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CENG 241 Digital Design 1 Lecture 6 Amirali Baniasadi [email protected] Decimal adder When dealing with decimal numbers BCD code is used. A decimal adders requires at least 9 inputs and 5 outputs. BCD adder: each input does not exceed 9, the output can not exceed 19 How are decimal numbers presented in BCD? Decimal 9 19 Binary 1001 10011 BCD 1001 (0001)(1001) 1 9 2 Decimal Adder Decimal numbers should be represented in binary code number. Example: BCD adder Suppose we apply two BCD numbers to a binary adder then: The result will be in binary and ranges from 0 through 19. Binary sum: K(carry) Z8 Z4 Z2 Z1 BCD sum : C(carry) S8 S4 S2 S1 For numbers equal or less than 1001 binary and BCD are identical. For numbers more than 1001, we should add 6(0110) to binary to get BCD. example: 10011(binary) = 11001(BCD) =19 ADD 6 to correct. 3 BCD adder Numbers that need correction (add 6) are: 01010 (10) 01011 (11) 01100 (12) 01101 (13) 01110 (14) 01111 (15) 10000 (16) 10001 (17) 10010 (18) 10011 (19) Decides to add 6? Adds 6 4 BCD adder Numbers that need correction (add 6) are: K Z8 Z4 Z2 Z1 0 1 0 1 0 (10) 0 1 0 1 1 (11) 0 1 1 0 0 (12) 0 1 1 0 1 (13) 0 1 1 1 0 (14) 0 1 1 1 1 (15) 1 0 0 0 0 (16) 1 0 0 0 1 (17) 1 0 0 1 0 (18) 1 0 0 1 1 (19) C = K + Z8Z4 +Z8Z2 5 Magnitude Comparators Compares two numbers, determines their relative magnitude. We look at a 4-bit magnitude comparator; A=A3A2A1A0, B=B3B2B1B0 Two numbers are equal if all bits are equal. A=B if A3=B3 AND A2=B2 AND A1=B1 AND A0=B0 Xi= AiBi + Ai’Bi’ ; Ai=Bi Xi=1 (remember exclusive NOR?) 6 Magnitude Comparators How do we know if A>B? 1.Compare bits starting from the most significant pair of digits 2.If the two are equal, compare the next lower significant bits 3.Continue until a pair of unequal digits are reached 4.Once the unequal digits are reached, A>B if Ai=1 and Bi=0, A<B if Ai=0 and Bi = 1 A>B = A3B3’+X3A2B2’+X3X2A1B1’+X3X2X1A0B0’ A<B = A3’B3+X3A2’B2+X3X2A1’B1+X3X2X1A0’B0 Xi=1 if Ai=Bi 7 Magnitude Comparators A3=B3 ? X3A2’B2 8 Decoders A decoder converts binary information from n input lines to a maximum of 2n output lines Also known as n-to-m line decoders where m< 2n Example 3-to-8 decoders. 9 Decoders: Truth Table X 0 0 0 0 1 1 1 1 Y 0 0 1 1 0 0 1 1 Z 0 1 0 1 0 1 0 1 D0 1 0 0 0 0 0 0 0 D1 0 1 0 0 0 0 0 0 D2 0 0 1 0 0 0 0 0 D3 0 0 0 1 0 0 0 0 D4 0 0 0 0 1 0 0 0 D5 0 0 0 0 0 1 0 0 D6 0 0 0 0 0 0 1 0 D7 0 0 0 0 0 0 0 1 10 Decoders: AND implementation 11 2-to-4 Decoder: NAND implementation Decoder is enabled when E=0 12 How to build bigger decoders? We can combine two 3-to-8 decoders to build a 4-to-16 decoder. Generates from 0000 to 0111 Generates from 1000 to 1111 13 Combinational Logic implementation A decoder provides the 2n minterms of n input variables. Any function is can be expressed in sum of minterms. Use a decoder to make the minterms and an external OR gate to make the sum. Example: consider a full adder. S(x,y,z) = Σ(1,2,4,7) C(x,y,z) = Σ (3,5,6,7) 14 Combinational Logic implementation 15 Encoders Encoders perform the inverse operation of a decoder: Encoders have 2n input lines and n output line. Output lines generate the binary code corresponding to the input value. 16 Encoders: Truth Table Outputs X 0 0 0 0 1 1 1 1 Y 0 0 1 1 0 0 1 1 Z 0 1 0 1 0 1 0 1 z=D1+D3+D5+D7 Inputs D0 1 0 0 0 0 0 0 0 D1 0 1 0 0 0 0 0 0 D2 0 0 1 0 0 0 0 0 D3 0 0 0 1 0 0 0 0 y=D2+D3+D6+D7 D4 0 0 0 0 1 0 0 0 D5 0 0 0 0 0 1 0 0 D6 0 0 0 0 0 0 1 0 D7 0 0 0 0 0 0 0 1 x=D4+D5+D6+D7 17 Priority Encoders Encoder limitations: If two inputs are active, the output is undefined. Solution: we need to take into account priority. What if all inputs are 0? Solution: we need a valid bit Input D0 D1 D2 0 0 0 1 0 0 X 1 0 X X 1 X X X D3 0 0 0 0 1 x X 0 0 1 1 Output y X 0 1 0 1 v 0 1 1 1 1 18 Priority Encoders: Map 19 Priority Encoders: Circuit 20 Multiplexers Multiplexer: selects one binary input from many selections example: 2-to-1 MUX 21 4-to-1 MUX Directs 1 of the 4 inputs to the output 22 Multi-bit selection logic Multiplexers can be combined with common selection inputs to support multi-bit selection logic 23 Implementing Boolean functions w/ MUX General rules for implementing any Boolean function with n variables: Use a multiplexer with n-1 selection inputs and 2 n-1 data inputs List the truth tabel Apply the first n-1 variables to the selection inputs of multiplexer For each combination evaluate the output as a function of the last variable. The function can be 0, 1 the variable or the complement of the variable. 24 Implementing Boolean functions w/ MUX 25 Implementing Boolean functions w/ MUX 26 Summary Reading up to page 154 27