|
這是小弟寫的文章 編講義用的 歡迎高手同行找碴:
1 A% j1 F: B' ~& l( Y4 K: A9 {0 a. i! I* H4 J! i3 ?4 j$ D# ]. }
This section highlights the importance of an accurate pre-determined DC compact model.\cite{dc!!!}+ d: p q9 I: s0 Z
For compact model developers, extraction of the DC parameters always needs to be carried out before the extraction of AC ones.
1 r% s& [2 u) U1 S. O% fFor devices measurement, DC measurement has to be carried out before AC measurement.
: u3 \ R( S; U& qFor modern IC designers, DC analysis needs to be carried out before AC analysis.\cite[4.6.4]{smith!}
1 b1 t# ]3 J! RFor SPICE-like simulators, DC simulation also needs to be swept first and then follows by AC simulation.\cite{nlz!}0 H; C: Y) i* S! g
From the device-level perspectives,' h; Q' T: B2 V6 D* }( m5 N
a set of equations which describe the terminal characteristics of DUT are written in a compact model.
1 r5 | T( Q# u$ ?4 y: f! W/ H& OThe equations are solved by SPICE-like simulators. `( P7 C" q7 D% S
The procedure of establishing equations, describing device's electrical characteristics, in a compact model
- ?( g( k/ S. R8 o/ h! }and have them solved by SPICE-like simulators is termed as: Device Characterization.8 x# V8 P# \# l$ O1 n8 J
The fully Characterizations are treated by two independent steps:
5 C. n# \, \% h0 g) r9 s% d(1)DC Characterization or DC parameters extraction1 s; g8 C9 w1 N2 p) h
(2)AC Characterization or AC parameters extraction* ^- x& B- z% e1 F
Characterizing a nonlinear electronics component always begins with
0 g3 H- N" t9 V: |) _! K9 L) QDC characterization and follows by the AC characterization.2 t3 r2 _8 G g3 F; W) u
Because the AC model is origined from the linearization of the DC model at an indicated operating point.+ r0 g4 s6 w4 B! c @
Accordingly, accuracy is highly required in DC characterization because of its piority in the procedure of parameter extraction .
T" {! h) L/ r, D! [# q0 _+ |6 p
1 E' I( q. _$ D# y2 w: uFrom the circuit-level perspectives,,- L4 T6 T5 O5 F" o5 O
Circuit analysis refers to solving a circuit with KVL and KCL./ x; z0 W+ U( B7 p6 j' A
To be more specific, it means solving out nodes voltages and branches currents of each element in a circuit.4 P" J. t" `- [2 `0 z* ?
As the source of stimulus can be systematically separated into DC sources and AC sources,* H. w+ @7 U5 M) A+ X. {
the unknown/solutions of the circuit are also separated into DC part and AC part.( N, a0 l: Q$ t j
Analyzing a circuit is treated in two independent meansA)DC analysis (B)AC analysis
- t2 w o: E, Z" h, VThe separation between DC analysis and AC analysis greatly simplified a complex circuit.9 p7 K4 b, z, p1 Y! k
DC analysis is being carried out before AC analysis.; I" m" b1 q/ J+ ] {
DC analysis determines the Q-point, including each node voltage and branch current.# t! Y$ `$ x( r$ K+ L. O0 D
AC analysis gives the frequency response, including bandwidth and gain.
9 J) a) n" |" c- u: gBefore performing AC Analysis, the DC operating point needed to be calculated from DC analysis first." b# c5 ~! ~3 l
This is to construct a linear small-signal model for the nonlinear component.
6 o6 d3 R7 D1 c0 K5 ?7 j' d/ uSo, the small signal (AC) response is highly dependent upon the presetting (DC) bias condition .5 x9 u& Q& u, L# ~, V4 P* b- f
# B+ j, h' {7 ^! W, lDC simulation in analog SPICE-like simulator, aims for computing the equilibrium points,& l- j4 k' F* I ]
which are the calculated DC node voltages and DC branch currents in a circuit.: T0 i. L& _* S r, p+ I1 H
They are the DC solutions of the DC equivalent equation/circuit.% R. E- X* [, H+ q8 @: r% C
A circuit will only reach its equilibrium if its stimulus is off
7 k; l: b) T! Dand the independent sources are remain constantly employed.+ z% ^9 v7 b$ B. J! n9 b* y) g9 [
4 a' m+ A3 O( j+ Q' B% G: j
3 ]+ Y5 X* S1 ]# ]- x3 M- ^: i
There is an important reason why a given electronics circuit always
, ]; `# i/ o8 \! x- ^need to be reduced into a DC equivalent circuit (large-signal model)! u2 r! _2 p! k! ]( @2 y( n1 s6 s" V
and followed by an AC equivalent one(small-signal model).$ X; i; Z( f. I! ]- l* Z
It has to do with the present of an active component in the circuit.- d) ~. B" f* H9 d3 @
The active component is a nonlinear element.So, it will have to be linearized.. L; M+ ~1 m; ^: }1 g
The employment of active elements, like transistors, make the circuit a nonlinear algebra system.
3 B& I& X! D1 ^& xNormally, the nonlinear equation can only be solved by means of iterative methods,) A6 b! Q- N; R) y0 a8 V3 @% L: R" x
such as Newton-Raphson algorithm \cite{nlz!}. This algorithm transforms the solutions
$ C9 a M! g' Kof the nonlinear equation into a sequence of linear equation.
! `) |( z# K+ S1 N6 p+ ^9 J m: V5 d; {$ ]: y& a3 E
* q: R+ m6 m! N" Z5 j
3 i% e+ z1 Y) ~5 Y, E! _( M
( m9 F9 L V; ~! C- ^
|
|