How many address bits are required for a 2K * 8 memory?

11 address bits
How many address bits are needed to operate a 2K * 8-bit memory? Explanation: For n address bits, the memory location will consist of 2n bits. Thus, for 2K, only 11 address bits are required, because 211 = 2K. 10.

How many chips are required to expand memory if available memory is 2K * 8 and required memory is 2K * 32?

(3) A RAM chip has 2K rows of cells to select (each row has 8 cells of 1 bit each which will always be selected together). So, we need 11 bits to select any of these 2K rows. (5) We use 5 address lines to select the RAM chips. So, decoder needs to be 5 to 32.

How many address lines are necessary on the chip of 2K byte memory?

11 address lines
Consider the EPROM. It is 2k in size. 11 address lines are needed to address all the addresses inside the EPROM. A similar calculation reveals that the 2K RAM also needs 11 address lines.

How many address lines are used in 4k memory?

12 address lines are require for 4k memory.

How do you calculate data line for memory?

If n=2, you can address 2 locations (0, 1, 2, and 3). As you can see, number of addressable locations = n^2. This means that n=log(1024) to the base 2. Thus, n=10.

What is the starting address of 8 KB memory chip if the last address is Ffffh?

The starting address of an 8K byte memory chip that ends at FFFFH is E000H. 8K is 8192 (8 * 1024) which is 2000H. Subtract 2000H from FFFFH, and add 1, and you get E000H.

Is 8K available?

Sadly, there’s more or less no 8K content available at the moment. In the meantime, 8K TVs make themselves useful by upscaling 4K, HD and even standard-def content. That means you can expect a gloriously cinematic experience right now, even though 8K content is far from mainstream.

What is 2K 4K 8K resolution?

2K: 2048 x 1080 pixels; 4K or Ultra HD: 3840 x 2160 pixels; 8K: 7680 x 4320 pixels; 10K: 10240 x 4320 pixels.

How many address pins are there in a 2K by 8 chip?

\\$\\begingroup\\$ Each 2k-by-8 ROM chip should have 11 address pins, A0 to A10, 8 data pins, and an output enable. To “parallel connect” the address lines, connect the A0 pins on each chip together. This will be the A0 address line for the entire circuit.

How many address lines are there in 128kx8 B Ram?

Here is an example of 128KX8 b RAM memory using memory components 8KX8 b and decoders DEC 3/8 (see attachment1). Here we have 16 memory components and 13 address lines for each component. Here, each connector of every decoder is connected to one select line of each memory component (there are total 16 connectors and 16 memory components).

How many number of 2kx8 ROM can be used in 4×4 decoder?

Use 2:4 Decoder and 16 2kx8 ROM in 4×4 manner. ie, for each decoder line enable 4 number of 2kx8 ROM will be enabled. Thanks for contributing an answer to Electrical Engineering Stack Exchange!

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.

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