Engineering Very simple logic circuit question

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The discussion revolves around designing a logic circuit where the output, HOLD, is LOW only when the input LOAD is HIGH and READY is LOW. Participants explore using NAND gates to achieve this, noting that connecting two NAND gates can function as an inverter, which is crucial for the desired output. A Karnaugh map is referenced for minimizing the expression, indicating that understanding this tool is important for solving the problem. The conversation highlights the asymmetry between the inputs and the necessity of treating them differently in the circuit design. Ultimately, a solution is found that confirms the correct conditions for the output based on the truth table.
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Homework Statement



Draw a logic circuit for the case where the output, HOLD, is LOW only if the input, LOAD, is HIGH and the input, READY, is LOW.

Homework Equations



Truth tables for AND, OR, NAND, NOR, XOR, XNOR are known (I'm not re-typing them here).

The Attempt at a Solution



logic.JPG


My drawing above uses an XNOR gate. However, I wasn't sure if it satisfies the original question because the output of an XNOR gate can be low for two conditions per the truth table below.

XNOR%20Truth%20Table.PNG


Is there any way to design a logic circuit to satisfy the top input being *only* high and the bottom input being *only* low? In other words, I need the XNOR functionality excluding the second row where A=0, B=1, and Output=0.
 
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No that would be wrong because of the reason you give.

Are you allowed to you use more than one gate? It can be done with two NAND gates for example.
 
There is nothing in the textbook that says I must use only one gate.
 
CWatters said:
It can be done with two NAND gates for example.

How would I chain two NANDs together to get the correct result?

Could I take the NAND output from the LOAD and READY and then push that output into an inverter?

Like this:

combo.JPG
 
JJBladester said:
How would I chain two NANDs together to get the correct result?

Could I take the NAND output from the LOAD and READY and then push that output into an inverter?

Like this:

combo.JPG

Inverting a NAND gives you an AND gate. You tell us, would that work? Are you just guessing? Have you studied Karnaugh maps yet?
 
I have studied Karnaugh maps. That is the chapter I'm on. You're right about the NAND + inverter equaling an AND gate. I see that now.

I know how to plug create a Karnaugh map and get a minimized expresison, but I don't know what to do with that expression. I'm not being lazy, LCKurtz. I'm just unsure of what to do in this problem and need some help. This whole topic is brand new to me.

Your response was a bit rude, but thanks for pointing me in the direction of the Karnaugh maps.
 
JJBladester said:
Your response was a bit rude...

I'm done helping you.
 
If you want to go by intuition, then you may want to notice that there is an asymmetry between the two inputs, which means that your circuit must treat the two input differently if you use standard gates as these all have symmetric inputs.
 
How would I chain two NANDs together to get the correct result?

A hint...

Nand gate truth table..

04124.png


What do you get if you connect both inputs to the READY signal?
 
  • #10
CWatters said:
A hint...

What do you get if you connect both inputs to the READY signal?

Connecting the READY signal to two NAND gate inputs yields a low. Also, this is the only case for a low in the AND gate (when the inputs are both high). Thanks CWatters... This is the push I needed.

I figured out the solution. It helped that I just finished a lab on combinational gates which I hadn't encountered prior to today.

My solution:

Two%2520NANDs.png


The truth table for this diagram shows that a low is possible *only* when LOAD is HIGH and READY is LOW.
 
  • #11
Yes that's exactly the circuit I was hinting at.

Connecting the READY signal to two NAND gate inputs yields a low

Connecting two inputs together turns a NAND gate into an inverter. You could also achieve the same effect by connecting one of the inputs to Vcc = Logic 1. That would have the slight advantage that the load on the READY signal would be just one input rather than two.
 
  • #12
CWatters said:
You could also achieve the same effect by connecting one of the inputs to Vcc = Logic 1. That would have the slight advantage that the load on the READY signal would be just one input rather than two.

Very interesting... Thanks for the insight CWatters. I don't know why LCKurtz was so frustrated with me. I wouldn't want him as a teacher.

I see from the truth table what you're talking about in terms of using a NAND gate as an inverter. I'm assuming that building a circuit strictly out of NAND gates would be much cheaper than out of AND, OR, inverter, etc. gates.
 
  • #13
JJBladester said:
I don't know why LCKurtz was so frustrated with me. I wouldn't want him as a teacher.

This is not acceptable on this forum. Another insult like that and this thread will be locked.
 
  • #14
JJBladester said:
I see from the truth table what you're talking about in terms of using a NAND gate as an inverter. I'm assuming that building a circuit strictly out of NAND gates would be much cheaper than out of AND, OR, inverter, etc. gates.

Depends. In the old days when everything was built using packages containing 3 or 4 identical gates you tried to use gates of the same type so only one package was needed. Saved cost and space on your printed circuit board. If you already had NAND gates in the design the chances are you had some that were unused so adding one more of the same type was sometimes free. These days your bit of logic might be burried away in a programable device alongside a 32bit microprocessor and a few hundred thousand or more programmable gates. The cost of adding an inverter or NAND gate might be just a few clicks of a mouse.
 

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