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Suggs named AFC Defensive Player of the Week

Ravens linebacker Terrell Suggs has been named the AFC Defensive Player of the Week after racking up three sacks and forcing a fumble in the Ravens' 16-6 win over the San Francisco 49ers on Thanksgiving night.

This is the second time that Suggs has won the weekly award this season -- he was honored after sacking Ben Roethlisberger three times in the Week 1 win over the Pittsburgh Steelers -- and the third time in his career. It was the fourth time this season that a Ravens player has been named the AFC Defensive Player of the Week.

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Collectively, the Ravens sacked 49ers quarterback Alex Smith nine times, which tied a franchise record.

"I would love to line up every day and get three sacks, I would love that. But it's the NFL, It's not going to happen like that," Suggs said after the Harbaugh Bowl. "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. So, I'm very fortunate and there's a lot to be thankful for on Thanksgiving."

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This weekly honor is one more thing Suggs has to be thankful for.

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