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## Full Adder Display

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**Topics**• A 1 bit adder with LED display • Ripple Adder • Signed/Unsigned Subtraction • Hardware Implementation of 4-bit adder**Implementation of a Full Adder**(carry-in)**Verilog Implementation**Use switches to input binary numbers—x, y, and z. z is the carry-in. Display the output on the LED. Press a button to determine which bit will be displayed. s represents the sum bit. c represents the carry-out bit. A mux is used to determine Whether s or c should be displayed.**Multiplexing 7-Segment Displays (Last Week)**If s[1:0]=00, then x[3:0]. If s[1:0]=01, then x[7:4]. If s[1:0]=10, then x[11:8]. If s[1:0]=11, then x[15:12]. Use Quad 4-to-1 mux Get values for an[3:0] from btn[3:0] so that only one LED is displayed.**Explanation of the Code**If btn[0] is pushed, t[0] is 0. If btn[1] is pusehd, t[0] is 1. So we can use t[0] as a selector bit for the MUX. t[ ] =s[] If the output of the MUX is a 0, a 0 Is displayed. If the output of the MUX is a 1, a 1 is displayed.**Implementation of a Full Adder**(carry-in)**Four-Bit Adder**C4 is calculated last because it takes C0 8 gates to reach C4. Each FA uses 2 XOR, 2 AND and 1 OR gate. A four-bit adder uses 8 XOR, 8 AND and 4 OR gate.**Hardware Simplification**input carry ) 2 gate delays for C3!**Four-bit adder with Carry Lookahead**Ripple adder uses 8 XOR, 8 AND and 4 OR gate. Lookahead implementation: 8 XOR, (4+6) AND, 1 2-input OR, 2 3-input OR.**Advantages**• C1, C2 and C3 do not have to wait for C1 and C2 to progate. • C3 is propagated at the same time as C1 and C2.**Topics**• Calculations Examples • Signed Binary Number • Unsigned Binary Number • Hardware Implementation • Overflow Condition**Unsigned Number**(2-bit example)**Unsigned Addition**• 1+2=**Unsigned Addition**• 1+3= (Indicates Overflow) (Carry Out) Overflow can be an issue in unsigned addition.**Unsigned Subtraction (1)**• 1-2= (1’s complement) (2’s complement)**Unsigned Subtraction (2)**• 2-1= Discarded**Summary for Unsigned Addition/Subtraction**• Overflow can be an issue in unsigned addition • Unsigned Subtraction (M-N) • If M≥N, and end carry will be produced. The end carry is discarded. • If M<N, • Take the 2’s complement of the sum • Place a negative sign in front**Signed Binary Numbers**• 4-bit binary number • 1 bit is used as a signed bit • -8 to +7 • 2’s complement**Signed Addition (70+80)**(Indicates a negative number) 70=21+22+26=2+4+64 80=24+26=16+64 10010110→01101001 →01101010 21+23+25+26=2+8+32+64=106 10010110↔-106 010010110 010010110↔ 21+22+24+27=2+4+16+128=150 Conclusion: There is a problem of overflow Fix: Use the end carry as the sign bit, and let b7 be the extra bit.**Signed Subtraction (70-80)**(Indicates a negative number) 70=21+22+26=2+4+64 80=24+26=16+64=01010000→10101111→10110000 11110110→00001001 →00001010 21+23=10 11110110↔-10 (No Problem)**Signed Subtraction (-70-80)**(Indicates a positive number! A negative number expected.) 70=21+22+26=2+4+64 80=24+26=16+64 101101010 →010010101 → 010010110 010010110 ↔21+22+24+27=2+4+16+128=150 101101010 ↔-150 Conclusion: There is a problem of overflow Fix: Use the end carry as the sign bit, and let b7 be the extra bit.**Observations**• Given the similarity between addition and subtraction, same hardware can be used. • Overflow is an issue that needs to be addressed in the hardware implementation • A signed number is not processed any different from an unsigned number. The programmer must interpret the results of addition and subtraction appropriately.**The Mode Input (1)**If M=0, = If M=1, = B0**The Mode Input (2)**If M=0, If M=1,**M=0 (Addition)**B3 B2 B0 B1 0**M=1 (Subtraction)**1 2’s complement is generated of B is generated!**Unsigned Addition**When two unsigned numbers are added, an overflow is detected from the end carry.**Detect Overflow in Signed Addition**Observe The caryinto the sign bit The carry out of the sign bit If they are not equal, they indicate an overflow.