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Berkeley ELENG 130 - Lecture 40

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Lecture #40Velocity SaturationMOSFET I-V with Velocity SaturationDrain Saturation Voltage VDSsatEXAMPLE: Drain Saturation VoltageSlide 6IDSsat with Velocity SaturationShort- vs. Long-Channel MOSFETVelocity OvershootSummary: NMOSFET I-VPMOSFET I-V with Velocity SaturationEE130 Lecture 40, Slide 1Spring 2007Lecture #40OUTLINEThe MOSFET:•Velocity saturationReading: Chapter 19.1EE130 Lecture 40, Slide 2Spring 2007Velocity saturation limits IDSsat in modern MOSFETSSimple model:sat is the electric field at velocity saturation:Velocity Saturationsatsatvvv1satsatv2for   satfor  < satSiin holesfor cm/s 106Siin sonfor electr cm/s 10866satvEE130 Lecture 40, Slide 3Spring 2007MOSFET I-V with Velocity SaturationLVVVmVVCLWIsatDSDSDSTGSeffoxeDS12LVIIsatDSDSDS1 channellongIn the linear region:EE130 Lecture 40, Slide 4Spring 2007Drain Saturation Voltage VDSsat•If satL >> VGS-VT then the MOSFET is considered “long-channel”. This condition can be satisfied when– L is large, or –VGS is close to VTLVVmVsatTGSDSsat11satsatv2EE130 Lecture 40, Slide 5Spring 2007Question: For VGS = 1.8 V, find the VDSsat of an NFET with Toxe = 3 nm, VT = 0.25 V, and WT = 45 nm, if L = (a) 10 m, (b) 1 um, (c) 0.1 m, and (d) 0.05 mSolution: From VGS , VT, and Toxe , n is 200 cm2V-1s-1. sat= 2vsat / n = 8 104 V/cm m = 1 + 3Toxe/WT = 1.2EXAMPLE: Drain Saturation Voltage11LVVmVsatTGSDSsatEE130 Lecture 40, Slide 6Spring 2007(a) L = 10 m: VDSsat= (1/1.3V + 1/80V)-1 = 1.3 V(b) L = 1 m: VDSsat= (1/1.3V + 1/8V)-1 = 1.1 V(c) L = 0.1 m: VDSsat= (1/1.3V + 1/.8V)-1 = 0.5 V(d) L = 0.05 m: VDSsat= (1/1.3V + 1/.4V)-1 = 0.3 V11LVVmVsatTGSDSsatEE130 Lecture 40, Slide 7Spring 2007Substituting VDSsat for VDS in the linear-region IDS eq’n. gives:For very short channel length: • IDSsat is proportional to VGS–VT rather than (VGS – VT)2• IDSsat is not dependent on LIDSsat with Velocity Saturation LVVILVVVVCmLWIsatTGSDSsatsatTGSTGSeffoxeDSsat1 channellong122TGSsatVVL    TGSoxesatTGSeffoxesatDSsatVVCvmWVVCmWI 2EE130 Lecture 40, Slide 8Spring 2007Short-channel MOSFET:• IDsat is proportional to VGS-VTn rather than (VGS-VTn)2• VDsat is lower than for long-channel MOSFET• Channel-length modulation is apparent0 1 2 2.5Vds (V)0.00.10.20.30.4Ids (mA/m)L = 0.15 mVgs = 2.5VVgs = 2.0VVgs = 1.5VVgs = 1.0VVds (V)Ids (A/m)L = 2.0 mVgs = 2.5 VVgs = 2.0VVgs = 1.5 VVgs = 1.0V0.00.010.020.03(a)(b)Vt = 0.7 VVt = 0.4 V0 1 2 2.5Vds (V)0.00.10.20.30.4Ids (mA/m)L = 0.15 mVgs = 2.5VVgs = 2.0VVgs = 1.5VVgs = 1.0VVds (V)Ids (A/m)L = 2.0 mVgs = 2.5VVgs = 2.0VVgs = 1.5VVgs = 1.0V0.00.010.020.03(a)(b)Vt = 0.7 VVt = 0.4 VShort- vs. Long-Channel MOSFETEE130 Lecture 40, Slide 9Spring 2007Velocity Overshoot•When L is comparable to or less than the mean free path, some of the electrons travel through the channel without experiencing a single scattering event  projectile-like motion (“ballistic transport”) The average velocity of carriers exceeds vsate.g. 35% for L = 0.12 m NMOSFET Effectively, vsat and sat increase when L is very smallEE130 Lecture 40, Slide 10Spring 2007Summary: NMOSFET I-V•Linear region:•Saturation region:Siin sonfor electr cm/s 1086satvLVVVmVVCLWIsatDSDSDSTnGSneffoxeDS12,satsatv2 LVVVVCmLWIIsatTGSTGSneffoxeDSsatDS122,EE130 Lecture 40, Slide 11Spring 2007PMOSFET I-V with Velocity Saturation•Linear region:•Saturation region:LVVVmVVCLWIsatDSDSDSTpGSpeffoxeDS12,


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Berkeley ELENG 130 - Lecture 40

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