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2019 NFL Free Agency: Buffalo Bills sign WR John Brown to three-year deal, per report

More speed for the Bills offense!

Free-agent wide receiver John Brown is signing with the Buffalo Bills, per several reports. NFL reporter Adam Caplan adds that the deal is for three years, $27 million dollars, with $11.7 million in guarantees and $10.1 million guaranteed at signing.

The lightning-fast Brown was a free-agent target for the Bills last offseason, but ultimately signed a one-year deal with the Baltimore Ravens. That season started hot, with 31 receptions for 586 yards and four touchdowns through the first eight games, but after Joe Flacco’s injury and Lamar Jackson’s promotion, he cooled off. Brown only caught 11 passes for 129 yards and a touchdown over the final eight games of the year as the Ravens committed to a handcuffed run-heavy scheme.

Freed from that system, Brown could flourish with the Bills and Josh Allen’s arm. Adding Robert Foster, the Bills now have two great deep threats to rotate on the field. He’ll need to stay healthy for another year to maximize that value. In 2017, quad, back, and toe injuries limited Brown to only ten games. In 2016, a tender hamstring held him back. 2018 was a clean sheet in that regard, so Brown’s worst may now be behind him.