13
\$\begingroup\$

I lately heard in a course that a main advantage of differential signals is that both signals show the same impedance, whereas a single-ended signal, which is measured against ground, has a high impedance at the point of measurement and a low impedance at ground (please refer to the picture below).

I understand why that is the case, but in which way does it matter? Why is this important? Is it because "environmental" noise affects both signals the same way?

enter image description here

\$\endgroup\$
15
\$\begingroup\$

The same impedance is good because capacitively coupled noise is inversely proportional to the impedance.

Usually differential signals are arranged so that both are subjected to the same noise. Consider twisted pair cable, for example. The noise then has the same magnitude and polarity, while the intended signal has opposite polarity. By taking the difference between the two signals at the receiving end, any values on both signals (the common mode signal) cancels.

If the impedance of the two signals is different, then the same noise coupled by the same capacitance to each line end up producing a different voltage in the two lines. Subtracting the voltages at the receiver now no longer cancels the noise completely.

\$\endgroup\$
8
\$\begingroup\$

Is it because "environmental" noise affects both signals the same way?

It is entirely for that very reason. Impedance balance to earth (or earth impedance balance) it is called and, in the presence of some interfering signal, both lines (wires) will be largely affected equally thus, when using a differential receiving amplifier, the common-mode signal due to the interference will be largely eradicated.

enter image description here

The top circuit (above) doesn't have a balanced earth impedance and will be more susceptible to line noise from an external interferer. The lower circuit does have good earth impedance balance and will be less susceptible.

See also this stack exchange Q and A and note the point I make in my answer about "Driving impedances (sending end) are matched".

\$\endgroup\$
1
  • \$\begingroup\$ BTW -- you can also see a balanced interface as a Wheatstone bridge, which can be handy from a circuit analysis standpoint. \$\endgroup\$ – ThreePhaseEel Feb 1 '17 at 23:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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