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Ben Roethlisberger wins AFC Offensive Player of the Week honors for Week 3

Big Ben won his 16th AFC Player of the Week award on Wednesday for his performance against the Tampa Bay Buccaneers

Pittsburgh Steelers v Tampa Bay Buccaneers Photo by Julio Aguilar/Getty Images

The national narrative for the last few years has been that Ben Roethlisberger is a fading version of the Pittsburgh Steelers quarterback who won two Super Bowls. On Wednesday, Big Ben was announced as the AFC Offensive Player of the Week for his efforts in the Steelers win over the Tampa Bay Buccaneers on Monday night.

While Roethlisberger might not be the man he was 10 years ago, very few of actually are either, and it is worth noting that this is the 16th time he has won this award. The most in team history, as per the team’s PR/Media Manager, Dom Rinelli.

Completing 30 of his 38 passes for a total of 353 yards and 3 touchdowns, Roethlisberger finished the game with a passer rating of 120.7. Currently second in the league with 1140 yards passing behind Ryan Fitzpatrick, the Pittsburgh quarterback is on pace to break his 2014 total for most yards in a season when he finished the year with 4,952 yards through the air.

Roethlisberger is the second Steelers’ player to win Player of the Week honors in 2018, after T.J. Watt won the award in Week 1 for his 3 sack performance against the Cleveland Browns.