FET emulation of tubes without trimmer?

[JDoyle]

Could it be forced into having a current controlled input?

The 3130, that is?

The input pins are CMOS gates, so no current is getting in there at all.

I think your best bet would be one of the compensation pins, though your are going to be dropping your current across a resistor in the long tailed pair, so that pretty much makes it a voltage input...

Pin 8, as it is connected directly to the collector of the VAS would seem to be your best choice, but in that case you are driving the CMOS output stage at the same time and it is impossible to isolate the two...

So... Sorry man, gotta say 'not really' to your question.

For current input you pretty much need a BJT with it's emitter connected to ground through something OTHER than a resistor, a CSS or directly, for example.

Take a look at Norton op amps (I'm pretty sure you and I have discussed them Stephen, in the ancient past , actually, I think YOU pointed them out to me!), which, along with single op amps, is another area I think has been overlooked...

Jay

[JDoyle]

BTW: take a look at the simple basic "Phase Splitter": here the cathode, emitter, or source IS an OUTPUT...

I'm not arguing that the source/cathode/emitter can't be an output - I've proven myself to be an idiot on many occasions for sure, but denying the existence a source follower would be a new low for me...

Obviously the source can be USED as an output, but that isn't my point.

My point is: without any DIRECT connection from the output to the input, you DO NOT have negative feedback.

Degenerative? Yes. Negative? No.

Regards,

Jay Doyle

[soulsonic]

Of course!

The tubescreamer and it's ilk has caused so so many arguements/discussions/debates about DUAL op amps, that I honestly think there is an entire realm being overlooked - SINGLE op amps.

In duals everything is already decided for you, offset adjustment, compensation, operating current, etc. But with singles, you normally get access to points of the circuit where you can seriously mess with what is going on.

Regards,

Jay Doyle

YES! I ordered a bunch of LM301s recently for exactly these reasons! The idea of being able to externally change things like bandwidth seems like a really cool way to fool around with emulating old devices, and your suggestions about "access" to different parts of the circuit has gotten alot of interesting ideas flowing in my head. Thanks again.

My point is: without any DIRECT connection from the output to the input, you DO NOT have negative feedback.

Degenerative? Yes. Negative? No.

That's always how I've understood it to be. It was explained to me that degenerative feedback is the result of loops of current throughout the circuit caused by inadequate bypassing. Sometimes it is used intentionally as a convenient way of controlling gain, but the "textbook" thing is to usually bypass K fully so that no degeneration happens. Of course, tone-changing circuits of all sorts manipulate degeneration in creative ways via different sizes of K caps and different amounts of bypassing used.
You can also compensate for degeneration without bypassing by using positive feedback. Circuits like that were used in circumstances where the use of condensers was undesirable for whatever reasons. It's an old technique, so maybe it was situations where reliability of the old failure-prone electrolytics might be an issue, or even something as simple as the large physical sizes of the old caps.

RCA book stuff.

[JDoyle]

Well, you could argue that it is in that the actions of the source resistor curtail the gain of the amplifier...having a negative effect on gain (if you like).

No need to argue that at all - I think that has been prooven pretty solidly.

I've actually seen it described as negative feedback in one or two textbooks.

No argument that the info is presented that way, I'm just disagreeing. I'm the type of person who when he read in grade school textbooks that 'George Washington never told a lie', thought 'I'm bet Martha Washington would disagree with that'.

Makes sense as the inclusion of a source resistor swamps the effects of the current-dependent, internal emitter-resistance, in turn improving linearity.

OK, here and below, as you noted in subsequent posts, you are mixing your transistor types. The statement above is true for BJTs, the swamping of emitter resistance, but not for JFETs. A source resistor is usually there for bias, not linearity. It raises the source voltage, which allows bias from a single polarity supply; the improvement in linearity is secondary and a result of the flattening of the load line.

As you well know, Jay, without collector-base feedback and with the source resistor decoupled/omitted, things go banana-shaped pretty quickly if the signal is above 10mV or so.

Completely agree with the above quote FOR BJTs.

It is, however, NOT the case for JFETs. The reason for the banana shape in a BJT stage like the one quoted above is because of the nonlinear, and incredibly low, input/emitter impedance because the input is a diode. And as you said, an emitter resistor swamps that impedance, linearizing the output by linearizing the INPUT.

But in a JFET the input impedance is essentially constant, a source resistor is used to facilitate single polarity supply biasing it has no effect on the operation of the input once the bias has been set.

In fact, the MOST linear (and also the highest gain) Q point for a JFET is Vgs=0. You just have to keep your input signal to 100 mV (note the order of magnitude difference in signal size when compared to BJTs) or less so as not to forward bias the gate-source diode.

Somewhere I have an old Siliconix book 'Designing with Junction Field Effect Transistors' that outlines 8 or so 'rules' for low distortion design - which in our world of through-the-looking-glass engineering means '8 rules to break to get distortion from a JFET' - and one of them was 'bias as close to Vgs=0 without the input signal forward biasing the gate-source diode'.

