Could someone check my calculations? I'm looking for the shortest time an Ethernet frame will occupy the wire. I make it about 5uS on 100Mbit as follows.

Minimum frame size in bytes is 46 payload + 14 header + 4 FCS = 64. Preamble (still needed on full duplex) of 8 octet times Total of 72 octets * 8 bits = 576 bits. Each bit time is 1/100,000,000 seconds. (IIRC the signalling rate is

125,000,000 bits with a 4B/5B encoding so we do get 100,000,000 data bits in a second.)Finally 576 * 1/100,000,000 = 5.76uS

On 10Mbit I make the figure ten times as much 57.6uS On Gbit I think they kept the basic timing budget from 100Mbit so make it's smallest-frame time about the same as for 100Mbit.

Corrections welcome. Does the above look about right?

-- James