Is it Possible to Solve This Problem - Equations and Building Circuits

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SUMMARY

The discussion centers on the feasibility of constructing three specific logical functions (F, G, H) using only four logic chips from a specified list. The functions are defined as F=(D#B)+(C#A), G=DCA'+DB'+CB'A', and H=DC'B+D'C'A+D'BA+C'BA. The available chips include 7400, 7402, 7404, 7408, 7410, 7411, 7420, 7421, 7425, 7427, 7432, and 7486. Participants conclude that achieving this with the given constraints is not possible, as the complexity of the functions exceeds the capabilities of four chips.

PREREQUISITES
  • Understanding of Boolean algebra and logic gates
  • Familiarity with the function of logic chips such as 7400 series
  • Knowledge of circuit design principles
  • Experience with protoboard assembly
NEXT STEPS
  • Research the capabilities and limitations of 7400 series logic chips
  • Learn about combinational logic design techniques
  • Explore methods for simplifying Boolean expressions
  • Investigate alternative circuit design approaches for achieving complex functions
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Electronics enthusiasts, circuit designers, and students studying digital logic who are interested in practical applications of Boolean algebra and logic gate configurations.

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# -> means XOR
' => means NOT
AB => means A and B
A+B => means A or B

Is it possible to build the following 3 functions
F=(D#B)+(C#A)
G=DCA'+DB'+CB'A'
H=DC'B+D'C'A+D'BA+C'BA

Using only 4 chips (Build all three functions using only 4 chips not 4 for each function)

I have 8 switches for the inputs and was going to use
Switch 1 for A
Switch 2 for B
Switch 3 for C
Switch 4 for D
Switch 5 for A'
Switch 6 for B'
Switch 7 for C'
Switch 8 for D'

I don't have any other switches. I'm just using a switch and flipping switch 8 to the opposite of switch 4 to get the NOT of D.

I can only use the following chips. At the end of the day there can only be 4 chips on my protoboard so if I use the same chip say twice than this counts as two chips.

7400, 7402, 7404, 7408, 7410, 7411, 7420, 7421, 7425, 7427,
7432, and 7486.

I'm getting that it's not possible and have spent several hours trying to do it. Thanks for any help that you can provide me!
 
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GreenPrint said:
# -> means XOR
' => means NOT
AB => means A and B
A+B => means A or B

Is it possible to build the following 3 functions
F=(D#B)+(C#A)
G=DCA'+DB'+CB'A'
H=DC'B+D'C'A+D'BA+C'BA

Using only 4 chips (Build all three functions using only 4 chips not 4 for each function)

I have 8 switches for the inputs and was going to use
Switch 1 for A
Switch 2 for B
Switch 3 for C
Switch 4 for D
Switch 5 for A'
Switch 6 for B'
Switch 7 for C'
Switch 8 for D'

I don't have any other switches. I'm just using a switch and flipping switch 8 to the opposite of switch 4 to get the NOT of D.

I can only use the following chips. At the end of the day there can only be 4 chips on my protoboard so if I use the same chip say twice than this counts as two chips.

7400, 7402, 7404, 7408, 7410, 7411, 7420, 7421, 7425, 7427,
7432, and 7486.

I'm getting that it's not possible and have spent several hours trying to do it. Thanks for any help that you can provide me!

What is the closest that you've gotten? Can you post what you have tried?
 
GreenPrint said:
I have 8 switches for the inputs and was going to use
Switch 1 for A
Switch 2 for B
Switch 3 for C
Switch 4 for D
Switch 5 for A'
Switch 6 for B'
Switch 7 for C'
Switch 8 for D'
Normally you would use a switch to provide D, and use a gate to generate D'.
 

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