Scott,
I think the issue is now traveling, so it is time to quit. Hopefully a few more folks will read the paper.
ES
1. Series derived, series applied
2. Parallel derived, parallel applied
3. Series derived, parallel applied
4. Parallel derived, series applied
2. Parallel derived, parallel applied
3. Series derived, parallel applied
4. Parallel derived, series applied
A friendly warning from your neighbourhood moderator team. Any posts that allude in any way to personal issues will be deleted and will lead to penalty points and bin time. That includes replies of personal nature to others' provocations.
You are right! I was thinking of way back when they were first becoming common! Certainly anyone doing pulse width modulation s in an amplifier power supply or light dimmer, can use a really good triangle wave.
While Bruno says degeneration gives the same loop gain, he nevertheless
has no problem differentiating it. I quote (hopefully without offending)
from page 12 and 13 of this feedback article:
What else has happened though? Degenerating the input pair has a second
consequence: the non-linear contribution of the input transistors is
reduced (fig 13)!
By how much? Well, after reducing C by about a factor 10, the same
rate-of-change demands only one tenth the current. Reducing I by a
factor 10 will reduce the 3rd harmonic in Vdiff by a factor 1000 and the
5th a staggering one hundred thousand times. Loop gain hasn’t fundamentally
changed but SID is all but eliminated entirely. The sonic improvement
must have been spectacular, but clearly this cannot be attributed to
reduced feedback, as feedback hasn’t been reduced! Most people now know
that degenerating the input stage is the recognized method of getting rid
of SID. Since SID can be eliminated independently from feedback, there are
no grounds for claiming a causal relation between the two. The experiment
was a resounding success, but the conclusions drawn from it were incorrect.
The correct conclusion is: SID can be eliminated without affecting loop
gain, so loop gain does not cause SID. SID is a circuit flaw pure and
simple, and degeneration of the input stage fixes it.
😎
has no problem differentiating it. I quote (hopefully without offending)
from page 12 and 13 of this feedback article:
What else has happened though? Degenerating the input pair has a second
consequence: the non-linear contribution of the input transistors is
reduced (fig 13)!
By how much? Well, after reducing C by about a factor 10, the same
rate-of-change demands only one tenth the current. Reducing I by a
factor 10 will reduce the 3rd harmonic in Vdiff by a factor 1000 and the
5th a staggering one hundred thousand times. Loop gain hasn’t fundamentally
changed but SID is all but eliminated entirely. The sonic improvement
must have been spectacular, but clearly this cannot be attributed to
reduced feedback, as feedback hasn’t been reduced! Most people now know
that degenerating the input stage is the recognized method of getting rid
of SID. Since SID can be eliminated independently from feedback, there are
no grounds for claiming a causal relation between the two. The experiment
was a resounding success, but the conclusions drawn from it were incorrect.
The correct conclusion is: SID can be eliminated without affecting loop
gain, so loop gain does not cause SID. SID is a circuit flaw pure and
simple, and degeneration of the input stage fixes it.
😎
degenerating the diff pair is adding local negative feedback, redistributing gain, diff input gm is reduced if you don't boost bias current
it is just that reducing Miller Cdom to compensate may not have much performance trade off despite this change giving higher TIS (VAS) output Z
it is just that reducing Miller Cdom to compensate may not have much performance trade off despite this change giving higher TIS (VAS) output Z
Interestingly, no one jumped in, although when it comes to say "you are wrong" to CH, JC, TL and even NP, many here are quick in responding. And I'm sure they have enough knowledge to answer my question.....
I repeat, it doesn't interest me whether it's a BJT, JFET or MOSFET (even idealized), but can anyone here at least with a simple model / math, tell me how much loop gain at 20kHz and at 100kHz has such a device with a 50 Ohms source resistor compared to one with 5 ohms (other values are OK too, but they have to be quite different between each others)?
Btw, what happened with at lest two posts of Charles Hansen, they seem to disappear?
Are posts being deleted here?
JCX, I think that we know that, as we have been doing it for the last 40 years. However, I would like to make another point:
Doing a simple Google search on Black and negative feedback, I got a long interview of him telling 'all' about the history of negative feedback.
First of all, he made NO mention of cathode degeneration. I believe this was obvious and sometimes then used for linearizing individual stages of amps that he worked on since the early 1920's. Mostly in his spare time, he tried other approaches. He gave up attempting to directly 'linearize' a tube, and concluded it was impossible to do so. He then spent a great deal of time, years, with feedforward solutions to circuit utilization, but found it almost impossible to keep adjusted properly, without constant supervision.
Finally, he came up with the multistage negative feedback concept, and was allowed to develop it is relative secrecy, away from Nyquist and Bode, among others, who he presumed would try to take his idea from him, or at the very least, predict that it was impossible. However, he had a stable negative feedback amplifier running in 1928.
Of course, the patent development did not happen until 1932, and it was frought with problems from both the American Patent Office and The British Postoffice, but by the time of the patent being processed, they had something like 70 amplifiers in operation and they could prove that it worked, by demonstrating them to interested parties.
Like many of us, Black was self motivated, not too intimidated by PhD's, and was driven (you know, the 10,000 hour rule) to spend his lunch hours and weekends studying everything that Western Electric offered. He also had a personal challenge, to make trans-USA long distance telephone service, practical. It was NOT considered very important by many people he worked with, and he had to go straight to the top at Western Electric, to be allowed to work on his amplifiers. He was also rather possessive that he did it, himself, usually over the objections of others, if they got a chance to comment, and he was rather touchy about others, like Bode, and Nyquist, who apparently were later given more credit than really due, although their work was useful in general, as well.
