The NFL announced Wednesday that Lamarr Houston was named the AFC Defensive Player of the Week.
The honor comes in the wake of the Oakland Raiders’ second win of the season with a overtime victory over the Jacksonville Jaguars. The defense held the Jaguars to 209 yards including 54 on the ground. Houston was the cause of the only turnover of the game, forcing a crucial fumble in overtime to set up the winning field goal.
The defensive end had a team-high and season-best eight tackles including five solo stops. He got his first sack of the season, one of two tackles-for-loss he had on Sunday. He made big plays in the fourth quarter and overtime to keep the Raiders in the game and helped them eventually best Jacksonville. His two quarterback hits and constant pressure helped to throw Blaine Gabbert and Chad Henne off the mark.
The honor marks a major step forward for an Oakland defense, which has struggled for most of the season. The 23 points the Raiders gave up were tied for the second least this season, matching how much they gave up the previous week in a loss to the Atlanta Falcons. The back-to-back strong performances by the defense bode well for a team playing two very winnable games in the next two weeks.
Houston needs to use this honor as momentum as the season hits the midway point and the games become more important. In fact, the entire Raiders team, especially the defense, should use this honor as a sign they can still wreak havoc on the rest of the league. Oakland has all the talent needed to cause problems in the AFC West and maybe sneak back in to win it, but it must use the win and national recognition to boost its morale and keep the level of play at a high level.