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Can anyone answer the question below? I have links to pictures of the question and answer. I don't understand the answer provided by the book. Why does it say in the answer, "The EPROM will assert whenever CS4 is asserted"? I thought EN had to be pulled low?

Also, in the answer, where do they get 100X XXXX from? What are those eight bits representing?

A microprocessor (μp) uses a 16-bit address bus and an 8-bit data bus. The address bus lines are labeled A15 to A0, where A15 is the most significant address bit and A0 is the least significant address bit. The microprocessor generates an active-low address strobe (AS) at the beginning of each memory access cycle to indicate that a valid address has been placed on the address bus. The AS signal is asserted for the entire memory access cycle.

In a certain application of the μp, the three most significant address bits (A15-A13) are used as inputs to a 3-of-8 decoder to generate chip-select signals (CS0 - CS7). The chip-select signals are used to enable an EPROM memory device whose address lines are tied directly to the lower 13 address lines (A12-A0) of the μp. The EPROM's output lines are enabled whenever its enable (EN) is pulled low. A figure of the microprocessor described above and the decoder truth tables are shown at the below link.

The range of addresses (expressed in hexadecimal) to which the EPROM in the figure will respond is most nearly:

A. 0000 to 1FFF

B. 4000 to 5FFF

C: 8000 to 9FFF

D: E000 to FFFF

Question: https://ibb.co/CvXr914

Figure and truth table: https://ibb.co/JdRz0XX

Answer: https://ibb.co/0BW3K27

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    \$\begingroup\$ What do you not understand? Try asking a specific question, instead of asking to write an essay to you how it works. \$\endgroup\$
    – Justme
    Commented Jan 22, 2022 at 17:30
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    \$\begingroup\$ What about the answer don't you understand? Be more specific. \$\endgroup\$
    – user16324
    Commented Jan 22, 2022 at 17:36

1 Answer 1

Reset to default
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Figure out the base address. What is the address with:

  • A[15..13] selecting CS4n asserted (that is, low)
  • A[12..0] all zero

Match that to one of the choices given.

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  • \$\begingroup\$ Thank you. CS4 asserted means that it is low? Or is it CS4n? What does CS4n mean? A[15..13]=100 to select CS4 low. But how do they get 100X XXXX in the answer? What are those 8 bits? A[15..8]? Why is it 8 bits? Because the data bus is 8 bits? \$\endgroup\$
    – pat10
    Commented Jan 24, 2022 at 21:20
  • \$\begingroup\$ In their document it’s just CS4. Industry convention adds some indication of active-low, such as CS4n, CS4_N, CS4#, CS4*, etc., as a reminder to the reader. But one goal of the question is that you understand that CS4 is in fact low-true, regardless of how it’s named. \$\endgroup\$
    – hacktastical
    Commented Jan 24, 2022 at 21:25
  • \$\begingroup\$ And yes, we infer that each addressed element is 8 bits (one byte), so the total address space is 64KB, and each CS segment is 8K. \$\endgroup\$
    – hacktastical
    Commented Jan 24, 2022 at 21:32
  • \$\begingroup\$ Okay, thank you. So A[15..13] = 100, and A[12..8] is don't care? \$\endgroup\$
    – pat10
    Commented Jan 24, 2022 at 21:51
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    \$\begingroup\$ The data bus is 8 bits wide, so the minimum addressable unit is 1 byte. There are 16 address bits, so total address space is 2^16, or 65536 bytes (64K). The top three address bits feed a 3-to-8 decoder, dividing the address space into eight 8K segments. One of those segments, CS4, selects an EPROM with 13 address bits, so 2^13 (8192), or 8K bytes. By itself, the EPROM has an extent of 0x0000 ~ 0x1FFF. CS4 maps the EPROM in an 8K segment starting at 0x8000 and has an extent of 8K, so the range is 0x8000 ~ 0x9FFF \$\endgroup\$
    – hacktastical
    Commented Jan 27, 2022 at 17:32

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