PITTSBURGH — Steelers quarterback Ben Roethlisberger was named AFC Offensive Player of the Week for his performance against the Tampa Bay Buccaneers.
Roethlisberger threw for 353 yards and three touchdowns against the Bucs – leading the Steelers to a 30-27 victory. It was the team’s first win of the season.
Through three games, Roethlisberger has 1,140 passing yards and 7 touchdowns. He’s thrown for over 300 yards in each of the team’s three games this season.
This is the 16th time in Roethlisberger’s career that he’s been honored as AFC Offensive Player of the Week.
This week, the Steelers play the Baltimore Ravens on Sunday Night Football on Channel 11.
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