0
\$\begingroup\$

I was reading about the series termination techniques.

I have seen many schematics with 33E series termination for clock signal. Can anybody please let me know why 33E is used & what happens if increase or decrease the resistor value & How the resistor value is calculated.

Thanks in advance.

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Do you understand why we need to terminate a high speed transmission between two points? \$\endgroup\$
    – Andy aka
    Commented Oct 12, 2020 at 9:15
  • \$\begingroup\$ No I have just started studying about this concepts, if you can explain it will be helpful \$\endgroup\$
    – Puneeth
    Commented Oct 12, 2020 at 11:32
  • \$\begingroup\$ This site is not geared up to answer open-ended question like that. I was just trying to establish a "yes" or "no". Because "no" is your answer I would encourage you to google this. There are good explanations out there but, because it is "no" I cannot make a reasonable answer for this site. \$\endgroup\$
    – Andy aka
    Commented Oct 12, 2020 at 11:46
  • \$\begingroup\$ Sure Thanks Andy aka \$\endgroup\$
    – Puneeth
    Commented Oct 12, 2020 at 14:52

1 Answer 1

1
\$\begingroup\$

I assume you mean 33 Ω

It's a 'reasonable' value, that gives reasonable results even if some of our assumptions are not quite right.

A typical line impedance for a trace on FR4 that's twice as wide as the substrate is thick, is 50 Ω. Logic boards often end up with line geometries in this ballpark, even if they're not designed to be exactly that. Even if very narrow traces are used, it's very difficult to get more than 100 Ω.

The driver will have its own output impedance, which is often in the 10 to 20 Ω ballpark.

For any given board design and driver part number, we can of course characterise the board and driver, and choose the correct value resistor that matches them.

However, most drivers are going to have an output impedance well below that of the tracks on most board designs, so raising the output impedance bit is guaranteed to improve things. 33 Ω is not a bad compromise until we have done the characterisation.

With a zero Ω output impedance driver and no resistor, and a series terminated line, the 'first transit' voltage at the far end of the line is twice the driver's output voltage. This may cause problems by turning on protection diodes in the receiver. The reflected signal will reflect off the driver, and could cause further switch events.

If we increase the resistor value too far, then the 'first transit' voltage at the far end of the line will be less than the driver's output high. A small reduction eats into the noise margin. A large reduction stops the link working. Reflections will eventually raise the far end voltage to the full voltage, but by then it's too late.

\$\endgroup\$
3
  • \$\begingroup\$ Thanks for the reply, since i am new to this termination techniques can you give a example and explain \$\endgroup\$
    – Puneeth
    Commented Oct 12, 2020 at 11:33
  • \$\begingroup\$ I was only going on your question, about the value of 33 ohms. "I don't understand the need for series termination" is a totally different question. Do you want to ask that one? \$\endgroup\$
    – Neil_UK
    Commented Oct 12, 2020 at 11:44
  • \$\begingroup\$ Neil_UK i got the answer, Thanks for the reply \$\endgroup\$
    – Puneeth
    Commented Oct 12, 2020 at 14:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.