ECE 875: Electronic Devices Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University ayresv@msu.edu

Lecture 35, 07 Apr 14 Chp 06: MOSFETs Channel Current IDS (n-channel p-substrate) Full IDS, use equation (23): Linear Saturation Intermediate regimes M VM Ayres, ECE875, S14

ID Lec 34: in charge sheet and constant mobility approximation: Add VFB to (19): VM Ayres, ECE875, S14

Lec 34: Example: What regime? Answer: Not clear but not needed. VT is a requirement that is fixed by the materials properties of the semiconductor and the insulator (oxide) What is the HW question? VM Ayres, ECE875, S14

Lec 34: Example: What regime? Answer: Not clear but not needed. VT is a requirement that is fixed by the materials properties of the semiconductor and the insulator (oxide) What is the HW question? Answer: find the channel doping. VM Ayres, ECE875, S14

Behavior regimes Saturation: VD ≈ VG - VT Linear: VD < VG - VT VM Ayres, ECE875, S14

Linear Regime: VM Ayres, ECE875, S14

In initial pinch: Qn(y=L) =0 Saturation Regime: In initial pinch: Qn(y=L) =0 VM Ayres, ECE875, S14

Saturation Regime: in initial pinch: Qn(y=L) =0 Lec 33: VM Ayres, ECE875, S14

Saturation regime: Example: where is VDsat in eq’n (20)? VM Ayres, ECE875, S14

Saturation regime: Answer: VDsat = Dyi(y=L) VM Ayres, ECE875, S14

Saturation Regime: in initial pinch: Qn(y=L) =0 K is a function of doping concentration and oxide capacitance: thickness and dielectric choice VM Ayres, ECE875, S14

Saturation Regime: in initial pinch: Qn(y=L) =0 (also dielectric choice) VM Ayres, ECE875, S14

Saturation regime: New: Note the influence of M(NA, dox) on VDsat, transconductance gm and IDsat VM Ayres, ECE875, S14

Intermediate Regime: VM Ayres, ECE875, S14

VT: NA VM Ayres, ECE875, S14

VM Ayres, ECE875, S14

VM Ayres, ECE875, S14

Use Fig. 22, Chp. 04 to find “a” value for fms: VM Ayres, ECE875, S14 Which line matches Pr. 6.07?

Use Fig. 22, Chp. 04 to find “a” value for fms: VM Ayres, ECE875, S14 Answer: Given: n+-poly on p-Si.

Why “a” value for fms : since fms is also a function of NA Why “a” value for fms : since fms is also a function of NA. Therefore Pr. 6.07 has an iterative solution VM Ayres, ECE875, S14