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Rams' Aaron Donald Wins 2018 NFL Defensive Player of the Year

For the third time in NFL history, a player has repeated as NFL Defensive Player of the Year.

Los Angeles Rams defensive tackle Aaron Donald took home the award Saturday night at the 2019 NFL Honors ceremony.

NFL

.@RamsNFL DT @AaronDonald97 is the 2018 Defensive Player of the Year! #NFLHonors https://t.co/JMVQRJmpxP

He joins Lawrence Taylor and J.J. Watt as the only three players to win the prize in successive years.

Donald recorded 59 tackles and an NFL-high 20.5 sacks during the regular season. His sack total came just two short of matching Michael Strahan for the most in a single season.