Download I. Inverters

MICROELECTRONICS LAB
(REPORT 2)
Your name
Your instructor’s name
Your class
The date
Grade:
/10
MOS Inverters Static and switching
characteristics
Comparing between the delay and power between different types of inverters
I.
II.
Inverters
A.
Enhancement load inverters
B.
Depletion load inverters
C.
CMOS inverters
Complex gate function
I.
Inverters
A. Enhancement load Inverter:
For the enhancement load inverter shown above you are required to:
1.
Write the net list for an enhancement load inverter with a W/L driver 8, W/L Load 1,
CL=1µF and a DC supply VDD=5V. (Minimum future length 0.8µ)
2.
The total power dissipation is equal to ………. (Verify from simulation)
3.
Plot the static power dissipation versus input voltage
4.
Draw the input and output waveforms versus time and extract from the graph tPHL and
tPLH.
5.
Plot the dynamic power dissipation versus time
6.
Apply one of the power reduction techniques, and explain in details how you were able to
reduce the power dissipation.
B. Depletion load Inverter:
For the depletion load inverter shown above you are required to:
1. Write the net list for the depletion load inverter with a W/L driver 8, W/L Load 1 and a DC
supply VDD=5V. (Minimum future length 0.8µ)
The total power dissipation is equal to ………. (Verify from simulation)
2.
3.
Plot the static power dissipation versus input voltage
4.
Draw the input and output waveforms versus time and extract from the graph tPHL
and tPLH.
5.
Plot the dynamic power dissipation versus time
6.
Apply one of the power reduction techniques, and explain in details how you were
able to reduce the power dissipation.
C. CMOS Inverter:
For the CMOS inverter shown above you are required to:
1. Design a Symmetric CMOS inverter where VDD=3.3V and CL=1µF
2.
The total power dissipation is equal to ………. (Verify from simulation)
3.
Plot the static power dissipation versus input voltage
4.
Draw the input and output waveforms versus time and extract from the graph tPHL and
tPLH.
5.
Plot the dynamic power dissipation versus time
6.
Apply one of the power reduction techniques, and explain in details how you were able to
reduce the power dissipation.
II.
Complex gate function
X=((A+B).C + D.E.F)(G.H+I))’
1. Show the schematic of the CMOS complex gate in terms of PMOS and NMOS and size
the transistors to have the worst case output resistance as 1µm/1µm, 2µm/1µm PMOS
inverter
2. Draw the input and output waveform versus time, and extract from the graph tPHL and
tPLH
3. Draw the dynamic power versus time (for worst case only)
APPENDIX
Resistive Load
.MODEL NMOS NMOS(LEVEL=3 TOX=1.0600E-08 VTO=0.393 UO=591.7)
Enhancement Load
.MODEL NMOS NMOS (LEVEL=3 VTO=1 GAMMA=0.38)
Depletion Load
.MODEL NMOSD NMOS (LEVEL=3 VTO=-2.8 GAMMA=0.38)
