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Bus Architecture

Bus Architecture. S 2. Access Select. Memory unit 4096x16. 111. S 1. S 0. address. 001. AR. 010. PC. 011. 16-bit Bus. DR. E. ALU. 100. AC. INPR. 101. IR. 110. TR. OUTR. clock. Instruction Format. 15. 14. 12. 11. 0. I. opcode. address.

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Bus Architecture

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  1. Bus Architecture S2 Access Select Memory unit 4096x16 111 S1 S0 address 001 AR 010 PC 011 16-bit Bus DR E ALU 100 AC INPR 101 IR 110 TR OUTR clock CSC321

  2. Instruction Format 15 14 12 11 0 I opcode address I = 0 means direct memory address I = 1 means indirect memory address CSC321

  3. The Control Unit Instruction Register (IR) 15 14 - 12 11 - 0 Other Inputs 3x8 Decoder 12 Control Unit D7 – D0 n I T15 – T0 4x16 Decoder Increment Sequence Counter Clear Master Clock CSC321

  4. Decoding the Instruction • We’ve seen how to fetch and decode instructions in RTL notation • We now need to look at how to execute each instruction T0 T1 T2 = 1, register or I/O = 0, memory reference = 1, I/O = 0, register = 1, indirect = 0, direct T3 T3 T3 T3 CSC321

  5. Register Instructions • These are all pretty simple • CLA D7I’T3B11: AC ← 0 • CLE D7I’T3B10: E ← 0 • CMA D7I’T3B9: AC ← AC’ • CME D7I’T3B8: E ← E’ • CIR D7I’T3B7: AC ← shr(AC), AC(15) ← E, E ← AC(0) • CIL D7I’T3B6: AC ← shl(AC), AC(0) ← E, E ← AC(15) CSC321

  6. Register Instructions • INC D7I’T3B5: AC ← AC + 1 • SPA D7I’T3B4: if (AC(15) = 0) then (PC ← PC + 1) • SNA D7I’T3B3: if (AC(15) = 1) then (PC ← PC + 1) • SZA D7I’T3B2: if (AC = 0) then (PC ← PC + 1) • SZE D7I’T3B1: if (E = 0) then (PC ← PC + 1) • HLT D7I’T3B0: S ← 0 • S is a flip-flop that starts/stops the master clock CSC321

  7. Register Instructions • Perhaps the most interesting thing about these instructions is the condition on which each is selected • D7I’T3Bi • The D7I’terms specify the type of operation (register) • Execution starts at time T3 since no additional operands need to be fetched from memory • The Bi term is the interesting one CSC321

  8. Register Instructions • Note how the opcodes were assigned hex (binary) bit patterns • Notice any patterns? • Very common practice in both hardware design and programming CSC321

  9. Memory Instructions • The condition on which each is selected comes from the decoding of the operand and starts at time T4 • DiT4 • The Di term specifies the particular operation • Execution starts at time T4 assuring the operand has been fetched from the effective address CSC321

  10. Memory Instructions • These are a bit more complex • AND operation • D0: AC ← AC ^ M[AR] • This is fine except for the fact that the operation must take place in the ALU and M[AR] cannot be routed there directly • Therefore, we must rework this functional RTL statement to reflect the actual hardware architecture CSC321

  11. Logical AND • We must first get M[AR] into the DR register • Then we can perform the AND operation D0T4: DR ← M[AR] D0T5: AC ← AC ^ DR, SC ← 0 • Requires two timing phases, T4 and T5 CSC321

  12. Arithmetic ADD • Similar in nature to the AND operation D1T4: DR ← M[AR] D1T5: AC ← AC + DR, E ← Cout, SC ← 0 • The result remains in the AC register • If we want to place it in memory we must perform a store (STA) instruction CSC321

  13. Load Accumulator • Similar in nature to the AND operation D2T4: DR ← M[AR] D2T5: AC ← DR, SC ← 0 • Recall that the AC is only accessible through the ALU • This is why one of the ALU functions was a transfer (without any arithmetic operation) CSC321

