1. ECE 875: Electronic Devices
Prof. Virginia Ayres
Electrical & Computer Engineering
Michigan State University

2. 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
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3. ID Lec 34: in charge sheet and constant mobility approximation:
Add VFB to (19):
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4. 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?
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5. 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.
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6. Behavior regimes Saturation: VD ≈ VG - VT Linear: VD < VG - VT
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7. Linear Regime:
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8. In initial pinch: Qn(y=L) =0
Saturation Regime:
In initial pinch: Qn(y=L) =0
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9. Saturation Regime: in initial pinch: Qn(y=L) =0
Lec 33:
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10. Saturation regime: Example: where is VDsat in eq’n (20)?
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11. Saturation regime:
Answer:
VDsat = Dyi(y=L)
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12. Saturation Regime: in initial pinch: Qn(y=L) =0
K is a function of doping concentration and oxide capacitance: thickness and dielectric choice
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13. Saturation Regime: in initial pinch: Qn(y=L) =0
(also dielectric choice)
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14. Saturation regime: New:
Note the influence of M(NA, dox) on VDsat, transconductance gm and IDsat
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15. Intermediate Regime:
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16. VT: NA
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17. VM Ayres, ECE875, S14

18. VM Ayres, ECE875, S14

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

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

21. 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 has an iterative solution
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