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Registered: ‎04-23-2020

Questions about Zynq's differential input voltage VID(Q-Qbar)

Hello. I am a Circuit engineer.

I am using a Zynq FPGA, I have set the Zynq Pin to DIFF_HSTL_I_18 and I will evaluate the differential input voltage in circuit board. When I checked the spec, I could check the VID level.
I drew a VID and the picture below is correct?

Zynq Specz7010.jpg

VID(Q-Qbar)

Q-Qbar.jpg

Eye Diagram(Differential probing)

EYE Diagram.jpg

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Registered: ‎01-22-2015

@JINS 

Welcome!

DIFF_HSUL_I_18 is a differential (not single-end) standard with circuit description shown by the following figure from Xilinx document UG471.
DIFF_HSTL_I_UG471.jpg

Note that Q and Qbar for DIFF_HSUL_I_18 are always positive voltages.  So, your first drawing is correct if you replace the word “Single-Ended” with the word “Differential”.

In your Eye Diagram, you should change “VID(Q-Qbar)” to “VID(Q-Qbar)/2”.

Note that VID=0.100V is an input minimum requirement for DIFF_HSUL_I_18.  However, the Zynq with VCCO=1.8V will typically output DIFF_HSUL_I_18 with Q  and Qbar swinging between VOH=VCCO-0.400V = 1.400V and VOL=0.400V.  So, with your oscilloscope, you will typically observe (Q-Qbar) = 1.400-0.400 = 1.000V.

Cheers,
Mark

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Visitor
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Registered: ‎04-23-2020

Thank you for the answer.

There is something I don't understand in the Eye diagram your answer.
When measuring Q-Qbar with a differential probe, it will be measured as shown in the figure below, and I think the eye diagram that all measured UI with differential probe are merged.
Therefore, I think VID(Q-Qbar) is correct, not VID(Q-Qbar)/2 in Eye Diagram.
If I am wrong, I would appreciate it if you could explain it in detail.

Thank you.

Q-Qbar_2.jpg

 

EYE Diagram_2.jpg

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Registered: ‎01-22-2015

@JINS 

If I am wrong, I would appreciate it if you could explain it in detail.
My mistake.  Your eye-waveform labels are correct.  When using a differential probe, sometimes VID(Q-Qbar) will be positive and sometimes it will be negative. 

Mark

 

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