  14. Store Accumulator • Similar in nature to the AND operation D3T4: M[AR] ← AC, SC ← 0 CSC321

  15. Branch Unconditionally • A branch is merely a modification of the program counter (PC) register D4T4: PC ← AR, SC ← 0 CSC321

  16. Branch and Save Return Address • BSA is the assembly language version of a subroutine call • It must store the address of the next instruction (which is in the PC since we incremented it after the fetch cycle) somewhere • It uses the effective address of the operand for this purpose • It then performs a branch to the subroutine address CSC321

  17. Branch and Save Return Address D5T4: M[AR] ← PC, AR ← AR + 1 D5T5: PC ← AR, SC ← 0 • This means that the subroutine actually starts one memory location after that specified in the operand • Consider an example… CSC321

  18. After time T5 (when the instruction is complete) the PC is here After time T3 the PC is here 0x20 0x20 0 BSA 0x50 0 BSA 0x50 0x21 0x21 0x50 0x50 0x21 0x51 0x51 SUBROUTINE CODE SUBROUTINE CODE 1 BUN 0x50 1 BUN 0x50 BSA instruction inserts this at time T4 Programmer inserts this command Branch and Save Return Address CSC321

  19. Increment and Skip if Zero • This is used for creating for loops • Typically, you will store a negative loop count prior to using an ISZ command • It is also used in coordination with a BUN instruction D6T4: DR ← M[AR] D6T5: DR ← DR + 1 D6T6: M[AR] ← DR, if (DR = 0) then (PC ← PC + 1), SC ← 0 CSC321

  20. Calculate loop count into AC Store loop count 0x20 0 STA 0xAA 0x21 Start of loop End of loop 0x50 0x51 0 ISZ 0xAA 0 BUN 0x21 -810 0xAA 0xFFFC Increment and Skip if Zero CSC321

  21. Input/Output Instructions • Three new flip-flops are introduced into the architecture to support input/output commands • FGI – 1 when information from the input device is available, 0 otherwise • FGO – 1 when the output device is ready to receive information, 0 otherwise • IEN – interrupt enable CSC321

  22. Input/Output Instructions • INP D7IT3B11: AC(0-7) ← INPR, FGI ← 0, SC ← 0 • OUT D7IT3B10: OUTR ← AC(0-7), FGO ← 0, SC ← 0 • SKI (used in a manner similar to the ISZ) D7IT3B9: if (FGI = 1) then PC ← PC + 1, SC ← 0 • SKO (used in a manner similar to the ISZ) D7IT3B8: if (FGO = 1) then PC ← PC + 1, SC ← 0 CSC321

  23. Input/Output Instructions • A problem arises when using the SKI and SKO instructions • Their purpose is to set up loop structures (similar to what we saw with the ISZ instruction) waiting for an input/output device to become available • This could cause large amounts of valuable time to be wasted CSC321

  24. Now for some more 8050 Assembly Language Programming • Subroutines ACALL sub … exit: jmp exit ; don’t let the main program run ; into the subroutine sub: … ret CSC321

  25. What Does ACALL Do? • Refer to page 17 of the programmer’s guide • Operation (PC) ← (PC) + 2 (SP) ← (SP) + 1 (SP) ← (PC7-0) (SP) ← (SP) + 1 (SP) ← (PC15-8) (PC10-0) ← page address a10 a9 a8 1 0 0 0 1 a7 a6 a5 a4 a3 a2 a1 a0 CSC321

  26. ACALL • Note that there is no parameter passage mechanism • How can I tell that? • Therefore, parameters and return values must be passed in memory or registers • You need to be careful about this • Is the memory used outside the subroutine? • Do the subroutines have to be reentrant? • etc. CSC321

  27. Programming Assignment • Previously you wrote code to determine if a number was prime • Place that code into a subroutine • Create a 2nd subroutine that is given a number in R0 and returns the next larger prime number in R1 • This subroutine should call the subroutine for determining if a number is prime CSC321

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