[SOLVED] CMOS design question

Status
Not open for further replies.

user7813

Newbie level 3
Joined
Feb 11, 2015
Messages
3
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Visit site
Activity points
24
I have an AND gate with 4 inputs. How could this be implemented in CMOS with minimal switching delay?

Code:
                            (+)
                             |
input 0 (inverted) -----+-- <--
                        |    |
input 1 ------------+------ -->
                    |   |    |
input 2 --------+---------- -->
                |   |   |    |
input 3 ----+-------------- -->
            |   |   |   |    |
            |   |   |   |    |
            |   |   |   |    +----+----+----+----+------ output
            |   |   |   |         |    |    |    |
            |   |   |   +------- -->   |    |    |
            |   |   |             |    |    |    |
            |   |   + --------------- <--   |    |
            |   |   |             |    |    |    |
            |   +------------------------- <--   |
            |                     |    |    |    |
            +---------------------------------- <--
                                  |    |    |    |
                                 (-)  (-)  (-)  (-)

There are at least 3 things, where I am not certain if it will work:
I) can I put 4 transistors in series, or does this require 4 times the voltage?
II) can I turn NAND into AND, simply by swapping (+) and (-)?
III) can I invert the input signal by changing the type of transistor?

no I have no chip design experience, but I have done some VHDL so far.
 

In very simple and top level terms, to have min. sw. delay use a min. no. of transistors in your design. Have you tried building various configs of the 4 i/p AND gate in your CAD tool and then measuring the delay in each case?
And how to build a 4 i/p AND gate using pMOS and nMOS transistors is covered in any basic graduate level VLSI test book.
 
Every resource I found in the internet says, that:
I) 4-input AND can only be done by 3 NAND/NOR gates in a tree-structure
II) to get a single AND, the only way is to combine NAND and inverter gates
III) there are constraints where n-type and p-type transistors can be used

Can you explain why?
 

Initially you had some set of questions, and now they have changed!
You mention in the above post "Every resource I found in the internet says, that:..........."
So first let me ask you an honest and direct question. What is it you really want? -- some problem you don't understand or do you want us to explain & qualify for something which you have already found in the internet?

Going with your original question, as I have understood, your problem is to design a 4 i/p AND gate that is having min. sw. delay. I have already mentioned to use the minimum no. of transistors.
If I were the designer, I would use the very basic approach of a 4 input NAND (4 pmos & 4 nmos) connected to a NOT/inverter (1 pmos & 1 nmos) gate. Voltage shouldn't be a big problem for just 4 inputs (experts should correct me if I am wrong here).

As for your other questions, I repeat, please study and *understand* your graduate level analog CMOS design or VLSI design textbook. Please go back and understand your basics.
 
I am no engineer. I only have fun in understanding how things are working.

Initially you had some set of questions, and now they have changed!
You mention in the above post "Every resource I found in the internet says, that:..........."

Agreed. First I read your post, and THEN I searched the internet

If I were the designer, I would use the very basic approach of a 4 input NAND (4 pmos & 4 nmos) connected to a NOT/inverter (1 pmos & 1 nmos) gate. Voltage shouldn't be a big problem for just 4 inputs (experts should correct me if I am wrong here).

What I understand now:
I) the number of transistors in series between (+)/(-) and output directly affects switching time
II/III) p-type must be on the (+) side, n-type on the (-) side

Can you confirm this?



PS: Now I use 3 parallel NAND/NOR/inverter trees and multiplex by a skewed clock to achieve the same result. However, the original question is still interesting to me.
 

Status
Not open for further replies.
Cookies are required to use this site. You must accept them to continue using the site. Learn more…