Roethlisberger wins AFC award

PITTSBURGH –- Steelers quarterback Ben Roethlisberger has been named AFC Offensive Player of the Week after throwing for 367 yards and four touchdowns in a 37-27 win over the Lions last Sunday.

It is the 10th time Roethlisberger has won the award and the first time since 2011.

Roethlisberger played his best game of the season against the Lions, and he was masterful in leading a 10-play, 97-yard touchdown drive that gave the Steelers the lead for good. Roethlisberger completed 29 passes for 45 yards in the game and finished with a season-high passer rating of 119.4.

He also led the Steelers to victory after a fourth-quarter tie or deficit for the 29th time in the regular season, and his 2,901 passing yards for the most in Steelers history after 10 games in a season dating back to 1960.

Next up for Roethlisberger is a return to his native Ohio where he usually plays well. The 10-year veteran is 16-3 in his career against the Browns and Bengals, and he has lost only one time in Cleveland (2009).