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Barrett named NFC Defensive Player of the Week

What a game he had against his former team.

Tampa Bay Buccaneers v Denver Broncos
Quarterback Jeff Driskel #9 of the Denver Broncos passes as he pressured by outside linebacker Shaquil Barrett #58 of the Tampa Bay Buccaneers.
Photo by Matthew Stockman/Getty Images

The National Football League today announced that Tampa Bay Buccaneers outside linebacker Shaquil Barrett was named NFC Defensive Player of the Week for Week 3. He becomes the seventh Buccaneers player to earn the award multiple times with Tampa Bay, after also winning NFC Defensive Player of the Week honors in Week 2 of 2019.

Against the Denver Broncos, Barrett led the Buccaneers with three tackles for loss and 2.0 sacks, while also recording six total tackles and a safety. His three tackles for loss tied for the most by any player in the NFL in Week 3, and his safety was the first for Tampa Bay since 2016.

Since joining the Buccaneers in 2019, the Colorado State product leads the NFL with 21.5 sacks and five multi-sack performances, while also ranking third in quarterback hits (39) and tied for third in forced fumbles (six) over that span.

Barrett joins Pro Football of Famers Derrick Brooks and Warren Sapp, as well as All-Pros Rondé Barber, Lavonte David, Hardy Nickerson and Simeon Rice as Buccaneers players to win the defensive award multiple times during their time with Tampa Bay.

Barrett is part of a defense that ranks fourth in the NFL in yards allowed (308.0 per game), third in rushing yards allowed (70.3 per game), and tied for seventh in points allowed per game (20.3), while tying for the third-most sacks in the NFL (12.0) and the second most tackles for loss (24).

(Courtesy of the Buccaneers Communications Department.)