WEBVTT 00:00.480 --> 00:05.680 Unwanted return current can cause several side problems like ground bounce, EMI and ringing. 00:05.840 --> 00:10.760 In this video I will cover what is return current, how it propagates and circulate down the transmission 00:10.760 --> 00:13.280 line, and what happens if return current gets disturbed. 00:13.480 --> 00:19.360 And finally, I will discuss what happens if we increase the frequency on the current distribution for 00:19.360 --> 00:23.320 a PCB track, along with a couple of demos of cadence spikes. 00:23.640 --> 00:24.760 So let's get started. 00:24.880 --> 00:30.440 In the first section of this video, we will try to visualize the return current, how it is created, 00:30.440 --> 00:34.920 what is the direction of propagation and direction of circulation for a return current? 00:35.120 --> 00:38.080 We can ask very first question why does current return? 00:38.320 --> 00:44.600 And its answer is very simple as per Kirchhoff law, which states that the current entering into a node 00:44.600 --> 00:50.720 or junction must be equal to the current flowing out, or algebraic summation of a current should be 00:50.720 --> 00:55.680 equal to zero, or we can say the current enter is equal to current exit. 00:55.880 --> 00:57.800 That means from q is equal to it. 00:58.360 --> 01:01.960 Every charge in and out should be equal to zero. 01:02.200 --> 01:07.550 Let's see the quick application of Kirchhoff's law in transmission line, and I amount of current going 01:07.590 --> 01:10.190 in and returning back to the source. 01:10.310 --> 01:12.430 That means it is flowing in loop. 01:12.870 --> 01:14.630 What do you think in a transmission line? 01:14.710 --> 01:18.230 How much time it take current to get back to the source? 01:18.750 --> 01:21.230 And the answer is it returns immediately. 01:21.310 --> 01:22.070 I will show you how. 01:22.230 --> 01:27.150 If you remember our discussion on zeroth and first order model of a transmission line, you can watch 01:27.150 --> 01:28.630 it by clicking over the I button. 01:28.630 --> 01:30.910 If you forgot about it. 01:30.910 --> 01:38.070 As soon as signal launch, it sees a bunch of tiny capacitor between signal and return plane underneath. 01:38.190 --> 01:42.590 We will just take one capacitor from the transmission line and try to understand the return current 01:42.590 --> 01:47.630 on it, and it will be similar for all other capacitor because it is a uniform transmission line. 01:48.310 --> 01:50.230 We all know the working principle of a capacitor. 01:50.230 --> 01:54.390 If the voltage across the capacitor is constant, there will be no current flow. 01:54.630 --> 02:00.230 But as signal starts travelling, there will be a change in voltage which will cause an amount of current 02:00.230 --> 02:01.630 will be flowing through it. 02:01.750 --> 02:05.590 That current flows till the transition is happening on a capacitor. 02:05.630 --> 02:09.650 Then signal will reach to another capacitor due to signals dynamic property. 02:09.730 --> 02:12.410 Let's understand how current is flowing through a capacitor. 02:12.530 --> 02:17.170 As you can see, capacitor on a transmission line is connected to signal and its return path. 02:17.410 --> 02:22.890 And due to voltage transition, positive charge will accumulate on first plate of a capacitor. 02:23.170 --> 02:26.690 And negative charge was already there on the other plate. 02:26.850 --> 02:32.530 Due to this, electrons will move toward the positive charge to neutralize it and result you will see. 02:32.570 --> 02:34.690 Current will flow through the capacitor. 02:34.930 --> 02:39.850 Same funda will work throughout the uniform transmission line, which will cause same amount of current 02:39.850 --> 02:40.130 flow. 02:40.130 --> 02:42.290 And we call it displacement current. 02:42.290 --> 02:44.130 As you can see on the above figure. 02:44.170 --> 02:48.290 Current is moving forward as well as returning back through coupling capacitor. 02:48.490 --> 02:50.530 In this case, current is moving forward. 02:50.530 --> 02:55.170 That means its direction of propagation will be forward and current is also moving clockwise. 02:55.170 --> 02:58.210 So that means its direction of circulation will be clockwise. 02:58.250 --> 03:02.570 Let's see couple of examples on this one when the direction of circulation will be anti-clockwise. 03:02.850 --> 03:04.330 First draw a transmission line. 03:04.330 --> 03:07.050 But this time we will give negative VDC voltage. 03:07.170 --> 03:11.120 Let's see on a single capacitor due to minus VDC transition. 03:11.280 --> 03:15.480 Negative charge will accumulate on first plate and positive on the second plate. 03:15.640 --> 03:20.080 An electron will flow downwards due to opposite electron flow. 03:20.320 --> 03:22.920 Direction of circulation will be anti-clockwise. 03:22.960 --> 03:27.320 Till now we have learnt how to visualise return current flow from coupling capacitors. 03:27.320 --> 03:28.040 Point of view. 03:28.240 --> 03:34.720 Now before going for simulation, let's see how electromagnetic field contribute to return current. 03:34.920 --> 03:40.360 So as soon as we pass an alternate current in an conductor, it creates an electric field around the 03:40.360 --> 03:43.560 conductor which will spread to the closest conductor. 03:43.560 --> 03:46.040 In this case, it will be our return plane. 03:46.360 --> 03:52.720 Then these electric fields give rise to electric potential difference between signal and return plane 03:52.720 --> 03:55.600 underneath and hence the current flows. 03:55.880 --> 04:01.680 This change in electric field will generate magnetic field around the conductor and for direction of 04:01.680 --> 04:04.000 magnetic field will follow right hand rule. 04:04.600 --> 04:06.520 Let's see a quick demo on piecewise. 04:06.720 --> 04:08.240 For that open Orcad capture. 04:08.920 --> 04:13.670 And here I have created a Spice model of transmission line, where I'm using three transmission line 04:13.710 --> 04:18.150 of 50 ohm characteristic impedance and 0.5 nanosecond time delay. 04:18.750 --> 04:23.190 I am terminating those transmission line with a 50 ohm resistor, and at the source side I am using 04:23.190 --> 04:28.590 a step response of one volt amplitude and its source resistor is 200 ohm. 04:29.190 --> 04:32.470 So before running the simulation, I just wanted to get the expected results. 04:32.630 --> 04:35.710 For that, I'm going back to our whiteboard. 04:35.710 --> 04:38.550 And here let's draw the voltage divider circuit. 04:38.990 --> 04:45.270 And it is very simple circuit where I'm giving a one volt step response which is connected to 200 ohm 04:45.270 --> 04:46.070 source resistor. 04:46.710 --> 04:49.630 And it is further connected to a 50 ohm transmission line. 04:51.550 --> 04:54.550 So here we'll apply Ohm's law which is V is equal to IR. 04:55.590 --> 04:56.710 And we know the voltage. 04:56.710 --> 04:58.670 We know the total resistance in series. 04:58.830 --> 05:00.990 Then current will be is equal to four milliamps. 05:02.590 --> 05:09.870 So expected results from the Spice simulation will be three waveform on each transmission line. 05:09.870 --> 05:15.380 Return current and delay between those waveform will be Will be 0.5 nanosecond and amplitude will be 05:15.380 --> 05:16.180 four milliamps. 05:16.380 --> 05:17.380 Let's see the results. 05:18.380 --> 05:21.820 And as you can see, green one is transmission line one. 05:22.060 --> 05:24.820 And on that it is going to four milliamps. 05:24.820 --> 05:31.580 And after 0.5 nanoseconds we'll see the current waveform of second transmission line. 05:32.220 --> 05:36.580 And after that the return current of third transmission line which is again after 0.5 nanosecond. 05:37.740 --> 05:41.260 Now let's see what are the factors that add discontinuity on the return path. 05:41.380 --> 05:43.900 As a designer, what are the precautions we should take? 05:44.380 --> 05:46.780 First one is slot on a return path. 05:47.220 --> 05:54.060 So in this example you can see the return current is flowing from point A to point B, and due to plane 05:54.060 --> 05:56.100 cutout there will be reflection. 05:56.340 --> 05:59.380 Second one is shared return path on a mixed signal board. 06:00.100 --> 06:02.380 In this example, let's suppose we are. 06:02.420 --> 06:07.340 We have a analog and digital circuit and a power supply on the analog section. 06:07.420 --> 06:13.860 So that means the current or return current will be flowing from digital toward the power supply will 06:13.860 --> 06:19.280 be shared with analog and which will cause interference in analog signals. 06:20.120 --> 06:25.640 So the right way to do it, we have to place power supply outside of analog and digital circuit, and 06:25.640 --> 06:27.360 its return path will be separated. 06:27.920 --> 06:32.920 Now let's see a quick demo of impedance discontinuity and its effect on return current. 06:33.520 --> 06:35.320 For that open sources. 06:35.680 --> 06:37.240 And here I did one change. 06:37.600 --> 06:41.520 That is I have replaced 50 ohm transmission line to 80 ohm characteristic impedance. 06:41.520 --> 06:43.840 That means there is impedance discontinuity. 06:43.880 --> 06:45.240 Let's run the simulation. 06:45.600 --> 06:51.680 And in simulation you can clearly see for first 50 ohm transmission line its current is four Ma. 06:52.120 --> 06:53.760 As we have calculated earlier. 06:54.600 --> 07:02.320 But again when it see a impedance discontinuity of 80 ohm the current drop to 3.5 milliamps. 07:02.640 --> 07:04.800 And again it will see a 50 ohm transmission. 07:04.800 --> 07:07.680 Line it again going for four milliamps. 07:08.040 --> 07:11.320 And there is few more reflection due to termination resistor. 07:11.640 --> 07:14.320 So as expected we are getting the same values here. 07:14.680 --> 07:17.280 And we got those values using Ohm's law only. 07:17.720 --> 07:22.830 Now another topic we're going to discuss is what is the effect of frequency on the return current and 07:22.830 --> 07:24.630 the distribution of return current. 07:24.870 --> 07:28.230 Let's talk about effect of frequency on return current distribution. 07:28.350 --> 07:32.270 So first we will see from skin depth point of view lower the frequency. 07:32.310 --> 07:36.990 Higher will be the penetration of charge in a conductor and vice versa. 07:37.550 --> 07:41.430 But we will try to understand from impedance point of view or resistance point of view. 07:41.430 --> 07:47.950 For low frequency, the ground current takes up the path of least resistance, but in case of high frequency, 07:47.950 --> 07:53.550 the ground current takes up the path for least inductance, and that's why it flow under the signal 07:53.790 --> 07:57.270 because of capacitive coupling between signal and its return path. 07:57.910 --> 07:59.310 Let's summarize this video. 07:59.550 --> 08:04.750 Till here we have learned what is return current, how displacement current propagate and circulate 08:04.750 --> 08:10.910 down the transmission line, along with few p spice demos of return current and effect of impedance 08:10.910 --> 08:12.390 mismatch on it. 08:12.390 --> 08:16.110 And we have seen the relation between frequency and current distribution. 08:16.230 --> 08:21.230 In the next video, we will discuss what happens when we place a power plane between signal and its 08:21.230 --> 08:21.990 return path.