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[問題求助] 關於大信號和小信號的差別

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21#
發表於 2007-9-25 23:25:05 | 只看該作者
我的認知
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小訊號分析  就是假設 訊號 在 固定增益 下做線性的訊號放大...
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實作上 這是太理想的理論5 x2 ^: G& R; Q* O) ~$ `8 T
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即使 維持在 saturate 工作  增益也不會是固定的...
22#
發表於 2007-9-28 16:37:17 | 只看該作者
我想gimayon你講的沒錯 就算同樣在saturation區域 小訊號分析也是會差很多
- A" N' N1 O' B* R& Q5 Z因為MOS的偏壓點都不一樣 增益也是會變動的
" O0 a3 Q  d9 U. C! n我補充一下 也就是因為這樣 在實作上 我們是不會把OP用在開迴路上面的5 X3 p+ G8 o; A) k
除了比較器外 幾乎所有的類比電路都會把OP接成負回授來使用的
" S) h$ R& @; K( M這樣就算OP的gain變化很大 只要gain夠大 對閉迴路增益而言變化還是不大/ @1 {# P/ `! b
實作上 在跑模擬時我們最後注重的是跑transient 就是在輸入端輸入一個step電壓 8 w8 X+ U. {: I/ c8 u' s5 R/ `
然後看output訊號的transient響應如何
/ R4 ~1 E7 K! E2 N/ i當然小訊號推導的公式是沒辦法反應出transient響應的 畢竟小訊號只是針對某一偏壓點才成立
9 f: z, i! A9 M) ]而step電壓應該算是一個大信號吧 即使所有的MOS都是在saturation
5 v$ L8 o/ M# ?0 Q' D而且一個是開迴路的觀念 一個是閉迴路的觀念
. z7 G# B( z) q3 c6 c" {我想你應該是要表達這觀點吧
% J! b  O" p; K/ E1 t" u) n不過話說回來 我覺得小訊號分析還是很重要啦 因為至少它可以定性的告訴我們每一顆MOS參數的影響
! n+ R6 F" }3 l7 M在DC gain error的分析也是一樣要靠小訊號 因為閉迴路穩態誤差是跟開迴路增益有關的
8 y8 s$ i0 S% G, m8 ]當然同樣電路作久了 要怎麼調MOS也都知道囉!

