Anyone good at maths? I need you

Assuming that the individual transistors scale with process (in reality they probably don't) means:

One 130um transistor be 26 times smaller on each side on a 5um process. So you would be able to fit 676 5um transistors in the space occupied by just 1 120um transistor.

so moving a die from 5um to 130um would make it 26 times bigger on each side or increase the area by 676 times.

So the 609mm2 die would then become 411684mm2 or a square about 64cm on each side!
 
Hodders said:
Assuming that the individual transistors scale with process (in reality they probably don't) means:

One 130um transistor be 26 times smaller on each side on a 5um process. So you would be able to fit 676 5um transistors in the space occupied by just 1 120um transistor.

so moving a die from 5um to 130um would make it 26 times bigger on each side or increase the area by 676 times.

So the 609mm2 die would then become 411684mm2 or a square about 64cm on each side!
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my previous working is wrong, I scaled the 207nm die by 26, basically scaled the 5900 die size by 26. This is the correct answer.
 
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rare said:
I


Pure interest really. I recall reading something about a modern CPU being made on an old process node would have resulted in the size of a large car.

I appreciate the above responses a great deal!
Click to expand...
Probably would need the engine of a large car to run it as I assume that it would be 26 x whatever the 5900 pulls.
 
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