I'll see if I can dig it out...

In fact, I'll respectfully submit that looking at Vgs as the locus of JFET distortion is a red herring. The variable that MOST determines the distortion content of a JFET stage is Vds (which, I admit, changes with Vgs, but the initial Vds point is the key). If you make sure to always keep Vds in the pentode region, you have the cleanest operation possible. It is when you let Vds slip down into the triode region that you start to get the majority of the distortion (excepting out and out clipping, of course).

This is also why a JFET produces mostly 2nd harmonic distortion, the distortion that occurs in the triode region only occurs as the signal swings toward ground at the drain, reducing Vds and bringing the operation into the triode region of the transfer curve. As it swings up towards V+, Vds increases and the operation moves into the pentode region and the output becomes more linear. This is asymmetric distortion - thus the 2nd harmonic. While in a BJT stage the distortion mechanism exists for the entire output swing, it's symmetric, thus the domination of 3rd harmonic in a BJT stage (and incidentally why you don't see the same type of transfer curve graphs in BJT datasheets that you see in Tube and FET datasheets).

With all of that said, I completely agree with Doug's post above. I think trying to get anything like 'tube sound' out of a single JFET is tilting at windmills...

Regards,

Jay Doyle

[StephenGiles]

Could it be forced into having a current controlled input?

The 3130, that is?

The input pins are CMOS gates, so no current is getting in there at all.

I think your best bet would be one of the compensation pins, though your are going to be dropping your current across a resistor in the long tailed pair, so that pretty much makes it a voltage input...

Pin 8, as it is connected directly to the collector of the VAS would seem to be your best choice, but in that case you are driving the CMOS output stage at the same time and it is impossible to isolate the two...

So... Sorry man, gotta say 'not really' to your question.

For current input you pretty much need a BJT with it's emitter connected to ground through something OTHER than a resistor, a CSS or directly, for example.

Take a look at Norton op amps (I'm pretty sure you and I have discussed them Stephen, in the ancient past , actually, I think YOU pointed them out to me!), which, along with single op amps, is another area I think has been overlooked...

Jay

Well, I was trawling through this
http://mypeoplepc.com/members/scottnoanh/birthofasynth/id28.html

...where Thomas Henry utilises pin 11 of a 4046 to achieve current control instead of the usual pin 9 for voltage control, and I thought perhaps there is a trick somewhere to give such an opamp a current controlled existence.

[gez]

Makes sense as the inclusion of a source resistor swamps the effects of the current-dependent, internal emitter-resistance, in turn improving linearity.

OK, here and below, as you noted in subsequent posts, you are mixing your transistor types. The statement above is true for BJTs, the swamping of emitter resistance, but not for JFETs. A source resistor is usually there for bias, not linearity. It raises the source voltage, which allows bias from a single polarity supply; the improvement in linearity is secondary and a result of the flattening of the load line.

My apologies for mixing things up there, it certainly didn't help.  However ( ). Yes, self-biasing is convenient, but that doesn't mean to say that the source resistor pays no part in linearising the amplifier.

Transconductance varies with drain current.  Tie the source to ground in a split supply, bias the gate from the negative part of the split and you'll see greater distortion that you would with the inclusion of a source resistor due to said variance in transconductance resulting from changing gate-source voltage.  The inclusion of a source resistor, with the typical self-biasing setup, means that gate-source voltage is relatively unaltered due to the follower action seen at the source, resulting in better linearity. Granted, output resistance is higher in JFETs, so things still aren't perfect, but it is a huge improvement.

Re the term negative feedback, we're just going to have to agree to disagree on that one.  There are plenty of authors who I admire that call it such, and in books that are definetly pitched at a level above that found in "grade school textbooks".  In a way, the input signal (or rather a copy of it seen at the source - an output) is being used to linearise itself, so...hmmm, somehow I still think you're not going to buy it! 

[alanlan]

My point is: without any DIRECT connection from the output to the input, you DO NOT have negative feedback.

vgs = vg - vs = vin - vs

if the output is taken from the source (source follower) then:

vgs = vin - vout

If that's not negative feedback I don't know what is.

Same argument follows if you take your output from the drain.

Degenerative = negative

Regenerative = positive

Simple.

[JDoyle]

My apologies for mixing things up there, it certainly didn't help.  However ( ). Yes, self-biasing is convenient, but that doesn't mean to say that the source resistor pays no part in linearising the amplifier.

Gez - I too apologize if you thought that I meant that the source resistor didn't play a part, of course it does! Actually, I kind of buried my agreement with this notion in the last sentence of the paragraph you quoted:

"the improvement in linearity is secondary and a result of the flattening of the load line."

Which leaves out everything you said, but I meant the same thing.  

