 This is a reflection, and it's exactly what I want. The voltage goes from 0, and then halfway to 3.3 volts, and then all the way up to 3.3 volts. And that middle part is about 6.6 nanoseconds, and this is exactly what I want to see. It's chiptips, chiptips, I have no music, and I can't sing! In an earlier video, I showed what would happen if you launched the edge of a signal onto a transmission line that was open at the other end. You would get a reflection from the back that would make it all the way to the front again. Now, I built this printed circuit board in order to find out if what I thought was true was actually true in practice. So what I have on one end is a 74LVC244 buffer, and it's buffering a clock signal that I'm generating from a frequency generator. Now, the edge of the buffer is something like, I think, 2 nanoseconds or less. In any case, that edge gets launched along this entire transmission line. Now, the printed circuit board is 40 centimeters long, but I added some squiggly traces in the middle to lengthen the effective length of the stick. So it's something like 60 centimeters to the end, and then if the signal bounces off the back, we get another 60 centimeters. Now, if light speed is 30 centimeters per nanosecond and half light speed, which is the speed of an electrical signal on a printed circuit board, is 15 centimeters per nanosecond, then it should take four nanoseconds to hit the end of the stick and another four nanoseconds to go back to the beginning of the stick. In addition, what I've done is I've source terminated the transmission line. So there's a 100 ohm resistor over here. Now, I know that in the earlier video, I said that the trace on a printed circuit board would be about 50 ohms. That's actually not true when I looked it up. It's actually somewhere between 90 and 120 ohms depending on your trace thickness, so I figured 100 ohms is a good bet. Now, in the source terminated case, what would initially happen is that you would get an initial voltage over here of half the actual driver voltage, because there's 100 ohms over here and 100 ohms in the transmission line. That's a voltage divider. It divides the voltage by 2, and that's in fact what you can see on the oscilloscope. Now, once the signal goes all the way to the end and reflects back, it doubles, and once that doubling hits the source, we should see the full voltage, which in fact we do see on the oscilloscope. And now we can make some measurements. So, if I take a look at, say, the start of where I launched the signal, and then I move this other cursor to the start of where it sees the reflected doubling signal, according to this I'm getting about 6.4, maybe 6.6 nanoseconds. Let me move this over a little bit. Yeah, it looks like it's about 6.6 nanoseconds. So what that means is, starting from no signal, the signal begins at zero. It travels all the way to the end for 3.3 nanoseconds, gets reflected, and then spends another 3.3 nanoseconds coming back, at which point the voltage starts to double. So that's 6.6 nanoseconds. Now, that's not the 8 nanoseconds that I thought it would be, and that's okay. In fact, it turns out to be 60% light speed instead of 50% light speed, so the electrical wave is traveling a little faster than I thought it would. But remember that the speed of an electrical wave on a transmission line is dependent on the epsilon sub R coefficient of the transmission line. And that means that instead of something like 4, it's maybe a little less, I don't know, 3.8 or something like that. Now, another thing that I've built into this printed circuit board is I have test points one-third down the line, two-thirds down the line, and all the way at the end of the line. And I also have footprints over here that I'm going to add some counterchips in order to test whether I get double clocking from anything. So I can actually probe one of these points. So let's take a look at one-third down the line. Now, what I expect to see is basically nothing until the wave hits this point. And then, of course, the wave goes all the way to the end and back again. Now, because this is two-thirds of the length, and the total length is 6.6, that, or rather, the total length is 3.3 nanoseconds, each of these segments is about 1.1 nanoseconds. That means that I should see 2.2 nanoseconds, 4.4 nanoseconds. So I should see a 4.4 nanosecond difference between when I first see the wave and then when I see the reflection of the wave. So let's find out if that's actually true. Let me carefully grab the signal. This isn't the best angle. This is really not the best angle. Sorry, I have to pull the printed circuit board away. Okay, so here we go, and let me move this cursor. Oops, I just lost it. And it looks like I'm seeing something like 4.7 nanoseconds. But the interesting thing is that, again, the voltage goes up to halfway and then it goes all the way up to the full voltage without a whole lot of bouncing going on after that. Now, let's take a