Electrical Engineering - Finfet

The following will be notes taken on the subject of FinFets

• Summary
• Classical Mosfet
• Transistor that operates by using an Electrical field to invert the channel allowing conduction from source to drain

• The Classical Mosfet Dimensions
• Include length the short distance in this picture, Width is the longer dimension
• Functionality
• Length reduction results in loss of control over the channel
• performance reduction

• Finfet
• Goal
• solve performance reduction caused by smaller sizes 
• Wrap the gate electrode around the channel

• Thin fin of sillicon acts as the channel and it is encased by the gate electrode
• Silicon fin surrounded by extension implant and poly oxide
• Premise
• di-electric and metal gate allows for a stronger e-field to be formed as it increases the dimensions of the gate effective using the height/thickness of the fin as the channel length allowing for a decrease in overall size when compared to a typical Mosfet

• Source and Drain can be wrapped in silicon germanium or silicon carbon stressors just like classical transistors
• Shown below is a representation of a finfet design used for production

• The gate electrode is uniform for ease of construction

• Clarification of terminology
• After seeing the design used for production, we can see that it is nearly identical to the construction of the tri-gate

• The only difference is that a tri-gate includes multiple sources and drains whereas the finfet description shows only a single source and drain but the process of construction seems identical, and terminology could be used interchangeably assuming you're fine with annoying Intel
• The original papers indicate that the original process of the finfet did not wrap around on top of the fin, lowering the overall z or height of the overall transistor but this is too complicated of a process for mass production so the standard finfet is equivalent to the tri-gate which does wrap around 

• Benefits
• Maintain performance which includes
• Conductivity when turned on
• Insulation when turned off
• Finfet in relation to a standard mosfet reduces electron tunneling effect when insulation is small
• Main issues are - weaker dielectric, small size
• Finfet significantly increases volume of dielectric reducing leakage currents
• increase switching speed
• due to lower size due to gate capacitance being smaller
• note - non-issue in modern constructions, interface delay between metals is more significant as we get to smaller scale
• Lower's Voltage requirements
• increases lifetime/efficiency of product per charge
• Company Promises
• estimation, 2-5% higher price in exchange for 37% speed increase and 90% reduction in leakage current

• Complications
•  Adds complexity to construction process
• possible reduction in yield
• can increase price significantly above 2-5% estimate for the company overall
• Depending on process
• high k-dielectrics are more expensive, but already required as transistor size decreases
• Increased modeling difficulty
• geomotries now are first order effects
• random fluctuations in manufacturing can now cause deviations in result
• impacts possible econcomic forecasts for the division
• Verification
• requires new software to ensure design rules for fin to fin spacing is followed

• History
• construction process has roots in 1990s

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Notes - Square Roots Mod n

The following are notes from Introduction to Cryptography with Coding Theory.

• Suppose we are told that x² ≡ 71 (mod 77) has a solution
• How do we find one solution/all solutions?
• Proposition
• if y has a square root mod p, then teh square roots of y mod p are +-x.
• If y has no square root mod p, then -y has a square root mod p, and the square roots of -y are +-x
• Example
• Lets find the square roots of 5 mod 11. Since (P+1)/4 = 3, we compute x ≡ 5³ ≡ 4 (mod 11). Since 4² ≡ 5 (mod 11), the square roots of 5 mod 11 are +-4 or 5,7 mod 11
• Let's try to find the square root of 2 mod 11. Since (p+1)/4 = 3, we compute 2³ ≡ 8 (mod 11). But 8² ≡ 9 ≡ -2 (mod 11), so we have found a square root of -2 rather than 2
• Example
• Composite Modulus square roots
• x² ≡ 71 (mod 77)
• x² ≡ 71 ≡ 1 (mod 7) and x² ≡ 71 ≡ 5 (mod 11)
• x ≡ +-1 (mod 7) and x ≡ +-4 (mod 11)
• The chinese remainder theorem tells us that a congruence mod 7 and a congruence mod 11 can be recombined into a congruence mod 77
• m₁ = 7, m₂ = 11, m = m₁m₂ = 77
• M₁ = m/m₁= 11, M₂ = m/m₂= 7
• M₁N₁ ≡ 1 (mod m₁₎ => 11 N₁ ≡ 1 (mod 7) => 4 N₁ ≡ 1 (mod 7) => N₁ ≡ 2 (mod 7)
• this will get us results of x ≡ +-15,+-29 (mod 77) 

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Notes - Primitive Roots

The following are notes from Introduction to Cryptography with Coding Theory.

• Consider the powers of 3 (mod 7)
• 3¹≡ 3, 3²≡ 2, 3³≡ 6, 3⁴≡ 4, 3⁵≡ 5, 3⁶≡ 1
• Note that we obtain all nonzero congruence classes mod 7 as powers of 3
• Generally when p is a prime, a primitive root mod p is a number whose powers yield every nonzero class mod p
• Primitive Roots
• Let g be a primitive root for the prime p
• Let n be an integer. Then gⁿ≡ 1 (mod p) if and only if n ≡ 0 (mod p - 1)
• If j and k are integers, then g^j ≡ g^j (mod p) if and only if j ≡ k (mod p - 1)

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Notes - Fermat's Little Theorem and Euler's Theorem

The following are notes from Introduction to Cryptography with Coding Theory.

