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Nand to nor circuit 
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#1
Apr2113, 06:15 AM

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I started Digital Logic this Spring at Uni and was generally thinking that is it possible to directly change a nand circuit to a nor circuit without making it again from the start,when i mean making a circuit I mean implementing it on paper not practically!
Like suppose we have a function in SOP form F=B'C'+A'C"+A'B' and I can implement it using NAND gates fine but what if I want to change that circuit directly into NOR how do I do that? 


#2
Apr2113, 06:27 AM

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#3
Apr2113, 06:41 AM

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#4
Apr2213, 12:08 AM

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Nand to nor circuit
You can implement the NAND function using only NOR gates.
De Morgan says (x.y)' = x' + y' But when an OR gate is not available, use NOR then invert its output using another NOR as an inverter. = ( (x' + y')' )' 


#5
Apr2213, 05:08 AM

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I know NAND=OR with inverted inputs and NOR=AND with inverted inputs 


#6
Apr2213, 06:49 AM

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Sorry, uknowwho. Of the three questions you have asked in this thread, I haven't been able to understand even one. I took a stab and answered what I thought could be a likely question. It seems I didn't guess right. 


#7
Apr2213, 07:54 AM

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#8
Apr2213, 09:41 AM

P: 25

Do I have to put invertors(bubbles) infront of the inputs and outputs of both the NAND's which were used to form the AND gate? You can see the orginal schemtic which I did 


#9
Apr2213, 09:43 AM

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#10
Apr2213, 10:01 AM

P: 25

Here's what I tried what you suggested



#11
Apr2213, 10:01 AM

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#12
Apr2213, 11:03 AM

P: 506

I think you need to revise your understanding of a couple of things:
The 'bubbles' that are part of the symbols for NAND and NOR gates are just part of the symbols for the respective components  you can't "draw them in" on inputs (otherwise you are specifying a different component which you haven't got access to and may not even be manufactured). If you want to invert a signal, you use an invertor, also called a NOT gate. NOT gates can be implemented by joining the inputs of either a NAND gate or a NOR gate, but you cannot implement a NOT gate with an AND or OR (or XOR) gate  look at the truth tables to understand why. This means that you can construct ANY logic circuit using only NAND gates, or only NOR gates, but not using only AND, OR or XOR gates. Your circuit with the NAND gates has two NAND gates connected in series, each with their inputs joined. What is the output from this section of circuit when the input is 1? And when the input is 0? If you want to build this circuit with NOR gates you might like to start with the fact that A'B' = (A + B)' and so can be implemented with a single NOR gate, although you will get to the same result by replacing the NANDs with 4 NORs and eliminating the series inverters. 


#13
Apr2213, 11:17 AM

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#14
Apr2213, 03:09 PM

P: 506

Well (i) I don't have any easy means of posting a sketch at the moment and (ii) I'm not going to do it for you anyway, you need to do it yourself. But I will help.
First of all there are 6 redundant gates in this circuit  eliminate them. Then replace every NAND gate with a NOR gate with each input and output connected through an inverter  or if you like, draw bubbles on each input and output (the output will therefore have two bubbles). Now you can simplify this circuit: wherever a bubble on an output connects to a bubble on an input, elminate them both. Note that you can only do this with the extra bubble on the outputs because one bubble is part of the circuitry of the NOR gate. Also where an inverter connects to a bubble, or a bubble to an inverter, eliminate them both. Your answer should have a total of 7 NOR gates, 2 of them wired as inverters. The effecient solution with NAND gates had a total of 9, 4 of them wired as inverters. This problem may be used to illustrate engineering efficiencies  if building a production run of this circuit using standard componenents, 9 NAND gates would require 3 packages (quad 2 input NAND) whereas 7 NOR gates would only require 2 (quad 2 input NOR). But the optimum solution would use 2 3input NORs and an inverter, which are available in a single package (CMOS 4000 or 4025). 


#15
May2613, 10:19 PM

P: 28

You might want to check out Logic Friday :)



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