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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:
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This section highlights the importance of an accurate pre-determined DC compact model.\cite{dc!!!}
( \1 n n: `# f0 ~& S! |9 S: xFor compact model developers, extraction of the DC parameters always needs to be carried out before the extraction of AC ones.
|8 ^/ h, q! k, s1 W+ rFor devices measurement, DC measurement has to be carried out before AC measurement.! n2 t9 o+ ~* u9 [7 l
For modern IC designers, DC analysis needs to be carried out before AC analysis.\cite[4.6.4]{smith!}
$ ]1 ^/ T0 y4 N% w# WFor SPICE-like simulators, DC simulation also needs to be swept first and then follows by AC simulation.\cite{nlz!}
, \ o( w% K1 G% `& U( S' U' E! KFrom the device-level perspectives,/ j1 [6 ]. O$ `' p* y( E
a set of equations which describe the terminal characteristics of DUT are written in a compact model.' @+ g7 J9 ~5 g7 @! k. O; u. d* R9 q
The equations are solved by SPICE-like simulators.
7 j# Z4 b: p3 D' iThe procedure of establishing equations, describing device's electrical characteristics, in a compact model
M m0 U4 ]6 b% F* Z, V- o2 land have them solved by SPICE-like simulators is termed as: Device Characterization.
( ^! l7 f8 g* vThe fully Characterizations are treated by two independent steps:
# q, W w+ e5 X5 e" a/ K) o(1)DC Characterization or DC parameters extraction% {6 \6 j4 ^9 m V
(2)AC Characterization or AC parameters extraction& z# o4 X! k' M1 y5 ]1 v
Characterizing a nonlinear electronics component always begins with
5 k0 z! K3 @% A% r- y* fDC characterization and follows by the AC characterization.
# b% @& q. x; p1 Z3 t0 _2 M, l; F3 kBecause the AC model is origined from the linearization of the DC model at an indicated operating point.4 |. ~% n* a' N& a- P1 Q
Accordingly, accuracy is highly required in DC characterization because of its piority in the procedure of parameter extraction .0 ^1 t6 @9 H6 d/ ?' y1 o
4 t8 z. h8 V5 H* x# VFrom the circuit-level perspectives,,
7 b5 }, @4 t4 H" B1 P/ O* M! G3 g9 {Circuit analysis refers to solving a circuit with KVL and KCL.
" y7 Z6 c: r/ S w! o" ITo be more specific, it means solving out nodes voltages and branches currents of each element in a circuit.! T I" L1 s: p& R4 Q- x" H
As the source of stimulus can be systematically separated into DC sources and AC sources,- ]+ x6 S6 C( ?1 Q! c
the unknown/solutions of the circuit are also separated into DC part and AC part.
, A' L; ]' s; }$ F/ }8 TAnalyzing a circuit is treated in two independent meansA)DC analysis (B)AC analysis
* V' B$ x" l5 B* q) ?3 rThe separation between DC analysis and AC analysis greatly simplified a complex circuit.
5 D0 g/ R+ n7 g. T& d- [) qDC analysis is being carried out before AC analysis.! }& d( Y5 w* [6 x5 C
DC analysis determines the Q-point, including each node voltage and branch current.
$ Y; k1 S1 V+ T& a0 E. k1 @AC analysis gives the frequency response, including bandwidth and gain.
) G3 ^8 s9 ?4 H- A; w$ S# I& ABefore performing AC Analysis, the DC operating point needed to be calculated from DC analysis first.+ `7 y# x1 @- t, K0 w
This is to construct a linear small-signal model for the nonlinear component.
$ T& q) Y/ g. o+ cSo, 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,
@1 _6 B0 c; k; B/ v+ a2 |2 jwhich are the calculated DC node voltages and DC branch currents in a circuit.
) ~5 I6 V5 T: G7 LThey are the DC solutions of the DC equivalent equation/circuit.
9 e8 j& ~7 s- r: {; lA circuit will only reach its equilibrium if its stimulus is off
0 |/ E1 z+ \; n4 Y; v3 iand the independent sources are remain constantly employed.
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/ f n; ~* e7 Z0 iThere is an important reason why a given electronics circuit always n: v( k4 m/ B4 @" Q( a
need to be reduced into a DC equivalent circuit (large-signal model)
( F& J! \- J% }+ }and followed by an AC equivalent one(small-signal model).
0 b$ p' | t$ Z6 Q' w" PIt has to do with the present of an active component in the circuit.5 W- Q! _/ s" B" f+ f
The active component is a nonlinear element.So, it will have to be linearized.
, \0 [" |& n" |, i9 @% _' v+ NThe employment of active elements, like transistors, make the circuit a nonlinear algebra system.6 D' e$ L, p/ g Z" S `) B( Z
Normally, the nonlinear equation can only be solved by means of iterative methods,
! j; z. a8 b$ k0 m( x# T; \1 m Nsuch as Newton-Raphson algorithm \cite{nlz!}. This algorithm transforms the solutions" l9 e$ X# w4 a. K* v( o6 f1 K
of the nonlinear equation into a sequence of linear equation.
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