Cincinnati Bengals wide receiver A.J. Green has been named the AFC Offensive Player of the Week, the NFL announced Wednesday.
Green, now in his fifth NFL season, recorded a career-high 227 receiving yards to go with 10 catches and two touchdowns in the Bengals' 28-24 win over the Baltimore Ravens. Even if he only played in the fourth quarter of the game, Green could have been a contender for this award after the last 15 minutes of play featured a thrilling back-and-forth affair between the two AFC North rivals.
Green was already having a great game with seven catches for 126 yards entering the final frame. But with the Bengals needing to rally twice in the final period, Green exploded for an additional 101 yards and two scores, including an 80-yard score after the Ravens had taken their first lead of the game.
After Baltimore retook the lead, the Bengals marched right back down the field before Green hauled in the game-winning seven-yard score. He now has 18 catches for 335 yards (18.6 ypc) and three scores through three games this season.
Green has also been nominated for the Castrol Edge Clutch Performer of the Week, which you can vote for him to win.