Lec19 - ECE 212 Lec 19 ECE 212 Digital Circuits II...

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ECE 212 Lec 19 ECE 212 Digital Circuits II Wednesday, 9 April 2008 HW Set 19 : Problem 1 No presenter. Problem 2 No presenter. Problem 3 Humberto A. Problem 4 Rhodes B. Problem 5 Andy B. Problem 6 Alyssa B. Lecture 19 Goals To understand the 6812 flag response. To understand the proper use of conditional branch instructions. To understand how to set or clear specified bits of a Byte in memory. To understand how to make a decision based on the value of a bit or bits of a Byte in memory.
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ECE 212 Lec 19 68HC12 Flags Condition Code Register (CCR) An 8-bit Special Purpose Register. Contains condition codes (Flags). C V Z N I H X S N = Negative Flag. Z = Zero Flag V = ( Signed number) Overflow Flag. C = Carry (or Borrow) Flag. Flag Response after Addition N = 8th bit of the result Z = ( all 8 bits of the result are zero ) V = Signed number overflow occurs C = The 9th bit of the sum A = a7 a6 a5 a4 a3 a2 a1 a0 (data) = + g7 g6 g5 g4 g3 g2 g1 g0 = ( z7' ·z6'·z5'·z4 '·z3 '·z2 '·z1 '·z0' ) = z7 = ( a7 ·g7 ·z7') + ( a7 '·g7' ·z7 ) = z8 z8 z7 z6 z5 z4 z3 z2 z1 z0 adda (data): The hardware always responds this way, regardless of the data type (unsigned or signed). Flag significance: depends the data type.
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ECE 212 Lec 19 Example ldaa #0xA9 adda #0xDC Problem: Determine the values of the Z, C, N, V flags after the adda instruction executes. Solution: A9 = 1010 1001 + DC = + 1101 1100 1 1000 0101 C V Z N I H X S 0 Result is not zero. Example ldaa #0xA9 adda #0xDC Problem: Determine the values of the Z, C, N, V flags after the adda instruction executes. Solution: A9 = 1010 1001 + DC = + 1101 1100 1 1000 0101 C V Z N I H X S 0 1
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ECE 212 Lec 19 Example ldaa #0A9h adda #0DCh Problem: Determine the values of the Z, C, N, V flags after the adda instruction executes. Solution: A9 = 1010 1001 + DC = + 1101 1100 1 1000 0101 C V Z N I H X S 01 1 Example ldaa #0A9h adda #0DCh Problem: Determine the values of the Z, C, N, V flags after the adda instruction executes. Solution: A9 = 1010 1001 + DC = + 1101 1100 1 1000 0101 C V Z N I H X S 1 0
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ECE 212 Lec 19 Flags and Arithmetic Overflow ( a7 ·g7 ·z7') + ( a7 '·g7' ·z7 ) = 1 A = a7 a6 a5 a4 a3 a2 a1 a0 (data) = + g7 g6 g5 g4 g3 g2 g1 g0 z8 z7 z6 z5 z4 z3 z2 z1 z0 Unsigned Data: Overflow iff Signed Data: Overflow iff z8 =1 iff C = 1 iff V = 1
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Lec19 - ECE 212 Lec 19 ECE 212 Digital Circuits II...

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