| Author |
Message |
neetinsingh
Joined: 18 Jun 2007
Posts: 51
Helped: 1
|
 25 Oct 2009 11:40 fan out fixing ???? |
|
|
|
|
| how fanout is met by replacing the cell with a higher driver cell?
|
|
| Back to top |
|
 |
skyfaye
Joined: 25 Feb 2008
Posts: 27
Helped: 2
|
25 Oct 2009 22:32 Re: fan out fixing ???? |
|
|
|
|
High fanout means high capacitive load and which means your signal transition is longer. To fix this problem, you can replace the driver with a stronger one or replicate the driver with each driver driving a subset of the load.
- Hung
|
|
| Back to top |
|
|
ebuddy
Joined: 15 May 2007
Posts: 28
Helped: 1
|
26 Oct 2009 3:51 Re: fan out fixing ???? |
|
|
|
|
| Large fan out will result big capacitive load, which slows down the transition time. Of course, the other factor effecting the Tr and Tf is the channel resistance of the driving MOS FETs. By using bigger MOS FETs, the channel resistance is reduced, and thus reduce the RC time constant. The result is improved transition time.
|
|
| Back to top |
|
|
Google
AdSense
|
26 Oct 2009 3:51 Ads |
|
|
|
|
|
|
| Back to top |
|
|
neetinsingh
Joined: 18 Jun 2007
Posts: 51
Helped: 1
|
27 Oct 2009 18:58 Re: fan out fixing ???? |
|
|
|
|
thanks hung n ebuddy 
|
|
| Back to top |
|
|
edaboy123
Joined: 28 Oct 2009
Posts: 4
|
29 Oct 2009 7:32 Re: fan-out fixing with higher drivver cell |
|
|
|
|
| Using bigger driver to drive high fan-out load will increase the load for pre-level. So it is not fact that the bigger for driver strength, the less for delay. So to reduce delay for a path, we need to consider a proper cell with proper size.
|
|
| Back to top |
|
|
