## Вход

embed
```Sequential Circuits
Flip Flops
1
Flip Flops
 A flip-flop is a one bit memory that is edge-triggered or edge-sensitive
memory element which means it active only at transitions; i.e. either from 0
 1 or 1  0
 The flip-flop is at a “set state” if it forces the output to logic 1 and it is at a
“reset state” if it forces the output to logic 0.
level
positive (rising) edge
negative (falling) edge
2
Set Reset Flip-Flop (SR FF)
• Two cross coupled NAND gates.
• Two inputs S and R.
• C is a control input (enable) to determine when
the state of the filp-flop can be changed.
Q
S
SR
C
R
FF
Q
S
Q
C
Q
R
C
S
R
Qn+1
0
X
X
No change
1
0
0
No change
1
0
1
0
Reset state
1
1
0
1
Set state
1
1
1
Undefined
3
Now we will study the case when we suppose we have two cases for the
present state and then apply the inputs.
SR
d ≡ don’t care
10
0d
1
0
d0
01
Excitation Table
Present state Qn
The State diagram
S
R
Next state Qn+1
0
0
0
0
0
0
1
1
0
1
0
0
0
1
1
0
1
0
0
1
1
0
1
1
1
1
0
x
1
1
1
x
No change
Reset state
Set state
Not allowed
4
To get the next state Q n +1
0
1
0
0
1
1
x
x
Then Boolean Expression is
Q n +1 = S + Qn
5
D flip-flop
• D flip-Flop is designed to eliminate the undefined
state in the SR FF When (S=R=1).
Q
D
D
C
D
Qn+1
0
X
No change
1
0
0
Reset state
1
1
1
Set state
C
FF
Q
D
Q
C
Q
6
To study the case when we consider that we have two cases for the present
state and then apply the inputs:
CD
11
0d,11
0d,10
Excitation Table
C
0
D
x
Qn
0
1
Qn+1
0
0
0
x
1
1
1
0
0
0
1
0
1
0
1
1
0
1
1
1
1
1
10
Hold
The State diagram
Reset
Set
7
1
1
1
1
Qn+1 = CD + Qn
When the FF is enabled, C = 1
Qn+1 = 1.D + 0.Qn = D
8
JK Flip-Flop
• It is similar to SR with J replaced by S and K replaced by R except for
undefined state.
J
K
Qn+1
0
0
No change
0
1
0
Reset state
1
0
1
Set state
1
1
Complement
Q
J
JK
C
K
FF
Q
Toggle
9
To study the case when we suppose that we have two cases for the present
state and then apply the inputs:
JK
10,11
Excitation Table
0d
J
K
Qn
Qn+1
0
0
0
0
0
1
0
1
0
0
1
1
0
1
0
0
Reset state
1
1
0
0
0
1
1
1
Set state
1
1
1
1
0
1
1
0
Toggle
d0
1
0
01,11
Hold
The State diagram
1
1
1
Qn+1 = Qn +
1
J
10
T Flip-Flop
It is equivalent to JK FF with input J= K=T
T
Qn+1
0
Qn
1
Q
T
Excitation Table
T
C
No change
FF
Q
Complement
Since it is equivalent to Jk, we can use the same characteristic equation to find
the next state:
Qn+1 = Qn + J
= Qn + T
11
```
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