Bengals QB Joe Burrow today was named AFC Offensive Player of the Week for his performance in the team's 31-17 win at the San Francisco 49ers in Week 8.
Burrow completed 28 of 32 passes for 283 yards and three TDs with zero INTs. His career-high 87.5 completion percentage led all NFL passers in Week 8 and was the second highest in a game in team history (minimum 20 attempts), while his 134.8 passer rating led all AFC QBs.
Burrow during the first half completed 19 consecutive passing attempts, marking the most consecutive completions in a game by any NFL passer this season, the most in Burrow's career and the second most in team history. Week 8 also was Burrow's ninth career game with a completion percentage of 75 or higher on at least 30 passing attempts, the second most such games within a player's first four seasons in NFL history.
Burrow finished off the Bengals' opening drive by recording his 90th career TD pass in his 49th game, which tied him as the sixth fastest QB in NFL history to pass for 90 TDs. He completed a pass to seven different receivers during the game, and additionally rushed for a season-high 43 yards on six carries with four resulting in a first down.
This is Burrow's sixth career AFC Offensive Player of the Week award. He received the honor twice (Weeks 4 and 16) during the 2021 season and three times (Weeks 7, 13 and 16) in 2022. He is the second Bengals player to be named AFC Offensive Player of the Week this season, following WR Ja'Marr Chase in Week 5.
Bengals players have combined for 15 weekly honors (Offensive, Defensive and Special Teams) since the start of the 2021 season, the second most in the NFL in that span.