Joe Burrow named Week 8 AFC Player of the Week after win against 49ers

CINCINNATI — Joe Shiesty is back, baby.

Joe Burrow has been named AFC Offensive Player of the Week for his week 8 performance against the San Francisco 49ers.

The Bengals quarterback played the best he has all season during Sunday's win in San Fran.

Burrow completed 28 of 32 passes for 283 yards, 3 touchdowns and zero interceptions, and he had a 134.8 passer rating, which led all AFC QBs. His 87.5% completetion rate led all NFL passers for the week.

During the game, Burrow showed impressive mobility in the pocket, escaping multiple sacks and gaining much needed yardage for the Bengals.

That game, which came after the Bengals' bye week, is the first game Burrow has looked comfortable since he injured his right calf in the preseason. Burrow had opened the season playing the worst he has in his career.

Burrow is now tied for the fifth-most games with at least 3 passing touchdowns in the first four seasons in the NFL.

The Week 8 honor also tied Burrow with Boomer Esiason for the only two Bengals quarterbacks to earn Offensive Player of the Week six or more times.

Prior to this week, Burrow also garnered the title in Week 4 and 16 of the 2021 season, as well as Week 7, 13 and 16 of the 2022 season.

Burrow is the second Bengals player to earn AFC Offensive Player of the Week this season. JaMarr Chase earned the title during Week 5 after his record-setting performance against Arizona.