PHYSICS CLASS 12TH -LOGIC GATES

1. Logic GatesLogic Gates
 Made by- sarthak yadavMade by- sarthak yadav
 Roll no-39Roll no-39
 Class 12 A2Class 12 A2
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2. Review of BooleanReview of Boolean
algebraalgebra
 Just like Boolean logicJust like Boolean logic
 Variables can only be 1 or 0Variables can only be 1 or 0
 Instead of true / falseInstead of true / false
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3. Review of BooleanReview of Boolean
algebraalgebra
 Not is a horizontal bar above the numberNot is a horizontal bar above the number
 0 = 10 = 1
 1 = 01 = 0
 Or is a plusOr is a plus
 0+0 = 00+0 = 0
 0+1 = 10+1 = 1
 1+0 = 11+0 = 1
 1+1 = 11+1 = 1
 And is multiplicationAnd is multiplication
 0*0 = 00*0 = 0
 0*1 = 00*1 = 0
 1*0 = 01*0 = 0
 1*1 = 11*1 = 1
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__
__

4. Review of BooleanReview of Boolean
algebraalgebra
 Example: translate (Example: translate (xx++yy++zz)()(xyzxyz) to a Boolean) to a Boolean
logic expressionlogic expression
 ((xx∨∨yy∨∨zz))∧∧((¬¬xx∧¬∧¬yy∧¬∧¬zz))
 We can define a Boolean function:We can define a Boolean function:
 F(x,y) = (F(x,y) = (xx∨∨yy))∧∧((¬¬xx∧¬∧¬yy))
 And then write a “truth table” for it:And then write a “truth table” for it:
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__ __ __
x y F(x,y)
1 1 0
1 0 0
0 1 0
0 0 0

5. Basic logic gatesBasic logic gates
 NotNot
 AndAnd
 OrOr
 NandNand
 NorNor
 XorXor 55

6. Converting betweenConverting between
circuits and equationscircuits and equations
 Find the output of the following circuitFind the output of the following circuit
 Answer: (Answer: (x+yx+y)y)y
 Or (Or (xx∨∨yy))∧¬∧¬yy 66
x+y
y
(x+y)y
____

7. Converting betweenConverting between
circuits and equationscircuits and equations
 Find the output of the following circuitFind the output of the following circuit
 Answer: xyAnswer: xy
 OrOr ¬¬((¬¬xx∧¬∧¬yy) ≡) ≡ xx∨∨yy 77
x
y
x y x y
_ __ _______

8. Converting betweenConverting between
circuits and equationscircuits and equations
 Write the circuits for the followingWrite the circuits for the following
Boolean algebraic expressionsBoolean algebraic expressions
a)a) xx++yy
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x x+y

9. Converting betweenConverting between
circuits and equationscircuits and equations
 Write the circuits for the followingWrite the circuits for the following
Boolean algebraic expressionsBoolean algebraic expressions
b)b) ((xx++yy))xx
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______________
x+y
x+y (x+y)x

10. Writing xor usingWriting xor using
and/or/notand/or/not
 pp ⊕⊕ qq ≡≡ ((pp ∨∨ qq)) ∧∧ ¬(¬(pp ∧∧ qq))
 xx ⊕⊕ yy ≡≡ (x + y)((x + y)(xyxy))
1010
x y x⊕y
1 1 0
1 0 1
0 1 1
0 0 0
x+y
xy xy
(x+y)(xy)
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11. Converting decimalConverting decimal
numbers to binarynumbers to binary
 5353 = 32 + 16 + 4 + 1= 32 + 16 + 4 + 1
= 2= 255
+ 2+ 244
+ 2+ 222
+ 2+ 200
= 1*2= 1*255
+ 1*2+ 1*244
+ 0*2+ 0*233
+ 1*2+ 1*222
+ 0*2+ 0*211
+ 1*2+ 1*200
= 110101 in binary= 110101 in binary
= 00110101 as a full byte in binary= 00110101 as a full byte in binary
 211= 128 + 64 + 16 + 2 + 1211= 128 + 64 + 16 + 2 + 1
= 2= 277
+ 2+ 266
+ 2+ 244
+ 2+ 211
+ 2+ 200
= 1*2= 1*277
+ 1*2+ 1*266
+ 0*2+ 0*255
+ 1*2+ 1*244
+ 0*2+ 0*233
+ 0*2+ 0*222
++
1*21*211
+ 1*2+ 1*200
= 11010011 in binary= 11010011 in binary 1111

