Digital circuit design

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6 Replies - 3641 Views - Last Post: 20 May 2010 - 11:29 AM

#1 NeoTifa  Icon User is online

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Digital circuit design

Posted 13 May 2010 - 06:49 PM

Okay, so remember when I tweeted about a 2-bit binary comparator I made? Well, now I'm trying to simulate a 6-bit binary comparator x__x. Basically, I'm using a hierarchal design, using individual 1-bit comparators, so I can make an n-bit comparator.

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So my input is 2 1-bit numbers to compare, so if A > B, then the top input gives a one, which leads to an OR gate. The A = B output leads to the EN (enable) of another comparator. At the end, they're all combined and it outputs the bigger number. What I want to know is.... what the fuck am I supposed to do with the last A = B?! XD This is what I have. Does this make sense? (My TA is gone for the night)

Thanks!

6-bit Or gate (they only take 2 inputs at a time)
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The output turning the A > B argument into a 6-bit output. (those are 2-to-1 MUX's btw)
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The whole circuit
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The part where I need the most help is in between the 6bitout and 6bitor.

This post has been edited by NeoTifa: 13 May 2010 - 06:51 PM


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#2 SpeedisaVirus  Icon User is offline

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Re: Digital circuit design

Posted 14 May 2010 - 10:52 AM

Ugh...hate digital design. Trying to deal with a bunch of asmd stuff right now :-/
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#3 NeoTifa  Icon User is online

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Re: Digital circuit design

Posted 14 May 2010 - 06:33 PM

:/ Well, how the fuck does that help me?! D:
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#4 SpeedisaVirus  Icon User is offline

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Re: Digital circuit design

Posted 15 May 2010 - 03:14 PM

Sheesh, sorry :dontgetit: I'm no superstar on it. Had to design on an exam which was miserable, but the notes I had helped helped some.
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#5 NeoTifa  Icon User is online

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Re: Digital circuit design

Posted 16 May 2010 - 06:15 PM

That's a step ahead of this lab (we learned it the next day). Just.... does my logic seem right concerning the last A = B?
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#6 NeoTifa  Icon User is online

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Re: Digital circuit design

Posted 19 May 2010 - 07:15 AM

Nevermind, talked to the professor. Turns out that A = B isn't necessary at the end, rendering the 2x1 MUX superfluous. Thanks anyways.
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#7 NeoTifa  Icon User is online

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Re: Digital circuit design

Posted 20 May 2010 - 11:29 AM

Mkay, I have another question. I'm trying to get ahead on my lab tonight, and this is what I have so far.

Quote

Hex Calculator
Consider a basic calculator that performs basic arithmetic operations on integers in 2s-complement representation. The output (display) of the calculator is an n-bit value that represents the current calculated value which itself will be stored in an n-bit register. The input is an n-bit value and a 2-bit function control (S1S0). During each clock period, the calculator executes exactly one of four possible options. When S = 00, the calculator ignores the input and continues to hold its current display value. When S = 01, the calculator clears the currently stored value to 0. When S = 10, the calculator replaces the value stored in the register with the sum of the currently held value and the input. When S = 11, the calculator also hold its current display value.

• (10 pts.) Design and simulate a 1-bit Hex Calculator. This device will have 3 inputs (Data input D0, and control inputs S1 and S0) and 1 output (Display output Z0). For storage, use a positive-edged D-type flip-flop.
• (10 pts.) Design and simulate the components necessary to extend your 1-bit Hex Calculator into an 8-bit Hex calculator. Design a high-level schematic that shows the implementation of your 8-bit Hex Calculator. Design, simulate, and test the necessary hierarchal modules for your design. These modules might include an 8-bit register, an 8-bit adder, a 2-to-1 8-bit multiplexer, and other devices that you feel may be necessary to implement your final solution.


Here is my work so far... Does this seem logical?

This post has been edited by NeoTifa: 20 May 2010 - 11:31 AM

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