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I am trying to transmit a 3.3V digital signal over 20 cm distance. The problem is the environment might be noisy because of high current wire near it. I read that current signal is more robust than voltage signal. Does that apply to my case? and why?

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  • \$\begingroup\$ The only way to change current is to inject or extract either charges or time, versus voltage which only needs a change in energy. \$\endgroup\$ Jul 1, 2017 at 3:39

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It's not whether the signal is coded as voltage or as current what makes it more or less robust, it's the power involved.

Receiver of voltage signals (e.g. CMOS gate input) have extremely high impedance. So a noise voltage source does not need to provide much power to disturb the signal. That's why you shouldn't use it for "long" (some 0.1m) distances, because changing magnetic fields through the area circumscribed by the signal and GND line can easily induce some volts; although not much power (BTW: that's why twisted pairs are used; it causes the noise voltages of each twist to roughly cancel out each other).

Current signals (e.g. 4-20mA) are working against some resistors at the receiver side that require quite a voltage to generate the current. That means noise (voltage or current) source need to provide considerable power to have an effect. This is what makes it more robust.

But as thobie mentioned there are other good ways of reducing noise interference than just by requiring more power:
E.g. using differential signaling. That way you can still use voltage signals with high impedance receivers (but need as second signal line).
That's what is used e.g. for USB.

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  • \$\begingroup\$ So if i use a termination resistor at receiver side for my case, for example 1kohm resistor to gnd, it would help my circuit? \$\endgroup\$
    – han
    Jul 13, 2017 at 11:53
  • \$\begingroup\$ Yes, because then a noise signal needs much more power to create the same noise voltage (compared to a high impedance receiver). Of course the transmitter side needs to be able to drive required power too. This increases signal to noise ratio (which is measured as ratio of power values). \$\endgroup\$
    – Curd
    Jul 13, 2017 at 20:24
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LVDS uses low impedance termination at the receiving end so the robustness comes also from the "current signal" + many other thing related to the design of the interface.

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  • \$\begingroup\$ Welcome to EE.SE. Your answer is short, uses an unexplained acronym and has no links to reference material and so is unlikely to be of any use. Consider improving it. Have a look at the quality of answers given to other questions on the site - particularly ones with high votes. \$\endgroup\$
    – Transistor
    Jul 1, 2017 at 6:27

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