
Home  Forums  Rules  Articles  diyAudio Store  Gallery  Wiki  Blogs  Register  Donations  FAQ  Calendar  Search  Today's Posts  Mark Forums Read  Search 
Solid State Talk all about solid state amplification. 

Please consider donating to help us continue to serve you.
Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving 

Thread Tools  Search this Thread 
2nd July 2012, 04:05 PM  #2691  
diyAudio Member
Join Date: Nov 2009
Location: algeria/france

Quote:
Quote:
Quote:
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 04:07 PM. 

2nd July 2012, 04:55 PM  #2692 
diyAudio Member

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
__________________
Whether we like to think of it this way or not, an audio engineer shares the professional goal of a magician  Richard Heyser Check out Linear Audio! Last edited by jan.didden; 2nd July 2012 at 05:15 PM. 
2nd July 2012, 05:50 PM  #2693  
diyAudio Member
Join Date: Feb 2003
Location: Jakarta

Quote:
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:' 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. 

2nd July 2012, 05:59 PM  #2694 
diyAudio Member

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
__________________
Whether we like to think of it this way or not, an audio engineer shares the professional goal of a magician  Richard Heyser Check out Linear Audio! 
2nd July 2012, 06:30 PM  #2695  
diyAudio Member
Join Date: Nov 2009
Location: algeria/france

Quote:
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. A Mort Subite or a Jenlain will do it perfectly... 

2nd July 2012, 06:38 PM  #2696  
diyAudio Member
Join Date: Feb 2003
Location: Jakarta

Quote:
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 2K75R) 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. 

2nd July 2012, 09:02 PM  #2697  
diyAudio Member
Join Date: Apr 2006
Location: Zagreb

Quote:
dado 

3rd July 2012, 12:29 AM  #2698  
Banned
Join Date: Jul 2011
Location: Portugal

Quote:
Quote:
I publishedit for educational purpose, to demonstrate the benefit of Current feedback. But this amp is playng music (fine, so fine) in my real system. I useit 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 neededit 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. Last edited by Esperado; 3rd July 2012 at 12:39 AM. 

3rd July 2012, 05:18 AM  #2699  
diyAudio Member
Join Date: Feb 2003
Location: Jakarta

Quote:
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. 

3rd July 2012, 05:27 AM  #2700  
diyAudio Member
Join Date: Feb 2003
Location: Jakarta

Actually I meant stability against oscillation, the hallmark of all high speed circuits.
Quote:
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 

Thread Tools  Search this Thread 


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Symetrical schematics are alike plague in Brazil, do you like them?  destroyer X  Solid State  151  1st July 2010 03:09 PM 
Symetrical out low Zout  karsten21  Tubes / Valves  5  1st February 2010 11:19 PM 
going balanced/symetrical  what benefits?  weissi  Solid State  15  20th October 2007 09:06 AM 
Symetrical field, is this?  Raka  MultiWay  6  14th September 2003 02:21 PM 
Nonsymetrical SMPS output  cm961  Parts  4  21st August 2003 11:27 PM 
New To Site?  Need Help? 