BEREA -- The awards keep coming for the Cleveland Browns former NFL Defensive Player of the Year.
Browns defensive end Myles Garrett earned the AFC Defensive Player of the Week for games played in Week 12, the National Football League announced Wednesday.
The award could be a precursor for another NFL Defensive Player of the Year Award after the season. Garrett currently has 18 sacks to lead the NFL and is on pace to shatter the NFL record for sacks in a season of 22.5.
In the 24-10 last week against the Raiders, Garrett had five tackles, three sacks, six quarterback hurries, four tackles for loss and two forced fumbles
On the season, Garrett has 44 tackles, 18 sacks and three forced fumbles. The Browns had 10 sacks, overall, in the game.
The three sacks in the Raiders game gave him 18 on the season. Garrett's first sack was his 16th sack of the season in the third quarter to tie his previously set franchise record. He added No. 17 and 18 in the fourth quarter. The Browns had five first half sacks before Garrett had his first.
Garrett had five sacks against the Patriots and four against the Ravens. Garrett needed 7.5 going into the game in the final seven games to match the NFL record of 22.5 set by Michael Strahan and T.J. Watt. He now needs 4.5 sacks with six games to play.
Garrett had the single season franchise record of 16 sacks, which he accomplished in 2021 and again in 2022. With 18, he now has a new franchise record.
This week's opponent, the 49ers, have 13 sacks as a team this season.
Against the Ravens, it was the sixth time Garrett has eclipsed 12 sacks in a season to break a record previously held by Hall of Fame rusher Lawrence Taylor.
Garrett's 18th sack through 11 games is the most in NFL history. Previously, Mark Gastineau had 17.5 in 1984 through 11 games.
Last year, Garrett won the same award the same week after a three sack performance over the Steelers in a win.