Jake Browning named AFC Offensive Player of the Week after history-making performance vs. Jags
In one of his first two professional starts, Browning recorded the greatest completion percentage of any quarterback in NFL history, 86.5.
On Wednesday, the Cincinnati Bengals quarterback Jake Browning was named the AFC Offensive Player of the Week. For his role in the Bengals’ 34-31 overtime victory over the Jacksonville Jaguars, Browning was named the winner.
Our QB correspondent @RyanDLeaf is back, and we are talking @jake_browning @Bengals pic.twitter.com/KIcsctTtj6
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Browning, starting for the first time in his NFL career, completed 32 of 37 passes (86.5%) for 354 yards and one score with no interceptions. In terms of passing yards, completion percentage, and completions, he led all NFL quarterbacks in Week 13.
In one of his first two career starts, Browning’s 86.5 completion % was the highest by a quarterback in the NFL and the fifth-highest in a Bengals game ever (minimum of 20 attempts).
After Jeff Blake passed for 387 yards in his second game against the Seattle Seahawks on November 6, 1994, Browning had the second-highest passing yards of any quarterback in team history after just two starts (354 yards). With a completion percentage of 85 or higher, Browning became the tenth quarterback in the Super Bowl era to pass for 350 yards or more.
In the third quarter, Browning connected with wide receiver Ja’Marr Chase for a 76-yard touchdown, the Bengals’ longest play from scrimmage this year. Browning converted 9/11 passes on third down.
Browning led Cincinnati into position for the game-winning field goal in overtime by going 4/5 for 36 yards.
Following quarterback Joe Burrow in Week 8 and wide receiver Ja’Marr Chase in Week 5, Browning is now the third Bengal to win the AFC Offensive Player of the Week award this season.
Since the beginning of the 2021 season, Bengals players have earned a total of 16 weekly awards (offensive, defensive, and special teams), which is the second-most in the NFL.
