Why are differential signals with opposite polarity?

[m4rtin]

If I read articles regarding differential signaling, then two signals are always with opposite polarity. Why is this needed if difference of two signals is all that matters? I mean for example if both signals are positive and noise causes both signals to increase by 1V, then the difference between two signals would remain the same because noise affected both signals simultaneously.

[suicidaleggroll]

You're right, the difference between the two signals is what matters.  If both signals are the same positive version of the same signal, when you difference them you get zero.

[Alex Eisenhut]

If I read articles regarding differential signaling, then two signals are always with opposite polarity. Why is this needed if difference of two signals is all that matters? I mean for example if both signals are positive and noise causes both signals to increase by 1V, then the difference between two signals would remain the same because noise affected both signals simultaneously.

Well, yes, but that's usually not wanted. That's a common mode voltage, common to both signals, and would cause unwanted current to flow.

[helius]

not just differential signals, but effectively all signals are defined by a difference (a subtraction). With single-ended signals the quantity that gets subtracted is ground or common. What differential signaling gets you is better CMRR, because you can a) use a dedicated "reference" for every signal, and b) lower EMI because the + and - paths are close together so their flux cancels. At the transmitter you drive the two wires in opposite polarity because that minimizes the impact of (and influence on) other signals.

[free_electron]

If I read articles regarding differential signaling, then two signals are always with opposite polarity. Why is this needed if difference of two signals is all that matters? I mean for example if both signals are positive and noise causes both signals to increase by 1V, then the difference between two signals would remain the same because noise affected both signals simultaneously.

You are reading the wrong articles

digital differential signals (lvds or pecl for example) can be going from 1.5 to 1.8 volt. while the opposite node goes from 1.8 to 1.5
In such case the signals are clearly both positive.

Actually, in most systems that use differential signals you will see that both signals are positive. Even systems that use analog circuitry may shift both branches if the diff pair up above the ground level ( called common mode biassing ). It makes the design of the power supply easier. Simply pivot the diff pair around 1/2 vcc or 1/2 vref on an a/d. saves a lot of headache and you get additional benefits : you're not running into the rails , and you stay away from the outer ends of the a/d range where the converters are typically behaving non-linearly.

[rs20]

In addition to other valid points made earlier, driving two signals with two opposite polarities also gives you twice the differential amplitude. For example, if you have a single-ended signal coming out of a 5V-powered driver, it can output anything from 0 to 5V (in an ideal world). If you have a differential driver, it can output from {A=0V, B=5V} to {A=5V, B=0V}, which is a differential signal from -5 to 5V, twice the swing. Doubling the signal means double the signal-to-noise ratio, which is an additional benefit over the fact that the receiver is immune to common mode noise.

You are reading the wrong articles

digital differential signals (lvds or pecl for example) can be going from 1.5 to 1.8 volt. while the opposite node goes from 1.8 to 1.5
In such case the signals are clearly both positive.

They're also clearly changing in the opposite direction...

Actually, in most systems that use differential signals you will see that both signals are positive. Even systems that use analog circuitry may shift both branches if the diff pair up above the ground level ( called common mode biassing ). It makes the design of the power supply easier. Simply pivot the diff pair around 1/2 vcc or 1/2 vref on an a/d. saves a lot of headache and you get additional benefits : you're not running into the rails , and you stay away from the outer ends of the a/d range where the converters are typically behaving non-linearly.

Yes, so exactly what the OP is describing, plus an uninteresting DC bias. Silly nomenclature quibbles aside...

[m4rtin]

You're right, the difference between the two signals is what matters.  If both signals are the same positive version of the same signal, when you difference them you get zero.

My idea was that both signals are positive, but for example 5V apart from each other. For example if Tx+_1 is +5V, then Tx+_2 is 0V at that time or if Tx+_1 is +10V, then Tx+_2 is 5V at the same time etc.

[rs20]

My idea was that both signals are positive, but for example 5V apart from each other. For example if Tx+_1 is +5V, then Tx+_2 is 0V at that time or if Tx+_1 is +10V, then Tx+_2 is 5V at the same time etc.

And what does the receiver do? Just look at Tx+_2 and ignore Tx+_1 because it's completely redundant information??

