Ramp measurment question

Hello Bob --
        My understanding is that you measure the 20% and 80% voltage points
on the *actual* waveform, i.e. the waveform produced by the specified 50ohm
load. As you pointed out, the output voltage swing is reduced by the
relativly heavy load and I bet there are drivers where the output, under a 50ohm
load, never even makes it to the 20% or 80% points of the no-load swing.

                Best Regards,
                Stephen Peters
                Intel Corp.
Received on Mon Feb 28 14:13:51 1994