EE 330 Lecture 15. Devices in Semiconductor Processes. Diodes Capacitors MOSFETs

Transcription

1 EE 330 Lecture 15 evices in Semiconuctor Processes ioes Capacitors MOSFETs

2 Review from Last Lecture Basic evices an evice Moels Resistor ioe Capacitor MOSFET BJT

3 Review from Last Lecture

4 Review from Last Lecture pn Junctions Physical Bounary Separating n-type an p-type regions f oping levels ientical, epletion region extens equally into n-type an p-type regions

5 Review from Last Lecture Analysis of Nonlinear Circuits (Circuits with one or more nonlinear evices) What analysis tools or methos can be use? KCL? KL? Superposition? Noal Analysis Mesh Analysis Two-Port Subcircuits oltage ivier? Current ivier? Thevenin an Norton Equivalent Circuits?

6 Review from Last Lecture ioe Moels (amps) ioe Characteristics (volts) (amps) ioe Characteristics (volts) ioe Characteristics (amps) (volts) Which moel shoul be use? The simplest moel that will give acceptable results in the analysis of a circuit

7 Review from Last Lecture ioe Equation ioe Moels t e 1 S (amps) ioe Characteristics (volts) ioe Characteristics if if (amps) (volts) ioe Characteristics Piecewise Linear Moels R if if (amps) (volts) 0 if 0 0 if 0

8 0 0 0 if if 0 ioe Moels ioe Equation 1 e t S Piecewise Linear Moels R When are the piecewise-linear moels aequate? When it oesn t make much ifference whether =0.6 or =0.7 is use When is the ieal PWL moel aequate? When it oesn t make much ifference whether =0 or =0.7 is use

9 aliate Moel Select Moel Example: etermine OUT for the following circuit 10K OUT 12 1 Solution: Strategy: 1. Assume PWL moel with =0.6, R =0 2. Guess state of ioe (ON) 3. Analyze circuit with moel 4. aliate state of guess in step 2 (verify the if conition in moel) 5. Assume PWL with = Guess state of ioe (ON) 7. Analyze circuit with moel 8. aliate state of guess in step 6 (verify the if conition in moel) 9. Show ifference between results using these two moels is small 10. f ifference is not small, must use a ifferent moel

10 Solution: 1. Assume PWL moel with =0.6, R =0 2. Guess state of ioe (ON) 10K OUT Analyze circuit with moel = 1. 14mA OUT 10K 4. aliate state of guess in step 2 To valiate state, must show >0 = OUT =1.14mA>0

11 Solution: 5. Assume PWL moel with =0.7, R =0 6. Guess state of ioe (ON) 10K OUT Analyze circuit with moel = 1. 13mA OUT 10K 8. aliate state of guess in step 6 To valiate state, must show >0 = OUT =1.13mA>0

12 Solution: 9. Show ifference between results using these two moels is small =1.14mA an =1.13 ma are close OUT OUT Thus, can conclue OUT 1.14mA

13 Example: etermine OUT for the following circuit 10K 0.8 OUT 1 Solution: Strategy: 1. Assume PWL moel with =0.6, R =0 2. Guess state of ioe (ON) 3. Analyze circuit with moel 4. aliate state of guess in step 2 5. Assume PWL with = Guess state of ioe (ON) 7. Analyze circuit with moel 8. aliate state of guess in step 6 9. Show ifference between results using these two moels is small 10. f ifference is not small, must use a ifferent moel

14 Solution: 1. Assume PWL moel with =0.6, R =0 2. Guess state of ioe (ON) 10K 0.8 OUT Analyze circuit with moel = OUT 10K 20 A 4. aliate state of guess in step 2 To valiate state, must show >0 = =20A>0 OUT

15 Solution: 5. Assume PWL moel with =0.7, R =0 6. Guess state of ioe (ON) 10K 0.8 OUT Analyze circuit with moel = OUT 10K 10 A 8. aliate state of guess in step 6 To valiate state, must show >0 = =10A>0 OUT

