CLEVELAND–For the third time this season, a Cleveland Browns player has won the AFC Player of the Week award.
Jason Campbell has earned the offensive honor, joining special teams teammates, Travis Benjamin and Spencer Lanning from previous games.
In a 24-18 victory vs. Baltimore on Sunday, Campbell tied a career high with three touchdown passes and completed 23 of 35 passes (65.7 %) for 262 yards with zero interceptions for a 116.6 passer rating.
Campbell’s Sunday afternoon was all the more impressive as he returned to the field after sustaining a rib injury in the first quarter. His QB play helped Cleveland snap an 11-game losing skid to the Ravens.
All three of Campbell’s touchdown passes came from the 20-yard line or closer, against a Ravens defense that entered the game No. 1 in the NFL inside the red zone.
His most impressive work on the night was with the game on the line in the fourth quarter. Protecting a 21-18 lead with 6:44 left in the contest, Campbell completed four of five passes for 42 yards, including a fourth-and-1 conversion, and rushed for a 12-yard first down during a 15-play, 67-yard drive that spanned 6:30 to set up a field goal with 14 seconds remaining and secure a Browns victory.
Campbell orchestrated a Browns scoring drive in each quarter, marking the first time that Cleveland has scored in all four quarters since a 30-7 win against Kansas City on Dec. 9.
Campbell has not thrown an interception all season, extending his overall streak to 90 attempts without getting picked off.
He became the first Browns quarterback to pass for at least three touchdowns with zero interceptions in a game since Brady Quinn in 2009 (Week 14).