Bengals QB Joe Burrow today was named AFC Offensive Player of the Week for Week 4 for his performance in the team's win on Thursday Night Football vs. the Jacksonville Jaguars.
Burrow threw for 348 yards and two TDs vs. Jacksonville, while also posting career-highs in both completion percentage (78.1) and passer rating (132.8) as he helped lead Cincinnati to a comeback win. Burrow led the Bengals to scores on all four of their second-half possessions (TD, TD, TD, FG), and his 10-play, 73-yard drive late in the fourth quarter bled 5:33 off the clock and set up a game-winning FG as time expired. In the second half alone, Burrow completed 17 of 20 passes (85 percent) for 253 yards and two TDs (152.1 rating).
This is Burrow's first career AFC Player of the Week award. He is the second Bengal this season to win an AFC Player of the Week award, joining rookie K Evan McPherson, who won it on special teams in Week 1. Additionally, WR Ja'Marr Chase won NFL Offensive Rookie of the Month for September.
