both sides and cannot be isolated on one side (as far as I know) to obtain a closed equation in the form: Ip = function( Vg, Vcc, Rp, mu )

This shows that in practice you don't get a PURE 1.5 exponent law.  In addition we know that Child's law is just an approximation, so the "real world" may differ even more.  *** The moral here is that the 1.5 exponent law is not as important as it appears at first sight. ***

If you want to know, I've run many simulations with different triode models comparing with a JFET circuit with different source resistors, and have come to the conclusion that a JFET with a source resistor chosen as Rs = K * Vp / Idss, where K lies between 1.5 to 2.0 is a much closer approximation in terms of the harmonic content of the resulting signal (as seen running an FFT analysis) than the 0.83 or 0.84 "magic value".  Of course this is only valid in the central third of the dynamic range, as the JFET won't follow the triode saturation characteristics.

So, do we need to change the paradigm on the 0.83 magic source value?  Absolutely not!  Based on listening tests I really like this 0.83 value on electric guitar, and I'm sure many other people do.  I've also done listening tests bypassing the source resistor in the JFET circuit to approximate to the JFET theoretical 2.0 exponent (and also adding some attenuation to compensate for the extra gain obtained when the capacitor is added, so as to compare things at a similar level).  In summary, I don't like much the result with the 2.0 exponent since it gets on the harsh side to my liking.  Moving in the opposite direction, I also liked the case where Rs = 1.0 * Vp/Idss, as it sounds fuller than the original guitar signal.

CONCLUSION:  The above is proof enough for ME that it is a waste of energy to pursue an EXACT triode emulation.  Instead, it pays-off to see which the basic distortion mechanisms are and try to get something similar that's pleasant to the ear.  As the saying goes, "there are many ways to skin a cat."

[PRR]

> is intended for music reproduction

No, he says he tried it on "pick-ups (lead, bass guitar, etc)":

"Several circuits were made in the form of tube simulator (Figs.8,9), active front-end electronics (Fig.6) for various pick-ups (lead, bass guitar, etc) and microphone amplifiers...."

> There is a mix knob ..., a mix of these two signals will no longer follow the 1.5 exponent rule anyway!

No, of course not..... but more knobs is more to play with.

> practical 12AX7 ... won't ... 1.5 exponent ... Why?

Because there is 100:1 electrostatic feedback inside the tube, and in most any useful rig we have large plate swing.

The opposite extreme is an infinite impedance load. Now the current is fixed, Child's Law does not visit us, gain is exactly Mu and distortion approaches zero.

In most practical cases the exponent is something less, gain is less, and distortion is in-between zero and the ~~5% of a hard-driven tube. In fact it is the load which reduces practical triode THD from 10% (Danyuk's fig 2) to ~~5% at full output.

> where K lies between 1.5 to 2.0 is a much closer approximation ... ... than the 0.83 or 0.84 "magic value".

Sure. Danyuk emulated the short-circuited triode. Useful triodes distort half as much. Your more-or-less 1.7 gives half the distortion of 0.84.

It is interesting that he did not emulate a tube triode's Mu. Would have been simple with that over-done servo-load.    

> we know that Child's law is just an approximation

Within it's assumptions, it is exact; and crops-up in other contexts. I suppose you mean that real tubes are approximations of Child's assumptions, particularly in being non-infinite (end effects). And (especially on 12AX7) the designers knew how to exploit those "violations" to enhance gain. (12AX7's zero-grid line is far from Child's Law; grid is "too close" to cathode, by design.)

There's other deviations. Tubes at very low current density act like BJTs: 1/Gm is proportional to current. At very high current, some tubes can show "resistance" losses in cathode, making them very linear (but low-gain). (And current flattens at high current, but with oxide cathodes you Never Go Way Up There.) Conversely the 1/Gm proportional to current law in BJTs degrades into a Child's Law at very high current density. Difference is tubes conduct poorly and are usually in Child turf, BJTs conduct well and shouldn't get into Child's territory.

The JFET is actually a very different device. Tubes and BJTs have a "fence across the path". JFET has no fence, it squeezes the sides of the path.

> *** The moral here is that the 1.5 exponent law is not as important as it appears at first sight. ***

We need The Answer. The answer to everything is 42. Use a 2N4242 working at 42V with 42 ohm resistor, the mice will be happy.

[tca]

You can also have the 3/2 law for any value of the operating point such that k takes the value

k=1/2*z^(-3/4)

where z is the normalized input voltage. If z=1/2 one gets k=2^(-1/4).