CMOS inverter
.MODEL NMOS NMOS
+TOX
LEVEL = 7
= 8E-9
+XJ
= 1E-7
+K1
= 0.4728294
+K3B
NCH
K2
= -10
= 1.830495E-5
DVT1W = 0
= 0.802594
DVT1
= 361.3464355
+UC
= 4.669186E-11 VSAT
= 0.2592162
UA
B0
+WR
+DWG
= 780.376869
=1
WINT
= 70.0489524
NLX
= 2.193565E-7
DVT2W = 0
DVT2 = -0.3
= -9.67751E-10 UB
= 1.898742E5
= 2.220067E-6
+KETA = -9.077245E-3 A1
+RDSW
= 0.4074928
K3
= 0.4276766
+U0
+AGS
VTH0
= 3.621074E-3
W0
+DVT0W = 0
+DVT0
= 2.2E17
=0
PRWG
= -1.034792E-8 DWB
A0
B1
A2
= 0.0713836
= 1.389179E-7
= 2.889157E-18
LINT
= 9.824117E-9
= 1.3381235
= 5E-6
= 0.3487525
PRWB
= -7.21866E-3
= 1.24319E-9
VOFF
= -0.0869274
+NFACTOR = 0.7366118
+CDSCD = 0
+ETAB
CIT
=0
CDSC
CDSCB = 0
= -7.988336E-3 DSUB
ETA0
= 0.0346215
= 0.3317286
PCLM
+PDIBLC1 = 4.161735E-3
PDIBLC2 = 1.19743E-5
+DROUT = 2.748338E-3
PSCBE1 = 7.42428E8
+PVAG
=0
DELTA = 0.01
+MOBMOD = 1
+KT1
= -0.11
+UA1
= 4.31E-9
+AT
UB1
WL
=0
=0
+LWN
LLN
+MJSW
= -5.6E-11
=1
=0
=0
CAPMOD = 2
= 2.58E-10
CGSO
= 1.012513E-3
CJSW
= 0.1518459
= 0.022
UC1
LW
=0
CJ
= 0.3510986
= -1.5
WWL
=1
CGDO
= 1E-12
UTE
WLN
=1
LWL
+XPART = 0.5
+MJ
=0
PSCBE2 = 1E-3
= 3.9
KT2
PB
= 2.58E-10
= 0.8
= 2.862666E-10 PBSW
CJSWG = 1.82E-10
+MJSWG = 0.1518459
CF
+PRDSW = -73.5674578
PK2
= 1.9546403
PDIBLCB = 0.1
RSH
= -7.61E-18
WWN
=1
+CGBO
=0
KT1L = 0
= 3.3E4
+WW
+LL
PRT
= 2.4E-4
=0
= 0.8
PBSWG = 0.8
PVTH0 = -0.0100437
= 3.087074E-3
WKETA = 3.003636E-3
+LKETA = 2.647195E-3
.MODEL PMOS PMOS
+TOX
LEVEL = 7
= 8E-9
+XJ
= 1E-7
+K1
= 0.3939412
K2
= 15.35112
W0
+K3B
NCH
+DVT0W = 0
+DVT0
= 1.4617205
= 8.52E16
= 0.0308482
= 1E-5
DVT1W = 0
DVT1
VTH0
K3
NLX
= -0.6831778
=0
= 1E-9
DVT2W = 0
= 0.3569339
DVT2 = -0.0368562
+U0
= 221.3795636
+UC
= 8.567678E-11 VSAT
+AGS
UA
= 0.389467
= 1.573901E-9
B0
+WR
= 4E3
=1
A0
= 1.999067
= 2.419633E-6
B1
= 5E-6
+DWG
= -2.578547E-8 DWB
+CDSCD = 0
+ETAB
+PDIBLC1 = 0
= 4.6592572
+KT1
= -0.11
+UA1
= 4.31E-9
+AT
= 3.3E4
+WW
=0
=0
+LWN
=1
+MJ
+MJSW
ETA0
= 0.4254421
PCLM
UB1
WL
LLN
= 1E-12
CJ
+MJSWG = 0.1676363
=0
CGSO
PB
= 2.91776E-10
=0
= 3.12E-10
= 0.8896731
PBSW
CJSWG = 4.42E-11
+PRDSW = -233.4720278 PK2
+LKETA = -0.0207051
=1
CAPMOD = 2
= 3.12E-10
CF
= -5.6E-11
=0
= 9.916255E-4
CJSW
= 0.1676363
LW
=0
CGDO
UC1
WWL
=1
LWL
= -1.5
WLN
=1
= 2.8
= 0.022
= -7.61E-18
WWN
= 0.392439
PSCBE2 = 5.04016E-10
KT2
=0
= 2.7235598
PDIBLCB = 4.528856E-3
UTE
KT1L = 0
= 2.4E-4
RSH
=0
= -0.1219424
= 0.0518465
PSCBE1 = 8E10
PRT
+XPART = 0.5
+CGBO
CDSC
DELTA = 0.01
+MOBMOD = 1
=0
VOFF
PDIBLC2 = 4.344554E-3
+DROUT = 5.604876E-3
+PVAG
LINT
=0
CDSCB = 0
DSUB
= 0.6320223
PRWB = 0.1688991
= 9.89001E-9
CIT
= 4.735705E-3
A2
= -0.2146377
= 1.569174E-7
+NFACTOR = 1.8063821
+LL
= 4.394989E-5
PRWG
WINT
= 5E-18
= 2E5
+KETA = -6.020293E-3 A1
+RDSW
UB
= 0.99
PBSWG = 0.99
PVTH0 = 0.0107102
= 1.861393E-3
WKETA = -6.345721E-3