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Question

What is smallest negative value expressed in 8 bit 2 complement from that when added to itself, the sum causes an overflow?

Answer 1

Smallest negative value expressed in 8 bit 2 complement from is $-128$ [ 1000 0000 ]

Now if we add it with itself : 

   1000 0000  $(-128)$

   1000 0000  $(-128)$

---

1 0000 0000  $(-256)$

 

Note: Here we can see overflow came because when we add two binary numbers there must be an overflow when : 

X   Y    Z 

0   0     1

1    1    0

Where X → Sign bit of first number 

            Y → Sign bit of second number 

    and, Z →  Sign bit of Resultant

Answer 2

The range of numbers that can be represented using 8 bits in 2's complement form is  -128 to +127

and hence the smallest negative value is -128

now if we add -128 with -128 itself then we will get -256 which can't be represented using 8 bits in 2's complement form, so overflow will occur.

So -128 is the smallest negative value expressed in 8 bit 2's complement from that when added to itself, the sum causes an overflow

 

I hope this is a correct explanation.