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Investigative activities designed to develop your understanding of combinational Logic
ASSIGNMENT OBJECTIVES
In this assignment, for LO4, you will need to show that you can:
LO4.1Evaluate digital electronic device families.
LO4.2aDesign combinational digital electronic circuits.
LO4.2b Design sequential digital electronic circuits.
LO4.3Test digital circuits by construction or by computer simulation.
Solution:
TASK 1: (LO4.1)
Solution –
- The Nominal values for the logic families
The nominal values for logic families
The typical value for the characteristics of the following logic families is as given below –(Maini, 2007), (121/D, 2000), (Instruments, 2002)
Characteristics |
LS TTL |
ALS TTL |
CMOS |
HCMOS |
V_{IL} |
0.8 V |
0.8 V |
1.5 V |
1.0 V |
V_{IH} |
2.0 V |
2.0 V |
3.5 V |
3.5 V |
I_{IL} |
0.4 mA |
0.1 mA |
1 uA |
1 uA |
I_{IH} |
20 uA |
20 uA |
1 uA |
1 uA |
V_{OL} |
0.5 V |
0.5 V |
0.5 V |
0.1 V |
V_{OH} |
2.7 V |
2.75 V |
4.5 V |
4.9 V |
I_{OL} |
8 mA |
8 mA |
0.4 mA |
4 mA |
I_{OH} |
0.4 mA |
0.4 mA |
0.4 mA |
4 mA |
Fan-out |
20 |
20 |
10 |
20 |
Noise Margin |
0.3 V |
0.3 V |
V_{NH}=1.4 V V_{NL}=1.4 V |
V_{NH}=1.4 V V_{NL}=0.9 V |
- Definition of the terms Fan-out and Noise Margin.
Fan-out –
The Fan-out is a term which defines maximum number of digital inputs that output of a single logic gate can provide. The most of the TTL gates can provide up to 10 other digital gate or device. So, the TTL gate is having fan-out of 10. (Definition of fan-out, 2017)
Noise Margin –
The Noise margin can be defined as the amount of noise that CMOS circuit can resist without any compromise in the working of the circuit. The Noise margin ensures that a signal that is logic ‘1’ with a definite noise added to it, is accepted as logic ‘1’ and not as logic’0’. This is mainly the distinction between value of signal and noise value. (VLSI System Design - An IIT Alumnus Initiative.)
The noise margin can be defined as –
TASK 2: (LO4.2a & LO4.3a)
A circuit whose output is HIGH when the inputs are –
A is high and B is low and either input C or input D is also HIGH.
- The Boolean expression that describes the performance of the required circuit,
The Boolean expression
- The expression in Sum-of-Products (SOP) form –
The SOP form
This can be easily get by the truth table –
Input A |
B |
C |
D |
Output (Y) |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
0 |
The k-map,
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
1 |
- By simulation software, we draw circuit diagram and simulate its operation,
The circuit schematic –
The output waveform –
Case A)
A=1, B=0, C=0, D=1
Case B)
A=1, B=0, C=1, D=0
Case C)
A=1, B=0, C=1, D=1
- A four-variable truth table and apply all combinations of input variables to the circuit and each time the logical state are noted –
The output for each possible input and its respective output
Input A |
B |
C |
D |
Output (Y) |
0 |
0 |
0 |
0 |
90 mV |
0 |
0 |
0 |
1 |
0.08 V |
0 |
0 |
1 |
0 |
0.08 V |
0 |
0 |
1 |
1 |
0.08 V |
0 |
1 |
0 |
0 |
0.08 V |
0 |
1 |
0 |
1 |
0.08 V |
0 |
1 |
1 |
0 |
0.08 V |
0 |
1 |
1 |
1 |
0.08 V |
1 |
0 |
0 |
0 |
0.08 V |
1 |
0 |
0 |
1 |
1.3 V |
1 |
0 |
1 |
0 |
1.3 V |
1 |
0 |
1 |
1 |
1.3 V |
1 |
1 |
0 |
0 |
0.08 V |
1 |
1 |
0 |
1 |
0.08 V |
1 |
1 |
1 |
0 |
0.08 V |
1 |
1 |
1 |
1 |
0.08 V |
- The circuit operation –
The Circuit operation
The circuit consist of A and B^{— }and B is passed through the inverter to get the inverted B. then the inverted B and A are passed to the AND gate. On the other path C and D are passed to the OR gate. The output of AND gate and output of OR gate is passed to the output AND gate to get the final output Y.
The final output will give High value only when (A is High and B is low) and either of C or D is High. This is clarified by the truth table, simulation, and k-map.
Opportunity for Distinction (D2) –
The all parts of Task 2 are completed.
- Boolean expression is defined for the required operation.
- The Sum of product is also determined.
- The circuit is designed and simulated. The simulation results for high output are noted as waveform and all others values are noted in the truth table.
- Four-variable truth table is made and applies all combinations of input variables to circuit and each time logical state are noted in the table.
