 Greetings, electrical friends. For the longest time, transmission lines were kind of a mystery to me, and after a couple of days of intensively studying things about transmission lines, they're still kind of a mystery to me, but less so. It's chiptips, chiptips, I have no music and I can't sing. A transmission line is two conductors separated by some non-conductor. So for example, this is a transmission line. Here's another transmission line. Here's a bunch of transmission lines. This is an Apple IIe and these are the slots on it and there are signals running back this way and there's a ground plane. So there's a ground plane, there's a trace, that's a transmission line. Here is another example of a transmission line. This is just some random board and there are some traces over here and there is no ground plane on this board, but there are ground traces. So of course there's going to be some sort of electrical field between a trace and ground or a trace and power or even between power and ground and that's a transmission line because it's two conductors separated by a non-conductor. So here's a diagram of a transmission line and what we're going to do is in order to see what happens on this transmission line kind of intuitively, we're going to say that the voltage source goes from zero volts all the way to one volt instantaneously. Now of course that never actually happens in practice. There's always some sort of a rise time. In addition, my voltage source has no source impedance, it's ideal and I keep all of these things in the drawer next to my massless frictionless pulleys. In addition, this wire will have no resistance in it. Now we also know that nothing can travel faster than the speed of light. So if this voltage goes from zero to one volt and this transmission line has a certain length, well the end of the transmission line, which I've left open circuited here, is actually not going to see that electrical field until some time later. Now the speed of light in a vacuum is something like one nanosecond per foot or one foot per nanosecond. So an electrical field can't travel more than one foot in one nanosecond. So if this were a one foot long transmission line then the electrical field would only be seen at the end of the transmission line one nanosecond after this voltage source goes to one volt. Now it's a little more complicated than that because when you have something between this conductor that's other than a pure vacuum then that's called a dielectric. And dielectrics have a certain characteristic called the dielectric constant, which we'll call epsilon sub R, and the dielectric constant tells us how much more slowly than the speed of light does an electric field travel down that medium. Now I have a table somewhere. Now I have a table of some typical dielectric constants, so we can see that for a vacuum the dielectric constant is one, that's just the definition of the dielectric constant for a vacuum, and an electrical field takes 33 picoseconds to travel one centimeter. In air the dielectric constant of air is extremely close to the dielectric constant of a vacuum, it's just a little bit bigger than one, and if you look at this equation over here this tells us how the dielectric constant relates to the speed of an electric field. What you do is you take the speed of light and divided by the square root of the dielectric constant. So of course the square root of a number very close to one is going to be a number very close to one, it's still going to be slightly greater than one, so C divided by a number slightly greater than one is just going to be slightly less than C, the speed of light. So for all intents and purposes the speed of an electric field through a transmission line separated by air is approximately 33 picoseconds per centimeter. Now the important thing that we're going to talk about is in a printed circuit board. So if you have say a ground plane and a trace separated by a printed circuit board that PCB is made up of FR4 impregnated with glass and 55% resin. Anyway the dielectric constant is measured to be 4.1. It's something close to four, the square root of four is two, so that means that the speed of the electric field traveling down a transmission line separated by a printed circuit board is going to be roughly half of that of the speed of light or 67 picoseconds per centimeter. That's a very important value and we'll refer to it a lot later. There's an interesting dielectric water, if you happen to have two conductors separated by water, well the dielectric constant is 78.2. That corresponds to 295 picoseconds every centimeter. That's extremely slow. But I don't really know of any dielectric made up of water, so we're not going to talk about that. But anyway that's how the particular material that separates the two conductors relates to the speed that the electric wave is going to propagate down that transmission line. So visually what's going on is this. We turn on the voltage source at time zero and it instantaneously goes up to one volt. So we got a field that goes between these two conductors and it's just a one volt potential between them and it starts traveling down the transmission line like this. Now as it travels down the transmission line, it also causes the electrons in the transmission line to move. Now one myth that I wanted to spell right away is that electrons