Rodgers Wins First Player of Week in Two Years
In what had to be one of the most predictable results in the history of the award, Green Bay Packers quarterback Aaron Rodgers was named NFC Offensive Player of the Week on Wednesday.
In Sunday’s victory over Oakland, Rodgers was 25-of-31 passing for 429 yards with five touchdowns and a max passer rating of 158.3. Rodgers added a rushing touchdown, as well, meaning he accounted for as many touchdowns as he threw incomplete passes. It was Rodgers’ first six-touchdown day since throwing for six against Chicago on Nov. 9, 2014.
“I think this was the most complete that I’ve played,” Rodgers said after the game. “I felt good about the timing. There were a lot of balls thrown on time. Obviously, the line played fantastic. We’ve had seven sacks in the last six games. It’s pretty remarkable for those guys the way they’re playing. Gave me a lot of clean pockets where I could just throw, and then a few times I had to hang in there or move slightly, we were able to complete those, and then guys made plays. That’s the way we want it to look.”
Rodgers owns the highest passer rating in NFL history but this was his first game to strike 158.3.
“I still don’t understand how they put that rating together, but it does sound pretty good,” Rodgers said.
According to the Packers, this marks the 17th time Rodgers has won the award – tops in franchise history. It was his first since leading a come-from-behind victory at Dallas in Week 5 of the 2017 season. He didn’t win the award last season, snapping an eight-year streak in which he won at least once.
Two weeks ago, running back Aaron Jones was named NFC Offensive Player of the Week for his four-touchdown performance at Dallas.
