Though his game ended on a bitter note, sustaining a knee injury that will keep him out of at least Sunday’s Wild Card game against the Buffalo Bills, T.J. Watt took home one award for his final regular-season performance. Watt was named the AFC Defensive Player of the Week for Week 18, posting two sacks in the Pittsburgh Steelers’ 17-10 win over the Baltimore Ravens.
Before his injury, Watt was having another great performance, and he finished the day with eight tackles (three for a loss), two QB hits, and two sacks. His two takedowns of QB Tyler Huntley gave him 19 for the season to lead the league for a third year, the first player in official NFL history to ever do so. Watt’s 96.5 sacks in his first three seasons are third-most by any player since 1982, trailing only Reggie White and DeMarcus Ware.
It’s the first time Watt has been named Defensive Player of the Week since 2021, though he was named the AFC Defensive Player of the Month for this past September.
Now, we’ll see if Watt brings home any additional hardware. He’s one of the frontrunners for Defensive Player of the Year but the award could go to Cleveland Browns DE Myles Garrett, the slight favorite by oddsmakers. Watt, however, has better stats across the board with more sacks, interceptions, QB hits, tackles for loss, pass deflections, and defensive touchdowns.
Though ruled out for this weekend, it’s possible Watt could return if the Steelers beat the Bills on Sunday. If Pittsburgh pulls off the upset, it will play at Baltimore in the Divisional Round.
