SN65CML100: About the SN65CML100DGKR OUTPUT ELECTRICAL CHARACTERISTICS

Hi Brian

Thank you for your prompt response.
In the data manual,I think DC coupling is used. If AC coupling is used,
How to determine the output range?
SN65CML100DGKR is powered by 3.3V voltage.

The connection composition is as follows:

LVDS sender (DC coupling ) -> SN65CML100DGKR (AC coupling )  -> CML receiver(internal bias voltage 1.05V connet with 50ohm).

Do I need to connect terminals（For example,50ohm pull up to 1.8V?） behind the SN65CML100DGKR and in front of the capacitor?

Thanks and Best Regards

Chunli Liang

Hi Brian

Because the CML Receiver is internally pulled up to 1.05V and the Input Voltage range of CML Receiver is 675mV~1725mV.
According to your answer (VOH is close to Vtt and Vol is Vtt-800mv), using the output of SN65CML100 and DC coupling method, if directly connected to CML Receiver, the Vol of SN65CML100 is 1050mV-800mv=205mV, It cannot meet the Input Voltage range(675mV~1725mV) of CML Receiver.
So, we want to use AC coupling to connect to CML Receiver. The input characteristics are determined by the 50 ohm pull-up of the 1.05V power supply generated internally by the CML Receiver IC. I think it can meet the Input Voltage range.
In addition, since it is a 1.05V power supply generated internally by the CML Receiver IC, I do not want to create another 1.05V power supply for the 50ohm resistor to the output of SN65CML100.
So, I would like to confirm if using a 1.8V power supply to pull up a 50ohm resistor to power the output of SN65CML100 has any negative effects?

 SN65CML100DGKR ( 50ohm pull up to 1.8V )  ->  AC coupling capacitor -> CML receiver(internal bias voltage 1.05V connet with 50ohm).

Thanks and Best Regards

Chunli Liang

For  CML device, 675mV~1725mV. is for single ended or differential?   for SN65CML100,  800mv is for differential voltage, single ended is 400mv., so Vol is 1050mv -400mv =650mv.

Since  SN65CML100DGKR  output is  ac coupled, 1.8V power supply  for  SN65CML100DGKR  will not affect  the result.

Best

Brian

Hi Brian

Thank you for the quick response.
For CML devices, 675mV~1725mV is the input voltage range.
Please refer to the blue box in the picture.

I think that for the VOL output of SN65CML100, if DC coupling is used, it cannot meet the single ended input voltage range of CML device.
If calculated based on the minimum output, VTT–1100=1050mV-1100mV=-50mV.
According to typical values, VTT–800=1050mV-800mV=250mV.
It also cannot meet the minimum input voltage of 675mV for CML devices.

I believe that for the differential output of SN65CML100, it meets the differential input requirements of CML device.
The electrical characteristics of the single ended signal above are the point that troubles me.

I'm really sorry, do you have any better suggestions for the output of SN65CML100 and the input of CML Receiver？

Thanks and Best Regards

Chunli Liang

Hi Chunlii:

    This is the single ended DC output from SN65CML100, Vtt is 1050mv, single ended Vod is 400mv. 

   You need to check with CML device  if  675mV~1725mV is single ended or differential. I think it's differential voltage.

Best

Brian

Hi Brian

Thank you for the quick answer.

I will check CML device.

In the image added on May 20th, [Table 161: Receiver DC Characteristics] records [Input Voltage range a or b].

So I think it's single ended voltage.

I think VIDTH is differential voltage.

Thanks and Best Regards

Chunli Liang