"2.6-Gbytes/s high-speed data transmission rate."

Ram foot print....256-Mbit =256 X 1,000,000 bits(256,000,000 bits) /8 = is equal to 32,ooo,ooo bytes.

For e.Digitals audio/video process form NAND, they only minimally need to size the cache of one read/write(or page).....for NAND that is 256bytes. At 8 bits per byte that would be 256 * 8 or 2, 048 bits or 2kb.

They do not need the 256,000,000 bits of space at the intermediate memory. for e.Digital that 256, 000,000 bits is best utilized within the flash foot print.

When they(e.Digital) say small foot print, the cache is what they are considering.

"dubbed the ECK2664JACN." is a whopper...... lots of refresh capacitors in that sucker....lol

doni