BEREA, Ohio — Fresh off his spectacular performance in the Browns' Week 8 upset win over the Ravens, quarterback Jameis Winston has been honored as the AFC Offensive Player of the Week.
Making his first start at QB since 2022 and first ever as a member of the Browns, Winston put on a dazzling performance against Baltimore, completing 27 of 41 passes for 334 yards and three touchdown passes. His 334 passing yards was the most by any NFL quarterback in Week 8. Winston's passer rating was a sterling 115.3.
With the Browns trailing 24-23 with 2:36 to play in the game, Winston led the offense on a nine-play, 69-yard drive that ended with a 38-yard touchdown pass to Cedric Tillman with just 1:04 to go. Winston was 6 of 9 for 74 yards on that winning drive.
The 30-year-old Winston was inserted as the Browns' starting quarterback after Deshaun Watson suffered a season-ending Achilles injury during the previous week's game against Cincinnati.
Winston had previously won player of the week honors in 2016 and 2019 while he was a member of the Tampa Bay Buccaneers. He is the first Browns quarterback to be named AFC Offensive Player of the Week since Baker Mayfield was honored in 2020. The last Browns player to win the award was Amari Cooper in Week 16 of 2023, after he set a Browns single-game record with 265 yards receiving in a win at Houston.
The 2-6 Browns will look to build on the momentum from the win over Baltimore this Sunday when they play host to the Los Angeles Chargers.