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# Question on LVDS Driver design

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#### suria3

##### Full Member level 5
lvds driver design

Hi guys,

I have a question here. I'm currently designing a LVDS driver. I have seen many papers about LVDS design. What I understand is the LVDS driver need to drive a 100ohm load with a 3.5mA current, so the swing will be 350mVpp. I understand that the 100ohm is the resistor at the receiver part. So, my question here is, in the LVDS driver design, do I need to include the 100ohm resistor on the chip itself? I have a 10kohm resistor as a dc extraction for the common mode feedback of LVDS to control on the common mode voltage but no 50ohm. So, please explain which one is suppose to be.

Thanks,
Suria

lvds driver output resistance

Hi,
If the outputs of differential signals are padp and padn, which are giving 350mV peak, the resistance between padp and padn should be 100ohms.
If your CMFB impedance is much higher vlaue, try making it 100 ohms resistance between the two outputs.
Thanks
-Bharat

lvds+dc coupled +edaboard

bsrivastava said:
Hi,
If the outputs of differential signals are padp and padn, which are giving 350mV peak, the resistance between padp and padn should be 100ohms.
If your CMFB impedance is much higher vlaue, try making it 100 ohms resistance between the two outputs.
Thanks
-Bharat

Hi,

Currently I'm simulating the LVDS driver by putting 100ohm resistor between the padn and padp (but this resistor wont be in the chip, the assumed 100ohm at the receiver). My common mode extracted resistor is 10Kohm each. Are you saying that even in the chip, the resistor between padn and padp must be put 100ohm across them?

Suria.

lvds has more ac impedance mismatch

No I do not think you should use 100 ohms resistors for common-mode sensing- effectively it means you have to double your current output for the same swing.
Are LVDS drivers source terminated as well? I think the answer to this question is your solution.

Bupesh

100 ohm resistor on lvds inputs

I think, we use 100 ohm specifically because we are assuming the distributed load offered by the transmission lines and not the lumped discreate element loads. so no need of 100 ohm resistor in the chip.

lvds driver source resistance

panditabupesh said:
No I do not think you should use 100 ohms resistors for common-mode sensing- effectively it means you have to double your current output for the same swing.
Are LVDS drivers source terminated as well? I think the answer to this question is your solution.

Bupesh

Hi,

The reason i put 100ohm between padp and padn is the assumption that the receiver is terminated by 100ohm. That 100ohm is just for the simulation purpose and won't be on the chip. So, I would say that my common mode extraction will have 10kohm resistor and there wont be any 100ohm resistor across them in the chip. My another question is, when it come to chip testing, do I need to put a 100ohm resistor across positive and negative output of the chip on the board level? Do, I need to terminate the LVDS output by putting a DC coupling capacitors at each output before connecting to the scope. I'm asking this because, the scope will also have 50ohm resistor to ground which will reduce my overall LVDS driver swing. So, which is the appropriate way to do?

Thanks,
Suria

lvds dc coupling

Hi
I would expect scope to be a high-impedance probe, and in case it is 100 ohms then I think it should be one with differential inputs. LVDS needs a current loop and the termination should between the positive and negative outputs. The termination on board will come useful if the output nodes are sensed with high impedance probes.
Ac- coupling will introduce its host of problems like droop, setting the common-mode at the receiver end etc - I would suggest dc-coupling to avoid all such complications.

Bupesh

impedance mismatch lvds

panditabupesh said:
Hi
I would expect scope to be a high-impedance probe, and in case it is 100 ohms then I think it should be one with differential inputs. LVDS needs a current loop and the termination should between the positive and negative outputs. The termination on board will come useful if the output nodes are sensed with high impedance probes.
Ac- coupling will introduce its host of problems like droop, setting the common-mode at the receiver end etc - I would suggest dc-coupling to avoid all such complications.

Bupesh

Hi ,

I attached a LVDS driver which i'm working on. If you notice there, there is 10kohm resistor for the dc extraction of the common mode feedback loop. I was informed by one of the designer that in order to match the 100ohm LVDS receiver resistor, there should be another 100ohm resistor on the chip as in the attachment to avoid any reflection and matching purpose. So, if that was the case , the total dc output impedance now is 50ohm. Now, to have the swing of 350mVpp of LVDS standard, the current is no more 3.5mA, but it is doubled to 7mA. So, i'm very confused now, which one it suppose to be. The 7mA current is no more follow the LVDS standard as i seen from papers and journals. Is it necessary the 100ohm internally for matching purpose. I really need to understand on this. I need help from the guys out there as well.

Thanks,
Suria.

lvds ac coupling circuit

Hello
I looked at some IEEE papers, and it seems source termination is rarely used (a few papers seem to be suggesting it). The source termination will help if the receiver impedance has a mismatch with the cable impedance. I think I have seen even one commercial product without source termination. Similarly, for DVI cable I have seen a paper recommending double termination - which we never really used.

Hope it helps
Bupesh

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