LVDS current drive question

[gumphzb]

For LVDS, the drive current is only 3.5mA to reach 600Mbps. How can this current drive the capacitance of trace? There is no way that the rising time can be faster than 1ns when the drive current is only 3.5mA. Please help me understand this. Thanks in advance.

 

[FvM]

LVDS is using a matched impedance 100 ohms transmission line. In this case, trace impedances don't count, pin capacitances however do, but they are sufficient low to achieve bandwidths in the GHz range. See https://en.wikipedia.org/wiki/Transmission_line

 

[chaerl]

In theory, the current in capacitor is:

I=C(dV/dt)

where:
dV = change in Voltage
dt = change in time

A correction factor is considered due typical rise time of the transmitter
where:

I=IOD=CL(VOD(20%to80%)/(tr-tLLHT))

VOD = Output Differential Voltage
IOD = Output Drive Current
tr = Rise or Fall Time during transition
tLLHT = Low-High Transition time (Correction Factor)
CL = Load Capacitance

You can use this formula to algebraically derived the parameters you needed.