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Input common mode voltage of AC coupled differential pair

sharpsharpie

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Hi,

I was reading OnSemi's AN8173 app note and couldn't understand why common mode voltage of AC coupled line is equal to VTERM.
My thought was each input of the receiver will swing between VTERM and VTERM-400mV, thus the common mode would be VTERM-200mV.
I think there's something I am missing understanding the circuit.
Can anyone explain why input swings are between VTERM+200mV and VTERM-200mV?
Thanks

Capture2.PNG
 

d123

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Hi,

Maybe they take Vterm to equal 0, so a swing either way is the +-200mV.
 

KlausST

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Hi,

There is C1 and C2 blocking any DC.

Without anything connected it is quite obvious that VTerm is on D amd /D.
Now if you provide a DC signal to the signal_input (left side of C1 and/or C2) then only at the very beginning you may be able to measure some non_VTerm DC voltage at the IC_inputs.
But RT1 and C1 form a RC = a first order high pass filter.
The same does RT2 and C2.
(Indeed the source impedance plays a role, too. Let's ignore it for now)
R x C gives the time constant tau at which DC is attenuated.

The result is:
After some tau (maybe some microseconds, maybe less) there will be only VTerm as DC remaining at the IC inputs.
At each input the signal swings VTerm +/- 200 mV.
--> VTerm is the common mode (DC) voltage

Klaus
 

FvM

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In addition, it should be noted that Vterm +/- 200 mV input voltage is only achieved if the input signal is DC-balanced respectively has a duty cycle around 0.5. Otherwise the common voltage would shift.

Fast serial interface standards like USB3, SATA or PCIe are using 8b/10b encoding to guarantee DC balanced signals.
 

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