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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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9 S+ t2 J' ?0 t. U; [This section highlights the importance of an accurate pre-determined DC compact model.\cite{dc!!!}
$ }) x/ H! b& i$ T2 Y6 d2 IFor compact model developers, extraction of the DC parameters always needs to be carried out before the extraction of AC ones.: t7 Z" w8 M$ M
For devices measurement, DC measurement has to be carried out before AC measurement.
# x1 k" M" a" L5 v& [For modern IC designers, DC analysis needs to be carried out before AC analysis.\cite[4.6.4]{smith!}
J c3 k, w, I- o5 z( j1 e( gFor SPICE-like simulators, DC simulation also needs to be swept first and then follows by AC simulation.\cite{nlz!}4 G4 f/ F( ^2 ]' P; |+ d+ H
From the device-level perspectives,' A4 \. R* J& y6 U: I
a set of equations which describe the terminal characteristics of DUT are written in a compact model.4 Y7 W% V9 f% W6 S0 c: Y# O
The equations are solved by SPICE-like simulators.8 @" C" k7 w9 a( j! V; N5 H1 Z
The procedure of establishing equations, describing device's electrical characteristics, in a compact model
( A3 H! W- N; P7 n/ f8 l! Fand have them solved by SPICE-like simulators is termed as: Device Characterization.
" N' O3 A% H' z7 B8 l( j0 {+ CThe fully Characterizations are treated by two independent steps:* u* U: v7 [ Y3 g
(1)DC Characterization or DC parameters extraction2 f1 C* W0 v: x5 P: [( c C
(2)AC Characterization or AC parameters extraction$ U2 j; F- N, Z: [" B
Characterizing a nonlinear electronics component always begins with
1 ^' @: M; O% i! u& y dDC characterization and follows by the AC characterization.
5 e+ ]0 s O0 d, c4 e- ? F" sBecause the AC model is origined from the linearization of the DC model at an indicated operating point.
8 E/ [' N: N- X5 Z* H, \0 R9 w+ LAccordingly, accuracy is highly required in DC characterization because of its piority in the procedure of parameter extraction .8 z- P: D0 d; s6 l3 d; p" V
8 e8 c `1 }& b1 x; ]1 P' CFrom the circuit-level perspectives,,
6 }9 E2 Q4 | _, N3 LCircuit analysis refers to solving a circuit with KVL and KCL./ Q# ^: ~, r# g5 |7 t7 P) `9 \
To be more specific, it means solving out nodes voltages and branches currents of each element in a circuit.. A6 B) @8 R' A6 I
As the source of stimulus can be systematically separated into DC sources and AC sources,: M) X" S n9 r" o0 {, ?
the unknown/solutions of the circuit are also separated into DC part and AC part.2 J+ C5 ]$ Y. I" |8 ^
Analyzing a circuit is treated in two independent means A)DC analysis (B)AC analysis
- b; B: L4 g/ ~. A9 ZThe separation between DC analysis and AC analysis greatly simplified a complex circuit.
6 z) ~3 R2 R' aDC analysis is being carried out before AC analysis.
) t( q# P3 o( o# W; v$ [DC analysis determines the Q-point, including each node voltage and branch current./ N0 i! V) L1 P
AC analysis gives the frequency response, including bandwidth and gain.
5 j8 ]* n# Z3 f0 I9 nBefore performing AC Analysis, the DC operating point needed to be calculated from DC analysis first.9 Z% H- q9 u# E+ P$ j
This is to construct a linear small-signal model for the nonlinear component.
6 r+ n( {) v" wSo, the small signal (AC) response is highly dependent upon the presetting (DC) bias condition .
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5 C C- i. B+ \% r m# qDC simulation in analog SPICE-like simulator, aims for computing the equilibrium points,% |: r |4 F+ X0 x
which are the calculated DC node voltages and DC branch currents in a circuit./ E/ x; Z7 Q. t2 d7 v) {
They are the DC solutions of the DC equivalent equation/circuit.& z6 x! Q& u1 z$ L; ^
A circuit will only reach its equilibrium if its stimulus is off
' s$ S7 }3 I$ _- ~; {and the independent sources are remain constantly employed.* S3 G) G! [% M! O, n
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& e0 {4 `) }0 j: AThere is an important reason why a given electronics circuit always: ^) w' ?( ^8 J) U: m( z
need to be reduced into a DC equivalent circuit (large-signal model)4 x2 |8 D9 k4 |6 {/ p+ H" q+ D
and followed by an AC equivalent one(small-signal model).
8 G6 k2 L7 z/ t( \+ J6 _: d3 cIt has to do with the present of an active component in the circuit.3 e, A7 _3 R8 C' z |( p
The active component is a nonlinear element.So, it will have to be linearized.
4 l* k3 L' ~. R& n& M9 p/ X7 m! `The employment of active elements, like transistors, make the circuit a nonlinear algebra system.
& Q* g: y' [! U) U1 n5 s. g/ gNormally, the nonlinear equation can only be solved by means of iterative methods,
2 `% |; W G2 m6 C- }' ksuch as Newton-Raphson algorithm \cite{nlz!}. This algorithm transforms the solutions6 ^* \" p Y% o. [! C
of the nonlinear equation into a sequence of linear equation.
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