- The circuit operation is discussed and explained.
Opportunity of Merit (M2) –
The Karnaugh Maps to check the SOP found is equivalent to original Boolean expression.
The k-map,
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
1 |
TASK 3: (L04.2b & L04.3b)
- Design of 4 bit asynchronous counter using JK flip flops to count between 2 and 9 –
State or Flow diagram,
The Sequence table –
Present State |
Next State |
0010 |
0011 |
0011 |
0100 |
0100 |
0101 |
0101 |
0110 |
0110 |
0111 |
0111 |
1000 |
1000 |
1001 |
1001 |
0010 |
The excitation table –
Present State |
Next State |
Flip Flop Input |
|||||||
JD |
KD |
JC |
KC |
JB |
KB |
JA |
KA |
||
0000 |
0010 |
0 |
X |
0 |
X |
1 |
X |
0 |
X |
0001 |
0010 |
0 |
X |
0 |
X |
1 |
X |
X |
1 |
0010 |
0011 |
0 |
X |
0 |
X |
X |
0 |
1 |
X |
0011 |
0100 |
0 |
X |
1 |
X |
X |
1 |
X |
1 |
0100 |
0101 |
0 |
X |
X |
0 |
0 |
X |
1 |
X |
0101 |
0110 |
0 |
X |
X |
0 |
1 |
X |
X |
1 |
0110 |
0111 |
0 |
X |
X |
0 |
X |
0 |
1 |
X |
0111 |
1000 |
1 |
X |
X |
1 |
X |
1 |
X |
1 |
1000 |
1001 |
X |
0 |
0 |
X |
0 |
X |
1 |
X |
1001 |
0010 |
X |
1 |
0 |
X |
1 |
X |
X |
1 |
1010 |
0010 |
X |
1 |
0 |
X |
X |
0 |
0 |
X |
1011 |
0010 |
X |
1 |
0 |
X |
X |
0 |
X |
1 |
1100 |
0010 |
X |
1 |
X |
1 |
1 |
X |
0 |
X |
1101 |
0010 |
X |
1 |
X |
1 |
1 |
X |
X |
1 |
1110 |
0010 |
X |
1 |
X |
1 |
X |
0 |
0 |
X |
1111 |
0010 |
X |
1 |
X |
1 |
X |
0 |
X |
1 |
The Karnaugh maps –
The k-map for J_{D} –
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
X |
X |
X |
X |
X |
X |
X |
X |
The k-map for K_{D} –
X |
X |
X |
X |
X |
X |
X |
X |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
The k-map for J_{C} –
0 |
0 |
1 |
0 |
X |
X |
X |
X |
X |
X |
X |
X |
0 |
0 |
0 |
0 |
The k-map for K_{C} –
X |
X |
X |
X |
0 |
0 |
1 |
0 |
X |
X |
X |
X |
1 |
1 |
1 |
1 |
The k-map for J_{B} –
1 |
1 |
X |
X |
0 |
1 |
X |
X |
1 |
1 |
X |
X |
0 |
1 |
X |
X |
The k-map for K_{B} –
X |
X |
1 |
0 |
X |
X |
1 |
0 |
X |
X |
0 |
0 |
X |
X |
0 |
0 |
The k-map for J_{A} –
0 |
X |
X |
1 |
1 |
X |
X |
1 |
0 |
X |
X |
0 |
1 |
X |
X |
0 |
The k-map for K_{A} –
X |
1 |
1 |
X |
X |
1 |
1 |
X |
X |
1 |
1 |
X |
X |
X |
1 |
X |
- The circuit simulation of sequence counter –
The sequence counter
The output of the above sequence detector
Clock |
Present state |
Next state |
1 |
0000 |
0010 |
0 |
0010 |
0010 |
1 |
0010 |
0011 |
0 |
0011 |
0011 |
1 |
0011 |
0100 |
0 |
0100 |
0100 |
1 |
0101 |
0110 |
0 |
0110 |
0110 |
1 |
0110 |
0111 |
0 |
0111 |
0111 |
1 |
0111 |
1000 |
0 |
1000 |
1000 |
1 |
1000 |
1001 |
0 |
1001 |
1001 |
1 |
1001 |
0010 |
0 |
0010 |
0010 |
1 |
0010 |
0011 |
0 |
0011 |
0011 |
Opportunity for Merit (M3) –
The report is produces as shown in this task or work.
Opportunity for Distinction (D2)
The circuit is design and demonstrated with the simulation circuit and its output.
Bibliography
121/D, D. (2000, Jan). LS TTL Data. Retrieved from Rev. 6: http://ecee.colorado.edu/~mcclurel/ON_Semiconductor_LSTTL_Data_DL121-D.pdf
Instruments, T. (2002, Sept). HCMOS Design Considerations. Retrieved from SCLA007A: http://www.ti.com/lit/an/scla007a/scla007a.pdf
Maini, A. K. (2007). Digital Electronics : Principles, Devices and Apllications. John Wiley & Sons.