Talking about diff signals, we hear that they carry equal & opposite voltage. How will I conclude that referring the figure below :-
What is the significance of common mode voltage in diff signals. Also, what are VoH & voL meant for??
Electronic – Differential Signals (LVDS)
common-modelvds
Related Solutions
In the circuit you showed one would generally make the pole for the differential signal smaller (~20 times) than the common mode pole. The reason for this is due to component tolerances. If the common mode filter for the one leg has a different pole frequency then this would affect the differential signal. This is because the filtering is no longer identical on both sides.
This could cause worse interference than not filtering, since the differential OP-AMP and ADC should have at least 50dB CMMR at low frequencies (if not higher).
If the differential pole is lower, then when the common mode filters start "interfering" there is no longer a differential signal of interest. This will prevent high frequency CM noise interfering with your measurements. This is important as CMMR drops off with frequency.
For common mode signals you get two wires: one with a signal and one with the inverse of that signal. If you add them you'll get zero, if you'd take the difference you'll get the signal.
What's the cool thing about this. If the two wires (the pair of wires) picks up some noise. Than the noise will be added on both wires. If you add the two signals you only keep the noise.. But if you take the difference you'll get the signal without the noise.
Signal is S (without noise)
Cable a => S
Cable b => -S
The Difference is S - (-S) = 2*S
Signal is S (with noise)
Cable a => S + noise
Cable b => -S + noise
The Difference is (S+noise) - (-S+noise) = S+S+noise-noise = 2S
So the noise will be cancelled.
example image:
When Common mode signals are used they are mostly transferred with twisted pair cables like this:
Best Answer
The common mode voltage is the average of the voltage of the two components of the differential signal. Ideally it doesn't vary when the signal is in a logical '1' or '0' state.
It is important to know the common mode voltage because different receivers might have different ranges of acceptable input common mode voltage and you want to know that your transmitter and receiver are compatible.
VOH and VOL are exactly what they are defined to be in your graph: the voltage levels that the signals switch between, depending whether the signal is a logical '1' or '0'.
Again, you might need to know these levels to determine if your differential transmitter and receiver are compatible. For example, you wouldn't want VOH to exceed the maximum instantaneous input voltage of your receiver.
The difference between VOH and VOL also tells you the amplitude of your signal. Sometimes you might need to check these values to know if the signal amplitude specified elsewhere is a differential or single-ended amplitude.
If you subtract off the common-mode voltage from each component signal, then they will be equal and opposite.
If the two components are \$V_A\$ and \$V_B\$, whenever \$V_A\$ is at \$V_{CM}+0.175\$ then \$V_B\$ is at \$V_{CM}-0.175\$ and vice versa.