# LVDS difference between Vpp and Vdiff

I am looking into LVDS and I see the terms Vpp and Vdiff being used. I understand how LVDS works but the terminology is a little bit confusing. For the below picture we can see Vcm = 1.2V and the maximum and minimum voltage swings by 1.35V and 1.05V. I assume that Vdiff = 1.35V - 1.05V = 0.3V and Vpp = 2 * Vdiff = 0.6V. Is that right?

And why do we have Vdiff and Vpp. For what do we need to know Vpp?

In the example below we see a LVDS circuit. There is a 350mV voltage drop across the resistor. Is this value the Vpp or Vdiff value?

Source: Texas Instruments (https://training.ti.com/lvds-overview)

• That's Vdiff because it's a snapshot in time. Commented Nov 23, 2020 at 13:22
• And you should give credit for the diagram you posted. Something as simple as "diagram from the TI LVDS Owner's Manual snla187" suffices, as that appears to be where it came from. Commented Nov 23, 2020 at 13:25

The differential signal voltage $$\v_{diff}\$$ is the difference between the positive signal and negative signal. It is a time-varying waveform.

The peak-to-peak voltage $$\v_{pp}\$$ is the difference between the maximum value of a signal (in this case $$\v_{diff}\$$) and its minimum value as the signal varies in time. It isn't a waveform. It's a parameter describing a waveform.

Vpp is the peak-to-peak swing of each signal relative to itself. That is,

• Vpp = Vmax - Vmin

This calculation removes the common-mode voltage.

Vdiff is the swing relative to the two signals in the pair. If the signals in the pair have equal Vpp swing, Vdiff = 2* Vpp

On a scope, you would observe a diff pair swing by using the sum of two channels with the (-) side inverted. In other words, for a signal pair (V+) and (V-), Vdiff is derived as follows:

• Vdiff = (V+) - (V-)

A differential probe does this for you, inverting the (V-) side using a differential amp.

So using your 300mV Vpp example, viewed this way a differential pair with a 300mV p-p swing on each signal will have, at most, +/-300mV between the two. But because we invert the (V-) side and sum to make Vdiff, its effective swing is twice that of Vpp:

• Logic '1': V+, V- = 1.35, 1.05V; Vdiff = +300mV
• Logic '0': V+, V- = 1.05, 1.35V; Vdiff = -300mV

The differential swing is therefore +/-300mV, or 600mV.

You'll notice that the common-mode offset (1.2V) is eliminated in the differential waveform - it cancels out.