Model mosp pmos kp2e 5 vto 077 gamma05 lambda05

Info icon This preview shows pages 18–22. Sign up to view the full content.

View Full Document Right Arrow Icon
.MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5 +LAMBDA=.05 TOX=41.5N CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PROBE V(2) V(3) V(5) V(6) .END First inverter : t r = 6.5 ns , t f = 7 ns , t PLH = 2.6 ns , t PHL = 3.2 ns Fourth inverter : t r = 8.2 ns , t f = 9.3 ns , t PLH = 4.7 ns , t PHL = 5.5 ns t PHL = R onn C ln 4 V DD - V TN V DD - 1 + 2 V TN V DD - V TN t PHL = 10 - 12 2 1 50 x 10 - 6 ( 29 5 - 0.91 ( 29 ln 4 5 - 0.91 5 - 1 + 2 0.91 ( 29 5 - 0.91 = 3.1 ns t PLH = R onp C ln 4 V DD + V TP V DD - 1 - 2 V TP V DD + V TP t PLH = 10 - 12 5 1 20 x 10 - 6 ( 29 5 - 0.77 ( 29 ln 4 5 - 0.77 5 - 1 + 2 0.77 ( 29 5 - 0.77 = 2.9 ns The first inverter matches the equations well. The 2:1 relation between rise/fall time and propagation delay holds as well. The propagation delay of the interior inverters is a factor of 2 slower than predicted by the formula because of the slow rise and fall times of the signals. *PROBLEM 7.44(b) - FIVE CASCADED MINIMUM SIZE INVERTERS VDD 1 0 DC 5 VIN 2 0 PULSE (0 5 0 0.1N 0.1N 30N 70N) * MN1 3 2 0 0 MOSN W=4U L=2U AS=16P AD=16P MP1 3 2 1 1 MOSP W=4U L=2U AS=16P AD=16P C1 3 0 1P *AS=4UM*W - AD=4UM*W * MN2 4 3 0 0 MOSN W=4U L=2U AS=16P AD=16P MP2 4 3 1 1 MOSP W=4U L=2U AS=16P AD=16P C2 4 0 1P * MN3 5 4 0 0 MOSN W=4U L=2U AS=16P AD=16P
Image of page 18

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
7-231 MP3 5 4 1 1 MOSP W=4U L=2U AS=16P AD=16P C3 5 0 1P * MN4 6 5 0 0 MOSN W=4U L=2U AS=16P AD=16P MP4 6 5 1 1 MOSP W=4U L=2U AS=16P AD=16P C4 6 0 1P * MN5 7 6 0 0 MOSN W=4U L=2U AS=16P AD=16P MP5 7 6 1 1 MOSP W=4U L=2U AS=16P AD=16P C5 7 0 1P .OP .TRAN 0.025N 70N .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5 +LAMBDA=.05 TOX=41.5N +CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PROBE V(2) V(3) V(5) V(6) .END First inverter : t r = 16 ns , t f = 7 ns , t PLH = 6.5 ns , t PHL = 3.0 ns Fourth inverter : t r = 17.5 ns , t f = 11 ns , t PLH = 10 ns , t PHL = 7.5 ns t PHL = R onn C ln 4 V DD - V TN V DD - 1 + 2 V TN V DD - V TN t PHL = 10 - 12 2 1 50 x 10 - 6 ( 29 5 - 0.91 ( 29 ln 4 5 - 0.91 5 - 1 + 2 0.91 ( 29 5 - 0.91 = 3.1 ns t PLH = R onp C ln 4 V DD + V TP V DD - 1 - 2 V TP V DD + V TP t PLH = 10 - 12 2 1 20 x 10 - 6 ( 29 5 - 0.77 ( 29 ln 4 5 - 0.77 5 - 1 + 2 0.77 ( 29 5 - 0.77 = 7.3 ns The first inverter matches the equations well. The 2:1 relation between rise/fall time and propagation delay holds as well. The propagation delay of the interior inverter is a factor of 2 slower than predicted by the formula because of the slow rise and fall times of the signals.
Image of page 19
7-232 7.45 B C 2 1 D B C D V DD 20 1 20 1 20 1 20 1 2 1 2 1 2 1 v o b ( 29 NMOS: 4 1 | PMOS: 40 1 7.46 B C D A D C V DD B A v O a ( 29 NMOS: W L = 4 2 1  = 8 1 PMOS: W L = 5 1 b ( 29 NMOS: W L = 3 4 ( 29 2 1  = 24 1 PMOS: W L = 3 5 1 = 15 1
Image of page 20

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
7-233 7.47 7.48 B C B C V DD Z = A+B+C A NMOS: PMOS: 1 3.75 2 1 B C D B C D V DD A Z = A+B+C+D NMOS: PMOS: 1 5 2 1 7.49 7.50 B C A C B A V DD Z = ABC NMOS: PMOS: 2.4 1 2 1 B C D A D C B A V DD Z = ABCD NMOS: PMOS: 3.2 1 2 1 7.51 Output Z is A multiplied by B.
Image of page 21
Image of page 22
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern