WEBVTT 00:00.520 --> 00:01.320 Hi everyone. 00:01.680 --> 00:04.480 Welcome to another episode of Steam Academy. 00:04.600 --> 00:10.000 My name is Avril and we often use the term differential impedance and differential pairs. 00:10.320 --> 00:16.360 And the advantage of using differential signals are low EMI, reduced ground bounce and crosstalk. 00:17.120 --> 00:22.080 Or if we wanted to send our signal on a longer transmission line, we use differential pairs. 00:22.640 --> 00:25.800 In this video, I'll give you a quick introduction on differential signals. 00:25.920 --> 00:31.720 Then we'll focus on differential impedance and what will be the effect of coupling on differential impedance 00:31.720 --> 00:34.880 using couple of simulations and models on cadence space. 00:35.000 --> 00:36.200 So let's get started. 00:37.880 --> 00:41.160 Let's start with a quick introduction of differential signals. 00:41.560 --> 00:47.920 Differential signals are simply pair of transmission lines with capacitive and inductive coupling between 00:47.920 --> 00:55.920 them, and carry true signal on one line and complementary of same signal on another line. 00:55.960 --> 01:00.680 Now let's talk about advantages of differential signals over single ended signals. 01:01.120 --> 01:07.950 First one is Total di by dt from output driver is greatly reduced compared to single ended signal, 01:08.470 --> 01:11.430 and the second one is in case of differential pairs. 01:11.430 --> 01:17.230 If noise is introduced in the signal, it will impact both of the signal at the same time due to coupling 01:17.230 --> 01:22.470 between those and, which will cancel out when comparator will extract signal from it. 01:22.550 --> 01:30.190 As you can see in the figure, and due to this noise prone behavior, we'll see less rail collapse, 01:30.230 --> 01:35.230 ground bounce, crosstalk, and switching noises will be reduced on differential pairs. 01:36.670 --> 01:42.150 Some very popular examples of differential pair signals are USB, HDMI and PCIe, etc.. 01:43.430 --> 01:48.510 Now in the next step, I'm just going to give you a quick demonstration of differential pair signals 01:48.510 --> 01:49.590 on Orcad capture. 01:49.630 --> 01:52.190 17.4 so let's open that. 01:52.910 --> 01:56.590 And this is the circuit that we're going to simulate at the driver side. 01:57.070 --> 02:03.270 We are generating a step response of five volt with certain time period and rise time and fall time. 02:04.030 --> 02:07.510 Then we have a two transmission line of 50 ohm each. 02:07.990 --> 02:15.390 I'm using a lossless transmission lines here, and at the receiver side I'm powering it with five volt 02:15.390 --> 02:22.990 VCC and -five volt V, and at the output side I am terminating it with 75 ohm resistor. 02:23.950 --> 02:26.670 Now what will be the expected results of this one. 02:26.870 --> 02:32.310 So at the input side there will be a five volt swing from 0 to 5 volt. 02:33.110 --> 02:40.350 Then at the differential signals or on our transmission line at Y will get a true signal. 02:40.470 --> 02:46.990 Or you can say a transmission of zero to positive certain voltage at the Z side will get the complementary 02:46.990 --> 02:47.870 of same signal. 02:47.990 --> 02:54.390 And this is a simple comparator which will compare these two and will get two times of swing of differential 02:54.390 --> 02:54.750 pair. 02:55.230 --> 02:55.630 Right. 02:55.670 --> 02:57.430 So this will be the expected output. 02:57.470 --> 02:58.710 Let's see the results. 03:00.070 --> 03:08.160 So as you can see the yellow line or the yellow waveform is our Vin which is going from 0 volt to 5 03:08.160 --> 03:11.780 volt, and Vout will be the two times of differential signal. 03:12.020 --> 03:13.740 And these are the differential signals. 03:13.860 --> 03:17.180 Blue one is true signal and red one is the complementary signal. 03:17.940 --> 03:19.620 I will attach this project in the description. 03:19.620 --> 03:21.740 You can download it and play along with this one. 03:23.460 --> 03:27.900 Now what will happen if some random noise is introduced in our differential pair signals? 03:28.220 --> 03:33.100 I am going to demonstrate noise prone behavior of differential signals in the next demo. 03:33.300 --> 03:36.100 So let's go back to Orcad capture 17.4. 03:37.060 --> 03:38.180 Here it is the circuit. 03:38.380 --> 03:40.260 So here I did only one change. 03:40.260 --> 03:44.180 I have introduced one volt random noise on our differential pair signals. 