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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:: ?. D6 m5 G/ ?* l$ e
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This section highlights the importance of an accurate pre-determined DC compact model.\cite{dc!!!}
: o7 o! I# f, x, c& K+ eFor compact model developers, extraction of the DC parameters always needs to be carried out before the extraction of AC ones. u; B$ c: ~& g. h
For devices measurement, DC measurement has to be carried out before AC measurement.& z$ y9 x2 ~. C
For modern IC designers, DC analysis needs to be carried out before AC analysis.\cite[4.6.4]{smith!}' n. u1 m5 R" h0 N: u9 V% b
For SPICE-like simulators, DC simulation also needs to be swept first and then follows by AC simulation.\cite{nlz!}
9 h D7 W! W# lFrom the device-level perspectives,* J$ X l4 ]9 K" E# T3 ^5 s
a set of equations which describe the terminal characteristics of DUT are written in a compact model.0 s% B" ?+ R, N% D0 d' Q z& G* ?
The equations are solved by SPICE-like simulators.+ C: o5 f# e, P' j( U6 V- \
The procedure of establishing equations, describing device's electrical characteristics, in a compact model
; X2 p2 o. a+ band have them solved by SPICE-like simulators is termed as: Device Characterization.
8 P2 @% `' e8 c5 x- P/ j; KThe fully Characterizations are treated by two independent steps:1 \/ z3 Z6 L1 C7 U; V
(1)DC Characterization or DC parameters extraction
, Z5 f( D2 N3 a3 `(2)AC Characterization or AC parameters extraction8 I) f# m+ o k' \3 H! v
Characterizing a nonlinear electronics component always begins with
1 K, {+ k% T6 s' yDC characterization and follows by the AC characterization.* n" j3 v0 C6 Y& f- ]
Because the AC model is origined from the linearization of the DC model at an indicated operating point.8 {* n; l* d6 Z6 x) \
Accordingly, accuracy is highly required in DC characterization because of its piority in the procedure of parameter extraction .' H- M+ @3 B9 i- @' Y* B5 R
, C+ X9 v) z: {" B4 O# V0 AFrom the circuit-level perspectives,,
+ w# S/ C' {8 _! B( v! {Circuit analysis refers to solving a circuit with KVL and KCL.
) |) J: ^+ K0 Q( g5 r6 s+ CTo be more specific, it means solving out nodes voltages and branches currents of each element in a circuit.9 x" P# T2 k) D, ]6 w K" t
As the source of stimulus can be systematically separated into DC sources and AC sources,
; N7 C# }: z3 U$ \; S- ~$ ]/ ^( E" u1 uthe unknown/solutions of the circuit are also separated into DC part and AC part.
8 n e6 _5 C' g% k) C( PAnalyzing a circuit is treated in two independent means A)DC analysis (B)AC analysis
' @* `/ E# V: k( _The separation between DC analysis and AC analysis greatly simplified a complex circuit.1 D- y8 a u% \" R. V" w
DC analysis is being carried out before AC analysis.2 z! t1 k/ ^& d9 ^9 A
DC analysis determines the Q-point, including each node voltage and branch current.
- D$ L. a. r& i# ]AC analysis gives the frequency response, including bandwidth and gain.3 B$ q( C7 E2 D1 h
Before performing AC Analysis, the DC operating point needed to be calculated from DC analysis first.- W. M3 l/ Z+ E8 p8 N i4 \
This is to construct a linear small-signal model for the nonlinear component.
( {3 o/ k* A. d# M1 V3 q2 KSo, the small signal (AC) response is highly dependent upon the presetting (DC) bias condition .
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- V+ f' H' E0 E1 }DC simulation in analog SPICE-like simulator, aims for computing the equilibrium points,
9 P4 ~' O9 g9 V1 D5 V6 a7 l' |which are the calculated DC node voltages and DC branch currents in a circuit.
" Z5 ?6 q( x/ `, K! SThey are the DC solutions of the DC equivalent equation/circuit.
) ?6 [5 `2 V U, U1 \# t8 {1 qA circuit will only reach its equilibrium if its stimulus is off
( g1 z" T9 l. J/ Land the independent sources are remain constantly employed.
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' R& u% F6 W: z6 O. FThere is an important reason why a given electronics circuit always
C& d j, O4 ]5 F2 a7 o* a# ineed to be reduced into a DC equivalent circuit (large-signal model)- ]) [+ W' X7 h
and followed by an AC equivalent one(small-signal model).
! h8 ]! x+ G5 V: r% j2 f6 [8 ]It has to do with the present of an active component in the circuit.& n( P( D- M& x5 r4 A4 }. K
The active component is a nonlinear element.So, it will have to be linearized.
+ ~! M2 l* f0 T6 E: FThe employment of active elements, like transistors, make the circuit a nonlinear algebra system., L) U$ C, A$ J, Z1 \8 ?
Normally, the nonlinear equation can only be solved by means of iterative methods,# S% W/ H! C0 z" z# Z( U' ~
such as Newton-Raphson algorithm \cite{nlz!}. This algorithm transforms the solutions R2 o& Q- X, G0 H5 O$ k$ a
of the nonlinear equation into a sequence of linear equation.5 f& A) H6 @$ `' P7 q' h- ~
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