Pittsburgh Steelers’ Roethlisberger named AFC Offensive Player of the Week for 18th time

PITTSBURGH — Pittsburgh Steelers quarterback Ben Roethlisberger has been named the AFC Offensive Player of the Week for the 18th time in his career.

This time around, Roethlisberger received the honor for his performance in Sunday’s 36-10 win over the Cincinnati Bengals at Heinz Field.

Roethlisberger completed 27 of 46 passes for 333 yards. He had four touchdowns with no interceptions.

After having been on the Reserve/COVID-19 List from Nov. 10 until the day before the game, not being able to practice all week didn’t seem to have much of an effect on him.

“I attribute it to the guys around me,” Roethlisberger said. “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.”