• Fermat's Little Theorem
• if p is a prime and p does not divide a, then
• a ^p-1≡ (1 mod p)
• Example
• 2¹⁰ = 1024 ≡ 1 (mod 11)
• can then evaluate 2⁵³ (mod 11)
• 2⁵³ = (2¹⁰)⁵2³ ≡ 1⁵2³  ≡ 8 (mod 11)
• We now require an analog for composite modulus n
• Define φ(n) as the number of integers a between 1 and n such that gcd(a,n) = 1
• Example
• φ(10) = 4 [1,3,7,9]
• Also can be deduced from chinese remainder theorem
• φ(n) = n Π_p|n (1-1/p)
• φ(p) = p - 1 where p is a prime number
• when n = pq where p and q are primes φ(pq) = (p - 1)(q - 1)
• Euler's Theorem
• If gcd(a,n) = 1, then a^φ(n) ≡ 1 (mod n)
• Example
• What are the last 3 digits of 7⁸⁰³
• same as working mod 1000
• 1000 = 2³5³
• φ(1000) = 1000(1-1/2)(1-1/5) = 400
• Example
• Compute 2⁴³²¹⁰  (mod 101)
• Note that 101 is prime, therefore 2¹⁰⁰  ≡ 1 (mod 101)
• Therefore (2¹⁰⁰)⁴³²2¹⁰ = 1⁴³²2¹⁰ ≡ 1024 (mod 101) ≡ 14

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Notes - The Chinese Remainder Theorem

The following are notes from Introduction to Cryptography with Coding Theory.

• Chinese remainder theorem is used to break a congruence mod n into a system of congruences mod factors of n
• Example
• Number satisfies x≡ 25(mod 42)
• means we can write x = 25 + 42k (k is some integer)
• we can rewrite 42 as 7*6
• x = 25 + 7(6k)
• x = 25 + 6(7k)
• Chinese Remainder Theorem
• suppose gcd(m,n) = 1. If an integer c is a multiple of both m and n, then c is a multiple of mn. given integers a and b there exists exactly one solution x (mod mn)to the simultaneous congruences
• x ≡ a (mod m)
• x ≡ a (mod m)
• x ≡ b (mod n)
• Lemma
• Let m,n be integers with gcd(m,n) = 1
• if an integer c is a multiple of both m and n, then c is a multiple of mn
• Example
• Solve x ≡ 3 (mod 7), x ≡ 5 (mod 15)
• 7*15 = 105
• list congruences
• 5,20,35,50,65,80
• 3,10,17,24,31,38,45,52,59,66,73,80
• x ≡ 80 mod 105
• solve
• nk ≡ a - b (mod m)
• k ≡ (a - b)i (mod m) [i is the multiplicative inverse of n]

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Notes - Pseudo-random Bit Generation

The following are notes from Introduction to Cryptography with Coding Theory.

• Ways to generate random bits
• sampling thermal noise from semiconductor resistor
• flipping coins
• Sampling natural processes however is very slow and its difficult to ensure that the enemy does not view the process
• Pseudo-Random generation
• example rand()
• takes a seed as an input produces output bitstream
• based on linear congruential generators
• x_n = a x _n-1 + b (mod m)
• number x₀ is the initial seed
• this is suitable for experimental purposes but not for cryptographic purposes because it is too predictable
• one way functions
• functions f(x) that are easy to compute but is computationally infeasible to determine y = f(x)
• therefore we can define a x_j = f(s+j) for j = 1,2,3
• DES, SHA
• Blum-Blum Shub pseudo-random bit generator
• quadratic residue generator
• generate two large primes p, q that are congruent to 3 mod 4
• then set n = pq and chose random integer x that is relatively prie to n
• BBS produces a sequence by
• x_j ≡ x²_j-1 (mod n)
• b_j is the least significant bit of x_j
• However BBS is slow to calculate due to size of primes

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Notes - Block Cipher

The following are notes from Introduction to Cryptography with Coding Theory.

• also known as the hill cipher
• choose an integer n, the key is an n x n matrix
• M =
• (1   2 3)
• (4   5 6)
• (11 9 8)
• Message is written as a series of row vectors
• abc becomes (0 1 2) which gets 
• (012)(1   2 3)
•         (4   5 6)
•         (11 9 8)
• mod 26
• (0,23,22)
• In order to decrypt we need the determinant of matrix M to satisfy
• gcd(det(M),26)=1
• Find the inverse of matrix M
•       (-14 11 -3)
• -1/3(34 -25  6)
•       (-19 13 -3)
• Since 17 is the inverse of -3 mod 26
•      (22 5  1 )
•      (6 17 24)
•      (15 13 1)
• can return plaintext by mulitplying inverse matrix by (0,23,22) mod 26
• In order to perform the block cipher we just divide the plaintext into blocks of n characters and if the ciphertext does not divide evenly, the blanks are left as 0 but the matrix multiplication is still done
• changing one letter changes n letters of plaintext therefore frequency decryption is very difficult
• Claude Shannon
• the fundamental foundations of cryptography include
• Diffusion
• means that changing one character of the plaintext, then several characters
• Confusion
• means that key does not relate in a simple way to the ciphertext
• each ciphertext should depend on several parts of the key

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