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While learning about the differential signals I saw this youtube video. It looks like this slide does not correctly shows the voltage levels of the complementary differential signals, as I have marked in the following image.

enter image description here

My understanding is that the 'Blue' colored complementary signal should be -2V to 0V signal. So when we take the difference of the two signals (Black colored and Blue colored) then we get the original 'Single Ended Signal'. Please correct me if my understanding is wrong here.

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  • \$\begingroup\$ Search "LVDS signalling voltages". There ARE true complementary differential signalling systems, but LVDS is not one of them; neither is this slide. Advantage : you only need one supply rail. \$\endgroup\$ – user_1818839 Sep 1 '20 at 10:29
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The slide is correct, the signal is differential.

The voltage levels do not have to be symmetric like: -1 V, +1 V

Voltage levels like 0V, + 2 V are perfectly acceptable.

In the botton trace there is simply a DC voltage (of + 1V) added to the differential signal. It was added to both signals (the black and the blue) so the differential signal: (black - blue) remains the same!

Suppose you measured these voltages compared to a +1 V DC voltage: voltmeter between + 1 V DC and each signal. Then you would measure +1 V and -1 V.

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\$\boxed{\text{Example}}\$

5 volt logic gates use 5 volts to represent logic level "1" and 0 volts to represent logic level "0". If you had a logic gate toggling its output you would see: -

101010101 etc..

And, if you inverted that signal you would see: -

010101010 etc..

But it will still use 5 volts and 0 volts to represent the logic levels. It won't suddenly apply a value of -5 volts to represent logic level "1".

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