Xavien Howard once again showed off his big-play ability in the Miami Dolphins' Thursday night victory against the Baltimore Ravens, and he's now been rewarded for it.
Howard was named Wednesday as the AFC Defensive Player of the Week for his performance in Miami's 22-10 victory at Hard Rock Stadium.
The biggest play of the game in the defensive battle came when Howard forced a fumble by wide receiver Sammy Watkins, recovered the loose ball and returned the fumble 49 yards for a fourth-quarter touchdown that gave the Dolphins a 15-3 lead.
Howard also had five tackles in the game. His touchdown was the only defensive score in the AFC in Week 10.
This is the first time a Dolphins player has won the award since Week 12 of last season when linebacker Kyle Van Noy was recognized for his three-sack performance in a home victory against the Cincinnati Bengals.
It's the third award for Howard, who also won it in 2017 (Week 14) and 2018 (Week 13).
Howard is the first Dolphins player to win AFC Defensive Player of the Week honors since Cameron Wake did it in 2013, 2014 and 2015, and he joins Sam Madison as the only Dolphins cornerback to be three-time winners of the award.