Hellooo new here. i went to desperate measures as its been impossible for me to find what to do via internet or book about this. professor is horrible and the book is worse at explaining this. anyway i hope someone can help me here i think its pretty simple.

well first i had to make a 4 to 1 multiplexer with 2 bit selector. i think i did it right.

and now i have to make a 4 to 1 multiplexer using an 8 bit bus. i searched the net and i can't find what 8 bit buses are and my book is retarded on it. please help. thanks

 

You could use 4 x CD4052 multiplexer IC's with some extra input logic.

See a data sheet at

http://www.fairchildsemi.com/ds/CD/CD4052BC.pdf

 

Hi,

An 8 bit bus is a colloection of 8 logical lines used to carry binary data. The lines can be named as D0, D1, D2, D3, D4, D5, D6, D7 (D = data), (lines 0 to 7). D0 = 1, D2 = 2, D3 = 4, D4 = 8, D5 = 16, D6 = 32, D7 = 64. With all lines on (high, logic 1), The data in binary will = 255 (1+2+4+8+16+32+64 = 255). So binary 00000000 = 0, 11111111 = 255, 00001111 = 15 (1+2+4+8) and so on.

What I think you are trying to do is decode part of the 8 bit bus and address the multiplexer. Am I right?

Daniel.

 

Hello FuuFuu

Do you know that multiplexers (MUX) are electronical switches. In a 4 to 1 MUX there are four inputs and one output. To select which input who shall bee connected to the output, you use a binary code on the "select inputs". To select 4 different inputs, you need two select lines S1 and S2. They can do four combinations, 00 01, 10 and 11. For 8 inputs you need 3 select lines and so on.

Doing a 4 to 1 MUX with a 8 bit bus seems to mee to bee confusing. What is the task for the MUX?

Gustav