Adders
DELAY IN HALF ADDER AND FULL ADDER
Delay - Definition:
Delay in an adder is the amount of time taken to produce a stable output (Sum and Carry) from the applied input.
Delay in a Half Adder:
Consider a half adder with $A, B$ as input and Sum, Carry as output.
In a half adder, when inputs A and B are applied, sum and carry will be computed in parallel. (i.e) sum generation and carry generation process are independent to each other. $$Sum=A\oplus B$$$$Carry=AB$$
Delay in a half adder is as follows,
Sum generation: $1$ EXOR gate delay to compute Sum. $$A\;to\;Sum\;\; (or)\;\; B\;to\;Sum$$
Carry generation: $1$ AND gate delay to compute Carry $(C_{out})$. $$A \;to\; C_{out}\;\; (or)\;\; B\; to\; C_{out}$$
Among them, EXOR gate delay is the largest. Hence in a half-adder, the maximum delay will be equal to one EXOR gate delay (A to Sum or B to Sum).
Practice Question
A half adder is implemented with one EXOR and one AND gate. The propagation delay of an EXOR gate is twice that of an AND gate. The propagation delay of an AND gate is $1.2$ microseconds.
What is the total propagation time taken for generating the sum and a carry?
Sum generation involves $1$ EXOR gate delay, which is equal to $1.2\;\mu s×2=2.4\;\mu s$
Carry generation involves $1$ AND gate delay, which is equal to $1.2\;\mu s$
Since Sum generation and Carry generation process are parallel, the total propagation time taken for generating sum and carry is the largest delay among them, which is $2.4\;\mu s$
Delay in a Full Adder:
Consider a full adder with $A, B, C_{in}$ as input and $Sum$, $C_{out}$ as output.
Case $1$: When all three inputs $A, B$ and $C_{in}$ are applied at the same time,
Sum generation: $2$ EXOR gate delay to compute Sum. $$A\;to\;Sum\;\; (or)\;\; B\;to\;Sum$$
Carry generation: $1$ AND gate $+$ $1$ OR gate delay to compute Carry $(C_{out})$. $$A \;to\; C_{out}\;\; (or)\;\; B\; to\; C_{out}$$
Case $2$: When the inputs $A, B$ are already available for a long time and $C_{in}$ is applied now.
What it really means???
Inputs A, B is already applied means, FIRST EXOR gate output is readily available (i.e.,) we can omit this EXOR gate delay while calculating delay for Sum generation.
Sum generation: $1$ EXOR gate delay to compute Sum.$$C_{in}\;to\;Sum$$ Carry generation$ AND gate $+$ $1$ OR gate delay to compute Carry $(C_{out})$. $$C_{in}\; to\; C_{out}$$
Practice Question
A full adder is implemented with two EXOR, two AND gate and OR gate. The propagation delay of an EXOR gate is twice that of an AND gate. The propagation delay of an AND gate is $1.2$ microseconds.
What is the total propagation time taken for generating the sum and a carry? (Assume all three inputs are applied at once)
$1$ EXOR gate delay is equal to $1.2\;\mu s×2=2.4\;\mu s$
Sum generation involves $2$ EXOR gate delay, which is equal to $2.4\;\mu s×2=4.8\;\mu s$
Carry generation involves $1$ AND gate delay$+$$1$ OR gate delay, which is equal to $1.2\mu s+1.2\mu s=2.4\;\mu s$
Since Sum generation and Carry generation process are parallel, the total propagation time taken for generating sum and carry is the largest delay among them, which is $4.8\;\mu s$
Practice Question
A full adder’s propagation delays are given below$$A\;to\;Sum:\;40ns$$$$B\;to\;Sum:\;40ns$$$$A\;to\;C_{out}:\;40ns$$$$B\;to\;C_{out}:\;40ns$$$$C_{in}\;to\;sum:\;35ns$$$$C_{in}\;to\;C_{out}:\;25ns$$
i) Find the maximum delay, when all three inputs are applied at once.
ii) Find the maximum delay, when A and B are applied long before $C_{in}$.
i) When all three inputs are applied at once, maximum delay in calculating sum and carry will be $$40\;ns$$
ii) When A and B are applied first, the first EXOR gate output will be readily available. So, we have to omit $1^{st}$EXOR gate while calculating delay (all $40\;ns$ delay can be omitted). When $C_{in}$ is applied, it will be utilized by $2^{nd}$ EXOR gate to calculate Sum.
So maximum delay here is $35$\;ns$