For the 18th time in his NFL career, Ben Roethlisberger has been named the AFC Offensive Player of the Week.

By: KDKA-TV News Staff

PITTSBURGH (KDKA) – Ben Roethlisberger has been named the AFC Offensive Player of the Week for the 18th time in his NFL career.

On Sunday against the Bengals, Roethlisberger completed 58% of his passes, going 27-for-46, for 333 yards and four touchdowns with no interceptions.

All of this came during a week when Roethlisberger did not practice due to being placed on the COVID-19 list on Tuesday, November 10, and was not activated until Saturday.

“I attribute it to the guys around me,” said Roethlisberger in an interview on the Steelers’ website. “I attribute it to the offensive line, no sacks. They gave me time to throw it against a lot of crazy looks and blitzes. I attribute it to the coaching staff for getting me ready to play, and all of us ready to play. Then, obviously, the skill guys, they made plays tonight. It wasn’t easy out there. The wind was blowing the football around a little bit, and they were able to concentrate and make plays out there when sometimes plays weren’t even there to be made.”

The Steelers are 9-0 for the first time in franchise history and will look to make it 10 wins in a row on Sunday against the Jacksonville Jaguars.