A. J. Bramlett Net Worth 2018: What is this basketball player worth?

A. J. Bramlett was a professional basketball player. Bramlett was born on January 10, 1977, in DeKalb, Illinois. Bramlett’s weight is 227 pounds. Bramlett’s height is 6 ft 10 in. This article takes a deep dive into A. J. Bramlett’s net worth.

A. J. Bramlett Career and Earnings/Salary

In high school, A. J. Bramlett played basketball for La Cueva High School (Albuquerque, New Mexico). Bramlett played basketball in college with Arizona Wildcats men’s basketball (1995-1999) and started playing professionally in 1999. In the 1999 draft, Bramlett was chosen number 39 in round 2 by the Cleveland Cavaliers. Bramlett left the game of basketball in 2008.

Bramlett played for the Dafni BC during 1999 as well as the Cleveland Cavaliers in 1999. Later, Bramlett played for the La Crosse Bobcats throughout 2000, the CB Valladolid over 2000-2001, the CE Lleida Bàsquet during 2001-2005, and the CB Sevilla during 2005-2006. Some of the highlights of A. J. Bramlett’s career included: Latvian Basketball League, List of NCAA Men’s Division I Basketball champions, and First-team List of All-Pac-12 Conference men’s basketball teams.

A. J. Bramlett Net Worth 2018

Player salaries range between $20,000 and tens of millions of dollars annually, depending on a number of considerations. Although on average, NBA players make a salary of almost $2 million, some with D-league crossover agreements are paid as little as $50,000. The salary cap for NBA players is approximately $100 million, although as of today nobody earns anywhere near that much money. The top players earn around $35 million. Outside of the National Basketball Association, yearly earnings can go below $20,000 annually for minor league teams.