Program Counter




The memory address register (AR) has 12 bits since this is the width of a memory address. The program counter (PC) also has 12 bits and it holds the address of the next instruction to be read from memory after the current instruction is executed. The PC goes through a counting sequence and causes the computer to read sequential instructions previously stored in memory. Instruction words are read and executed in sequence unless a branch instruction is encountered. A branch instruction calls for a transfer to a nonconsecutive instruction in the program. The address part of a branch instruction is transferred to PC to become the address of the next instruction. To read an instruction, the content of PC is taken as the address for memory and a memory read cycle is initiated. PC is then incremented by one, so it holds the address of the next instruction in sequence.

Two registers are used for input and output. The input register (INPR) receives an 8-bit character from an input device. The output register (OUTR) holds an 8-bit character for an output device.

Program Counter



Frequently Asked Questions

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Ans: Computer instructions are normally stored in consecutive memory locations and are executed sequentially one at a time. The control reads an instruction from a specific address in memory and executes it. It then continues by reading the next instruction in sequence and executes it, and so on. view more..
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Ans: It is sometimes convenient to use the address bits of an instruction code not as an address but as the actual operand. When the second part of an instruction code specifies an operand, the instruction is said to have an immediate operand. view more..
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Ans: In this chapter we introduce a basic computer and show how its operation can be puter specified with register is defined by its internal transfer registers, statements. the limirlg The otganization and control of structure, the comand the set of instructions that It uses. The design of the computer is then carried out in detall. Although the basic computer presented in this chapter is very small compared to commercial computers, It has the advantage of being simple enough so we can demonstrate the design process without too many complications. view more..
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Ans: The memory address register (AR) has 12 bits since this is the width of a memory address. The program counter (PC) also has 12 bits and it holds the address of the next instruction to be read from memory after the current instruction is executed. The PC goes through a counting sequence and causes the computer to read sequential instructions previously stored in memory. Instruction words are read and executed in sequence unless a branch instruction is encountered. view more..
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Ans: The basic computer has eight registers, a memory unit, and a control unit. Paths must be provided to transfer information from one register to another and between memory and registers. view more..
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Ans: The input data and output data of the memory are connected to the common bus, but the memory address is connected to AR. Therefore, AR must always be used to specify a memory address. By using a single register for the address, we eliminate the need for an address bus that would have been needed otherwise. The content of any register can be specified for the memory data input during a write operation. Similarly, any register can receive the data from memory after a read operation except AC . view more..
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Ans: The basic computer has three instruction code formats, as shown in Fig. 5-5. Each format has 16 bits. The operation code (opcode) part of the instruction contains three bits and the meaning of the remaining 13 bits depends on the operation code encountered. view more..
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Ans: Before investigating the operations performed by the instructions, let us discuss the type of instructions that must be included in a computer. A computer should have a set of instructions so that the user can construct machine language programs to evaluate any function that is known to be computable. The set of instructions are said to be complete if the computer includes a sufficient number of instructions in each of the following categories: view more..
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Ans: The timing for all registers in the basic computer is controlled by a master clock generator. The clock pulses are applied to all flip-flops and registers in the system, including the flip-flops and registers in the control unit. The clock pulses do not change the state of a register unless the register is enabled by view more..
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Ans: The last three waveforms in Fig. 5-7 show how SC is cleared when D3T4 = I. Output D3 from the operation decoder becomes active at the end of timing signal T2• When timing signal T4 becomes active, the output of the AND gate that implements the control function D3T4 becomes active. This signal is applied to the CLR input of SC. view more..
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Ans: A program residing in the memory unit of the computer consists of a sequence of instructions. The program is executed in the computer by going through a cycle for each instruction. Each instruction cycle in turn is subdivided into a sequence of subcycles or phases. In the basic computer each instruction cycle consists of the following phases: view more..
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Ans: Initially, the program counter PC is loaded with the address of the first instruction in the program. The sequence counter SC is cleared to 0, providing a decoded timing signal To. After each clock pulse, SC is incremented by one, so that the timing signals go through a sequence T0, T1, T2, and so on. The rnicrooperations for the fetch and decode phases can be specified by the following register transfer statements. view more..
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Ans: The timing signal that is active after the decoding is T3• During time T,, the control unit determines the type of instruction that was just read from memory. The flowchart of Fig. 5-9 presents an initial configuration for the instruction cycle and shows how the control determines the instruction type after the decoding. The three possible instruction types available in the basic computer are specified in Fig. 5-5. view more..
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Ans: Register-reference instructions are recognized by the control when 07 = 1 and I = 0. These instructions use bits 0 through 11 of the instruction code to specify one of 12 instructions. These 12 bits are available in IR(0-11). They were also transferred to AR during time T2• view more..
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Ans: In order to specify the rnicrooperations needed for the execution of each instruction, it is necessary that the function that they are intended to perform be defined precisely. Looking back to Table 5-2, where the instructions are listed, we find that some instructions have an ambiguous description. This is because the explanation of an instruction in words is usually lengthy, and not enough space is available in the table for such a lengthy explanation. We will now show that the function of the memory-reference instructions can be defined precisely by means of register transfer notation. view more..
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Ans: This is an instruction that performs the AND logic operation on pairs of bits in AC and the memory word specified by the effective address. The result of view more..
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Ans: This instruction adds the content of the memory word specified by the effective address to the value of AC. The sum is transferred into AC and the output carry C,., is transferred to the E (extended accumulator) flip-flop. The rnicrooperations needed to execute this instruction are view more..
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Ans: This instruction transfers the memory word specified by the effective address to AC. The rnicrooperations needed to execute this instruction are view more..




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