The Miami Dolphins earned a win on Sunday over the Las Vegas Raiders, and they wouldn’t have been able to do so without the impressive performance by cornerback Jalen Ramsey.
Ramsey was targeted seven times on Sunday, allowing just two receptions for 23 yards. He also recorded a pass breakup and two interceptions, including one that sealed that game for Miami.
For his efforts, Ramsey was awarded AFC Defensive Player of the Week honors.
Who else but 5️⃣?
Congrats @jalenramsey on being named AFC Defensive Player of the Week! 🤟 pic.twitter.com/NnistIf6j2
— Miami Dolphins (@MiamiDolphins) November 22, 2023
This is the second time in his career that the cornerback has received this award, with the first time coming back in 2016 when he was with the Jacksonville Jaguars.
Cornerback Xavien Howard was the last Dolphin to win the honor back in 2021 when he had five tackles and a forced fumble that he picked up and returned for a touchdown against the Baltimore Ravens.
Miami is racking up the Player of the Week award this season, as he’s the fourth Dolphin to do so this season. Tua Tagovailoa, De’Von Achane and Raheem Mostert all received the offensive version of the same honor earlier in the year.
