Making a 3*3 block in Bejeweled

Question

More puzzle but still math:
You surely know the rules of Bejeweled (or comparable games): Start in a stable state. Flip two adjacent cells. Everywhere where $>2$ equal symbols are adjacent horizontally or vertically, they vanish and everything on top on it falls down. (If more matches, delete at the same time.) Repeat until no more matches are made and the field is stable again.
While playing Klaxxon (comparable rules but diagonals count too) I always wondered if I can theoretically make a $3*3$ block or if the $5*5$ board was too small to prepare a chain reaction. The same question for Bejeweled: can you make a $i*j$ block and what $m*n$ playground area do you need to prepare it. (One number may even be infinite - I can easily do an infinite vertical stack one cell wide.) So, here is how I do the $3*3$ block. The X are "random" nonmatching cells. Swap X and A. Then the A vanishes, then B, then C, then D makes the desired block. (If I didn't botch the setup - always check if no other cells match!)
XDXXCX
XCXXXX
XDXDCX
XCDCXX
CDCDCC
XXDCXX
CXCDXC
XXDCXX
XBXXBC
XBXXBX
AAXAAX
XBAXBX
Can you improve on $12*6$? (Bonus if you use as few colors as possible, super bonus if no debris remains.) More interesting, can you even give a generic recipe for making an $i*j$ block?