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Flip-Flops and LatchesAn SR latch is shown in figure13.3. The latch Truth table is shown in the following table.
The two inputs, S and R denote ``set'' and ``reset'' respectively. The latch has memory, and
the present output is dependent on the state of the latch. Thus the output at instant,
denoted by is dependent on output at instant, denoted by .
Figure 13.3: Construction of a latch from NOR gates
Students should verify the veracity of the truth table from the figure 13.3.
S R
1 0 1 0
0 1 0 1
1 1 0 0
0 0
Note that in state, both and are 0, which seems absurd. Thus,conventionally, the state is said to be ``not allowed''.
A similar latch, known as latch is constructed using NAND gates (as opposed to NOR
gates for latch). The students should again check that the working of the latch coheres
with that of the truth table.
Figure 13.4: Construction of a latch from NAND gates
0 1 1 0
1 0 0 1
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0 0 1 1
1 1
To avoid ``race'' between the inputs, to have a control on when the input affects the latch,
the circuit13.5 is often implemented.
Figure 13.5: Circuit to avoid ``race'' condition
The inputs have an effect on the latch only when , otherwise, the previous state is
maintained. The input may be a clock, so that whatever transitions in and take place
before the clock changes to do not affect the outputs, and only when the inputs have
become stable is the system affected.
Digital CircuitsBoolean Operators:Single Input Single Output: Operation A and B take binary (0,1) values.
eg. NOT operation12.1.
Figure 12.1: A NOT gate
Multiple Inputs Single Output:
Eg 1. OR operation 12.2.if either of ``or'' is
The truth table is as follows:
AB
C
0 0 0
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01
1
10
1
11
1
Truth table for OR gate
Figure 12.2: An OR gate
Eg 2. AND operation is shown in figure 12.3
Figure 12.3: An AND gate
AB
C
00
0
01
0
10
0
11
1
Truth table for AND gate
These were basic gates which are implemented using transistor and other devices.
The transistor implementation is shown if figure 12.4
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Figure 12.4: Implementation using a BJTIC based gates are shown in figure12.5
Figure 12.5: IC based gates
Other functions
NAND: Not + AND
Figure 12.6: NAND gate
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AB
C
00
1
0 1 1
10
1
11
0
Truth table for NAND gate
NOR gateNot+OR gate
Figure 12.7: NOR gate
AB
C
00
1
0
1
0
10
0
11
0
Truth table for NOR gate
X-OR gate
Exclusive-OR gate
Figure 12.8: X-OR gate
A B C
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00
0
01
1
10
1
11
0
Truth table for X-OR gate
X-NOR gate
Exclusive NOR
Figure 12.9: X-NOR gate
AB
C
00
1
01
0
10
0
11
1
Truth table for X-NOR gate.
Using NAND gates
NOT
Figure 12.10: Realizing a NOT gate using a NAND gate
OR The following statements are calledDeMorgan's Theorems and can be easily verifiedand extended for more than two variables.
(12.
1)
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(12.
2)
(12.
3)(12.
4)
In general:(12.
5)
Thus :(12.
6)
Now it is easy to see that , which can be checked from the truth tableeasily. The resulting realization of OR gate is shown in12.11
Figure 12.11: Realization of OR gate by NAND gates
AND gate
Figure 12.12: Realization of AND gate by NAND gates
X-OR gate
(12.7)
Clearly, this can be implemented using AND, NOT and OR gates, and hence can beimplemented using universal gates.
Figure 12.13: X-OR gate
X-NOR gate
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(12.8)
Again, this can be implemented using AND, NOT and OR gates, and hence can be
implemented using universal gates, i.e., NAND or NOR gates.
Figure 12.14: X-NOR gate
Boolean ExpressionsA general realization of a Boolean expression is shown in 12.15
Figure 12.15: Realization of a Boolean Expression: shown as a black box
Example:In a car, we have the following components:
ADay-night sensor: Day-1, Night-0
B
Lamps on: On-1, Off-0
CIgnition on: On-1, Off-0
D
Warning light for lamps-onIn this case, the truth table for the logic D would be
A B C D
0 0 0 00 0 1 1
0 1 0 1
0 1 1 0
1 0 0 0
1 0 1 0
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1 1 0 1
1 1 1 1
Therefore,
, which can be written as in the sum of product form. We arrive at this bylooking at the combinations when the outout is one.
