Bengals QB Joe Burrow today was named AFC Offensive Player of the Week for Week 7 for his performance in the team's 35-17 win over the Atlanta Falcons on Sunday.
Burrow completed 34 of 42 passes (80.1 percent) for 481 yards and three TDs for a passer rating of 138.2. He led the NFL in Week 7 in completions, passing yards and passer rating, and his passing yardage total was the third-most in a single game in Bengals history. Burrow also ran three times for 20 yards with a one-yard TD in the third quarter. He became the third player in NFL history with at least 475 passing yards, three TD passes and one rushing TD in a single game, joining Norm Van Brocklin (1951) and Billy Volek (2004).
Each of Burrow's three TD passes were for 30 or more yards, including a 60-yarder to WR Tyler Boyd two minutes into the first quarter. Burrow now has 12 TD passes of 50 or more yards since entering the league in 2020, four more than any quarterback in that span. It also tied Pro Football Hall of Famers Dan Marino and Fran Tarkenton for the most TD passes of over 50 yards by a player in his first three seasons in NFL history.
Sunday marked Burrow's fifth career game with over 400 passing yards, surpassing Marino (four) for the most such games by a player in his first three seasons in NFL history.
This is Burrow's third career AFC Offensive Player of the Week award. He won it in Week 4 of the 2021 season vs. Jacksonville, and again in Week 16 against Baltimore. Bengals players have combined to win three weekly awards this year, with Burrow joining DE Trey Hendrickson (AFC Defensive Player of the Week, Week 3) and K Evan McPherson (AFC Special Teams Player of the Week, Week 4).