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This chapter provides an intuitive understanding of semiconductor devices, covering their operation, basic equations, and models for manual analysis and SPICE simulation. It discusses the critical features of p-n junctions in integrated circuits and highlights the significance of energy bands, charge neutrality, and current flow in different regions of a diode. Furthermore, it delves into the challenges posed by sub-micron technology, avalanche breakdown, and the behavior of MOS transistors under various operational conditions, ultimately providing insight into future trends in digital circuit design.
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Digital Integrated CircuitsA Design Perspective Jan M. Rabaey Anantha Chandrakasan Borivoje Nikolic The Devices July 30, 2002
Goal of this chapter • Present intuitive understanding of device operation • Introduction of basic device equations • Introduction of models for manual analysis • Introduction of models for SPICE simulation • Analysis of secondary and deep-sub-micron effects • Future trends
B A Al SiO 2 p n Cross-section of pn -junction in an IC process A Al A p n B B One-dimensional representation diode symbol The Diode Mostly occurring as parasitic element in Digital ICs
What is the most important fact(feature) about semiconductor? • 왜 반도체인가? 전도도가 도체와 절연체의 중간이라서? • Ans; 자유전자와 hole 의 공존(에너지 대역 상에서 서로 다른 곳에) • 왜 free electron 과 hole 이 존재하게 되는가? • Fermi 준위가 Band Gap 을 통과하는 물질이 반도체이다. Special names were given to two neighboring E-bands; Others are simply nameless; Conduction band and Valence band as 1st lane and 2nd lane in the 2-lane highway system. • Bound electron has quantized (discrete) E-levels; Perturbation of neighboring atoms in infinitum would produce E-continuum; Periodic structure of crystal lattice yields some screening of the perturbing effects of indirect neighbors yielding E-bands.
@반도체 에서 Energy Gap 이 생기는 이유 • 수소 단원자(Single hydrogen atom) 모델; discrete energy levels • 두 개의 수소 원자 모델; Split Energy levels • N 개의 수소 원자 모델; N 개의 split levels • What happens if N becomes so large?; Energy band and then Energy continuum. • What happens if that is the case with single crystal?; Energy gap appears due to screening effects
Questions to ask now. 1. Why are Q- and Q+ of the same magnitude? 2. Why BV gets smaller with higher doping? 3. How are potential drop divided between n- and p-regions in the depletion region? 4. How does current flow in each region, depletion region, neutral region, and region in between? Depletion Region
@PN junction • Charge neutrality; Qp=Qn ->q*Np*Xp=q*Nn*Xn • Va=Vp+Vn • Express Vp, Vn with N, X. • Express BV as a function of Ecrit. • Explain why BV goes low with higher N(Np or Nn).
Forward Bias Lp ; Diffusion Length Forward bias is typically avoided in Digital ICs
Reverse Bias The Dominant Operation Mode
Junction Capacitance You can derive this either from C=Es/Xd or from C=dQ/dV Why does LGJ(Linearly-Graded Junction) show less steep Cj Vs. Vd behavior?
Small-signal vs. Large-signal Capacitance • Small-signal; C=Es/Xd, where Xd is a function of V(bias voltage) • Large-signal; verage capacitance is defined by Ceq=(Qj(V1)-Qj(V2))/(V1-V2)
Secondary Effects 0.1 ) A ( 0 D I –0.1 –25.0 –15.0 –5.0 0 5.0 V (V) D Avalanche Breakdown For high-doped PN diode, this can be Zener breakdown Due to BTBT(Band-to-Band Tunneling)
|V | GS A Switch! An MOS Transistor What is a Transistor?
