Hi guys,

i am working on a very simple circuit that take in a 3 bit unsigned value ABC and output a 5 bit binary number S0-S4 that is 5 times of the input value.

This my truth table :

A B C S4 S3 S2 S1 S0  V
0 0 0  0  0  0  0  0  1
0 0 1  0  0  1  0  1  1
0 1 0  0  1  0  1  0  1
0 1 1  0  1  1  1  1  1
1 0 0  1  0  1  0  0  1
1 0 1  1  1  0  0  1  1
1 1 0  1  1  1  1  0  1
1 1 1  X  X  X  X  X  0 (Invalid)

SOP expressions i got are :

S4 = A
S3 = B + A.C
S2 = A.C' + A'.C = A XOR C
S1 = B
S0 = C

I have a 74LS20 NAND gate and a 74LS83 4-bit Parallel adder only, how do i actually start connecting this circuit?

 

As this seems more like a homework question then a general one, I'll feed you the answer in pieces.

The problem is not how to write logic expressions but how to apply the hardware you actually have on hand.

It was once quite useful when functions came on discrete chips, now with micro controller and FPGAs it's less useful but still something good to know.

Your hint is you have an adder to do multiplication. Somewhere your brain should start coming up with addition terms to satisfy y = 5 * x. Here's a useful one:

y = 4*x + x

Think of how to do THAT with the adder.

 

hi thanks for the kind response and guide,

I know that its 4*X + X = Y
I also know there's a relation of shifting 2 bits. like 00 to the left.

I dont know how to do it with adder as i dont know how to start connecting an adder.

 

and i know i need to do a multiply by 4 circuit. I will connect the MSB of my Input to the MSB of my output since its the same, which results in the soultion with number of inputs.

 

You don't use an adder to multiply.

You have a number on 3 wires, and you need to shift the value by two. You got that part correct. Now how could one shift a number on some wires by two....

Well, maybe by shifting the wires themselves?

Last hint for now: Look at the outputs S0 and S1, then ask how x and 4x influence these number. That should show you how to hook up the added.