Pittsburgh Steelers defensive end Stephon Tuitt has been named AFC Defensive Player of the Week for his performance in the team’s win over the Baltimore Ravens last weekend.
Tuitt was dominant on Sunday, recording nine total tackles, two sacks, three tackles-for-loss and three quarterback hurries.
This is the first time Tuitt has been awarded the weekly honor. He is the second Steeler defender to win the award. Outside linebacker T.J. Watt was named AFC Defensive Player of the Week in Week Two.
Selected in the second round of the 2014 NFL Draft out of Notre Dame, Tuitt has spent his entire seven-year career in Pittsburgh. He has 29.5 career sacks and 44 tackles-for-loss.