Homework Solutions.pdf - ECE3030 Spring 18 Homework%1...

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ECE3030 Spring 2018 Homework 1 Due: January 23, Tuesday, 2018 at the beginning of the class. No late homework will be accepted. TypeGwritten solution is preferred Total points = 100 (2% of the final grade) Boltzman Constant (k B = 1.38 x 10 G23 J/K) 1. What are the basic physical components of a von Neumann machine? (10 point) 2. What are the fundamental requirements a physical medium must satisfy to be able to represent a bit? (10 point) 3. Consider the following energy barrier with E b =3000k B . (40 point) Boltzman Constant (k B = 1.38 x 10 G23 J/K) a. Compute the transition probability from S to D (p SGD )at temperature (T) of 300K. b. Computer the temperature at which p SGD =10 G4 . c. Compute the value of E d that will make p SGD =10p DGS at T=300K, where p DGS is the transition probability from D to S 4. Consider a circuit of 100 million (10 8 ) devices. The circuit failure probability at a temperature of 300K should be 0.01. (40 point) Boltzman Constant (k B = 1.38 x 10 G23 J/K) a. Compute the required minimum energy barrier height of the devices. b. Assuming switching energy is equal to the barrier height, and all devices switch once in every nanosecond (10 G9 ), compute the total power dissipation of the circuit.
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Solution 1. Central Processing Unit, Memory (or Storage), and Interconnect 2. Distinguishability and Conditional Change of State 3. Energy barrier and transition probability (a) ! "#$ = exp * + , - . = exp /000, - /00, - = exp −10 = 4.5×10 #7 (b) ) ! "#$ = exp * + , - . => ln ! "#$ = − * + , - . ; target p SGD = 10 G4 => ; = − < = ln ! "#$ > ? = − 3000> ? (−4ln 10 ) > ? = 3000 4 ln 10 = 325D (c) ! $#" = exp * + F* G , - . => H IJK H KJI = exp * + , - . + * + F* G , - . = exp * G , - . We want: H IJK H KJI = 10 ; therefore, we need < M = > ? ;NO 10 = 690> ? , at T=300K 4. Part (a) Circuit failure probability = p circuit Device failure probability = transition probability = p dev Number of devices = N dev ! RSTRUSV = 1 − 1 − ! MWX Y GZ[ => ln 1 − ! MWX = NO 1 − ! RSTRUSV \ MWX = ln 0.99 ×10 #] => 1 − ! MWX = exp NO 0.99 ×10 #] ! MWX = 1 − exp NO 0.99 ×10 #] = 10 #^0 = exp < = > ? ; < = = −> ? ;NO 10 #^0 = 10> ? ;NO 10 = 9533×10 #_/ J Part (b) `abcd = \ MWX < = ebfghℎfOj gfkc = 10 ] ×9533×10 #_/ 10 #l m = 9533×10 #n m = 9.5km
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ECE3030 Spring 2018 Homework 2 [Lecture slides 4 and 5] Due: Feb 1, Thursday, 2018 at the beginning of the class. No late homework will be accepted. TypeFwritten solution is preferred Total points = 100 (2% of the final grade) Boltzman Constant (k B = 1.38 x 10 F23 J/K) 1. Consider a nFtype silicon with N D = 10 22 1/m 3 ; Assume, N c = N v =10 25 1/m 3 , n i =10 16 1/m 3 , Boltzman Constant (k B = 1.38 x 10 F23 J/K), and Temperature = 300K [50 point] a. Draw the energy bandFdiagram showing the position of E C , E V , and E F , and compute the difference (in terms of k B T) between E F , E C, and E V , assuming E G =42k B T b. Compute the hole concentration in the device. 2. Consider the following MOSFET with pFtype source, nFtype channel, and pFtype drain. Assume doping (pFtype) density in source and drain are same N source =N drain =0.1N V . The doping (nFtype) density in the channel is N channel =10 F5 xN C . [50 point] a. Draw the energy band diagram showing the position of E C , E V , and E F in all regions.
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