John Johnson was released by the Browns in March, marking an end to his tenure with his second career team. A reunion with his first is now in order. The veteran safety has agreed to a one-year deal that will return him to the Rams.
Johnson spent his first four seasons in Los Angeles, quickly establishing himself as a key member of the team’s secondary. He started 48 of his 54 games with the Rams and played every snap of the 2020 season. That made him a hot commodity on the open market when he reached free agency, and he arrived in Cleveland with plenty of expectations.
The Boston College product signed a three-year, $33.75M with the Browns in 2021, and provided playmaking in his debut season. Johnson recorded three interceptions and five pass breakups while remaining a full-time starter. Things did not go as well last year, though, one in which he was moved closer to the line of scrimmage. That decision resulted in a 100-tackle campaign for the third time in his career, but only one interception and a career-worst passer rating allowed in coverage (82.8).
The Browns saved $9.75M by designating Johnson as a post-June 1 release, something which came about after it was clear they would look to trade him in a cost-shedding move. Cleveland signed Juan Thornhill and Rodney McLeod in free agency to work alongside incumbent Grant Delpit on the backend. Thornhill in particular will be counted on to replace Johnson’s production.
The Rams have a number of relatively inexperienced options at the safety position (among many others after embracing a youth movement this offseason). Jordan Fuller can be expected to once again shoulder a starting role after being limited to just three games last year, but Johnson could find himself alongside the former sixth-rounder with the first-team defense if he can regain his previous Los Angeles form. Doing so would help the Rams improve on their highly disappointing performance in 2022 while boosting his free agent stock ahead of the next offseason in the process.
More must-reads: