The Intel 8085 ("eighty-eighty-five") is an 8-bit microprocessor produced by Intel and introduced in 1976. It is a software-binary compatible with the more-famous Intel 8080 with only two minor instructions added to support its added interrupt and serial input/output features. Intel 8085 can read or write 8 bits at a time. Also, memory is used for storing data and results, 8 bits in each memory location. Like two memory locations and four memory locations, if there are eight memory locations, 3-bit address is needed to specify the location of interest. If address is 000, location 0 is selected, etc. and if address is 111, location 7 is selected. On A15-8 the 8085 sends out the MS byte of address. These are o/p tri-state (a state of high impedance) signals used as higher order 8 bits of 16-bit address. Their pin numbers are 28 to 21, respectively in the pin diagram of 8085 microprocessor. These signals are unidirectional and are given from 8085 to select memory or I/O devices. With 16 address pins the 8085 is capable of addressing any memory location out of 216 = 26×210 = 64×1024 = 65,536 locations.
Fig: Pin diagram of 8085
Let us consider this following example and also the voltage levels at AD15, AD14, …, AD8 pins. If we consider instruction MVI E, ABH then it means that ABH will be moved or copied to the register E. And as a result, the previous value of E will get over written.
|2000||1E||MVIE, ABH||E ← ABH|
|2001||AB||ABH as operand|
This instruction will have seven T-states as shown below.
Summary − So this instruction MVI E, ABH requires 2-Bytes, 2-Machine Cycles Opcode Fetch and Memory Read) and 7 T-States for execution as shown in the timing diagram.