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Baltimore Ravens outside linebacker Terrell Suggs has been named the AFC Defensive Player of the Week after his performance against San Francisco on Thanksgiving.
Suggs sacked 49ers quarterback Alex Smith three times, which was part of a nine-sack night by the Ravens defense, tying a franchise record.
Baltimore (8-3) dominated San Francisco’s offense in the 16-6 victory, and Suggs played a key role in harassing Smith. The Ravens held the 49ers to 170 total yards, and San Francisco didn’t have a play longer than 20 yards.
“When you’ve got a unit that’s playing in sync with each other, that’s when you get a lot of sacks, they start to become contagious,” Suggs said after the game.
All three of his sacks came in the second half, and the three-sack night ended a three-game streak by Suggs without a sack.
“It has been a little while, but it’s the NFL,” Suggs said after the game. “There’s a lot of guys with a lot of sacks with a team that’s not very good and a team that doesn’t have a chance with the playoffs. But here I am, I have some sacks and my team is 8-3 and we’re going on a playoff run.”
This is the second time this season that Suggs has been named AFC Defensive Player of the Week, and the fourth time a Ravens linebacker has won the award this season.
Suggs won the award after Baltimore’s 35-7 win over Pittsburgh in Week 1, where he had three sacks and two forced fumbles. Ray Lewis and Jarret Johnson have also won the award this year.
Suggs now has 49 tackles, nine sacks, three forced fumbles and two interceptions on the year.