Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

[SOLVED] Floorplan: Aspect ratio

Status
Not open for further replies.

rohit_singh1

Junior Member level 3
Junior Member level 3
Joined
Jan 28, 2011
Messages
29
Helped
1
Reputation
2
Reaction score
1
Trophy points
1,283
Activity points
1,495
Hi All,

I understand the definition of aspect ratio w.r.t. floorplan, but I am finding it difficult to understand the following example that I found in a book:


Normally, METAL1 is used up by the standard cells. Usually, odd
numbered layers are horizontal layers and even numbered layers are
vertical. So for a 5 layer design, AR = 2/2 = 1.
b. Example2: For a 6 layer design, AR = 2/3 = 0.66

How/why did you get those numbers (AR = 2/2 , AR = 2/3) is my question.

Thanks!
 
Last edited:

metal1,3,5 horizontal
metal2,4,6 vertical

metal1 is not usable, so if you have 5 layers, for routing resource, you have metal 2,3,4 5 available, i.e. the resource on horizontal direction (2 metal layers) and vertical direction (2 metal layers) is the same, and AR = 2/2 =1; if you have 6 layers, you have metal 2,3,4,5,6, which means vertical direction (2,4,6) has more resource that horizontal(3,5). If you make horizontal direction wider, ie. AR = 2/3, then you make the routing resource roughly the same for both directions. Make sense?
 
Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top