travel at the speed of light. They don't. First of all electrons have mass and nothing that has mass can even approach the speed of light unless you've got a linear accelerator or a supercollider or something like that. Since we don't have a supercollider, if we have an electron, here's a little electron, and we've got a difference in voltage potential. So for example here it could be one volt and here it could be zero volts which is what's going to happen. So initially there's zero volts all along the transmission line. And then when you turn on the voltage source you get one volt over here and this part of the transmission line is still zero. So because of that potential difference you're going to have charges moving. So the charges are going to move this way and I guess they're also going to move this way. And that's in this part of the transmission line. In this part of the transmission line of course the electrons aren't moving because the field hasn't gotten to that part of the transmission line. Now what exactly is the speed of that electron? Well it's not the speed of light, it's called the drift speed of the electron. And in a tiny copper wire of about diameter two millimeters that drift velocity is something like 23 micrometers per second. It's extremely slow. That's of course because a field has no mass but the electron does have a mass and as the field goes by the electron it's sort of like wafts the electrons down the conductor sort of. So we do get a current and the current is related by of course Ohm's law. So V equals I R. Now instead of the resistance I'm going to talk about the impedance Z sub zero. Z sub zero is the characteristic impedance of a transmission line. Now if you just imagine a regular wire you know that a wire has some sort of resistance to it. And the longer the wire the more resistance you're going to have. Not quite with a transmission line. A transmission line has a certain characteristic impedance which is not a DC resistance. It's actually a resistance to a change in flow of electrons or flow of electric field. And the other interesting thing about a characteristic impedance is that it doesn't depend on the length of the transmission line. It's an impedance that is characteristic of just a piece of a transmission line no matter how long it is if it's a centimeter long or a kilometer long. But one final interesting thing about the characteristic impedance is that it doesn't depend on frequency. Now that's not quite true because it does depend on the dielectric constant and the dielectric constant does change a bit with frequency but not that much so you could consider the characteristic impedance to be constant with respect to frequency. Okay so this is the relationship that we're looking at so we know that the current is going to be the voltage potential difference divided by Z0. Now for a transmission line on a printed circuit board Z0 is about 50 ohms. It's plus or minus but we're just going to call it 50 ohms. So if I have my one volt potential over here and this part of the transmission line is zero volts the difference in potential in that segment of the transmission line is one volt so I should see a current of 20 milliamps flowing that way and I guess flowing the other way over here. Again this part of the transmission line does not see that current and this is different from you know just a static analysis of a circuit where the current has to be the same everywhere throughout a wire. So now what's going to happen as the wave travels down the transmission line at half the speed of light it keeps going and the electrons start moving this way and we start seeing that 20 milliamp current progressively going down the line. Okay now what happens when the one volt wave hits the very end of the transmission line? Well one of the problems with this is that you can't have electrons floating off of the end of the conductor so you have to have a zero milliamp current at the end of the conductor. But we've got electrons flowing this way so what happens is there is kind of a reaction at the end of the transmission line which forces electrons to flow this way in order to counteract the electrons that are flowing this way so that net there's no current at the end of the transmission line so there's no electrons falling off the end. Now the only way for that to happen is for there to be an equivalent electric field being reflected off the end of the transmission line here let's cover that up of exactly the same amount of electric field that's approaching the end of the transmission line. Now why is that? Well if you look at this this is now two volts okay because the electric field goes this way and then it gets reflected back so it's so it adds on to itself so I have two volts over here and one volt over here that's a potential voltage difference going this way of one volt which means of course that we're going to get a current going this way of 20 milliamps now 20 milliamps going this way and 20 milliamps going that way is going to exactly cancel out. So of course that wave has to propagate down the transmission line in the other direction at half the speed of light and again we see zero milliamps over here but because this reflection hasn't yet gotten to this part of the transmission line this transmission line is still seeing a 20 milliamp current going this way. So the electric