Ravens linebacker Patrick “Peanut” Onwuasor has cracked into the awards, as he has been named the AFC’s Defensive Player of the Week. It’s his first NFL award.
The third-year linebacker had two sacks and forced a key fumble late in Saturday’s 22-10 Ravens win in Los Angeles when he punched the ball out of tight end Antonio Gates’ arms. He also had nine tackles.
The former undrafted Portland State product has emerged as a key player in Baltimore’s top-ranked defense and has been playing his best football in recent weeks. Onwuasor has 3.5 sacks and two forced fumbles over the past four games.
“Patrick has really stepped it up, taking it to another level the last two, three, four games – both on defense and special teams,” Head Coach John Harbaugh said last week. “So, credit to him. It’s important for us.”
After converting from being a college safety to NFL linebacker, Onwuasor has always shown himself to be a big hitter. He’s gradually improved next to Pro Bowler C.J. Mosley, and has become an excellent blitzer.
Onwuasor split time, and starts, with rookie Kenny Young at the start of the year but is now taking the lion’s share of the load. He had 30 defensive snaps (of 63 total) in Saturday’s win, compared to 12 for Young. Onwuasor has started the past eight games.