Rams DT Aaron Donald Named 2018 Defensive Player Of The Year

Rams DT Aaron Donald won the 2018 Defensive Player of the Year award Saturday night at NFL Honors.

.@RamsNFL DT @AaronDonald97 is the 2018 Defensive Player of the Year! #NFLHonors pic.twitter.com/JMVQRJmpxP

— NFL (@NFL) February 2, 2019

Here are the vote totals for Defensive Player of the Year award:

1. Rams DT Aaron Donald – 45
2. Bears OLB Khalil Mack – 5

Donald, 27, is a former first-round pick of the Rams back in 2014. He was in the final year of his four-year, $10.136 million rookie contract when the Rams picked up his fifth-year team option for the 2018 season, which cost them around $6.892 million.

Los Angeles signed Donald to a record six-year, $135 million extension back in August, which included a $40 million signing bonus and $86,892,000 guaranteed. He is set to earn base salaries of $9,108,000 and $17,000,000 over the next two years. 

In 2018, Donald appeared in all 16 games for the Rams and recorded 59 total tackles, 20.5 sacks, two fumble recoveries, four forced fumbles and one pass defense. Pro Football Focus rated him as the No. 1 overall interior defender out of 118 qualifying players.