The MOS Transistor Polysilicon Aluminum
MOS Transistors -Types and Symbols D D G G S S Depletion NMOS Enhancement NMOS D D G G B S S NMOS with PMOS Enhancement Bulk Contact
The Threshold Voltage Potential drop for surface inversion Why has NMOS appeared Later than CMOS, i.e., P-well CMOS? Ans; Positive Qss. PMOS, p-well CMOS, NMOS, N-well CMOS, twin-well CMOS
-4 x 10 6 VGS= 2.5 V 5 Resistive Saturation 4 VGS= 2.0 V Quadratic Relationship (A) 3 VDS = VGS - VT D I 2 VGS= 1.5 V 1 VGS= 1.0 V 0 0 0.5 1 1.5 2 2.5 V (V) DS Current-Voltage RelationsA good ol’ transistor
Pinch-off Transistor in Saturation
A model for manual analysis parabola
-4 x 10 2.5 VGS= 2.5 V Early Saturation 2 VGS= 2.0 V 1.5 Linear Relationship (A) D I VGS= 1.5 V 1 VGS= 1.0 V 0.5 0 0 0.5 1 1.5 2 2.5 V (V) DS Current-Voltage RelationsThe Deep-Submicron Era Mobility is inversely proportional to the vertical electric field, i.e., gate-source bias
5 u = 10 sat ) s / m ( n u x = 1.5 x (V/µm) c Velocity Saturation VDsat=Lvsat/mobility Constant velocity Constant mobility (slope = µ)
@Perspective Vdsat is Vds which launches Ecrit in the channel Explain why Vdsat is low with short-channel MOSFET. I D Long-channel device V = V GS DD Short-channel device V V - V V DSAT GS T DS
-4 x 10 -4 x 10 6 2.5 5 2 4 1.5 (A) 3 (A) D D I I 1 2 0.5 1 0 0 0 0.5 1 1.5 2 2.5 0 0.5 1 1.5 2 2.5 V (V) V (V) GS GS ID versus VGS linear quadratic quadratic Long Channel Short Channel
-4 -4 x 10 x 10 2.5 6 VGS= 2.5 V VGS= 2.5 V 5 2 Resistive Saturation VGS= 2.0 V 4 VGS= 2.0 V 1.5 (A) (A) 3 D D VDS = VGS - VT I I VGS= 1.5 V 1 2 VGS= 1.5 V VGS= 1.0 V 0.5 1 VGS= 1.0 V 0 0 0 0.5 1 1.5 2 2.5 0 0.5 1 1.5 2 2.5 V (V) V (V) DS DS ID versus VDS Long Channel Short Channel
G S D B @A unified modelfor manual analysis Vmin selects which of the three operating modes Come into play, linear, pinch-off,or velocity saturation. Channel length modulation Channel pinch-off velocity saturation
-4 x 10 2.5 VDS=VDSAT 2 VelocitySaturated 1.5 Linear 1 VDSAT=VGT 0.5 VDS=VGT Saturated 0 0 0.5 1 1.5 2 2.5 @Simple Model versus SPICE 모델과 식의 차이; Vsat onset 이 gradual 하기 때문 (A) D I (Channel pinch-off) V (V) DS
-4 x 10 0 -0.2 -0.4 (A) D I -0.6 -0.8 -1 -2.5 -2 -1.5 -1 -0.5 0 V (V) DS A PMOS Transistor VGS = -1.0V VGS = -1.5V VGS = -2.0V Assume all variables negative! VGS = -2.5V
Subthreshold Conduction • Subthreshold or weak inversion region • Diffusion current due to concentration gradient saturates with Vd of > 0.1 V. • Slope factor; S=n(kT/q)ln(10)
Polysilicongate Source Drain W x x + + n n d d Gate-bulk L d overlap Top view Gate oxide t ox + + n n L Cross section The Gate Capacitance
Gate Capacitance Cut-off Resistive Saturation Most important regions in digital design: saturation and cut-off
Gate Capacitance Accumulation region Inversion with source/drain electron Cgc=Cgcb+Cgcs+Cgcd Cgb=0 Capacitance as a function of the degree of saturation Capacitance as a function of VGS (with VDS = 0)
@Extracting the SPICE parameter by Measuring the Gate Capacitance Apply const I, and measure V(t) to obtain C.
Diffusion (Junction) Capacitance Channel-stop implant N 1 A Side wall Source W N D Bottom x Side wall j Channel L Substrate N S A