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Tre’Davious White named AFC Defensive Player of the Week

Those skills learned in his goalie academy are really paying off!

Buffalo Bills cornerback Tre’Davious White was named the AFC Defensive Player of the Week after recording two interceptions during Buffalo’s (10-4) playoff-clinching, 17-10 win over the Pittsburgh Steelers in Week 15.

White, who on Tuesday earned the first Pro Bowl accolade of his young career, was the anchor of a stout Bills defense that forced five turnovers and held the Steelers (8-6) to just 226 net yards of offense—the fewest yards in a game for Pittsburgh since the 2013 season opener.

This is White’s second time winning the AFC Defensive Player of the Week award, as he was also honored in Week 7’s win over the Miami Dolphins. White becomes the first Buffalo player to earn multiple Defensive Player of the Week accolades in the same season since 2013 (Mario Williams).

White also made four tackles and broke up two passes in the win over the Steelers. He allowed five catches (on nine targets) for 50 yards, picking off two Devlin “Duck” Hodges to raise his season-long total to six, tied for the league lead in INTs. White has also broken up 17 passes, tied for third in the NFL.

White and the Bills travel to Foxborough to face their AFC East rivals—the New England Patriots (11-3)—at 4:30 p.m. Saturday. If Buffalo wins out and the Patriots lose out (New England plays the Dolphins in Week 17) the Bills would win the AFC East championship.

As is, the Bills have already clinched their second playoff berth in the last three years, and Buffalo can finish no worse than fifth in the AFC.