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23#
發表於 2007-9-30 20:11:17 | 只看該作者
版主功力遠勝那些幹 n 年 讀死書的  "資深工程師"
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佩服佩服!!
24#
發表於 2007-10-1 10:08:20 | 只看該作者
別這樣講啦 只是個人一些的經驗提出來交流而已
7 O) a4 J7 Y3 p  H) r比起一些真正的高手 自己功力還差得遠勒!
: l8 Q" X" Z! _" W0 M5 C6 M/ s, a不過希望大家藉這個版 彼此互相討論心得 這樣大家都能進步!
" j3 l) I5 ]2 ^( y2 @- I2 U  ^: a0 y節省自己摸索 嘗試錯誤的時間 藉由討論把類比設計的觀念弄得更懂!
25#
發表於 2007-10-9 01:32:25 | 只看該作者
其實我建議你因該拋除大信號和小信號之間的差別
- b) T! m/ X$ ], r% E& D8 _對我來說一般都是先了解電路的大訊號特性(例如I-V曲線)
! S& E) Y" v4 Q( {/ G9 t4 y! F然後才決定我的直流操作點 然後在該點上的斜率就是我得小訊號特性了$ |. p. ?$ B3 l% i5 A$ [- f/ X" A
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實際我再設計時還會分析一下(例如I-V曲線)再直流操作點附近是否線性
. [3 _: H! k5 r4 u# q如果非線性的話  則會造成小訊號在動作時的非線性  也就是一般所謂的THD
26#
發表於 2008-7-24 20:48:29 | 只看該作者
看了此貼,收獲頗多,向大家學習中。
27#
發表於 2016-12-4 20:24:04 | 只看該作者
這是小弟寫的看一看您應該會比較清楚) b. t" X3 b7 K" Y$ F- _
DC versus AC& U5 `+ m5 {! Z+ m+ H: a, f6 I$ M
=======================
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<電路上為何要先設置直流 才看交流>
This section highlights the importance of an accurate pre-determined DC compact model.\cite{dc!!!}
For compact model developers, extraction of the DC parameters always needs to be carried out before the extraction of AC ones.For devices measurement, DC measurement has to be carried out before AC measurement.
, J- R: _: s5 K1 C2 XFor modern IC designers, DC analysis needs to be carried out before AC analysis.\cite[4.6.4]{smith!} For SPICE-like simulators, DC simulation also needs to be swept first and then follows by AC simulation.\cite{nlz!}In other words, before performing all other simulations such as Transient, S-parameter, AC, DC simulation stands as the initial first step.
From the device-level perspectives,
& B& v3 t4 P" T$ w7 h, `  @) Da set of nonlinear differential equations which describes the terminal characteristics of DUT are written in a compact model.The equations are solved by SPICE-like simulators.
% D' {$ ?4 z9 m* HThe procedure of establishing equations, describing device's electrical characteristics, in a compact model and have them solved by SPICE-like simulators is termed as: Device Characterization. The fully Characterizations are treated by two independent steps: (1)DC Characterization or DC parameter extraction(2)AC Characterization or AC parameter extraction.Characterizing a nonlinear electronics component always begins with DC characterization and follows by the AC characterization.Because the AC model is originated from linearization of the DC model at an indicated operating point.
; k/ V/ l6 {7 N. e+ O( hAccordingly, accuracy is highly required in DC characterization because of its priority in the procedure of parameter extraction.
From the circuit-level perspectives,Circuit analysis refers to solving a circuit with KVL and KCL.To be more specific, it means solving out nodes voltages and branches currents of each element in a circuit.As the source of stimulus can be systematically separated into DC sources and AC sources,
( f. ]8 D, @- w8 b! eThe unknown/solutions of the circuit are also separated into DC part and AC part. Analyzing a circuit is treated in two independent meansA)DC analysis (B)AC analysis
$ \& t0 @; V+ vThe separation between DC analysis and AC analysis greatly simplified a complex circuit.DC analysis is being carried out before AC analysis.DC analysis determines the Q-point, including each node voltage and branch current. % B2 j/ W# o  D- i' b! I
AC analysis gives the frequency response, including bandwidth and gain.Before performing AC Analysis, the DC operating point needed to be calculated from DC analysis first.
1 ]6 A! g2 k/ j0 K& W1 V5 f  \This is to construct a linear small-signal model for the nonlinear component. So, the small signal (AC) response is highly dependent upon the preset (DC) bias condition.
" f; C6 S: W: e' [1 M. nDC simulation in analog SPICE-like simulator, aims for computing the equilibrium points, which are the calculated DC node voltages and DC branch currents in a circuit.3 e) {4 T! a0 J: I3 v6 i
They are the DC solutions of the DC equivalent equation/circuit. A circuit will only reach its equilibrium if its stimulus is off and the independent sources are remained constantly employed.
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28#
發表於 2016-12-4 20:24:16 | 只看該作者
There is an important reason why a given electronic circuit always needs to be reduced into a DC equivalent circuit (large-signal model) and followed by an AC equivalent one (small-signal model). It has to do with the presence of an active component in the circuit. The active component is a nonlinear element. So, it will have to be linearized. The employment of active elements, like transistors, make the circuit a nonlinear algebra system. Normally, the nonlinear equation can only be solved by means of iterative methods, such as Newton-Raphson algorithm \cite{nlz!}. This algorithm transforms the solutions of the nonlinear equation into a sequence of linear equation.
# Q/ [8 k1 D$ w2 i5 @9 g6 BLikely, in linear electronic circuits, the superposition principle skillful separates all the electrical characteristics, voltages and currents, into the DC part and the AC part. The AC signals are superimposed on the DC signals.The two independent quantities, AC quantities and DC quantities, can be determined by means of two independent means in circuit analyses. Firstly, DC analysis is carried out in order to determine DC bias points, which are the DC quantities $V_{E}$ $V_{B}$ $V_{C}$ $I_{E}$ $I_{B}$ $I_( N6 x  `4 C8 L: x
{C}$.The DC analysis are performed when the original circuit is simplified into the equivalent DC circuit. Where all the passive elements are remained in the circuit, but all the reactive elements such as: capacitors and inductors are removed.Secondly, the AC analysis is performed on the equivalent AC circuit, which is the so called: the small-signal equivalent circuit.The equivalent AC circuit is executed by removing all the DC sources and replacing transistors with the small-signal models.
29#
發表於 2017-1-22 06:58:25 | 只看該作者
本帖最後由 陳璟賢@FB 於 2017-1-22 07:38 AM 編輯
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gimayon 發表於 2007-8-28 01:38 AM
: O8 a) ?" w7 ~: _5 U畢業了就把小訊號仍了吧
$ p. B) v1 s  e7 D: V9 v實作上  沒有無限小這種鬼東西!!
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小訊號是指整個訊號的一段線性區間9 g! V, H( \3 r/ u0 K/ C
看的出來你扔了
# {/ K! t; r5 O) f9 P1 P, V連看個datasheet都有困難8 d/ B& K0 M3 p" ?. K. D2 _
Split-pi topology就能解決的事也在問
+ G, S+ {- h. V: G5 t1 h就不要在這邊誤導人了
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30#
發表於 2017-1-22 07:13:51 | 只看該作者
本帖最後由 陳璟賢@FB 於 2017-1-22 07:49 AM 編輯 5 ?5 w8 d  {! p. @8 ~, O, J
gimayon 發表於 2007-9-25 10:41 PM) z$ q" N% `1 P$ T. c
單單以訊號能否使 mos 工作在 saturate 判斷 是否為 小訊號 有失公允
( ?2 d6 w) W- E以模擬 OP AC GAIN 為例% Q) [& S8 _( C6 e
兩端輸入各為 ...