look at what happens if I go 2 thirds of the way down the line. And again, because the distance from that point to the end of the line is 1.1 nanoseconds, I should see a 2.2 nanosecond reflection. So let's see if that's true. And I'm getting something a little less than 2.7. It looks like maybe 2.5 or so. That's pretty close. Maybe it's not exactly 1.1 nanoseconds. Maybe it's more like 1.2. Now, I also have a test point all the way at the end of the line. Now, at the end of the line, you should basically see no delays between the end of the line and the end of the line. So you should see the voltage pretty much jump all the way up to its full voltage. So let's see what happens there. And indeed, that's what we see. Now, at this point, there may be an objection. So if I pull up the wave again, you can see that the voltage goes from 0 to, let's see what the measurement is. It looks like it's about 1.8, and then the top voltage is 3.3. Now, this is the chip that I'm planning to use for the counter. It's a 74 LVC 161. Now, you can see that its low-level voltage is 0.8, while its high-level voltage is 2. Well, the problem is that 1.8 is smack between the low and the high thresholds. So am I not going to get a bad clocking behavior? Because I'm sitting for a few nanoseconds in the middle of that dead zone. The answer is no, and the reason for that is this input transition rise and fall rate. So because I'm using a 3.3 volt supply, the maximum rise and fall rate is 10 nanoseconds per volt, or putting it the other way, 0.1 volts per nanosecond. Now, between 0.8 and 2 is 1.2 volts, and at 0.1 volts per nanosecond, that means that I have to go at least as fast as 12 nanoseconds, or faster. So as long as I can make this transition within 12 nanoseconds, I'm safe. And in fact, since this entire stick is something like 6.6 or 7 nanoseconds, it doesn't matter how long I spend within that transition region anywhere on this printed circuit board. Now, I have here a second printed circuit board that has a termination resistor of 0 ohms, so basically it's just a jumper. So I wanted to see what would happen in this case. This would be equivalent to the case where I have a completely closed circuit on the source end and a completely open circuit on the back end. Now, it isn't going to be completely closed circuit, or short circuit, because there is some resistance in the output driver itself. So what I expect to see is a lot more reflections going back and forth. So let's see if that's the case. So here I'm probing the very beginning of the transmission line, and you can see that in fact there are a lot of reflections going on. So if I measure what that top value is, it looks like it's going up to about 3.7 volts. So that's not quite the doubling that I would have expected. There's another reflection further down that looks like it's even higher. This is at 4 volts, and the settling voltage here looks like it's about 3 volts. So in fact it bounces up to something like 3.6 volts, down to 3 volts, up to 4 volts, down to what is this? About 3.6 volts again, then it comes down to about 2.9, and eventually it's going to settle out around 3.3 volts. But you can see that there's nowhere near the clean type of signal that you would see with a properly terminated transmission line. Now that's at the source end. Let's see what happens one-third down the way. This is obviously worse. So let's go to the beginning of the signal. Let's call it over there, and let's pull in this part of the signal. So we can see that as the wave makes its way past that point, it goes up to something like 3.2 volts. So in other words, it's pretty much the full voltage. However, when the reflection comes back, it goes up to 5.4 volts. That's pretty crazy. Then it goes back down to something like 2.4 volts, and then it goes up to 4 volts, and then it's going to bounce up and down until, again, it settles at 3.3 volts. That's really no good. Let's take a look at what happens two-thirds down the line. Boy, that's pretty bad. So again, it starts at zero. Rapidly goes up to 3.2, bounces a little bit, and then goes up to 5.8 volts, and then it sort of slowly drifts down to about 2.2 volts, down to about 2 volts, goes up to 4.4 volts. Now, this is the interesting thing right over here. This is 2 volts. This is just about the high threshold. Now, if it had gone below 2 volts or if maybe there's a little bit of noise on the line over here, I could see a case where we might get double clocking. So let's see what happens at the end of the transmission line now, and I kind of expect that it's going to jump from zero all the way up to some high voltage. Okay, that's maybe not quite what I expected, but it does jump to something like 6 volts. So it does pretty much jump to the high voltage right away, and it goes down to, wow, it's 1.9 volts. So that's below the high threshold. So you could very well get some double clocking going on over here, or at least some undefined behavior or some behavior