Steelers’ Ben Roethlisberger named AFC Offensive Player of the Week |
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Joe Rutter
Steelers quarterback Ben Roethlisberger throws a pass against the Tampa Bay Buccaneers during the first half Monday, Sept. 24, 2018, in Tampa, Fla.

For the first time in his 15-year NFL career, Ben Roethlisberger has passed for 300 yards in each of his first three games in a season.

For his most recent effort, Roethlisberger was named AFC offensive player of the week.

Roethlisberger was honored Wednesday after completing 30 of 38 passes for 353 yards and three touchdowns in the Steelers’ 30-27 win over the Tampa Bay Buccaneers on Monday night.

It was the 16 th time in his career that Roethlisberger received the AFC’s top weekly honor.

Roethlisberger ranks second in the NFL with 1,140 passing yards. He has thrown seven touchdown passes against four interceptions.

Roethlisberger said he’d gladly trade the award for an improved record. The win was the first of the season for the Steelers, who take a 1-1-1 record into their AFC North game Sunday night against the Baltimore Ravens.

“I’d rather be 3-0 and not throw for 300,” Roethlisberger said. “For me, it’s always about winning and losing, but I also think that it’s the direction offenses are going, games are going. You look at the number of points that are being scored. You have to keep up with the Joneses, if you will.”

Joe Rutter is a Tribune-Review staff writer. You can contact Joe at [email protected] or via Twitter @tribjoerutter.

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