LVDS pairs need 100 ohms differential impedance. However, I am having trouble achieving that in my design.

My design requires that the pairs travel over very thin flexible PCB, about 50mm total transmission length. The substrate is 50um, with a 50um coverlay. Using Saturn's PCB Toolkit, it seems to be extremely difficult to achieve 100R impedance without making the tracks too thin to manufacture, or the S/H ratio going out of spec.

However, I can achieve 80R impedance without too much difficulty.

Saturn PCB differential impedance LVDS

My question is: Can I just use 80R tracks, and terminate them with a 80R resistor?

As far as I understand it, as long a every part of my transmission line has the same impedance, I shouldn't get any reflections. The only down side will be a reduced voltage at the receiver, but since I should have very few other losses in the system, that should still be plenty of voltage swing for the receiver.

  • \$\begingroup\$ What is the driver device? \$\endgroup\$
    – The Photon
    Oct 30, 2018 at 18:18
  • 1
    \$\begingroup\$ LVDS is current out, so with lower load, you'll get lower voltage at the receiver. Check the specs of the receiver for your transmission distance and speed, it may be OK. You might be able to increase the tx current output a little to compensate. \$\endgroup\$
    – Neil_UK
    Oct 30, 2018 at 19:12
  • \$\begingroup\$ @ThePhoton - In this case, it's a Lattice Crosslink FPGA, but this question is relevant to many occasions when we're using LVDS over flex. \$\endgroup\$ Oct 31, 2018 at 14:22
  • \$\begingroup\$ The reason to ask is that FPGAs in particular might not use a "true" current-mode LVDS driver. For example, the IOs on your FPGA "can be used in conjunction with off-chip resistors to emulate LVDS output drivers." (sys/IO Usage Guide) That implies to me that you can't trust them to be truly fixed current outputs like Neil and Kevin have mentioned LVDS drivers should supply. \$\endgroup\$
    – The Photon
    Oct 31, 2018 at 16:01
  • \$\begingroup\$ On the other hand, that might be better for you, if you can convince them to produce a full LVDS output swing with reduced load impedance. \$\endgroup\$
    – The Photon
    Oct 31, 2018 at 16:02

1 Answer 1


You might see if your flex vendor can provide 0.075 mm trace width. This capability is fairly common for rigid board vendors now, and my understanding is that flex has historically been able to support smaller features than rigid (but I don't design flexes regularly).

With 0.075 mm traces it's pretty easy to get 100 ohm differential:

enter image description here

That said, if your driver IC can handle an 80 or 85 ohm load, I don't see any reason not to use that instead. You'll get lower loss and probably more accurate characteristic impedance with the wider traces.

  • \$\begingroup\$ LVDS drivers are constant current so they won't care about driving a lower impedance. As the OP states the voltage at the receiver will be lower. I have had a similar issue with a 16 layer board limited to overall ~80mil. The required trace width was about 3.7mil - below the 4mil min our PCB vendor would guarantee. The terminators were internal. We pushed the vendor to make the board at 3.7mil and we had no problems. This was at 2.7Gbps. \$\endgroup\$ Oct 31, 2018 at 0:01

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