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The minimum number of NAND gates required to implement A+AB'+AB'C ? Plz explain aslo...
Dear studentMinimal expression is ANo. Of nand gates required is 0Regards
Dear student Above ans is incorrect There are two addition Terms and Three inputsInputs are a,b,cfor b to b’ conversion one NAND gate is requiredFor adding two numbers 3 NAND Gates are requiredTherefore split the term as Y = a + ab’ + ab’c asZ = b’ —- One NAND GateZ1 = ab’ is an and operation . this requires 3 NAND gates of 2 inputsZ2 = ab’ + ab’c = Z1(1+c) = Z1 since 1 + c in digital is 1theretofore Y becomes a + Z1 Totally 4 NAND gates of two input is the Minimal requirementThere are other different methods too for implementation of this function
There are two addition Terms and Three inputs
Inputs are a,b,c
for b to b’ conversion one NAND gate is required
For adding two numbers 3 NAND Gates are required
Therefore split the term as Y = a + ab’ + ab’c as
Z = b’ —- One NAND Gate
Z1 = ab’ is an and operation . this requires 3 NAND gates of 2 inputs
Z2 = ab’ + ab’c = Z1(1+c) = Z1 since 1 + c in digital is 1
theretofore Y becomes a + Z1 Totally 4 NAND gates of two input is the Minimal requirement
There are other different methods too for implementation of this function
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