. d' j2 E9 h+ D! v# t  c1 Y/ X6 D首先你是講雙輸出還是單輸出?1 _3 @5 T, Q" O& O1 \  o9 ]
雙輸出要固定輸出直流準位不然浮接spice怎麼算gain?8 X% n+ L5 b7 X) H) q
2mv 3mv 可以差到40db? 我保證你有電晶體掉到線性區甚至cut off
+ U/ K/ i; _) J( X/ ~9 x你不會模擬就說小訊號錯?
9 r* b; e* z1 i) i你的模擬檔 或 電晶體尺寸要不要PO上來給人笑一下阿
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31#
發表於 2017-1-24 02:56:08 | 只看該作者
本帖最後由 陳璟賢@FB 於 2017-1-24 03:35 AM 編輯 & l- N: e  ^6 b- S- o; J* R
gimayon 發表於 2007-9-25 11:25 PM
, S8 d( z$ M2 W) p5 k6 B我的認知
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小訊號分析  就是假設 訊號 在 固定增益 下做線性的訊號放大...

5 d) l  w* v4 [5 Z- t1 _2 b小訊號模型從來就沒有要表達說gm不會變
9 I: I( A$ X7 N4 s它是幫助分析出哪些電晶體的gm、ro貢獻增益
/ j  e6 l( S6 }哪幾個寄生電容跟電阻形成極點、零點0 ?! y+ t$ w8 g9 T% r+ e: B) j# o
隨便用個CS放大器去掃VGS對gm的變化(VGS當成輸入)( e, D; W- N3 k
就可以知道不同VGS(Vt~VDD),gm會差到幾倍!& z4 S, j( ]6 H0 Q  L8 c/ x
幾分鐘就能驗證的事情你在這話唬爛
& O1 R6 |5 _4 k還是說你連這都不會模擬?- |2 d3 X. s, g! s
輸入變化2mv~3mv會差到40db 100倍?
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OP gain會差到40db 要麻你輸出偏壓點變了
. i$ ^+ I- Y. P  t% r, T8 y+ H你知道偏壓點變了 相當於Rout在變嗎?6 Z, k: ]: m$ Q
或者你電晶體根本沒有操作在飽和區
* U$ u9 J9 M( s# u連模擬電路的基本工都沒有. e$ {$ {* @( ?' y& p2 j# H
牽拖gm變化差到40db!再來講小訊號模型可以丟掉' `; x& I* `) o& p( r
自己觀念錯還敢嗆人讀死書
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我看你在EETOP論壇 2014年了,都還在問大學專題程度的問題8 x% B( w1 V! A/ c( L% F% j
或者直接伸手討答案、討code! 給你做再多年也是毫無長進$ R9 w6 r7 v8 @( H5 I
不要出去屁說自己工程師幹幾年!因為你根本連解決問題,發現問題的能力都沒有4 S  j" G% e; Q0 U1 ?

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. N+ G9 M- {1 c+ s" ?" n補充內容 (2017-1-24 03:39 AM):
- N& L- ]: w7 ?: M% L' Z: J' v基本功
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