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Generation of Vcm in LVDS driver

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garvind25

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Hi,

I was trying to make a LVDS driver. In this connection, I am referring to the paper here. I am unable to understand how VCM voltage is getting generated. I hope someone can help me.

As per the LVDS standards, the nominal current through switching bridge is 3.5 mA (Iout). The LVDS standards also require a VCM (nominally of 1.25v) to be generated by the driver. As per the paper,to generate this Vcm, the circuit uses two 10K resistors as voltage divider. But as per my calculation, (pls. see the image below), the Vcm is coming out to be 0.174v [(3.5mA X ((10k+10k)||100))/2].

lvds_Vcm.jpg

Where am I going wrong/ what am I missing here. Any guidance is welcome.
Looking forward to your replies.

Thanks and Regards,
Arvind Gupta.
 

There would be a feedbacl loop for which the VDD/2 divider
is only the reference. Details of such things are seldom shown
in datasheets / app notes but can sometimes be found in
"back in the day" circuit design papers (from when LVDS was
anything like innovative).

The average of a signal pair would be fed back to a poor-boy
op amp against that 1.2V reference, to as to get the current
sources to balance on that pivot point.
 

Well...I could not understand the above explanation. How is Vcm=1.25 v even getting generated with the help of resistors? Pls. elaborate on the answer. Other than that, can anyone refer to me a simpls LVDS driver and receiver ckt. pls?

Thanks,
Arvind Gupta

- - - Updated - - -

I forgot to add this in my previous post: It is mentioned in the paper that Iout or the bias current for the bridge is fixed at Vod(nominal)/50. I suppose Vod(nominal) would be 0.35v. Then Iout comes to 0.35/50=7 mA. The specifications of LVDS says that Iout should be 3.5mA. So how come this mismatch?


Thanks and Regards,

Arvind Gupta.
 
Last edited:

how come this mismatch?
You overlooked that the hypothetical LVDS driver uses a termination resistor Rt_t, thus it needs to source 7 mA to get 3.5 mA at the receiver.
 

The swing is 350mV with 100mA current going through 100 ohm resistor.

The middle point is 1.25V, so the swing is between 1.075-1.425V.
 

Not sure what you're talking about. Transmitter termination resistor is 100 ohm in case of doubt (to match the cable impedance). 10k resistor has nothing to do with termination, it's used in the voltage divider for common mode voltage feedback.
 

[(3.5mA X ((10k+10k)||100))/2]=3.5mA X100/2=175mV, which is half of the total swing.
 

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