12. Converting binaryConverting binary
numbers to decimalnumbers to decimal
 What is 10011010 in decimal?What is 10011010 in decimal?
1001101010011010 = 1*2= 1*277
+ 0*2+ 0*266
+ 0*2+ 0*255
+ 1*2+ 1*244
+ 1*2+ 1*233
++
0*20*222
+ 1*2+ 1*211
+ 0*2+ 0*200
= 2= 277
+ 2+ 244
+ 2+ 233
+ 2+ 211
= 128 + 16 + 8 + 2= 128 + 16 + 8 + 2
= 154= 154
 What is 00101001 in decimal?What is 00101001 in decimal?
00101001 = 0*200101001 = 0*277
+ 0*2+ 0*266
+ 1*2+ 1*255
+ 0*2+ 0*244
+ 1*2+ 1*233
++
0*20*222
+ 0*2+ 0*211
+ 1*2+ 1*200
= 2= 255
+ 2+ 233
+ 2+ 200
= 32 + 8 + 1= 32 + 8 + 1
= 41= 41
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13. A note on binaryA note on binary
numbersnumbers
 In this slide set we are only dealing withIn this slide set we are only dealing with
non-negative numbersnon-negative numbers
 The book (section 1.5) talks about two’s-The book (section 1.5) talks about two’s-
complement binary numberscomplement binary numbers
 Positive (and zero) two’s-complement binaryPositive (and zero) two’s-complement binary
numbers is what was presented herenumbers is what was presented here
 We won’t be getting into negative two’s-We won’t be getting into negative two’s-
complmeent numberscomplmeent numbers
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14. How to add binaryHow to add binary
numbersnumbers
 Consider adding two 1-bit binary numbersConsider adding two 1-bit binary numbers xx andand yy
 0+0 = 00+0 = 0
 0+1 = 10+1 = 1
 1+0 = 11+0 = 1
 1+1 = 101+1 = 10
 Carry isCarry is xx ANDAND yy
 Sum isSum is xx XORXOR yy
 The circuit to compute this is called a half-adderThe circuit to compute this is called a half-adder 1414
x y Carry Sum
0 0 0 0
0 1 0 1
1 0 0 1
1 1 1 0

15. The half-adderThe half-adder
 Sum =Sum = xx XORXOR yy
 Carry =Carry = xx ANDAND yy
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x y Carry Sum
0 0 0 0
0 1 0 1
1 0 0 1
1 1 1 0

16. Using half addersUsing half adders
 We can then use a half-adder to computeWe can then use a half-adder to compute
the sum of two Boolean numbersthe sum of two Boolean numbers
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1 1 0 0
+ 1 1 1 0
010?
001

17. How to fix thisHow to fix this
 We need to create an adder that can take aWe need to create an adder that can take a
carry bit as an additional inputcarry bit as an additional input
 Inputs:Inputs: xx,, yy, carry in, carry in
 Outputs: sum, carry outOutputs: sum, carry out
 This is called a full adderThis is called a full adder
 Will addWill add xx andand yy with a half-adderwith a half-adder
 Will add the sum of that to theWill add the sum of that to the
carry incarry in
 What about the carry out?What about the carry out?
 It’s 1 if either (or both):It’s 1 if either (or both):
 xx++yy = 10= 10
 xx++yy = 01 and carry in = 1= 01 and carry in = 1
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x y c carry sum
1 1 1 1 1
1 1 0 1 0
1 0 1 1 0
1 0 0 0 1
0 1 1 1 0
0 1 0 0 1
0 0 1 0 1
0 0 0 0 0