We can alternatively, express this in the product of sums form by looking at the
combinatins when the outout is low as
Using SOP and POS, it can be implemented as follows:
Next, we will try to reduce the number of gates by combining terms suitably.
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We can get the above by clubbing the s in the k-map shown.
Now, if we club the zeroes together in the k-map,
Check that we get the same expression by simplifying the product of sums expression (byusing (X+Y)(X+Z)=X+YZ)
Multiplexer
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The truth table for the multiplexer is as follows:
0 0
0 1
1 0
1 1
Multiplexers (MUX)For logic function realizations, instead of logic gates, Multiplexers can also be used
Consider a boolean function f={1,2,6,7}. Here input variables are A,B,C.
Multiplexer schematic for multiplexer is shown in figure 13.1
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Figure 13.1: ( ) multiplexer implementation of logic function
Depending on the contorl input combination specfic input is connected to the single output
of multiplexer.
(13.1)
When multiplexer is used to implement the above function. We connect boolean logic
'1' at the inputs corresponding to control inputs ABC= 1, 2, 4, and 6. For all other inputboolean logic '0' is connected.
In case we take a multiplexer we can make as control input and then determinewhat should be connected at the inputs of multiplexer as shown below.
(13.2)
(13.3)
(13.
4)
(13.
5)
Multiplexers (MUX)For logic function realizations, instead of logic gates, Multiplexers can also be used
Consider a boolean function f={1,2,6,7}. Here input variables are A,B,C.
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Multiplexer schematic for multiplexer is shown in figure 13.1
Figure 13.1: ( ) multiplexer implementation of logic function
Depending on the contorl input combination specfic input is connected to the single outputof multiplexer.
(13.1)
When multiplexer is used to implement the above function. We connect boolean logic'1' at the inputs corresponding to control inputs ABC= 1, 2, 4, and 6. For all other input
boolean logic '0' is connected.
In case we take a multiplexer we can make as control input and then determinewhat should be connected at the inputs of multiplexer as shown below.
(13.
2)
(13.
3)
(13.
4)
(13.5)
The realization is shown in figure 13.2 using a Mux.
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Figure 13.2: Multiplexer ( ) Implementation
Flip-Flops and LatchesAn SR latch is shown in figure13.3. The latch Truth table is shown in the following table.The two inputs, S and R denote ``set'' and ``reset'' respectively. The latch has memory, and
the present output is dependent on the state of the latch. Thus the output at instant,
denoted by is dependent on output at instant, denoted by .
Figure 13.3: Construction of a latch from NOR gates
Students should verify the veracity of the truth table from the figure 13.3.
S R
1 0 1 0
0 1 0 1
1 1 0 0
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0 0
Note that in state, both and are 0, which seems absurd. Thus,
conventionally, the state is said to be ``not allowed''.A similar latch, known as latch is constructed using NAND gates (as opposed to NOR
gates for latch). The students should again check that the working of the latch cohereswith that of the truth table.
Figure 13.4: Construction of a latch from NAND gates
0 1 1 0
1 0 0 1
0 0 1 1
1 1
To avoid ``race'' between the inputs, to have a control on when the input affects the latch,
the circuit13.5 is often implemented.
Figure 13.5: Circuit to avoid ``race'' condition
The inputs have an effect on the latch only when , otherwise, the previous state is
maintained. The input may be a clock, so that whatever transitions in and take place
before the clock changes to do not affect the outputs, and only when the inputs have
become stable is the system affected.
Sequential circuits
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In the above circuit, we have the problem of multiple tranistions when the clock is active.
Master Slave Flip-Flop (S-R)
When , , and are both 1. Therefore, it is an undefined condition. Thiscan be eliminated by proper feedback.
for the above circuit, the truth table is
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1 1
0 1 0
1 0 1
0 0
The problem with the circuit shown above is that when clock =1, the feedback will causeoscillatinons and when clock goes zero, the predicting the ouput state is difficult. On the
other hand, master slave configuration does not allow oscillation.
Edge triggered Flip-Flop
The above diagram shows a positive edge triggered flip-flop. The truth table is as follows
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0 1 0 1
1 0 1 0
0 0
1 1
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