Suppose that the transmitter transmits Tx+_1 = +5V, and Tx+_2 = 0V. I suppose this represents the transmitter sending a "0V" signal. Now,

Scenario A: The transmitter decides to send a "5V" signal, so it sends Tx+_1 = +10V, and Tx+_2 = 5V. Fine, the receiver can suppose that a "5V" signal is being sent.
Scenario B: Someone turns on a vacuum cleaner in the vicinity, or something else switches on on the same PCB, inducing a noise current in the two wires carrying Tx+_1 and Tx+_2, causing those wires to briefly increase by 5V. So the receiver sees Tx+_1 = +10V, and Tx+_2 = 5V briefly, which looks exactly like the transmitter is sending a "5V" signal. This is obviously an error. Bad. Etc.

This is the key point: noise induced on twisted pair or side-by-side traces on a PCB is common mode noise, which means that it appears almost equally on both wires. If you're just sending the same signal on both wires, your signal is common mode as well, which means that the receiver cannot tease apart the signal from the noise. Adding a DC offset to one of the signal achieves nothing useful. If you're sending the signal as a differential signal (Tx+ swings up while Tx- swings down, even though the polarity of both might remain positive at all times), then it's trivial to eliminate the noise and recover the signal: subtract Tx- from Tx+, and almost all the noise will cancel out.

And, as already explained, a differential transmitter sends exactly opposite currents through the two wires, which means that the radiated emissions basically cancel out, which is another nice bonus.

[Mechatrommer]

i thought its something with cancelling capacitive coupling hence achiving higher data rate transmission and signal integrity, but i need a backup to say so, this link can be more complete in explanation i think fwiw... http://www.pcbcarolina.com/images/presentation_-_hartley_-_diff_pairs.pdf

[m4rtin]

Suppose that the transmitter transmits Tx+_1 = +5V, and Tx+_2 = 0V. I suppose this represents the transmitter sending a "0V" signal. Now,

Scenario A: The transmitter decides to send a "5V" signal, so it sends Tx+_1 = +10V, and Tx+_2 = 5V. Fine, the receiver can suppose that a "5V" signal is being sent.
Scenario B: Someone turns on a vacuum cleaner in the vicinity, or something else switches on on the same PCB, inducing a noise current in the two wires carrying Tx+_1 and Tx+_2, causing those wires to briefly increase by 5V. So the receiver sees Tx+_1 = +10V, and Tx+_2 = 5V briefly, which looks exactly like the transmitter is sending a "5V" signal. This is obviously an error. Bad. Etc.

To be honest, I still don't get it Let's say that transmitter decides to send a "5V" signal and sends this signal on Tx+_1 wire at +10V and on Tx+_2 wire at 5V. Now there is a noise somewhere in the middle which will affect both Tx+_1 and Tx+_2 wires in a way that voltage will increase to +15V on Tx+_1 and +10V at Tx+_2. As receiver cares only about difference of signals, then it shouldn't matter if it receives +10V/+5V or +15V/+10V, i.e. receiver will see the signal at +5 volts in both cases.

[rs20]

To be honest, I still don't get it Let's say that transmitter decides to send a "5V" signal and sends this signal on Tx+_1 wire at +10V and on Tx+_2 wire at 5V. Now there is a noise somewhere in the middle which will affect both Tx+_1 and Tx+_2 wires in a way that voltage will increase to +15V on Tx+_1 and +10V at Tx+_2. As receiver cares only about difference of signals, then it shouldn't matter if it receives +10V/+5V or +15V/+10V, i.e. receiver will see the signal at +5 volts in both cases.

Very simply: in the model that you are proposing, what do you think the transmitter should send in these two cases?:
a) "sending a signal at +5V"
b) "sending a signal at +0V"

And what mathematical operation is the receiver performing to determine what was sent? I'm really unclear as to what you're proposing anymore. I thought in your OP you were proposing to send the same (polarity?) signal on both wires, but now you're talking about subtracting the two signals at the other end to recover the signal, which clearly isn't compatible with that.

[Zero999]

What sort of signal are you talking about?

If it's digital, you can use a transceiver IC such as the SN7516.
http://www.ti.com/lit/ds/symlink/sn75176a.pdf