Ravens Terrell Suggs named AFC Defensive Player of the Year

KANSAS CITY, MO - Ravens linebacker Terrell Suggs has been named the AFC Defensive Player of the Year.

The honor comes from the NFL 101 Awards which are voted on by media from around the country. The 42nd Annual NFL 101 Awards will salute Suggs and other players and coaches from around the NFL at their annual gala in Kansas City in early March. The black-tie awards event was found in 1969.


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Through nine season, Terrell Suggs has always made a mark on the Ravens defense, but this may have been one of the best seasons of his career. The outside linebacker recorded an AFC-high 14 sacks and forced a team-high seven fumbles. He also broke the franchise record for sacks which now stands at 81.5 in his career.


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Suggs was also named to his fifth Pro Bowl.

Suggs and the rest of the Ravens defense finished the regular season ranked #3 as a unit in the NFL helping the Ravens to a 12-4 record, their third AFC North Division title, and their first ever sweep of the AFC North.

The target for Suggs in Sunday's AFC Championship game, New England quarterback Tom Brady, was also honored by the NFL 101. Brady was given the AFC Offensive Player of the Year nod. Brady lead the AFC in passing attempts (611), completions (401), completion percentage (65.6), yards (5,235), touchdowns (39), and passer rating (105.6). Those numbers added up to one of the top ranked offensive teams in the NFL and led the Patriots to a 13-3 record and AFC East Division title. This is the second straight season and third time Brady has been named AFC Offensive Player of the Year.

A panel of 101 sportswriters and broadcasters who cover the NFL voted on each award at the end of the regular season.

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