Talk:Asymptotic gain model

Page contents not supported in other languages.
From Wikipedia, the free encyclopedia

Revision[edit]

I made the first revision of this article, but I'm not very satisfied with it. The model was found in course material from Holland; firstly a cource given by Catena Microelectronics, then a "book" by Verhoeven and others. Both these originated fact-wise from a PhD thesis by Ernst Nordholt, who later wrote the Catena course material. I wrote the article to aid my own understanding of the subject, hoping that others would later fill in.

Things that need to be checked or clarified: Is the term asymptotic gain model used elsewhere, or was it invented by Nordholt? How does the terminology and model relate to the works of Black? How is the model useful for design?

The "book" by Verhoeven and others was later made into a real book (ISBN 1-40-207590-1), which is now for sale. I don't have it. Hopefully the dimension problems have been solved. --HelgeStenstrom 13:31, 24 Sep 2004 (UTC)

Rewrote the article[edit]

Sorry but I couldn't really salvage the previous article so I rewrote it. The equations and symbols are all different, but I'm sure they refer to the same thing. I based my article on "Feedback Amplifier Principles" by Rosenstark and various other IEEE papers, especially those by Middlebrook. There's still alot content that needs to be added (e.g. an article on return ratio and an example would be nice), so help would be appreciated. Roger 20:07, 14 May 2007 (UTC)[reply]

Is this statement necessary?[edit]

"Besides their usefulness for complicated circuits, numerical simulations are helpful in checking formulas. When a lot of algebra is involved, doubts arise that resulting formulas are error free. A way to check algebraic formulas is to put them into a spreadsheet and compare the results with a numerical simulation."

Good advice, but equally applicable to any circuit theorem. -Roger (talk) 19:40, 25 January 2008 (UTC)[reply]

Hi Roger; Maybe you can help me with this. The reason I put this here is that I found finding T, G_0 and G_00 was a real pain, due possibly to my propensity for algebraic error. On the other hand, I found using the small-signal circuits in SPICE to implement the the listed steps was awfully easy. So I found the errors in my algebra this way. The uniqueness of this example is probably the need to follow a sequence of steps - its not just y=x^2.

Brews ohare (talk) 19:47, 25 January 2008 (UTC)[reply]

Yep, the math does get messy, real fast. I've found the best way to check results is by doing a symbolic analysis with SapWin. That way you end up with an expression you can compare with your own and know for sure if yours is right. -Roger (talk) 20:39, 25 January 2008 (UTC)[reply]

Current vs. transR gain[edit]

I think my perspective was not understood. Finding the current gain first is the simple way to go, and then make connection with the transR gain later. There is no reason to make transR gain the primary choice. The short-circuit current gain allows determination of gain for any output variable choice. Brews ohare (talk) 23:31, 25 January 2008 (UTC)[reply]

I retained your improvements in phrasing.

Brews ohare (talk) 23:33, 25 January 2008 (UTC)[reply]

The short circuit current gain may simplify things, but it isn't accurate to define G_inf and G_0 to be current gains while the overall gain is called a transresistance gain. Plus the additional explanations necessary mostly complicate things. A better approach would be to make it a current amplifier (a more natural choice, but would require new figures). -Roger (talk) 23:41, 25 January 2008 (UTC)[reply]
I'm guessing you mean rewriting Example 1, the single transistor amplifier the same way? I don't think that is necessary, because for that amplifier voltage output is a natural. For the two-transistor case, that could be done too, by keeping the VF stage in the feedback loop, as discussed. I don't think the use of symbol G for gain suggests we are aiming for a transR gain, do you?

Brews ohare (talk) 00:19, 26 January 2008 (UTC)[reply]

Example 1 seems fine. All the gains are transresistance. In example 2 it is stated that the aim is to find the transresistance gain of the amplifier. Yet the direct and asymptotic gains are both found as current gains. All the gains should be of the same kind.
I understand what you did, but its technically incorrect and requires extra explanation and potential confusion for a reader. If you don't like the changes I made, then one suggestion is to state that the aim is to find the current gain (which would probably need new figures).
The section "Gain using alternative output variables" is probably unnecessary too since its just an exercise in applying Thevenin/Norton's theorem and is applicable to any amplifier. -Roger (talk) 00:29, 26 January 2008 (UTC)[reply]
Hope this is now put to rest. I have clarified the preambles. I don't see a need to re-do the figures. If you think changes are needed, please advise.Brews ohare (talk) 01:58, 26 January 2008 (UTC)[reply]

Brews ohare (talk) 01:26, 26 January 2008 (UTC)[reply]

It needs to be stated somewhere clearly what kind of gain we're interested in. The statement:
In Figure 5, the output variable is voltage across RL, which leads to the transresistance gain vL / iS. Another possibility is to take the output as the current in RL, which leads to the current gain of the amplifier with load RL, namely iL / iS. Another possibility is the output current βiB (the short-circuit load current), which leads to the short-circuit current gain of the amplifier, namely βiB / iS. The last choice leads simply to the other two, so initial focus is on finding this short-circuit current gain...
isn't necessary this early on. I think the best approach is to treat the amplifier as a current amplifier (and state so) and perhaps at the end mention that Thevenin/Norton's Theorem means that we can also treat it as a voltage amplifier. Also I don't think we need to elaborate on the loading effect. This is a common occurrence and doesn't need to be elaborated on for every amplifier. -Roger (talk) 02:11, 26 January 2008 (UTC)[reply]
Hi Roger:
See Asymptotic_gain_model#Examples. How does that look? There is no presumption that G is transR.
The next opportunity to raise the issue is Asymptotic_gain_model#Gain_G0_with_T_.3D_0, which you have objections to. Any suggestions?

Brews ohare (talk) 02:36, 26 January 2008 (UTC)[reply]

Hi Brews. This elaboration is unnecessary and potentially confusing:
In Figure 5, the output variable is voltage across RL, which leads to the transresistance gain vL / iS. Another possibility is to take the output as the current in RL, which leads to the current gain of the amplifier with load RL, namely iL / iS. Another possibility is the output current βiB (the short-circuit load current), which leads to the short-circuit current gain of the amplifier, namely βiB / iS. The last choice leads simply to the other two, so initial focus is on finding this short-circuit current gain:
Better to just call it a current amplifier and perhaps point to Thevenin/Norton for the equivalent gains. -Roger (talk) 03:10, 27 January 2008 (UTC)[reply]
Did current amplifier bit, changed figure; pointed to loading article, but left explicit application in - I think that is clearer than having the reader try to figure out how to use the other articles - Brews ohare (talk) 00:50, 28 January 2008 (UTC)[reply]

Middlebrook's General Feedback Theorem[edit]

The Asymptotic Gain Model is very closely related to the General Feedback Theorem (GFT) developed by R. David Middlebrook. A short introduction and a number of relevant links are available in the corresponding section of my webpage http://www.geocities.com/frank_wiedmann/loopgain.html.

Frank —Preceding unsigned comment added by 212.144.152.110 (talk) 19:01, 21 July 2008 (UTC)[reply]