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Yannick Ngakoue named Week 7 AFC Defensive Player of the Week

After a standout game against the Philadelphia Eagles, Yannick Ngakoue has been named the AFC Defensive Player of the Week for his play in Week 7's matchup.

The defensive end finished Sunday's contest with four tackles (two solo), two tackles for loss, two sacks, two quarterback hits and two passes defensed on his way to registering his 12th career multi-sack game and second this season.

His two QB hits and two sacks contributed to the total of 10 hits and 4.0 sacks he has recorded so far on the season, second-most on the team behind the other half of the Raiders' power edge duo, Maxx Crosby.

Stat line:

Two sacks

His two sacks tied for second-most in the AFC and tied for third in the league in Week 7.

Two passes defensed

His two passes defensed were the most among defensive ends in Week 7 and marked a new single-game career high for the six-year vet. He was also the only player in the league with at least two sacks and two passes defensed for the week.

Two tackles for loss

He tied for first in the AFC and tied for second in the NFL among defensive ends in Week 7 with his two tackles for loss.

Four total tackles

His four total tackles tied for third-most among defensive ends in the AFC for the week.

Two QB hits

His two quarterback hits tied for second in the AFC and tied for fourth in the NFL in Week 7.

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