Washington Redskins linebacker Keenan Robinson has been named the NFC Defensive Player of the Week for Week 7. The Redskins posted an excellent team defensive effort against the Titans, and much of that success can be attributed to Robinson. He recorded a career-high 14 tackles, the most by a Redskins player since Perry Riley also recorded 14 tackles in Week 1 last season. Behind Robinson’s strong game, the Redskins were able to hold the Titans to just 236 yards, their lowest total of the season. Robinson is the first Redskins player to win this award since cornerback DeAngelo Hall did so back in 2011.
The 6’3 240 pound linebacker was highly regarded coming out of the University of Texas in the 2012 NFL Draft where the Redskins selected him in the 4th round. It took a while for Robinson to get significant playing time for the Redskins, as he was stuck behind future Hall of Famer London Fletcher for the past few seasons. However, after Fletcher retired this past offseason, Robinson won the job outright and has been performing very well since. He has recorded 58 tackles through the first 7 games this season and has been the primary play caller for the Redskins defense.
The Redskins will be looking for more strong play from Robinson in the near future. They will face the Dallas Cowboys in Week 8 on Monday Night Football.