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Old 2nd July 2012, 03:05 PM   #2691
wahab is offline wahab  Algeria
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Quote:
Originally Posted by Esperado View Post
Wahab, please, play with my simulations here, Crescendo revisited

You will see the differences between CFB and VFB *on the same amp*.
VFB ---> NFB
Bandwich: 200Khz ---> 3Mhz
Slew rate: 220v/µs ---> 1200v/µs
hd: 0.0023% ---> 0. 0002%
IM: x ---> x /10

As you can see, and as said L.C, improvement is in all the domains (except DC stability). The major improvement is listening pleasure.
Hi , Esperado , i ll give it a try as soon as possible.

Quote:
Originally Posted by Bigun View Post
I'm afraid that according to my understanding (I don't claim to have a perfect understanding of anything in this world) most, if not all of this, is completely wrong and will be confusing.
There are things that are counter intuitive , mind you....

Quote:
Originally Posted by janneman View Post
So you can't say that the gain of one stage only decreases distortion of that stage.

jan
I said that it cancel distorsion of that stage as well as the one of following stages but not the distorsion of preceding stages...

Say an amplifier with two chained stages with gains of A and B respectively.

Let s call their caracteristic functions F(x) and G(x) respectively,
x being the input signal and a function of time.

The first stage caracteristic function can be summarized as :

F(x) = ax + d , a being the stage gain and d a function of x that is the stage distorsion.

Identicaly , the second stage caracteristic function will be :

G(x) = bx + e , b is the stage gain and e its distorsion.


Let H be the caracteristic of the chained stages , wich can be devellopped as :

H(x) = G(F(x)) , that is G as a function of F(x)

= G(ax + d)

= B(ax + d) + e

= ABx + Bd + e

Now we apply NFB in such a way that the amp has CLG of 1 , that is , we divide
the final caracteristic function by the factor AB wich is the total available gain.


This yield the CLG caracteristic function , let s call it K(x).

K(x) = H(x) / (AB)

= (ABx + Bd + e) / AB

= x + d/A + e/AB

As you can see , the first stage distorsion , d , is reduced only by the
loop gain provided by the first stage , while the second stage distorsion , e ,
is reduced by both the first and second stage loop gains.

Last edited by wahab; 2nd July 2012 at 03:07 PM.
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Old 2nd July 2012, 03:55 PM   #2692
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I don't think that is correct. If you have two functions in series F(x) and G(x), the total fransfer function is the product of the two, namely G(x)*F(x), which in your example should come out to (ax+d)*(bx+e), no?

Edit: I do think actually you are right; the error of the last function e is not processed by the first function (ax+d).
Wow! That is really interesting. I need to digest it furtehr.

jan
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Last edited by jan.didden; 2nd July 2012 at 04:15 PM.
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Old 2nd July 2012, 04:50 PM   #2693
Jay is offline Jay  Indonesia
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Quote:
Originally Posted by janneman View Post
Edit: I do think actually you are right; the error of the last function e is not processed by the first function (ax+d).
It is a very simple Mathematics, Jan. I thought you cared with the understanding/interpretation of the Math, not the Math itself.

The math presented by Wahab is correct (of course), but the interpretation of it can be right can be wrong. Grammatically I disagree with the logic:'

Quote:
Originally Posted by wahab View Post
= x + d/A + e/AB

As you can see , the first stage distorsion , d , is reduced only by the
loop gain provided by the first stage , while the second stage distorsion , e ,
is reduced by both the first and second stage loop gains.
The Mathematical expression does not say anything about a reduction of stage distortion, rather, it says that the final distortion is "dominantly" determined by the earlier (here the first) stage distortion. This is logical because first stage distortion will be "augmented" by the gains of stage 2, stage 3, stage 4 and so on.
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Old 2nd July 2012, 04:59 PM   #2694
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Yes I agree with you and of course Wahab is correct in that view - it is a different way of expressing it. We might have a langauge thing here.

Wahab is correct, as you made clear, that the first stage distortion is dominant. I took his statement too literally; of course the stage distortion is not modified as such.
But, to be honest, I didn't realise this dominance of the 1st stage so in that respect I learned something from this!
Thank you and thank you Wahab - you can have that beer now

jan
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Old 2nd July 2012, 05:30 PM   #2695
wahab is offline wahab  Algeria
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Quote:
Originally Posted by janneman View Post
I don't think that is correct. If you have two functions in series F(x) and G(x), the total fransfer function is the product of the two, namely G(x)*F(x), which in your example should come out to (ax+d)*(bx+e), no?

Edit: I do think actually you are right; the error of the last function e is not processed by the first function (ax+d).
Wow! That is really interesting. I need to digest it furtehr.

jan
Hi , Jan

That s not a transfer function but a caracteristic function
wich is in time domain.

The Laplace transform of a caracteristic function is a transfer function ,
that is , it is mapped in a complex frequency domain.

I myself need to look further as it s not a rigorous demonstration
mathematicaly speaking , since i deliberatly simplified the things
by downplaying non dominant terms.

Indeed , IIRC , some people that were in this forum already pointed
what is explained in theses lines.


Quote:
Originally Posted by janneman View Post
Thank you and thank you Wahab - you can have that beer now

jan
A Mort Subite or a Jenlain will do it perfectly...
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Old 2nd July 2012, 05:38 PM   #2696
Jay is offline Jay  Indonesia
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Quote:
Originally Posted by dadod View Post
so maybe thia amp works better with 4.7 V zener instead of 3.9 V.
dado
With the tendency for the output to have positive DC, and with both FB paths have equal 1K resistance, it is the lower input transistor that will have the tendency to get too low Vbe.