16 Solution: 9. Show ifference between results using these two moels is small =10A an =20A are not close OUT OUT 10. f ifference is not small, must use a ifferent moel Thus must use ioe equation to moel the evice OUT OUT 0.8- = 10K = e S t K OUT 0.6 Solve simultaneously, assume t =25m, S =1fA Solving these two equations by iteration, obtain = an OUT =18.60μA

17 Use of Piecewise Moels for Nonlinear evices when Analyzing Electronic Circuits Process: Observations: 1. Guess state of the evice 2. Analyze circuit 3. erify State 4. Repeat steps 1 to 3 if verification fails 5. erify moel (if necessary) o Analysis generally simplifie ramatically (particularly if piecewise moel is linear) o Approach applicable to wie variety of nonlinear evices o Close-form solutions give insight into performance of circuit o Usually much faster than solving the nonlinear circuit irectly o Wrong guesses in the state of the evice o not compromise solution (verification will fail) o Helps to guess right the first time o Moel is often not necessary with most nonlinear evices

18 Use of Piecewise Moels for Nonlinear evices when Analyzing Electronic Circuits Process: 1. Guess state of the evice 2. Analyze circuit 3. erify State 4. Repeat steps 1 to 3 if verification fails 5. erify moel (if necessary) What about nonlinear circuits (using piecewise moels) with time-varying inputs? 1K out 80sin500t 1 1K Same process except state verification (step 3) may inclue a range where solution is vali

19 Example: etermine OUT for N =80sin500t 1K 80sin500t N 1 out 1K Guess 1 ON (will use ieal ioe moel) 1K N 1 out 1K OUT = N =80sin(500t) ali for >0 Thus vali for N > 0 N 1K

20 Example: etermine OUT for N =80sin500t 1K 80sin500t N 1 out 1K Guess 1 OFF (will use ieal ioe moel) 1K N 1 out 1K OUT = N /2=40sin(500t) ali for <0 2 Thus vali for N < 0 N

21 Example: etermine OUT for N =80sin500t 1K 80sin500t N 1 out 1K Thus overall solution N 80 OUT 80sin 500t for N 0 40sin 500t forn 0 t OUT 80 t -40

22 Use of Piecewise Moels for Nonlinear evices when Analyzing Electronic Circuits Process: 1. Guess state of the evice 2. Analyze circuit 3. erify State 4. Repeat steps 1 to 3 if verification fails 5. erify moel (if necessary) What about circuits (using piecewise moels) with multiple nonlinear evices? 1K 20 out Guess state for each evice (multiple combinations possible) 4K

23 Example: Obtain OUT 1K 20 out K

24 Example: Obtain OUT 1K 20 out Guess 1 an 2 on 4K 1K 20 out OUT =-20 ali for 1 >0 an 2 > mA 0 4K mA 0 1K 4K Since valiates, solution is vali

25 Types of ioes pn junction ioes Signal or Rectifier Pin or Photo Light Emitting LE Laser ioe Metal-semiconuctor junction ioes Zener aractor or aricap Schottky Barrier

26 Basic evices an evice Moels Resistor ioe Capacitor MOSFET BJT

27 Capacitors Types Parallel Plate Fringe Junction

28 Parallel Plate Capacitors con 2 con 1 A 1 A 2 C insulator A = area of intersection of A 1 & A 2 One (top) plate intentionally C A size smaller to etermine C

29 Parallel Plate Capacitors f C Cap unit area where C C C ε A C ε A

30 Fringe Capacitors C C ε A A is the area where the two plates are parallel Only a single layer is neee to make fringe capacitors

31 Fringe Capacitors C

32 Junction Capacitor Capacitance C C 1 jo φ p C A A B n n for FB φ 2 φ B 0.6 n ; 0.5 B C epletion region Note: is voltage epenent -capacitance is voltage epenent -usually parasitic caps -varicaps or varactor ioes exploit voltage ep. of C

