Unlocking the Power of NAND Gates:
NAND gate:
This type of gate is a NOT-AND gate, which is equivalent to an AND gate
then a NOT gate. If any of the inputs are low, the output of the NAND gate
is high. An AND gate with a tiny circle on the output serves as the sign.
The little circle is a representation of inversion.
|
A |
B |
~(A.B) |
|
0 |
0 |
1 |
|
0 |
1 |
1 |
|
1 |
0 |
1 |
|
1 |
1 |
0 |
Diagram:
Objective:
It can be fairly determined that every Boolean function can be
implemented with NAND gates if we are able to show that the logical
operations AND, OR, XOR and XNOR can be implemented with NAND
gates.
IC Number:
The 7400
series is one of the widely used ICs that contains NAND gates.
1: Press
on “P” at top Left corner .
2: Type
7400 and press Enter Key.
3: Final
Result
(3)
Task 1 :
Verification of AND function
Observation:
· By giving input ‘0’ and ‘0’ the output was ‘0’.
· By giving input ‘0’ and ‘1’ and output was ‘0’.
· By giving input ‘1’ and ‘0’ and output was ‘0’.
· By giving input ‘1’ and ‘1’ and output was ‘1’
· Above steps verify that it’s an AND gate.
|
A |
B |
F=A.B |
|
0 |
0 |
0 |
|
0 |
1 |
0 |
|
1 |
0 |
0 |
|
1 |
1 |
1 |
Observation Table:
|
A |
B |
A.B |
M=~(A.B) |
~(M) |
|
0 |
0 |
0 |
1 |
0 |
|
0 |
1 |
0 |
1 |
0 |
|
1 |
0 |
0 |
1 |
0 |
|
1 |
1 |
1 |
0 |
1 |
Note (Bar or ~ on A.B (
AND gate uning NAND:
Task 2 :
Verification of OR function
Observation:
· By giving input ‘0’ and ‘0’ the output was ‘0’.
· By giving input ‘0’ and ‘1’ and output was ‘1’.
· By giving input ‘1’ and ‘0’ and output was ‘1’.
· By giving input ‘1’ and ‘1’ and output was ‘1’
· Above steps verify that it’s an OR gate.
|
A |
B |
F=A+B |
|
0 |
0 |
0 |
|
0 |
1 |
1 |
|
1 |
0 |
1 |
|
1 |
1 |
1 |
Observation Table:
|
A |
B |
X=~(A) |
Y=~(B) |
X.Y |
~(X.Y)=X+Y |
|
0 |
0 |
1 |
1 |
1 |
0 |
|
0 |
1 |
1 |
0 |
0 |
1 |
|
1 |
0 |
0 |
1 |
0 |
1 |
|
1 |
1 |
0 |
0 |
0 |
1 |
Step 1: Take Complement of A and Store in X.
Step 2: Take Complement of B and Store in Y.
Step 3: Take AND of X, Y.
Step 4: Complement of ~A.~B gives A+B.
OR gate uning NAND:
Task 3 :
Verification of XOR function
Observation:
· By giving input ‘0’ and ‘0’ the output was ‘0’.
· By giving input ‘0’ and ‘1’ and output was ‘1’.
· By giving input ‘1’ and ‘0’ and output was ‘1’.
· By giving input ‘1’ and ‘1’ and output was ‘0’
· Above steps verify that it’s an XOR gate.
|
A |
B |
A⊕B |
|
0 |
0 |
0 |
|
0 |
1 |
1 |
|
1 |
0 |
1 |
|
1 |
1 |
0 |
Observation Table:
|
A |
B |
C=NAND |
X=A.C |
Y=B.C |
~X |
~Y |
M=~X.~Y |
~M |
|
0 |
0 |
1 |
0 |
0 |
1 |
1 |
1 |
0 |
|
0 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
|
1 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
|
1 |
1 |
0 |
0 |
0 |
1 |
1 |
1 |
0 |
Step 1 : Column C store the Values
According to NAND operation .
Step 2 : Apply AND operation on A
and C , store them in Column X.
Step 3 : Apply AND operation on B
and C , store them in Column Y.
Step 4 : Take complement of X and Y
.
Step 5 : Again apply AND operation on Complement of X and Y, store
them in Column M.
Step 6 : Take Complement of M after that we get XOR by using only
NAND.
Task 4 : Verification of XNOR function
Observation:
· By giving input ‘0’ and ‘0’ the output was ‘1’.
· By giving input ‘0’ and ‘1’ and output was ‘0’.
· By giving input ‘1’ and ‘0’ and output was ‘0’.
· By giving input ‘1’ and ‘1’ and output was ‘1’
· Above steps verify that it’s an XNOR gate.
|
A |
B |
~(A⊕B) |
|
0 |
0 |
1 |
|
0 |
1 |
0 |
|
1 |
0 |
0 |
|
1 |
1 |
1 |
|
A |
B |
C=NAND |
X=A.C |
Y=B.C |
~X |
~Y |
M=~X.~Y |
|
|
0 |
0 |
1 |
0 |
0 |
1 |
1 |
1 |
|
|
0 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
|
|
1 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
|
|
1 |
1 |
0 |
0 |
0 |
1 |
1 |
1 |
|
|
M=~M |
~M |
|
||||||
|
0 |
1 |
|
||||||
|
1 |
0 |
|
||||||
|
1 |
0 |
|
||||||
|
0 |
1 |
|
||||||
Step1 : Column C store
the Values According to NAND operation .
Step2 : Apply AND operation on
A and C , store them in Column X.
Step 3 : Apply AND operation on B
and C , store them in Column Y.
Step 4 : Take complement of X and Y
.
Step 5 : Again apply AND operation on Complement of X and Y, store
them in Column M.
Step 6 : Take Complement of M.
Step 7 : Again Complement of M after that we get XNOR by using only
NAND.
XNOR gate uning NAND:
|
|
NAND |
NOR |
|
NOT |
1 |
1 |
|
AND |
2 |
3 |
|
OR |
3 |
2 |
|
XOR |
4 |
5 |
|
XNOR |
5 |
4 |
The table shows how many gates are needed to build a certain gate
using NAND or NOR. I-e
I need 2 NAND gates to established
AND gate or 3 NOR gates!
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