Seahawks wide receiver DeAndre “DeAndre” Hopkins is a player, and a winning one at that.
The NFL announced Tuesday that Hopkins, the reigning Defensive Player of the Year, will earn $1 million this season in a contract extension.
The extension, announced by the league, comes less than a week after Hopkins won the 2016 Super Bowl.
Hopkins was the NFL’s MVP after catching four touchdown passes against the New England Patriots and has been the most consistent and productive receiver in the league.
The six-year, $75 million deal comes at a time when Hopkins is on pace to earn a total of $24 million in 2017.
It’s the fourth consecutive year he’s earned a contract of at least that size.
Hopkins, who is scheduled to become an unrestricted free agent next month, is expected to earn about $14 million in 2019.
The Seahawks’ decision to extend Hopkins is the latest example of the league expanding its deal with its biggest players, which began with the NFL Players Association’s $100 million deal in 2016.
The NFL also announced Tuesday the Seattle Seahawks will be the NFL and the NFLPA’s only teams to be awarded the 2018 Super Bowl title.
The Seattle Seahawks have been crowned the champions for four straight seasons.
The Seahawks, who won the Super Bowl in 2019, have now won the NFL title in five of the last six years.