Ben Roethlisberger named AFC Offensive Player of the Week

(Photo by Justin K. Aller/Getty Images)

Ben Roethlisberger has been named AFC Offensive Player of the Week for his performance against the Cincinnati Bengals on Sunday where he completed 27 of 46 passes (58.7 percent) for 333 yards, four touchdowns, no interceptions, and had a rating of 110.1 in the Steelers' 36-10 win.

It's the 18th time Roethlisberger has won the honor in his career. Roethlisberger didn't even practice all week after he was placed on the reserve/COVID-19 list on Tuesday, Nov. 10 for being deemed as a close contact with Vance McDonald, who the team announced the day before tested positive for the virus. Roethlisberger was removed from the reserve/COVID-19 list on Saturday, along with Vince Williams, Jaylen Samuels and Jerald Hawkins. All four players participated in an extended walkthrough on Saturday.

The last time Roethlisberger won AFC Offensive Player of the Week was for his Week 10 performance against the Carolina Panthers in 2018 when he completed 22 of 25 passes (88%) for 328 yards, five touchdowns, no interceptions and had a perfect quarterback rating of 158.3.

Roethlisberger is the fifth player from the Steelers to be honored for a league award this season. Stephon Tuitt was named AFC Defensive Player of the Week after a dominant outing against the Baltimore Ravens in Week 8. Chase Claypool was named AFC Offensive Player of the Week for his four-touchdown presentation against the Philadelphia Eagles in Week 5, T.J. Watt won AFC Defensive Player of the Week for his Week 2 performance against the Denver Broncos and was named AFC Defensive Player of the Month for September.

Roethlisberger also was voted as the FedEx Air Player of the Week for Week 10.