Why Does My 2:4 Decoder Solution Differ from the Book?

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Discussion Overview

The discussion revolves around the design of a 2:4 decoder circuit and the differences between a participant's solution and the one provided in a textbook. Participants explore the implications of their designs, focusing on Boolean logic, truth tables, and circuit implementation.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion regarding the textbook solution for a 2:4 decoder and requests clarification.
  • Another participant suggests constructing a Boolean truth table for the 2-bit circuit to understand the logic better.
  • A participant notes that their implementation of the decoder differs in the handling of output Y0, questioning the rationale behind the textbook's approach.
  • It is proposed that the textbook solution may prioritize ease of understanding and clarity in mapping the addressing sequence compared to the participant's design.
  • One participant argues that the textbook design aims to balance loading on inverted and non-inverted signals, which could affect output characteristics and propagation times.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the superiority of either design. Multiple competing views remain regarding the rationale behind the textbook solution versus the participant's approach.

Contextual Notes

Participants express uncertainty about the implications of their circuit designs, particularly concerning output characteristics and propagation times, without resolving these issues.

anhnha
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Hi.
I am confused a bit about the solution for 2:4 decoder in the picture below.
Please help me with the question in the figure.
Thanks.

attachment.php?attachmentid=69132&stc=1&d=1398579539.jpg
 

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Think of the binary equivalent, 2 bits has how many possible conditions? The circuit is a translation of the Boolean logic. work out your boolean truth table. of a 2 bit circuit, then draw the same circuit with and or gates.
 
Hi.
I understand all possible conditions. I didn't draw Y1, Y2, Y3 because they are same in terms of implementation.
My confusion about Y0.
I am wondering why the author used that solution not mine.
 
most likely ease of understanding the address bus relation. Notice that he's taken care to add both conditions of A0 and A1 in separate lines where you did an equivalent with use of strictly high inputs? In the first case its easier for someone new to understand. Its easier to correlate the truth table to desired conditions.

edit: he's mapped the addressing in the following sequence for visual ease.

00
01
10
11

You've mapped
11 with an inverted condition later on then
01
10
11 or at least I hope you did as you only showed one decoded gate
 
Last edited:
Thanks. I see no difference in functioning between my solution and the key in the book.
I also can derive the circuit given in the solution.
However, I can't figure out why the author used the key solution not the one I get above.
 
I would say that the book design is such to keep the loading on inverted and non-inverted signals equal. Also if you look at the book solution, the output is virtually identical for each pin. This could also be to keep the propagation times equal in all outputs. There are numerous ways to solve a problem, but sometimes you need to see the complete solution and keep things in sync and balanced as possible for other reasons.

Your output stage using 2 transistors to pull the signal to rail, will cause a different voltage output when a current is supplied on a "high" output. This results in different output characteristics for different pins, again not a good thing. You really want the outputs and prop times equal for the outputs.
 
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