that comes and goes. Now, I'm going to zoom out so that we can get more of an idea of what that waveform at the end looks like. Okay? And this is what you would describe as ringing. So it goes up to some high volt, some high voltage, which is, oops, how did I not actually sure how I moved that over? Okay, well, anyway, let's go and bring the first one down here. Okay, so this goes all the way up to 6 volts, and then it goes all the way down to 1.7 volts, goes back up to 4.3 volts, goes down to 2.9 volts, goes up to about 3.8 volts, down to about 3.2 volts, and then eventually it settles out at, well, it looks like about 3.4 volts or so. Okay, maybe my voltage supply is 3.4 and not 3.3, but the point is that that's a terrible signal. And of course, if my transmission line were any longer, these peaks would be even higher. Now, what's the difference between these peaks, say? Just for fun, the difference in time is something like 16 nanoseconds or so. Okay, I'm honestly not sure what that means. Perhaps that is, let's see, yeah, I'm not really sure what that time is, but what it means is we start at, say, the high over here, 6 volts, and then I suppose that this low would be the time that it takes to go from here back to here, possibly. Or maybe the low is when the wave makes it all the way to the end, and the next peak is when it comes back. Although 16 nanoseconds, yeah, given that this is, given that a round trip is 6, I would say that this is about 2 round trips that it's making. So it's probably, the low voltage is probably when the wave hits the source, and then the high is when the wave comes back and adds to itself again. I guess that makes sense. So that would be, you know, 1 round trip, 2 round trips maybe? I don't know. Anyway, the point is that this is not what you want. So this is the zoomed out view of what happens at the end of the transmission line. So let me swap out the properly terminated transmission line and see what we get. Okay, and let's go to the end of the transmission line. Well, that's certainly a lot better. You can see that it's extraordinarily stable, no ringing. In fact, it looks like a perfect termination to me. Let's go down two thirds down the line. So again, there's a tiny little jog over there, and that's simply because the wave is reflecting off the back end of the transmission line, which is exactly as expected. Of course, if I go one third down the line, that little halfway point is going to be, I'm not hitting my ground point properly here. So that halfway point is going to be a little bit bigger. But in any case, this is a very nicely terminated transmission line. So the idea is now that, of course, if I have chips all the way, you know, at various points on the transmission line, they should get clean clocks, at least clean clocks relative to their required transition times. Now, what I really wanted to show was that with an improperly terminated transmission line, like this one with the zero ohm resistor, if I stuck a counter all the way at the end, it would start seeing all those extreme bumps in voltage. Now, one of the things that we can look at is the maximum input voltage that is allowed. Now, this is an LVC, which means that it can actually take TTL voltages, so that means that the maximum voltage it is allowed to see on any input is 5.5 volts. But you saw that we were exceeding that at many points. So that's one danger with an improperly terminated transmission line, at least with this one, is that we would probably end up damaging these chips, or at least reducing their lifespan or making them operate inconsistently. And the other thing, of course, is the high-level input voltage and the low-level input voltage combined with the required transition speed. So we did see that at some points we dipped below 2.0 volts, which is not really good. It would be even worse if we dipped below 0.8 volts and then came back up, but we didn't see that actually happening. And the other thing was that the signal did sort of bounce around for what seemed to me like a lot longer than the required maximum 12 nanoseconds. So it exceeded 12 nanoseconds of transition time, which is really not great for consistent operation. So I could put a clock here and just sort of let it go and see if I can find double clocks or inconsistent clocks or weird things that happen. But I don't think I'm going to do that for this video. Maybe, you know, at some point I'll actually do that experiment. But what I really wanted to show was the voltage waveforms and what a properly terminated transmission line can do for you. And for me, because when I'm making my RISC-5 processor that has 32 registers and they're all going to be lined up on a bus, well, it's going to be probably something like this long. Maybe a little shorter, maybe a little longer. But the point is that at least now I know how to properly terminate that bus. Anyway, thanks for watching. Okay, so here we go and let me move this cursor. Oops, I just lost it.