Write an 8085 assembly language program to simulate the throw of a die using an interrupt.We have a counter for this program, which counts the ranges from 1 to 6, and again repeats the sequence of count in a matter which is endless in an infinite loop. The die throwing getting the head and tail probability gets simulated by pressing the key ‘Vect Intr’ on the keyboard. The branching of 8085 branches to RST7.5 ISS. Here, the current value of the counter gets is displayed in the data field, which controls and returns to the main program for the continuation ... Read More

 Let us say that the 8085 which is interrupted because of RST6.5 pin and has been branched to the ISS for the pin RST6.5. Then, except the DI instruction at the beginning of this ISS, all the interrupts gets disabled except TRAP. So, even if RST7.5 pin is in activated state in the middle of the execution of RST6.5 ISS, the interruption of 8085 will not occur due to RST7.5. Actually a higher priority interrupt is RST7.5, but the lower priority interrupt ISS cannot be interrupted by it of RST6.5. We solve this problem by specifically having the instruction which ... Read More

Intel 8255 is a peripheral interface (PPI) chip which is programmable. It is used for the connection of peripheral devices and interfacing. We call Peripheral device also as Input Output device. We use Input Output ports for the connection of Input Output devices. Hence 8255 is a programmable Input Output port chip. It is a 40 pin chip available for dual line packaging. Power supply of +5 Volt DC is needed for its working. It consists of two programmable Input Output ports having of size 8 bits and two programmable Input Output ports of size 4 bits. We call them ... Read More

A non-maskable interrupt is a Trap Interrupt which implies that whenever this pin gets activated, the 8085 always gets interrupted even if the state of 8085 is in DI. The input of Trap input is level sensitive and edge sensitive. Hence the Trap line always makes a transition from 0 to 1, and remains in state 1 until the end of the execution of an instruction for the interruption of 8085. A vectored-interrupt in 8085 is a TRAP. The starting address of 8085 is known by itself the of the ISS as 4.5 * 8 = 0024H. Hence we name the ... Read More

Before having a discussion regarding the demerits or merits of I/O mapped I/O and memory-mapped I/O, let us have a generic discussion regarding the difference between I/O mapped I/O and memory mapped I/O.In Memory Mapped Input Output −We allocate a memory address to an Input-Output device.Any instructions related to memory can be accessed by this Input-Output device.The Input-Output device data are also given to the Arithmetic Logical Unit.Input-Output Mapped Input Output −We give an Input-Output address to an Input-Output deviceOnly IN and OUT instructions are accessed by such devices.The ALU operations are not directly applicable to such Input-Output data.So as ... Read More

RST7.5 pin is an input which is edge-sensitive. Peripherals uses it for sending a pulse, rather than a sustained high level, for the interruption of the processor. Internal to 8085 we have a flip-flop which gets connected to the interrupt pin RST7.5. We set this flip-flop to 1, when a positive-going edge occurs on the input RST 7.5. The waveform of pin RST7.5 and output Q of RST7.5 flip-flop is shown in the Fig.Internal interrupt signal RST7.5 has a priority higher than the internal interrupt signals of RST6.5, RST5.5 and INTR. As we can see from fig the RST7.5 internal interrupt ... Read More

 Both the pins RST5.5 and RST6.5 pins are inputs which are level sensitive. RST6.5 is of higher priority than RST5.5 but the pin RST5.5 is of higher priority than INTR. RST5.5 and RST6.5 have similar functions. The point to be noted that these pins must remain high till the 8085 checks all the internal interrupt signals at the end of the instructions. As we can easily see from the Fig. We activate the RST5.5 and RST6.5 internal interrupt signals if and only if when the external interrupt pins are in logic 1 state;Flip-flop IE is in logic 1 state;SIM instructions ... Read More

We have assumed that the interrupt system gets enabled by using the EI instruction, and the signals which have higher priority are not in active state.In the penultimate clock cycle of the last machine cycle of every instruction, the 8085 senses all the internal interrupt signals.If the INTR internal signal which is at logic 1, the 8085 enters to a machine cycle which is called interrupt acknowledge (INA) machine cycle.The interrupts from the Input Output port gets acknowledged by the 8085 by the activation of INTA* pin in the T2 state of the machine cycle INA where INTA* is a ... Read More

Intel 8085 consists of a RESET_IN* pin which is an active low input pin. We RESET 8085 by placing a logic 0 on this pin at least for 0.5μs, after that the power is supplied to Vcc pin of 8085. Moreover, in practice we place the RESET_IN* in logic 0 state for a few milliseconds. A typical reset circuit which we use in ALS 8085 kit, is shown in the following Fig.The moment when the power supply is switched on, the Vcc pin gets +5V power here the RESET_IN* pin stays in logic 0 state for a time dependency on the ... Read More

Before having a discussion regarding the demerits of I/O mapped I/O and merits of memory-mapped I/O, let us have a generic discussion regarding the difference between I/O mapped I/O and memory mapped I/O.In Memory Mapped Input Output −We allocate a memory address to an Input-Output device.Any instructions related to memory can be accessed by this Input-Output device.The Input-Output device data are also given to the Arithmetic Logical Unit.Input-Output Mapped Input Output −We give an Input-Output address to an Input-Output deviceOnly IN and OUT instructions are accessed by such devices.The ALU operations are not directly applicable to such Input-Output data.So as ... Read More