INDIANAPOLIS - The NFL today announced that Indianapolis Colts defensive tackle DeForest Buckner has been named the AFC Defensive Player of the Week for Week 15. This marks the first time in his career that Buckner has earned Player of the Week honors.
In the Colts' 27-20 win against the Houston Texans, Buckner totaled four tackles (three solo), 2.0 tackles for loss, a career-high 3.0 sacks and one forced fumble. He is one of just four players in the NFL this season to register at least four tackles, 2.0 tackles for loss, 3.0 sacks and one forced fumble in a single game. Buckner's 3.0 sacks are tied for the third-most by any player in a single game this season.
In 2020, Buckner has played in 13 games (12 starts) and has compiled 52 tackles (33 solo), 8.0 tackles for loss, 7.5 sacks, one pass defensed, two forced fumbles and one fumble recovery. Among defensive tackles, he ranks in the top five in tackles (tied-third), solo tackles (tied-second), sacks (second) and forced fumbles (tied-third). Buckner is part of a Colts defense that ranks seventh in the NFL in yards per game (334.1 avg.) and fifth in rushing yards per game (98.1 avg.) this season. In 2020, Indianapolis ranks third in the league in takeaways (24) and is fifth in interceptions (15).
Buckner is the fifth Indianapolis player to earn AFC Player of the Week honors this season, joining cornerback Xavier Rhodes (AFC Defensive Player of the Week for Week 3), linebacker E.J. Speed (AFC Special Teams Player of the Week for Week 10), kicker Rodrigo Blankenship (AFC Special Teams Player of the Week for Week 11) and cornerback Kenny Moore II (AFC Defensive Player of the Week for Week 14).