[tca]

Ip = [ Vg + (Vcc - Rp * Ip) / mu ] ^ 1.5

Notice that Ip is on both sides and cannot be isolated on one side (as far as I know) to obtain a closed equation in the form: Ip = function( Vg, Vcc, Rp, mu )

Just square both sides of the equation and then use the general formula of roots for a cubic equation (http://en.wikipedia.org/wiki/Cubic_function).

.. it pays-off to see which the basic distortion mechanisms are and try to get something similar that's pleasant to the ear.  As the saying goes, "there are many ways to skin a cat."

I couldn't agree with you more!

[tca]

Is there a particular reason for the Fetzer valve circuit to be working in common drain (with only current gain)? I haven't seen any reason in Dimitri's paper for this. Probably because of the "equivalent" triode valve circuit? Is this the only reason?

Thanks.

P.S.

This is a stupid question... see below!

[PRR]

> reason for the Fetzer valve circuit to be working in common drain

?? The standard Fetzer is common-source:  http://www.runoffgroove.com/oldfetzer.html

Dimitri's plan is smothered in added amplification so FET gain is not needed; however it too is wired common-source.

[tca]

?? The standard Fetzer is common-source:  http://www.runoffgroove.com/oldfetzer.html

You are obviously right... sorry about that question. I was thinking about a buffered version of the schematic!

[tca]

I've written a small page explaining how the exact value of the constant k, which determines the value of the source resistor, is obtained.
http://www.diale.org/triode.html

Any comments are welcome.

Cheers.

[Davelectro]

I'm kinda lost here. What "K" should one aim for in a first stage of a Fetzer-based preamp (for example)?

[PRR]

TCA: Thanks for your essay

I'm not qualified to proof-read your math. (In a recent post I arrived at "1" when the truth was "2".)

I do know that "real" device equations may have one general trend, and many "minor" trends due to end-effect or other practical construction details.

> The basic equations that governs the behavior of a JFET, in the saturated regime

FWIW:

This is so only for "(infinitely) long channel" JFETs.

Some production JFETs are "long-enough channel" to follow the equation closely.

A short-channel JFET has lower output resistance, but also higher gain in low-impedance loads. Short-channel JFETs are popular in RF circuits where all loads are low-impedance and gain is essential.

We often can not know which type "our" JFET is. Back around 1980 the manufacturers published some clues in their data-sheets but the JFET market has been stagnant for years and there is little new data, and not much old data.

If you have curves, you can compare with Silconix's examples:


From: Designing with Field-Effect Transistors, 2nd edition, Siliconix Inc, Ed Oxner

It should be noted that vacuum triodes also have "long/short-channel" properties. WE300B is very linear, exponent less than 3/2 over the useful range. 12AX7 is very non-linear in some areas.

And a vacuum triode's "3/2" law is, in most simple circuits, strongly linearized by load resistance and Mu.

And this K-fudge _approximates_ a 3/2 law only over a limited range. Perhaps good-enough for 10% swings and 0.5% THD. Not-the-same for guitar players who routinely slam 30% swings and >5% THD.

> What "K" should one aim for

I suspect start with >0.5 but <1.0, then trim to taste.

[Eb7+9]

it helps to look at the right set of transfer curves ... have a look here:

http://www.lynx.bc.ca/~jc/transferCurvature-TubeSimulation.html

my approach to achieving similar Plate-like wave bending goals based on biasing a jFET anywhere inside
its Vp range while providing full gain/curvature (ie., by avoidance of Source resistance)

Dimitri's plan is smothered in added amplification so FET gain is not needed

it's not about gain but more about gain curve ...

the op-amp based trans-resistance amplifier converts Drain current to output voltage (linearly)
and fixes Drain voltage steadily via NFB in op-amp // acting like an AC ground for the Drain

the  intent is that the wide-range (large-signal) non-linear Grid Voltage to Drain current transfer function
is now mirrored out and converted to (usalble) voltage ... more importantly, the whole non-linear
current transfer function now gets preserved in shape as it is made to become a voltage-voltage function ...
by providing y = ax type of transform in the trans-R amp (essentially shape-order preserving)

the resistance in the Source circuit simply drops the power factor, from a 2 (un-degenerated jFET) to 1.5
(un-degenerated triode) ... this shows the principle at play (jFET current-transfer test bed) and shows the ability to
hit a power function curve somewhat lower than 2 (variability)

most musicians, tho, would be happy to have as much curvature/bending (= dynamic harmonics) as possible ...

[tca]

I'm kinda lost here. What "K" should one aim for in a first stage of a Fetzer-based preamp (for example)?

I would say: 2^(-1/4)=0.84 plus/minus 50%.