18. The full adderThe full adder
 The “HA” boxes areThe “HA” boxes are
half-addershalf-adders
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x y c s1 c1 carry sum
1 1 1 0 1 1 1
1 1 0 0 1 1 0
1 0 1 1 0 1 0
1 0 0 1 0 0 1
0 1 1 1 0 1 0
0 1 0 1 0 0 1
0 0 1 0 0 0 1
0 0 0 0 0 0 0
s1
c1

19. The full adderThe full adder
 The full circuitry of the full adderThe full circuitry of the full adder
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20. Adding bigger binaryAdding bigger binary
numbersnumbers
 Just chain full adders togetherJust chain full adders together
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...

21. Adding bigger binaryAdding bigger binary
numbersnumbers
 A half adder has 4 logic gatesA half adder has 4 logic gates
 A full adder has two half adders plus a OR gateA full adder has two half adders plus a OR gate
 Total of 9 logic gatesTotal of 9 logic gates
 To addTo add nn bit binary numbers, you need 1 HAbit binary numbers, you need 1 HA
andand nn-1 FAs-1 FAs
 To add 32 bit binary numbers, you need 1 HATo add 32 bit binary numbers, you need 1 HA
and 31 FAsand 31 FAs
 Total of 4+9*31 = 283 logic gatesTotal of 4+9*31 = 283 logic gates
 To add 64 bit binary numbers, you need 1 HATo add 64 bit binary numbers, you need 1 HA
and 63 FAsand 63 FAs
 Total of 4+9*63 = 571 logic gatesTotal of 4+9*63 = 571 logic gates 2121

22. More about logic gatesMore about logic gates
 To implement a logic gate in hardware,To implement a logic gate in hardware,
you use a transistoryou use a transistor
 Transistors are all enclosed in an “IC”, orTransistors are all enclosed in an “IC”, or
integrated circuitintegrated circuit
 The current Intel Pentium IV processorsThe current Intel Pentium IV processors
have 55 million transistors!have 55 million transistors!
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23. Flip-flopsFlip-flops
 Consider the following circuit:Consider the following circuit:
 What does it do?What does it do? 2323

24. MemoryMemory
 A flip-flop holds a single bit of memoryA flip-flop holds a single bit of memory
 The bit “flip-flops” between the two NANDThe bit “flip-flops” between the two NAND
gatesgates
 In reality, flip-flops are a bit moreIn reality, flip-flops are a bit more
complicatedcomplicated
 Have 5 (or so) logic gates (transistors) per flip-Have 5 (or so) logic gates (transistors) per flip-
flopflop
 Consider a 1 Gb memory chipConsider a 1 Gb memory chip
 1 Gb = 8,589,934,592 bits of memory1 Gb = 8,589,934,592 bits of memory
 That’s about 43 million transistors!That’s about 43 million transistors!
 In reality, those transistors are split into 9In reality, those transistors are split into 9
ICs of about 5 million transistors eachICs of about 5 million transistors each
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25. HexadecimalHexadecimal
 A numerical rangeA numerical range
from 0-15from 0-15
 Where A is 10, B is 11,Where A is 10, B is 11,
… and F is 15… and F is 15
 Often written with aOften written with a
‘0x’ prefix‘0x’ prefix
 So 0x10 is 10 hex, orSo 0x10 is 10 hex, or
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 0x100 is 100 hex, or0x100 is 100 hex, or
256256
 Binary numbers easilyBinary numbers easily
translate:translate: 2525

26. DEADBEEFDEADBEEF
 Many IBM machines would fill allocatedMany IBM machines would fill allocated
(but uninitialized) memory with the hexa-(but uninitialized) memory with the hexa-
decimal pattern 0xDEADBEEFdecimal pattern 0xDEADBEEF
 Decimal -21524111Decimal -21524111
 Makes it easier to spot in a debuggerMakes it easier to spot in a debugger
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