03:45.180 --> 03:48.500 And let's see what would the expected results. 03:48.580 --> 03:50.820 So at the input side there will be no change. 03:50.900 --> 03:52.940 It will shrink from 0 to 5 volt. 03:53.380 --> 03:58.780 Now on differential pair signals we'll see one volt of random noise on both of them. 03:59.620 --> 04:05.980 And at the receiver side if everything is fine with the differential signal there should be a proper 04:06.020 --> 04:07.770 clean signal at the receiver end. 04:09.050 --> 04:10.210 So let's see the results. 04:11.650 --> 04:16.330 So blue one is our input signal which has swing from 0 to 5 volt. 04:17.050 --> 04:23.050 And yellow is the output signal which is pretty much clean after introducing this amount of noise. 04:24.370 --> 04:25.930 Let me know in the comment section if you can. 04:25.970 --> 04:32.010 Able to send a some amount of noise from differential pair or transmission lines to Vout. 04:33.210 --> 04:37.210 Now let's move to another section and let's talk about differential impedances. 04:40.210 --> 04:45.890 Now if you recall from our transmission line discussions how we have derived impedance for a single 04:45.890 --> 04:48.130 ended signal at a particular instant. 04:48.130 --> 04:54.450 Using the first order model of a transmission line, there L naught was inductance per unit length, 04:54.490 --> 05:00.010 C naught was capacitance per unit length, and using that we have derived z naught is equal to square 05:00.010 --> 05:01.370 root of l naught by c naught. 05:02.730 --> 05:09.050 Now same thing we can represent in the form of stack view or front view where we Where we have signal 05:09.050 --> 05:15.730 at the top layer, return plane or return path at the bottom layer, and then z naught will be the instantaneous 05:15.730 --> 05:18.650 impedance seen by the signal propagating down the line. 05:19.610 --> 05:25.610 But in case of differential pair signals, there will be two identical transmission lines parallel to 05:25.610 --> 05:26.210 each other. 05:26.490 --> 05:31.610 And as we know them, as a true signal and complementary signals travelling down the transmission line 05:31.610 --> 05:34.050 will see same instantaneous impedance Z naught. 05:35.090 --> 05:40.650 Because as we know, the characteristic impedance is independent of voltage, either it is positive 05:40.650 --> 05:41.810 or negative voltage. 05:43.090 --> 05:49.650 Anyways, so in this case differential impedance is the equivalent impedance between these signals, 05:50.250 --> 05:53.770 which will be z naught plus z 0 or 2 times of z naught. 05:53.810 --> 05:59.890 That means if we route a two 50 ohm transmission lines parallel to each other, then its z equivalent 05:59.890 --> 06:03.050 or z differential will be 100Ω, right? 06:03.210 --> 06:05.850 So here we can see a very popular statement. 06:05.850 --> 06:12.600 In ideal world differential impedance of two uncoupled transmission line will be two times single ended 06:12.600 --> 06:14.280 impedance as seen by the driver. 06:15.760 --> 06:17.760 But we don't live in ideal world. 06:17.760 --> 06:23.600 So let's talk about what are the two things that complicate the differential impedance in real world. 06:24.120 --> 06:29.080 And the first one is the impact of coupling of transmission lines or two signals. 06:30.400 --> 06:35.480 And second is common signal and its generation and control on differential pairs. 06:36.400 --> 06:42.040 In this video we will focus on coupling impact on differential impedance, and we will see how to estimate 06:42.040 --> 06:45.360 differential impedance in case of coupling or no coupling. 06:48.000 --> 06:50.440 Now here the first thing we can ask what is coupling. 06:50.760 --> 06:56.680 So if we bring two stripline or microstrip traces close to each other, the fringe electric and magnetic 06:56.680 --> 06:57.840 field will overlap. 06:57.840 --> 07:00.560 That means the coupling between the traces will increase. 07:01.480 --> 07:08.000 Now here when I'm talking about coupling, that means either there can be capacitive coupling or mutual 07:08.000 --> 07:09.040 inductance coupling. 07:10.560 --> 07:16.120 Now to Now, to understand the effect of coupling on z differential, we have to imagine a zero order 07:16.120 --> 07:20.640 model for two transmission lines where we have a true signal and complementary signal. 07:21.360 --> 07:25.120 The capacitive coupling between true signal and return path is C1. 07:25.440 --> 07:27.680 And for complementary signal it's C2. 