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Ben Roethlisberger wins AFC Offensive Player of the Week award for game against Panthers

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Big Ben won player of the weeks honors for the second time this season for his efforts in Week 10.

Carolina Panthers v Pittsburgh Steelers Photo by Justin K. Aller/Getty Images

A Pittsburgh Steelers player winning player of the week honors continues to be a biweekly award for the team in 2018 after Ben Roethlisberger was named AFC Offensive Player of the Week on Wednesday for his efforts against the Carolina Panthers in Week 10.

Roethlisberger won the award for the second time this season and 17th time in his career after recording a perfect game against Carolina. Big Ben finished the contest with the maximum quarterback rating of 158.3 for completing 22 of 25 passes for 328 yards and 5 touchdowns. He also became the only player in NFL history to have finished a game with a perfect quarterback rating four times.

With 2,888 passing yards on the season, Roethlisberger currently ranks fourth in the league, as Patrick Mahomes leads the NFL with 3,150 yards. His 21 touchdowns also find him in a tie for fourth place, while Mahomes is again the leader with 31. An average of 320.9 yards per game is enough to rank him second behind Matt Ryan (335.0).

The Steelers’ quarterback is one of three Pittsburgh players to win a weekly award in 2018 and this is the fifth time a member of the team has been honored this season. T.J. Watt won defensive player of the week honors in Week 1 and Week 5, while Roethlisberger won the offensive award in Week 3 and James Conner won it in Week 8. No NFL team has won more individual weekly honors that the Steelers so far this year.