If you want to see what a real engineer does to make a serious breakthrough in his career, I recommend that you study the development of negative feedback from the inventer, himself.
Doing a simple Google search on Black and negative feedback, I got a long interview of him telling 'all' about the history of negative feedback.
First of all, he made NO mention of cathode degeneration. I believe this was obvious and sometimes then used for linearizing individual stages of amps that he worked on since the early 1920's. Mostly in his spare time, he tried other approaches. He gave up attempting to directly 'linearize' a tube, and concluded it was impossible to do so. He then spent a great deal of time, years, with feedforward solutions to circuit utilization, but found it almost impossible to keep adjusted properly, without constant supervision.
Finally, he came up with the multistage negative feedback concept, and was allowed to develop it is relative secrecy, away from Nyquist and Bode, among others, who he presumed would try to take his idea from him, or at the very least, predict that it was impossible. However, he had a stable negative feedback amplifier running in 1928.
Of course, the patent development did not happen until 1932, and it was frought with problems from both the American Patent Office and The British Postoffice, but by the time of the patent being processed, they had something like 70 amplifiers in operation and they could prove that it worked, by demonstrating them to interested parties.
Like many of us, Black was self motivated, not too intimidated by PhD's, and was driven (you know, the 10,000 hour rule) to spend his lunch hours and weekends studying everything that Western Electric offered. He also had a personal challenge, to make trans-USA long distance telephone service, practical. It was NOT considered very important by many people he worked with, and he had to go straight to the top at Western Electric, to be allowed to work on his amplifiers. He was also rather possessive that he did it, himself, usually over the objections of others, if they got a chance to comment, and he was rather touchy about others, like Bode, and Nyquist, who apparently were later given more credit than really due, although their work was useful in general, as well.
If you want to see what a real engineer does to make a serious breakthrough in his career, I recommend that you study the development of negative feedback from the inventer, himself.
The usual feedback formula is expressed in terms of A (open loop gain) and b (or beta) the feedback. For degeneration, which samples output current, you can get an approximate result by using device gm for A, and degeneration resistor value for b. The result is the closed-loop gain expressed as a transconductance.sottomano said:
Any frequency lags within the device need to be included in A, and any external capacitance across the resistor needs to be included in b. Then you may need to apply a correction if the device has significant base/gate current.
Dave, would it be like:
Vout=Ve= Ie*Re
Ie=(Vin-Vout)*Gm
thus Vout= (Vin-Vout).(Gm*Re)
reworked to Vout/Vin=Acl=(Gm*Re)/(1+Gm*Re), for DC and ignoring base current, no load?
Which approaches '1' for Gm*Re >> 1.
Half way correct?
jan
Vout=Ve= Ie*Re
Ie=(Vin-Vout)*Gm
thus Vout= (Vin-Vout).(Gm*Re)
reworked to Vout/Vin=Acl=(Gm*Re)/(1+Gm*Re), for DC and ignoring base current, no load?
Which approaches '1' for Gm*Re >> 1.
Half way correct?
jan
Yes, you have arrived at the same point by essentially deriving the feedback equation for this particular case.
Simply putting A=gm and b=Re into CLG=A/(Ab+1) makes it much clearer that this is feedback.
Simply putting A=gm and b=Re into CLG=A/(Ab+1) makes it much clearer that this is feedback.
Ok thanks. But I still seem to not get it. Lets say gm=1000, Re=1 Ohm. So Acl is 1000/1001=0.999
Then I add a Re of 100. Gm=1000, Re=100, Acl=0.99999.
So adding more degeneration increases Acl? Ie less feedback? And what about loop gain?
Then I add a Re of 100. Gm=1000, Re=100, Acl=0.99999.
So adding more degeneration increases Acl? Ie less feedback? And what about loop gain?
More degeneration i.e. higher value of Re, increases feedback, so you get closer to '1'. In the limiting case with an open emitter, Acl = 1.
Edit: I believe Gm for a bipolar at 10mA is around 0.4 , which means for Re=10 ohms an Acl of 4/5, and for Re = 100 ohms, 40/41.
jan
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I was away for a certain time but is it true that best cancelation of distortion happens in BJT´s when Gm is reduced by half, this done by emitter degeneration ? Is this from maximum Gm ? That would make substantial bias necessary to begin with. Imagine 100mA in the input stage of a power amp. Maybe Dan D`Agostino would do that.
Thanks again. But....I know (knew...) that Loop Gain = Aol -Acl. Ie, more feedback=more loop gain, and with same open loop gain, less gain of the stage.
So here we have the case that loop gain is reduced with more feedback. This is very confusing!
Adding more degeneration reduces transconductance, but over a bigger resistor so voltage gain to the emitter increases slightly. You have to define carefully what you mean by output: voltage or current?sottomano said:
Best cancellation of third-order distortion occurs when Re=1/(2gm), so effective gm is reduced to 2/3rds. Different values would probably give cancellation of particular other odd-order distortions. There is not, as far as I know, a value which cancels all of them and you can't cancel any even-order distortions. The best you can do is have enough feedback to reduce all of them. Fairly easy with a small-signal BJT, and routinely used (e.g. LTP degeneration). I suppose you could use an LTP (which cancels even-order) combined with a little degeneration (to cancel third-order), but you would still have all the higher odds.Joachim Gerhard said:
Thanks DF. I looked up the BC550 for example and it has the most gain at 10mA so it is not that extreme. Still i am surprised on how little bias many designs starve the input transistors. OK, again feedback fixes that substantially.
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