Hi everyone.
Welcome to another episode of Steam Academy.
My name is Avril and we often use the term differential impedance and differential pairs.
And the advantage of using differential signals are low EMI, reduced ground bounce and crosstalk.
Or if we wanted to send our signal on a longer transmission line, we use differential pairs.
In this video, I'll give you a quick introduction on differential signals.
Then we'll focus on differential impedance and what will be the effect of coupling on differential impedance
using couple of simulations and models on cadence space.
So let's get started.
Let's start with a quick introduction of differential signals.
Differential signals are simply pair of transmission lines with capacitive and inductive coupling between
them, and carry true signal on one line and complementary of same signal on another line.
Now let's talk about advantages of differential signals over single ended signals.
First one is Total di by dt from output driver is greatly reduced compared to single ended signal,
and the second one is in case of differential pairs.
If noise is introduced in the signal, it will impact both of the signal at the same time due to coupling
between those and, which will cancel out when comparator will extract signal from it.
As you can see in the figure, and due to this noise prone behavior, we'll see less rail collapse,
ground bounce, crosstalk, and switching noises will be reduced on differential pairs.
Some very popular examples of differential pair signals are USB, HDMI and PCIe, etc..
Now in the next step, I'm just going to give you a quick demonstration of differential pair signals
on Orcad capture.
17.4 so let's open that.
And this is the circuit that we're going to simulate at the driver side.
We are generating a step response of five volt with certain time period and rise time and fall time.
Then we have a two transmission line of 50 ohm each.
I'm using a lossless transmission lines here, and at the receiver side I'm powering it with five volt
VCC and -five volt V, and at the output side I am terminating it with 75 ohm resistor.
Now what will be the expected results of this one.
So at the input side there will be a five volt swing from 0 to 5 volt.
Then at the differential signals or on our transmission line at Y will get a true signal.
Or you can say a transmission of zero to positive certain voltage at the Z side will get the complementary
of same signal.
And this is a simple comparator which will compare these two and will get two times of swing of differential
pair.
Right.
So this will be the expected output.
Let's see the results.
So as you can see the yellow line or the yellow waveform is our Vin which is going from 0 volt to 5
volt, and Vout will be the two times of differential signal.
And these are the differential signals.
Blue one is true signal and red one is the complementary signal.
I will attach this project in the description.
You can download it and play along with this one.
Now what will happen if some random noise is introduced in our differential pair signals?
I am going to demonstrate noise prone behavior of differential signals in the next demo.
So let's go back to Orcad capture 17.4.
Here it is the circuit.
So here I did only one change.
I have introduced one volt random noise on our differential pair signals.
And let's see what would the expected results.
So at the input side there will be no change.
It will shrink from 0 to 5 volt.
Now on differential pair signals we'll see one volt of random noise on both of them.
And at the receiver side if everything is fine with the differential signal there should be a proper
clean signal at the receiver end.
So let's see the results.
So blue one is our input signal which has swing from 0 to 5 volt.
And yellow is the output signal which is pretty much clean after introducing this amount of noise.
Let me know in the comment section if you can.
Able to send a some amount of noise from differential pair or transmission lines to Vout.
Now let's move to another section and let's talk about differential impedances.
Now if you recall from our transmission line discussions how we have derived impedance for a single
ended signal at a particular instant.
Using the first order model of a transmission line, there L naught was inductance per unit length,
C naught was capacitance per unit length, and using that we have derived z naught is equal to square
root of l naught by c naught.
Now same thing we can represent in the form of stack view or front view where we Where we have signal
at the top layer, return plane or return path at the bottom layer, and then z naught will be the instantaneous
impedance seen by the signal propagating down the line.
But in case of differential pair signals, there will be two identical transmission lines parallel to
each other.
And as we know them, as a true signal and complementary signals travelling down the transmission line
will see same instantaneous impedance Z naught.
Because as we know, the characteristic impedance is independent of voltage, either it is positive
or negative voltage.
Anyways, so in this case differential impedance is the equivalent impedance between these signals,
which will be z naught plus z 0 or 2 times of z naught.
That means if we route a two 50 ohm transmission lines parallel to each other, then its z equivalent
or z differential will be 100Ω, right?
So here we can see a very popular statement.
In ideal world differential impedance of two uncoupled transmission line will be two times single ended
impedance as seen by the driver.
But we don't live in ideal world.
So let's talk about what are the two things that complicate the differential impedance in real world.
And the first one is the impact of coupling of transmission lines or two signals.
And second is common signal and its generation and control on differential pairs.
In this video we will focus on coupling impact on differential impedance, and we will see how to estimate
differential impedance in case of coupling or no coupling.
Now here the first thing we can ask what is coupling.
So if we bring two stripline or microstrip traces close to each other, the fringe electric and magnetic
field will overlap.
That means the coupling between the traces will increase.
Now here when I'm talking about coupling, that means either there can be capacitive coupling or mutual
inductance coupling.
Now to Now, to understand the effect of coupling on z differential, we have to imagine a zero order
model for two transmission lines where we have a true signal and complementary signal.
The capacitive coupling between true signal and return path is C1.
And for complementary signal it's C2.
Now when we put these two tracks close to each other, there will be a mutual coupling between these
two, which I am representing here with C1 two.
Either you can use this capacitive model or you can use the inductive model for the same.
Now let's talk about case one.
What if coupling between two signal increased.
What will be the effect on z differential.
Now when we are increasing the coupling that means the mutual capacitive coupling will increase.
That means its capacitance value will increase.
So C1 two will go higher.
Now overall charge required to charge these two capacitor C1 and C1 two.
Let's imagine the true signal only.
We need high amount of charge, right?
That means the current requirement will increase.
Then the characteristic impedance Z naught, which is inversely proportional to C equivalent.
And I required will decrease.
And we know that that z differential is directly proportional to the characteristic impedance of each
signal.
So the overall differential impedance will go low.
So let's conclude this in case one.
What if coupling increased z differential will go low.
Now let's move to case two.
What if we separate two traces.
What will be the effect on z differential.
So as you can see if we separate two traces c one to our mutual capacitive coupling will reduce.
So that means the characteristic impedance which is inversely proportional to C12 will go high.
And z differential is directly proportional to characteristic impedance.
So z differential will also increase.
So let's conclude this point.
If we separate two traces, the z differential will increase.
Similarly, you can estimate the impact on z differential as you reduce the height or change the value
of dielectric material.
Now let's see a quick demo of how we can estimate the differential pair impedance using cross section
editor.
For that, open PCB editor 17.4.
And this is the project for which we are going to calculate differential impedance.
To open cross section editor you have to go to setup and click over cross section.
Now here you have to select primary tab and double click over Signal Integrity.
Now you have to go to the right side.
And here you will find two.
Option one is differential spacing and one is differential znaught.
In case if it is showing none here you have to just click over here and select edge.
Now here I'm going to calculate or estimate the differential An impedance for layer L1.
Just let me know in the comment section why I'm not calculating it for top layer.
Anyways, so here you can see the width I have set ten mil and I want to calculate the differential
impedance.
Let's set the target impedance of 90 ohm and choose the field to be recalculated.
I have selected line spacing here.
So as you can see for ten mil width, if we put four mil spacing between two tracks, the differential
impedance between those will be 90 ohm.
All right.
And same thing you can calculate for the width.
Now this was all about coupling between two tracks and the effect of coupling on differential pairs.
Now in the next video I will talk about what will be the effect on differential impedance of common
signal.
So now you can just apply and click over okay.
All right.
So see you in the next video.