Patriots’ Gronkowski gets $4.3M raise, highest paid TE in NFL

EAST RUTHERFORD, N.J. — Rob Gronkowski of the New England Patriots has all the incentives to become the NFL’s highest paid tight end.

Agent Drew Rosenhaus said Thursday the Patriots have added $4.3 million in incentives to Gronkowski’s contract for the next two seasons. The deal includes $1 million in per game bonuses and $3.3 million in incentives for catches, playing time and touchdowns.

Gronkowski has the potential to make $12.3 million this season and $13.3 next year when his base salary jumps to $9 million.

If he hits all the incentives, Gronkowski would be the NFL’s highest paid tight end. He is fourth among tight ends in base salary, trailing Green Bay’s Jimmy Graham ($10 million), Kansas City’s Travis Kelce ($9.36 million) and Washington’s Jordan Reed ($9.35 million).

Most of the incentives are reachable if the 29-year-old says healthy.

The nine-year veteran would receive incentives of $1.1 million for 70 catches, playing 80 percent of the offensive snaps, nine touchdown catches and 1,085 yards receiving. However, he can collect on no more than three incentives.

Gronkowski had 69 catches for 1,084 yards and eight touchdowns in 14 games last season.

Gronkowski’s deal was first reported by ESPN.

This is the second straight year the Patriots have added incentives to Gronkowski’s contact before the start of the season.

Gronkowski has not caught a pass in limited action in the preseason. He warmed up with the Patriots before Thursday night’s preseason finale against the Giants, but he was not expected to play.