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Vlsi Lab-3

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Introduction:

This experiment deals with the layout design of a CMOS Inverter using Microwind. CMOS Inverter: 25m (cmos02.rul) EDA Tool: Microwind 3.1 Full Version Transistor Sizing: NMOS- 4/3(WIDTH/LENGTH) PMOS- 13/3(WIDTH/LENGTH) MOS Model: Very Simple Level-1 Supply Voltage: 5.0V I/O Voltage: 5V

Objectives
To manually design the mask layout of a CMOS Inverter To check the design for design rule errors To check the functionality of the inverter using simulation with the built-in simulator To extract netlist from the inverter layout for SPICE and perform simulation

Physical Design (Layout) (Mask-Level Layout):

Layout Simulation:
Voltage vs.Time (Transient Analysis): Output voltage vs. Time & Input Voltage vs. Time:

Voltage vs. Voltage (DC Analysis): Output voltage vs. Input Voltage:

Current vs. Voltage (DC Analysis): Inverter Current vs. Inverter Input Voltage:
DURING PULL-DOWN, THE INVERTER CURRENT IS NMOS DRAIN CURRENT.

DURING PULL-UP, THE INVERTER CURRENT IS PMOS DRAIN CURRENT.

Extracted SPICE netlist:


* IC Technology: CMOS 90nm, 6 Metal Copper - strained SiGe - LowK * VDD 1 0 DC 5.00 Vinput 6 0 DC 0 PULSE(0.00 5.00 0.10N 0.01N 0.01N 0.10N 0.22N) * * List of nodes * "N2" corresponds to n2 * "output" corresponds to n3 * "input" corresponds to n6 * * MOS devices MN1 0 6 3 0 N1 W= 0.20U L= 0.15U MP1 1 6 3 2 P1 W= 0.70U L= 0.15U * C2 2 0 0.638fF C3 3 0 0.465fF C5 1 0 0.393fF C6 6 0 0.023fF * * * n-MOS Model 1 : * low leakage .MODEL N1 NMOS LEVEL=1 VTO=0.34 UO=500.000 TOX= 1.2E-9 +GAMMA=0.500 PHI=0.150 +CGSO=100.0p CGDO=100.0p +CGBO= 60.0p CJSW=240.0p * * p-MOS Model 1: * low leakage .MODEL P1 PMOS LEVEL=1 VTO=-0.32 UO=190.000 TOX= 1.2E-9 +GAMMA=0.400 PHI=0.150 +CGSO=100.0p CGDO=100.0p +CGBO= 60.0p CJSW=240.0p * * Transient analysis * * (Pspice) .TEMP 27.0 .TRAN 0.1N 0.50N .PROBE .END

Discussion:
A CMOS inverter is designed in microwind. The width of PMOS(Wp) is 13 and the width of the NMOS(Wn) is 3(which is less than 13/3). To make the the inverter symmetric we chose Wp>3Wn. After examining the graph we see that without the switching time there is no current in the inverter.

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