RS flip flop

1
RS flip flop
Nand Gate
2
RS flip flop
0
R

Q
p
q

P
0
S
Let R and S initially be logic 0. This results in
Q and P being at a logic 1 (from the truth table)
Nand Gate
3
RS flip flop
0
1
R

Q
p
q

1
P
0
S
If Q and P are at logic 1 then. q and p must be
also !
Nand Gate
4
RS flip flop
0
1
R

Q
p
q

1
P
0
S
So if R 0, p1 then Q must be 1 And S 0, q1
then P must be 1 (from the truth table) therefore
Nand Gate
THIS IS A STABLE STATE
5
RS flip flop
0
1
R

Q
p
q

1
P
0
S
Nand Gate
Now lets change S to logic 1
6
RS flip flop
0
1
R

Q
p
q

1
P
1
S
Nand Gate
Now S 1, q1 so P must change .
7
RS flip flop
0
1
R

Q
p
q

0
P
1
S
Nand Gate
To a logic 0 But this means p must change
8
RS flip flop
0
1
R

Q
p
q

0
P
1
S
So now we have R 0, p 0 so Q1 S 1, q 1 so
P0
Nand Gate
9
RS flip flop
0
1
R

Q
p
q

0
P
1
S
So now we have R 0, p 0 so Q1 S 1, q 1 so
P0
Nand Gate
THIS IS A STABLE STATE
10
RS flip flop
0
1
R

Q
p
q

0
P
1
S
Nand Gate
Now lets change R to logic 1..
11
RS flip flop
1
1
R

Q
p
q

0
P
1
S
So now we have R 1, p 0 so Q1 S 1, q 1 so
P0
Nand Gate
THIS IS A STABLE STATE
12
RS flip flop
1
1
R

Q
p
q

0
P
1
S
Nand Gate
Now lets change S to logic 0..
13
RS flip flop
1
1
R

Q
p
q

0
P
0
S
Nand Gate
This means P goes to logic 1
14
RS flip flop
1
1
R

Q
p
q

1
P
0
S
Nand Gate
So now p must be logic 1..
15
RS flip flop
1
1
R

Q
p
q

1
P
0
S
Nand Gate
So now Q must be logic 0..
16
RS flip flop
1
0
R

Q
p
q

1
P
0
S
Nand Gate
So now q must also be logic 0..
17
RS flip flop
1
0
R

Q
p
q

1
P
0
S
So now we have R 1, p 1 so Q0 S 0, q 0 so
P1
Nand Gate
THIS IS A STABLE STATE
18
RS flip flop
1
0
R

Q
p
q

1
P
0
S
Nand Gate
Now lets change S back to logic 1..
19
RS flip flop
1
0
R

Q
p
q

1
P
1
S
Nand Gate
THIS IS A STABLE STATE
20
RS flip flop
RS
qp
Plotting a Karnaugh map for Q and P on the same
map..
21
RS flip flop
RS
qp
K-Map for QP Which are the Stable States ?
22
RS flip flop
RS
qp
K-Map for QP showing Stable States
23
RS flip flop
The RS flip flop is said to be in a Stable State
when qpQP !
24
Simple isnt it !