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這是小弟寫的文章 編講義用的 歡迎高手同行找碴:0 j) s2 X' Q; Y- D' `! F; X/ w
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This section highlights the importance of an accurate pre-determined DC compact model.\cite{dc!!!}
) E5 b/ {# E! R7 Q2 n ]. y$ X# HFor compact model developers, extraction of the DC parameters always needs to be carried out before the extraction of AC ones.' Q5 d, H* [# Y8 s
For devices measurement, DC measurement has to be carried out before AC measurement.
) C" w# N1 U4 x1 [# q) oFor modern IC designers, DC analysis needs to be carried out before AC analysis.\cite[4.6.4]{smith!}3 ]4 O5 D9 G$ V" J6 E9 t
For SPICE-like simulators, DC simulation also needs to be swept first and then follows by AC simulation.\cite{nlz!}! G# k+ Y' C( N) `* K& W. y# t# o
From the device-level perspectives,
+ d9 ]% Y {) O, s+ M6 b; x& }$ _a set of equations which describe the terminal characteristics of DUT are written in a compact model.
8 h9 `$ h. k/ [. O2 B# `( [The equations are solved by SPICE-like simulators.
( B) w) [3 q; XThe procedure of establishing equations, describing device's electrical characteristics, in a compact model
% H$ ?1 Y' p7 n# ?9 n2 s, }4 N' S' nand have them solved by SPICE-like simulators is termed as: Device Characterization.4 F6 n" @( ]9 l
The fully Characterizations are treated by two independent steps:
( _1 M/ I+ X& A4 q(1)DC Characterization or DC parameters extraction
& N4 }# s7 U2 t1 a9 h* t(2)AC Characterization or AC parameters extraction
- |9 f* h6 [) Q0 m3 `5 I6 |7 m8 |" yCharacterizing a nonlinear electronics component always begins with8 F( W6 F$ [% B$ u" K3 {
DC characterization and follows by the AC characterization.
# d) T/ }# E9 F, s a# h: BBecause the AC model is origined from the linearization of the DC model at an indicated operating point.$ e! z) U2 T3 _6 k. ?! x. z
Accordingly, accuracy is highly required in DC characterization because of its piority in the procedure of parameter extraction .- t; n% H: D( C: c' F; ^% M; b" q
0 N' Q7 g% }/ V. s$ l* F/ qFrom the circuit-level perspectives,,
- v1 z; q* l% ?% d" uCircuit analysis refers to solving a circuit with KVL and KCL.
$ g4 N# C# W. U9 hTo be more specific, it means solving out nodes voltages and branches currents of each element in a circuit.6 x H( ^; [1 }2 y; s: \
As the source of stimulus can be systematically separated into DC sources and AC sources,
+ f9 d$ ?" E+ a4 P2 Lthe unknown/solutions of the circuit are also separated into DC part and AC part.
9 w* L( y5 v# R, K' h+ aAnalyzing a circuit is treated in two independent meansA)DC analysis (B)AC analysis
% ?5 p/ d9 u( FThe separation between DC analysis and AC analysis greatly simplified a complex circuit.7 o- F" y {' E+ C* B5 D
DC analysis is being carried out before AC analysis.
7 i2 e* x# J7 W9 O4 U; I* y2 DDC analysis determines the Q-point, including each node voltage and branch current.' j, T( d- }: p m0 b) g
AC analysis gives the frequency response, including bandwidth and gain.
( b3 p" L$ V- s1 T/ c6 mBefore performing AC Analysis, the DC operating point needed to be calculated from DC analysis first.9 \- j! C" E8 W2 W6 @2 b' T
This is to construct a linear small-signal model for the nonlinear component.( v( T! F+ m$ R
So, the small signal (AC) response is highly dependent upon the presetting (DC) bias condition .
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DC simulation in analog SPICE-like simulator, aims for computing the equilibrium points,
0 u9 \4 ^2 s! `, k) B( F8 k7 dwhich are the calculated DC node voltages and DC branch currents in a circuit.6 q6 K, k% D) x2 z8 @, D) c0 P5 n
They are the DC solutions of the DC equivalent equation/circuit.+ {* B, b( u7 m
A circuit will only reach its equilibrium if its stimulus is off
1 V: H" W1 g. D- T2 o& Wand the independent sources are remain constantly employed.6 F/ H M' [: U: o7 \/ y1 z% I
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There is an important reason why a given electronics circuit always0 @" ~# m! |8 h& i- L( m
need to be reduced into a DC equivalent circuit (large-signal model)$ Y3 D9 U S" x
and followed by an AC equivalent one(small-signal model).
/ D" ~8 r L1 K# TIt has to do with the present of an active component in the circuit.
# O* M x7 t! o& PThe active component is a nonlinear element.So, it will have to be linearized.
* Z* ~5 J$ M: _The employment of active elements, like transistors, make the circuit a nonlinear algebra system." x. u3 _; W% }4 V0 E
Normally, the nonlinear equation can only be solved by means of iterative methods,
5 M" P/ K$ v, ~; Isuch as Newton-Raphson algorithm \cite{nlz!}. This algorithm transforms the solutions) X$ [! W& m7 E; W
of the nonlinear equation into a sequence of linear equation.
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