There are many ways to increase this Vbe. Easy way is to increase the current thru the 1K resistors (set by the 3K3 of the CCS).

Using higher zener voltage will have the same effect to the CCS (i.e. to increase current).

Increasing the FB resistors (e.g doubling the resistance with 2K||75R) will also solve the issue.

But all of these will put the circuit in worse stability. This is probably the reason why this topology is not popular, despite its good sound character.
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Old 2nd July 2012, 08:02 PM   #2697
dadod is offline dadod  Croatia
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Quote:
Originally Posted by Jay View Post
With the tendency for the output to have positive DC, and with both FB paths have equal 1K resistance, it is the lower input transistor that will have the tendency to get too low Vbe.

There are many ways to increase this Vbe. Easy way is to increase the current thru the 1K resistors (set by the 3K3 of the CCS).

Using higher zener voltage will have the same effect to the CCS (i.e. to increase current).

Increasing the FB resistors (e.g doubling the resistance with 2K||75R) will also solve the issue.

But all of these will put the circuit in worse stability. This is probably the reason why this topology is not popular, despite its good sound character.
Yes, that is correct, but I've got to high distortion 0.3% at 1 kHz and 0.5% at 20 kHz and very bad FFT, when 3.9 V was used. I don't get it, some people like high distortion or what. By the way, with 4.7 V phase marhin I've got is 47 degree. Am I doing something wrong with this simulatio??
dado
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Old 2nd July 2012, 11:29 PM   #2698
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Quote:
Originally Posted by Jay View Post
So how do you define your minimum stability (requirement) in simulation?
For Dc ? i don't care it in simul, but in real life.
Quote:
Originally Posted by Jay View Post
Is it simulated performance? What is the real number, not simulated numbers? I believe that it is very difficult to achieve simulated performance without careful and skillful fine tuning because from the nature of this "current" feedback it is easy that the input transistor will have too low Vbe.
I saw that you have trimmed the current source (lower ccs has 3K32, upper ccs has 3K3) and also the base voltage of VAS transistor with 30 Ohm above zener. Could you give a hint how do you do the trimming process?
Yes this distortion is in simul, i don't have powerful enough distortiometer to measure so low levels in real life. In real life, i use crossed fingers when power on, then my ears and an oscilloscope. ;-)
I published-it for educational purpose, to demonstrate the benefit of Current feedback.

But this amp is playng music (fine, so fine) in my real system. I use-it to power a JBL motor in a spherical waves horn made in plain wood, between 1KHz and 20 KHz
The resistance you are talking about is just a fine tuning for 0Vdc at the output. I needed-it because my protection circuit is tuned to be very 'aware' of DC and HF. The real values are not exactly the same on both channels.

I am waiting L.C propose his parts (if i can afford them) to compare on a musical point of view.
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Last edited by Esperado; 2nd July 2012 at 11:39 PM.
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Old 3rd July 2012, 04:18 AM   #2699
Jay is offline Jay  Indonesia
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Quote:
Originally Posted by dadod View Post
I don't get it, some people like high distortion or what... Am I doing something wrong with this simulatio??
Hehehe nobody likes high distortion especially when it is third order. The only objection I have with this circuit is the relatively higher 3rd order distortion as compared to 2nd order. I would prefer higher overall distortion but from 2nd order.

You didn't do something wrong I believe. Here is a trick:

The FB resistor (1K) is connecting the input transistor's emitor and the output.

You want the input transistor to have Vbe of at least 0.6v (approximation). Base voltage (Vb) is assumed zero and so is output voltage (Vout). So to get 0.6v at the emitters (Ve) you want a certain current thru the FB resistors (This is what the CCS is for).

V = 0.6V
R = 1K
I = calculate! ---> 0.6mA.

CCS current == current thru FB resistor + current thru transistor

Note that the current thru the transistor is also affected by the collector resistor (1K here, which is a good value). This current is around 0.5mA. So more or less you want the CCS to give out 0.6mA+0.5mA = 1.1mA.

You may want to set both CCS to output 1.2mA. You will find that the higher the current, the lower the distortion. That's why you got ("too") low distortion when you use higher zener voltage which affect the base voltage of the CCS, giving very high current I believe.

But I suspected that it would be better (e.g. for the stability) if top CCS gives different current than the bottom CCS. With positive voltage at the output you want the top CCS to have slightly higher current. You can try 1.2mA for the top and 1.1mA for the bottom CCS.

That's just my guesswork, you will have to troubleshoot the correct setting after you build it (if you intent to). In Christophe's circuit, top CCS has 3K3 and bottom CCS has 3K32. This means that the top CCS has slightly more current (I guess). But unfortunately the base voltage of the ccs also determines the current. I guess that you need to work out the 3K3 resistors on the CCS and the 30R resistor at the base of the VAS transistor.

You owe to yourself to build any of the SSA amp. Better is the more complex one.
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Old 3rd July 2012, 04:27 AM   #2700
Jay is offline Jay  Indonesia
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Quote:
Originally Posted by Esperado View Post
For Dc ? i don't care it in simul, but in real life.
Actually I meant stability against oscillation, the hallmark of all high speed circuits.

Quote:
Originally Posted by Esperado View Post
The resistance you are talking about is just a fine tuning for 0Vdc at the output. I needed-it because my protection circuit is tuned to be very 'aware' of DC and HF.
Ah, so it will save the speaker when there is oscillation.

Quote:
Originally Posted by Esperado View Post
I am waiting L.C propose his parts (if i can afford them) to compare on a musical point of view.
Unfortunately he didn't use lateral output. I guess the CSA will use the BIGBT output. But hopefully your amp will outperform the CSA so I don't need to build the CSA
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