Re the term negative feedback, we're just going to have to agree to disagree on that one.  There are plenty of authors who I admire that call it such, and in books that are definetly pitched at a level above that found in "grade school textbooks".

Again, I apologize, this time for the grade school reference, I was trying to come up with an analogy and that was the first one I came up with...

A better one, considering our agreement to disagree, AND the fact that it is based on a definition is:

'I have textbooks that say Pluto is a planet.'

In a way, the input signal (or rather a copy of it seen at the source - an output) is being used to linearise itself, so...hmmm, somehow I still think you're not going to buy it!  

Well, it is still missing that key 'fed back from the output to the input' bit I think is necessary.

Plus, considering the inherent distortion of the device, is the signal at the source REALLY a copy of it?

So you know, I think our opinions converge in the case of a BJT where the seperation of input and output is the resistance of the base-emitter diode - there still isn't a direct feedback connection, but instead of mega or even terraohms worth of resistance between the gate and source of a JFET or MOSFET, respectively, we're talking a hundred or so ohms, not to mention the fact that the input signal literally exits at the emitter of a BJT, instead of merely being 'sensed' by a FET...

Agree to disagree. Definitely.

Though I wonder who Harold Black would agree with. I've never read that paper...

Regards,

Jay Doyle

[gez]

'I have textbooks that say Pluto is a planet.'

And I have comics that say he was a dog!

[JDoyle]

vgs = vg - vs = vin - vs

if the output is taken from the source (source follower) then:

vgs = vin - vout

If that's not negative feedback I don't know what is.

Same argument follows if you take your output from the drain.

Degenerative = negative

Regenerative = positive

Simple.

Then in my opinion, you don't know what it is...

In neither your math, nor your dictionary definitions (which are wrong, type 'define: degenerative' into Google and the word 'negative' doesn't even APPEAR on the page, nevermind form the core of the definition; same goes for the Webster's dictionary I have sitting on my desk), do you show that (or how) ANY of the signal is fed from the output back to the input.

It sure is simple when you redefine everything - including what negative feedback is...

[JDoyle]

And I have comics that say he was a dog!

WAS?!?!?!?!?

When did he die??

[gez]

Damn, it's always the detail I get wrong... ( )

[alanlan]

Then in my opinion, you don't know what it is...

Oh really?  and I always thought I had a grasp on the subject.  Thanks for putting me right - I'm very grateful.

P.S. instead of entering the single word "degenerative" into google, try typing it along with "negative" and "feedback" and then you can put the rest of the electronics world right on the subject.

[johngreene]

I thought everybody knew that negative feedback is just a more transparent form of degenerative feedback.   

degenerative feedback has more mojo.

--john

[alanlan]

and to add a quote:
"However, unlike the resistor in the drain circuit, any current in Rsource causes a voltage across it. The voltage across it raises Vsource and therefore lessens the voltage difference between the gate and source. Rsource causes a voltage that opposes the change in Vbias - it's giving us negative feedback."

from http://www.geofex.com/article_folders/mosboost/mosboost.htm

[Fp-www.Tonepad.com]

You can email me, i may have some info.

[DougH]

I don't see what the big deal is- it's just a question of semantics. Either way, negative or degenerative, the overall effect is to reduce gain.

[earthtonesaudio]

Thought I would bump this again: http://www.colorado.edu/physics/phys3330/phys3330_fa06/pdfdocs/AN102FETbiasing.pdf
Says you can minimize device-to-device gain variations (page 4-6) using the graphs and some calculations, but I tried and came up with weird numbers.  Anyone else tried successfully?

[R.G.]

Says you can minimize device-to-device gain variations (page 4-6) using the graphs and some calculations, but I tried and came up with weird numbers.  Anyone else tried successfully?

I read that years ago, and decided I could not make it work reliably.

There's one of those engineering limits things going on here. If there was a simple way to make JFETs have much less unit-to-unit variation by how you biased them, just by using a couple of resistors, they would be used in more circuits as something other than switches or input followers.

It's important to remember that there are two kinds of things that are hard to do/understand/etc. Some things are hard because you're inexperienced or have just never seen the trick; once you learn the trick to it, they become easy. And some things are hard because they really are hard, not because there's a trick. I understand that Gordon Liddy - no, you won't remember him, you're too young - used to amaze people at %^&*tail parties by laying a burning cigarette in the palm of his hand and letting it smoulder there, without even wincing.  Invevitably, someone would say to him "Wow. That's incredible. What's the trick?" His reply was "The trick is not caring."

[puretube]

There`s a current project in Elektor April/May 2008,
where a different approach is taken:
Error-Correction, instead of negative feedback (to minimize nonlinearities, though...).

I can imagine that some people here will find some hair to split on that subject, too!   

Author


(Note to self: Error-Injection, to increase/induce nonlinearities...)

[EDIT]: a certain Mr. Graeme is cited in that article...

more stuff...