33 Capacitance Junction Capacitor C Cj0A C C 1 jo φ A B n for FB φ 2 φ B 0.6 n ; 0.5 B oltage epenence is substantial

34 Basic evices an evice Moels Resistor ioe Capacitor MOSFET BJT

35 n-channel MOSFET Poly n-active Gate oxie p-sub

36 n-channel MOSFET Source Gate L rain W L EFF Bulk

37 n-channel MOSFET Poly Gate oxie n-active p-sub epletion region (electrically inuce)

38 n-channel MOSFET Operation an Moel S GS BS B G Apply small GS ( S an BS assume to be small) epletion region electrically inuce in channel Terme cutoff region of operation =0 G =0 B =0

39 n-channel MOSFET Operation an Moel S GS BS B G ncrease GS ( S an BS assume to be small) epletion region in channel becomes larger =0 G =0 B =0

40 n-channel MOSFET Operation an Moel S GS BS B G =0 G =0 B =0 Moel in Cutoff Region

41 n-channel MOSFET Operation an Moel S Critical value of GS that creates inversion layer terme threshol voltage, T ) BS B GS G ( S an BS small) ncrease GS more nversion layer forms in channel nversion layer will support current flow from to S Channel behaves as thin-film resistor R CH = S G =0 B =0

42 Trioe Region of Operation G S R CH S GS B BS = 0 For S small R G CH L W W μcox L 0 B 1 GS T COX GS T S Behaves as a resistor between rain an source Moel in eep Trioe Region

43 Trioe Region of Operation G R CH GS B BS = 0 R CH For S small L W 1 GS T COX Resistor is controlle by the voltage GS Terme a oltage Controlle Resistor (CR)

44 n-channel MOSFET Operation an Moel S GS BS B G ( S an BS small) ncrease GS more nversion layer in channel thickens R CH will ecrease Terme ohmic or trioe region of operation R CH = S G =0 B =0

45 n-channel MOSFET Operation an Moel S GS BS B G ( BS small) ncrease S nversion layer thins near rain =? G =0 no longer linearly epenent upon S Still terme ohmic or trioe region of operation B =0

46 Trioe Region of Operation G S GS B BS = 0 R For S larger CH L W 1 GS T COX G μc B OX 0 W L GS T 2 S S Moel in Trioe Region

47 n-channel MOSFET Operation an Moel S GS BS B G ( BS small) ncrease S even more nversion layer isappears near rain Terme saturation region of operation Saturation first occurs when S = GS - T =? G =0 B =0

48 Saturation Region of Operation G S GS B For S at onset of saturation BS = 0 or or μc μc OX W L W L equivalently OX equivalently GS GS T T 2 S GS S 2 T GS T G μcoxw 2L 0 B GS T 2

49 n-channel MOSFET Operation an Moel S GS BS B G ( BS small) ncrease S even more (beyon GS - T ) Nothing much changes!! Terme saturation region of operation =? G =0 B =0

50 Saturation Region of Operation G S GS B BS = 0 For S in Saturation G μcoxw 2L 0 B GS T 2 Moel in Saturation Region

51 Moel Summary G S GS B 0 GS T W μc S OX GS T S GS S GS T L 2 T W 2 μcox GS T GS T S GS T = =0 G B 2L BS = 0 Cutoff Trioe Saturation This is a piecewise moel (not piecewise linear though) Note: This is the thir moel we have introuce for the MOSFET (eep trioe special case of trioe where S is small ) R CH L W 1 GS T COX

52 Moel Summary G GS S B BS W μc BS = 0 0 μc OX OX W L 2L = = 0 G B GS GS T T 2 2 S S GS GS GS T T T S S GS GS T T Observations about this moel (evelope for BS =0): = f, 1 GS S = f, G 2 GS S = f, B 3 GS S This is a nonlinear moel characterize by the functions f 1, f 2, an f 3 where we have assume that the port voltages GS an S are the inepenent variables an the rain currents are the epenent variables

53 En of Lecture 15