Buccaneers quarterback Jameis Winston has been named NFC Offensive Player of the Week for Week 4. This is the second time Winston has won the NFC Offensive Player of the Week honors (Week 1 in 2016), and he also becomes the seventh Tampa Bay offensive player and fifth quarterback to win the award multiple times.

In Sunday’s 55-40 win over the previously undefeated Rams in Los Angeles, Winston completed 28-of-41 passes (68.3 percent) for 385 yards with a season-high four touchdowns and just one interception. Winston posted a 120.5 QB rating.

Bucs QB Jameis Winston – Photo by: Getty Images

The four TD passes were the most by any quarterback in Week 4, and it was the fifth time Winston had at least four touchdown passes in a game, surpassing Brad Johnson for the most games in team history. Averaging 9.4 yards per attempt for the week, Winston joined Patrick Mahomes as the only players this season to maintain more than nine yards per attempt in a game, while recording at least 40 pass attempts.

Winston, who is in a fifth-year option contract year, is currently tied for fourth in the NFL with nine passing touchdowns on the season, while ranking fifth with 1,167 yards passing and sixth with 8.4 yards per attempt. Winston and the Bucs travel to New Orleans on Sunday to face the 3-1 Saints in an NFC South showdown.

Winston is Tampa Bay’s first NFC Offensive Player of the Week honoree this season and the second player to win one of the league’s weekly awards, after outside linebacker Shaquil Barrett earned NFC Defensive Player of the Week honors in Week 2. Winston is the first Buccaneers offensive player to win the award since Ryan Fitzpatrick won in back-to-back weeks to begin the 2018 season.