Floorplan Suresh

7/30/2019 Floorplan Suresh

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1.) How do you determine Die (core) area given total chip area 8um x 12 um? Hardblockages 25%, IO area = 6%, macro area = 30%. Come up with the aspect ratio of thisdesign?

Total area of Die=96 um^2IO area=5.76um^2Core area=Die area-IO area

=96-5.76=90.24um^2

Total IO area=5.76IO area per side=5.76/4=1.44IO area on top side=12X0.12=1.44IO area on left side=8X0.18=1.44Height of the IOs on top and bottom based on the area per side=0.18umHeight of the IOs on left and right based on the area per side=0.12um

Height of core=8-0.18=7.82 umWidth of core=12-0.12=11.88 umAspect ratio of core=7.82/11.88=0.658

Or

Total area of the core=90.24um^2=9.49X9.49umAspect ratio=9.49/9.49=1

2.) What constraints you are going to take care at floor plan stage?

Placement blockages

Routing blockages

Orientation of standard cell and macros

Utilization

Aspect Ratio

Offset to IO boundary

IO placement order and offset

Placement Guide, fence and region

3.) Insert well taps cells (process guideline pitch =15un) from library in staggered manner? Ifwell tap not there use BUF 1.

In the library there are no well tap cells, so I have used TBUFX1 instead of well tap cell.

It has been inserted in the core with a pitch of 15 micro meters. You can insert this byusing GUI option as well as command mode.


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4.) Create halos of 2.5un on all memories? Merge halos where ever required.

The halos are created for all the macros where ever its required. For some macros the

halos created only on top & right side and for other macros its created bottom & rightside. The halos are merged where the two macros are coming closer to each other, in that

location no need provide halos for two macros. The screen shot has been attached for

merging of halos,


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5.) Why soft, hard, partial blockages are used? Define a partial blockage around one of the

memory and try to place 10 BUF1 into it?The hard blockage used where you dont want to place any standard cell in that area. It

wont allow standard cell to be placed during placement and optimization stage.The soft blockage used only the tool will add some standard cell during optimization

stage and not during placement stage.The partial blockage used to give percentage of that area (blockage area) allowed placingstandard cell during placement stage. If you give partial blockage as 40%, the tool will

allow only 40 % of the blockage area to be used or cells are placed in 40% of the area of

the blockage. The remaining 60% of the area will be free.


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6.) Create routing channel b/w any two macros? Show with the right calculation of channellength?

There are two macros sitting adjacent to each other. Those macros are MEM2_128X32

and MEM2_128X32.

Number of pins in Macro (MEM2_128X32): 148Number of pins in top side: 134Number of pins in right side: 14Number of pins in Macro (MEM2_128X32):148Number of pins in top side: 134Number of pins in right side: 14The first macro pins are oriented towards the core. So pins of first macro you dont needto consider while calculating channel between two macros.

For the second macro also 134 pins are only comes in between two macros, so for

channel calculation you consider this 134 pins only. The remaining 14 pins of this macro

will come other side of the macro (comes core side not in-between two macros), it wont

count for channel length calculation.The length of channel between two macros should able to route 134 pins.Number of routing layers available for routing the macro pins: 6The metal layers 2 to 7 you can use it for signal routing, metal 8 & 9 used for power

routing and metal 1 is used for power rail routing.You can use 6 metal layers for routing of pins of the macro in the channel.You need to route these pins of macro through horizontal and not in vertical, you arehaving routing blockage over the macros in vertical. So only consider horizontal routing

pitch of metals.

Routing pitch of METAL3: 0.2Routing pitch of METAL5: 0.2Routing pitch of METAL7: 0.2Horizontal routing layers available: 3Channel length: (134 * 0.2)/3

=26.4/3= 8.8 micro meter

So you can take channel length as 9 micro meters.


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7.) Report congestion and what % of congestion you see and which layer is higher?

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Floorplan Suresh