Pittsburgh Steelers defensive end Stephon Tuitt had a great game against the Baltimore Ravens in Week 8 and it’s resulted in the team’s former second-round draft pick being named the AFC Defensive Player of the week for Week 8 of the 2020 season.
Tuitt finished Sunday’s win against the Ravens with two sacks for a combined loss of 15 yards, nine tackles, including eight solo stops, three tackles for a loss, and three quarterback hurries.
For the season, Tuitt now has six sacks, 25 total tackles, six tackles for loss and 17 quarterback hits. More importantly, Tuitt has remained healthy all season after having his 2019 season cut short by pectoral injury in Week 6.
This marks the second time that Tuitt has won the AFC Defensive Player of the Week award. He also won it during the 2016 season in Week 11.
This is the second time this season a Steelers defensive player has won this ward as outside linebacker T.J. Watt was previously named the AFC Defensive Player of the Week for Week 2.