XIO2001: Calculating Vrx_cm_acp from Datasheet

Darren (FAE)

Part Number: XIO2001

[1] In the datasheet, there is an equation for calculating V_rx_cm_acp as follows.

A customer I have has pointed out that the absolute value result unit will be [V], not [V^2], so when you divide by V, you end up with no units.
This doesn't make any sense. Is the equation correct?

[2] VRX-DIFFp-p from the Datasheet lists a value of 1.2V max.

Under testing, a customer has confirmed REFCLK+ and REFCLK- are both around ±400mV, but calculating the p-p value from the equation in the datasheet, they end up with 1.6V ~ 1.7V

Is there any problem when the differential signal p-p is larger than 1.2V? Is there any chance the IC could be destroyed/malfunction?

Regards,

Darren

over 4 years ago

0 Brian Zhou over 4 years ago

TI__Guru*** 146515 points

1: where did you get that equation? this this what I saw

2; it should be ok if a little over 1.2v

0 Darren (FAE) over 4 years ago in reply to Brian Zhou

TI__Expert 5483 points

Hi Brian, the image you attached didn't show up (you can't CTRL+V an image in-line, you actually have to save it to desktop, then upload it manually into the post)

The equation I re-wrote above for Vrx-cm-acp comes directly from the datasheet p19 Section 6.8 PCI Express Differential Reference Clock Input Ranges(1)
For REFCLK+ and REFCLK- where the MAX value listed is 140mV.

The MAX value listed is in [mV] - but the equation listed results in no units.

Could you confirm please?

So 1.6V ~ 1.7V should not be an issue? They are nervous about operating the device outside of datasheet recomendations. Could you help me look into why the MAX value is 1.2V? What circuitry could be damaged by having a larger differential voltage for the clock signal? Just to be sure, but is the equation for #2 also correct?

Thanks!

0 Darren (FAE) over 4 years ago in reply to Darren (FAE)

TI__Expert 5483 points

Hi Brian,

Any update on the above?

0 Brian Zhou over 4 years ago in reply to Darren (FAE)

TI__Guru*** 146515 points

1: it should be an error. The one on top of p19 should be correct

2: you can see Vih_se on p19. it can handle up to 3.3v

0 Darren (FAE) over 4 years ago in reply to Brian Zhou

TI__Expert 5483 points

1: In the equation on top of p19, replace Vrxp with Vrefclk+ and Vrxn with Vrefclk-, right? Thanks!

2: I'm still a little confused. Vih_se is for single-ended. Since the customer is using differential, and the voltage for Vrefclk+ or Vrefclk- is ±0.4V, the differential voltage is 2*|Vrefclk+ - Vrefclk-| =1.6V or 1.7V.

The voltage at the pin is ±0.4V, so within the pin absolute max of Vdd = 3.3V.

But the question is, in differential mode, why is the MAX listed as1.2V...? Does the device go into some kind of non-lineaerity behavior if the differential input p-p voltage is larger than 1.2V?

You have mentioned it should be fine to have up to 1.7V differential voltage (±450mV at REFCLK+/-)
Is this still true...?

Thanks!

0 Brian Zhou over 4 years ago in reply to Darren (FAE)

TI__Guru*** 146515 points

1: yes

2: you mean single ended is -450mv to 450mv when used as differential?

0 Darren (FAE) over 4 years ago in reply to Brian Zhou

TI__Expert 5483 points

➡The voltage applied at REFCLK+ pin is ±400mV.
➡The voltage applied at REFCLK- pin is ±400mV.

Thus, the maximum voltage difference between the two is:
|REFCLK+ - REFCLK-| = |400mV - -400mV| = 800mV; which equals a difference of 800mV.

But the equation VRX-DIFFp-p from Section 6.8 in the datasheet, shows the "differential input peak-to-peak voltage" equation as being:
2*|REFCLK+ - REFCLK-| = 1.6V.

Is the equation correct? If it is correct, then is 1.6V "differential input peak-to-peak voltage" acceptable? The DS states 1.2V max. What problems can occur by going above the 1.2V max spec?