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Lab 1 - 2 to 1 MUX

Date post: 11-Apr-2015
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Design of a 2 in, 1 out MUX using Cadence
7
A X1 X0 Y 0 0 0 0 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 0 1 1 1 1 I started with first writing down the Boolean equation that describes the 2-input mux on the gate level diagram. Using Demorgan's theorem, I can simplify the Boolean equation. The truth table is easily found by plugging in all possible values of A, X0, and X1 into the inputs and recording Y. 1 - Truth table and reduced Boolean equation. Lab1 Page 1
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Page 1: Lab 1 - 2 to 1 MUX

A X1 X0 Y

0 0 0 0

0 0 1 0

0 1 0 1

0 1 1 1

1 0 0 0

1 0 1 1

1 1 0 0

1 1 1 1

I started with first writing down the Boolean equation that describes the 2-input mux on the gate level

diagram.

Using Demorgan's theorem, I can simplify the Boolean equation.

The truth table is easily found by plugging in all possible values of A, X0, and X1 into the inputs and

recording Y.

1 - Truth table and reduced Boolean equation.

Lab1 Page 1

Page 2: Lab 1 - 2 to 1 MUX

I decided against using symbols because of the simplicity of the circuit. Any larger than this, and I'm sure it

would be wise to work at the gate level .

Each gate was made with CMOS transistor logic.

2 - Transistor Level Schematic (Virtuoso)

Lab1 Page 2

Page 3: Lab 1 - 2 to 1 MUX

Converting the netlist with schm2sim.pl:

>perl schm2sim.pl

stepsize 50analyzer A X1 X0 Yvector in A X1 X0set in 000sset in 001sset in 010sset in 011sset in 100sset in 101sset in 110sset in 111s

Finally, using IRSIM with the input script and the CMOS description file found on the TA page.

>irsim cmos.prm sch.sim -sim.cmd

The following irsim script was used to generate inputs:

cmd.bat

The logic generated correctly describes the 2-input mutiplexer.

3 - Logic Simulation (irsim)

Lab1 Page 3

Page 4: Lab 1 - 2 to 1 MUX

First step was to calculate the capacitive load for the final NAND gate. The specification call for the mux

driving 32 minimum size CMOS inverters, and a 100fF wiring cap.

Estimating the capacitive load of one CMOS inverter using only the gate capacitances for both the NMOS and

PMO S device :

Solving for final load by multiplying Cinv by 32 and adding the 100fF wiring cap:

Calculate the current required to switch the load in the allotted 600ps propagation delay.

4 - Hand Calculations

Lab1 Page 4

Page 5: Lab 1 - 2 to 1 MUX

tphl tplh tp pstat pdyn

Hand 600ps 600ps 600ps 0 700uW

Spice 598ps 650ps 598ps 720uW

5 - Table of Calculations

Lab1 Page 5

Page 6: Lab 1 - 2 to 1 MUX

After first simulating the circuit with the orginal calculated Wp and Wn, I realized that not taking into

account some of the parasitics caused me to underestimate the current required to charge the load.

By increasing my Wn to 660nm and my Wp to 2.22um for the last nand gate to take into account the

parasitics, I was able to meet the 600ps specification for rise time.

6 - Simulation Waveforms

Lab1 Page 6

Page 7: Lab 1 - 2 to 1 MUX

The layout design was completed successfully and passed the CRC and all design constraints.

7 - Layout Design

Lab1 Page 7


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