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I'm trying to see how a long transmission line behaves by playing with its parameters in LTspice. The input signal is differential, but the output does not seem to be.

So I created the following simplified circuit to try to understand what is happening:

enter image description here

Here are the outputs:

enter image description here

Why is the common-mode voltage 0 V at the output?

Why is Vout- stuck at 0 V and not oscillating around 0 V too (like Vout+)?

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  • \$\begingroup\$ part of your schematic is missing in the picture. Is vout- connected to ground? Try plotting (Vout+) - (Vout-) \$\endgroup\$
    – Thomas C.
    Commented Feb 23, 2023 at 15:24
  • \$\begingroup\$ No it is not. It's just another resistor \$\endgroup\$
    – HatimB
    Commented Feb 23, 2023 at 15:29
  • \$\begingroup\$ I am not sure the picture views results are ok (?). \$\endgroup\$
    – Antonio51
    Commented Feb 23, 2023 at 16:56

2 Answers 2

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You should be getting a 'floating nodes' error from that schematic, not sure why you're not.

The T-Line in LTspice doesn't have any DC connection from one side to the other, you might think it would work like a piece of coax but it doesn't.

Try adding a large value resistor (R=1E12) from each side of the output to ground. That appeared to solve the problem when I simulated it.

t-line ltspice

To feed the common mode voltage through you can connect the resistors across each side of the T-Line like this:

t-line2 ltspice

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  • \$\begingroup\$ I don't get any "Floating node error". The trick you suggested works. I get a differential signal at the output now, but without the common mode voltage. Do you have another trick to get the DC component too? LTSpice has a "TransmissionLineInverter" simulation example for a coax cable, but I can't understand the logic behind it. \$\endgroup\$
    – HatimB
    Commented Feb 23, 2023 at 16:48
  • \$\begingroup\$ @HatimB Ah, okay... return the two large resistors to Vin+ and Vin- instead of ground. Added to answer. \$\endgroup\$
    – GodJihyo
    Commented Feb 23, 2023 at 17:00
  • \$\begingroup\$ I tried a solution closer to what you suggested before posting the question, but not this one. Your solution works! Thank you very much. \$\endgroup\$
    – HatimB
    Commented Feb 23, 2023 at 17:40
  • \$\begingroup\$ LTspice's topology checker is enabled by default. It will throw a warning (shown in the error log) that the node is floating and then will automatically add a conductance to GND based on whatever the gfloat options parameter is set to (1p by default). You can disable topology check by setting .options topologycheck=0, and then the circuit will get all sorts of singular matrix errors and undefined/infinite values for nodes. \$\endgroup\$
    – Ste Kulov
    Commented Feb 23, 2023 at 19:04
  • \$\begingroup\$ How can be, in the first picture Vout- and Vout+, around 0 V? \$\endgroup\$
    – Antonio51
    Commented Feb 24, 2023 at 8:40
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Out of curiosity, I made a simulation (very weird, in fact).

Here is what I get (Td = 5 ns) ...

At input

enter image description here

At output

enter image description here

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  • \$\begingroup\$ I don't think your impedances are matched. \$\endgroup\$
    – Ste Kulov
    Commented Feb 23, 2023 at 20:30
  • \$\begingroup\$ Right. but OP schematic picture and others are not also (to see "something", I changed only transmission line time delay). \$\endgroup\$
    – Antonio51
    Commented Feb 24, 2023 at 8:49

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