The NFL Network began unveiling their annual Top 100 players in the league on Sunday night, and it didn’t take long for a Buccaneer player to be announced.
Long time defensive standout Lavonte David came in at No. 100.
Kicking of the #NFLTop100 countdown at 100…@Buccaneers LB Lavonte David! pic.twitter.com/NNQviBzbec
— NFL Network (@nflnetwork) July 27, 2020
The list is complied by fellow NFL players who take an off-season poll where they vote on their fellow players based on their performance for the most recent NFL season. In the eight-year history of the rankings, only Tom Brady has been voted No. 1 more than once (2011, 2017, 2018).
David had yet another superlative season in 2019, finishing the year with 122 combined tackles (81 solo), one sack, three forced fumbles, one fumble recovery, seven pass breakups and an interception while starting all 16 games. This marks David’s fourth appearance on the list, but first since 2016.
Four other Bucs will be in the Top 100 and will be announced over the next three nights.