How many address lines are required to interface 8k byte Eprom and 1k byte RAM with microprocessor 8085?

8085 has 16 address lines (A0 – A15), hence a maximum of 64 KB (= 216 bytes) of memory locations can be interfaced with it. The memory address space of the 8085 takes values from 0000H to FFFFH.

What is the maximum number of 4096 8 Eprom chips we can interface with 8085?

How many address lines in a 4096*8 EPROM CHIP? 12 Address lines.

How many memory chips are connected with 8086?

Interface two 4K x 8 EPROMS and two 4K x 8 RAM chips with 8086.

Why do we need interfacing in microprocessor?

When we are executing any instruction, we need the microprocessor to access the memory for reading instruction codes and the data stored in the memory. The interfacing circuit therefore should be designed in such a way that it matches the memory signal requirements with the signals of the microprocessor.

What is the address of the last memory location of an 8K Eprom when the starting address is 8000h?

Find the starting address. Will someone Please tell how it is calculated? The starting address of an 8K byte memory chip that ends at FFFFH is E000H. 8K is 8192 (8 * 1024) which is 2000H.

How many address lines are needed for 2 MB?

Answer: It requires 21 address lines to address two megabytes of memory.

How many address lines are needed for 4096 8 Eprom chip interfacing?

How many address lines in a 4096 x 8 EPROM CHIP? Ans: 12 address lines.

How many address lines are required to decode 8K memory?

8k= 2^13. So 2^13 address lines are required.

What is the maximum memory addressing capacity of 8086?

1 Mb
Address Bus − 8085 has 16-bit address bus while 8086 has 20-bit address bus. Memory − 8085 can access up to 64Kb, whereas 8086 can access up to 1 Mb of memory.

What is memory interfacing in 8086 microprocessor?

• The general procedure of static memory interfacing with 8086 is briefly described as follows: 1. Arrange the available memory chip so as to obtain 16- bit data bus width. The upper 8-bit bank is called as odd address memory bank and the lower 8-bit bank is called as even address memory bank.

How to design an interface between 8086 CPU and EPROM?

Design an interface between 8086 CPU and two chips of 16K x 8 EPROM and two chips of 32K x 8 RAM. Select the starting address of EPROM suitably. The RAM address must start at 00000H. Solution. The last address in the map of 8086 is FFFFFH.

How many chips are required to interface 8086 with 32K RAM?

It is required to interface two chips of 32K x 8 ROM and four chips of 32K x 8 RAM with 8086, according to the following map. Show the implementation of this memory system. Solution. Let us write the memory map of the system as shown in Table.

How many memory locations can be interfaced with 8085?

8085 has 16 address lines (A0 – A15), hence a maximum of 64 KB (= 216 bytes) of memory locations can be interfaced with it. The memory address space of the 8085 takes values from 0000H to FFFFH. Ex: Interface a 6264 IC (8K x 8 RAM) with the 8085 using NAND gate decoder such that the starting address assigned to the chip is 4000H.

How many 4K x 8 memory chips are there in a computer?

The memory system in this example contains in total four 4K x 8 memory chips. The two 4K x 8 chips of RAM and ROM are arranged in parallel to obtain 16-bit data bus width. If A 0 is 0, i.e. the address is even and is in RAM, then the lower RAM chip is selected indicating 8-bit transfer at an even address.