07:28.440 --> 07:33.880 Now when we put these two tracks close to each other, there will be a mutual coupling between these 07:33.880 --> 07:36.480 two, which I am representing here with C1 two. 07:37.520 --> 07:41.680 Either you can use this capacitive model or you can use the inductive model for the same. 07:42.560 --> 07:44.600 Now let's talk about case one. 07:44.600 --> 07:46.880 What if coupling between two signal increased. 07:46.960 --> 07:49.720 What will be the effect on z differential. 07:50.960 --> 07:56.320 Now when we are increasing the coupling that means the mutual capacitive coupling will increase. 07:56.320 --> 07:59.320 That means its capacitance value will increase. 08:00.640 --> 08:02.480 So C1 two will go higher. 08:03.320 --> 08:10.440 Now overall charge required to charge these two capacitor C1 and C1 two. 08:10.480 --> 08:12.160 Let's imagine the true signal only. 08:12.950 --> 08:15.710 We need high amount of charge, right? 08:15.830 --> 08:17.870 That means the current requirement will increase. 08:19.350 --> 08:25.910 Then the characteristic impedance Z naught, which is inversely proportional to C equivalent. 08:26.030 --> 08:28.470 And I required will decrease. 08:29.150 --> 08:33.950 And we know that that z differential is directly proportional to the characteristic impedance of each 08:33.950 --> 08:34.390 signal. 08:34.750 --> 08:37.310 So the overall differential impedance will go low. 08:38.030 --> 08:39.910 So let's conclude this in case one. 08:39.910 --> 08:43.430 What if coupling increased z differential will go low. 08:44.710 --> 08:45.990 Now let's move to case two. 08:46.710 --> 08:48.630 What if we separate two traces. 08:48.710 --> 08:50.950 What will be the effect on z differential. 08:52.270 --> 08:57.990 So as you can see if we separate two traces c one to our mutual capacitive coupling will reduce. 08:58.910 --> 09:03.230 So that means the characteristic impedance which is inversely proportional to C12 will go high. 09:03.790 --> 09:07.470 And z differential is directly proportional to characteristic impedance. 09:09.030 --> 09:10.910 So z differential will also increase. 09:11.470 --> 09:13.070 So let's conclude this point. 09:13.510 --> 09:18.540 If we separate two traces, the z differential will increase. 09:20.180 --> 09:26.580 Similarly, you can estimate the impact on z differential as you reduce the height or change the value 09:26.580 --> 09:27.900 of dielectric material. 09:28.660 --> 09:35.100 Now let's see a quick demo of how we can estimate the differential pair impedance using cross section 09:35.100 --> 09:35.620 editor. 09:37.060 --> 09:39.300 For that, open PCB editor 17.4. 09:39.300 --> 09:43.500 And this is the project for which we are going to calculate differential impedance. 09:44.420 --> 09:48.540 To open cross section editor you have to go to setup and click over cross section. 09:50.500 --> 09:55.060 Now here you have to select primary tab and double click over Signal Integrity. 09:56.140 --> 09:57.740 Now you have to go to the right side. 09:57.980 --> 09:59.300 And here you will find two. 09:59.300 --> 10:04.340 Option one is differential spacing and one is differential znaught. 10:04.380 --> 10:09.940 In case if it is showing none here you have to just click over here and select edge. 10:11.140 --> 10:17.140 Now here I'm going to calculate or estimate the differential An impedance for layer L1. 10:17.620 --> 10:20.980 Just let me know in the comment section why I'm not calculating it for top layer. 10:21.940 --> 10:29.020 Anyways, so here you can see the width I have set ten mil and I want to calculate the differential 10:29.020 --> 10:29.780 impedance. 10:30.220 --> 10:36.860 Let's set the target impedance of 90 ohm and choose the field to be recalculated. 10:37.340 --> 10:39.340 I have selected line spacing here. 10:42.060 --> 10:50.420 So as you can see for ten mil width, if we put four mil spacing between two tracks, the differential 10:50.420 --> 10:53.260 impedance between those will be 90 ohm. 10:53.300 --> 10:53.820 All right. 10:53.820 --> 10:55.860 And same thing you can calculate for the width. 10:57.460 --> 11:03.820 Now this was all about coupling between two tracks and the effect of coupling on differential pairs. 11:05.780 --> 11:10.580 Now in the next video I will talk about what will be the effect on differential impedance of common 11:10.580 --> 11:11.100 signal. 11:12.860 --> 11:15.700 So now you can just apply and click over okay. 11:16.300 --> 11:16.780 All right. 11:16.780 --> 11:18.060 So see you in the next video.