the reflected electric wave starts traveling back and back and back and as it travels back and back and back it's canceling out the current that's on this transmission line and eventually it reaches the beginning of the transmission line. So now the transmission line is completely at two volts and there's no current anywhere in the transmission line but there's a problem because the moment the reflected wave which is now adding up to two volts hits the beginning of the transmission line well we have another boundary condition that says that this end of the transmission line must be at one volt. So the only way for that to happen is for this wave to get reflected off the beginning of the transmission line like this. So it's going to reflect except it has the opposite sign so if we add up all of this we see one volt, two volts, one volt. So now this part of the transmission line remains at one volt. And again because there is now a potential difference here is one volt and here is two volts we get an additional 20 milliamps flowing this way that adds to the 20 milliamps flowing this way and the other 20 milliamps flowing this way which became zero milliamps and now we've got another 20 milliamps flowing this way. Now we've got 20 milliamps flowing this way and we've got this wave traveling down the line like this at half the speed of light and the current is going this way at 20 milliamps. The current on this part of the transmission line is zero still and when the wave hits the end of the transmission line the same thing has to happen. We've got 20 milliamps flowing this way but you can't have electrons getting sucked into the end of the wire. So there has to be some reaction over here to cancel that out and to cancel that out we get another wave, move this paper down, we get another wave that has equal sign that's going to reflect off the end of the transmission line. And if we look at that and if we add this up this is one volt, one volt, minus one volt, minus one volt we've got zero volts over here, we've got one volt over here so the potential difference is one volt that way which means that we're going to get 20 milliamps going that way which exactly counteracts the 20 milliamps going this way again resulting in zero milliamps at the end of the line. That's great. So the reaction works and the reflected wave travels back at half the speed of light again canceling out the current flowing on the transmission line until it hits the end again. So we've got zero volts over here but we're required to have one volt over here so what happens well we're back to the original situation where we have this and we get this one volt wave having to travel down the transmission line. So let's graph that, let me get all my little pieces of paper here, so let's graph what happens at the end of the transmission line over here. So we're going to graph the voltage at the end and this is time so of course let's suppose that, let's see, okay so light travels at one foot per nanosecond, half of that means one foot per two nanoseconds and let's suppose that this is a one foot transmission line so that's two nanoseconds so it takes two nanoseconds for the wave to go from the beginning to the end so at time zero we're going to have the voltage at the end of the transmission line being of course zero because the electric field hasn't had time to get to the end of the transmission line and it doesn't have time until two nanoseconds later so this goes like this like this like this like this and then the moment that the electric field hits the end of the transmission line well it can't be one volt it has to be, where's my little piece of paper here we go it has to be two volts in order to make sure that the current at the end of the transmission line is never non-zero it has to be zero so the moment that the electric field hits the end of the transmission line we get two volts on the end of the transmission line so let's say two volts and it goes like this okay so now the reflected wave goes back to the end of the line or rather to the beginning of the line and that takes two nanoseconds so at the end of the transmission line we're still seeing two volts now of course again the moment that this wave hits the beginning of the transmission line we have to have a reaction in order to cancel out that voltage to make the voltage equal to the voltage source and that travels down the line and it takes two nanoseconds to travel down the line so finally at the end over here we get this reflected wave hitting the end of the transmission line which is immediately causing a reaction like this so we see zero volts at the end of the transmission line so let's draw that out so here's another two nanoseconds and now the transmission line at the end goes back down to zero and that pattern is going to repeat because now this wave travels back takes two nanoseconds and then we're back to this situation where this wave takes another two nanoseconds to get to the end and so on and so forth so four nanoseconds here and so on so there's two nanoseconds this is at six nanoseconds this is at 10 nanoseconds so what happens is at the end of the transmission line you're going to see this bounce up and down and up and down and up and down and it'll go on forever again it goes on forever with no loss because these are all ideal components so basically the energy just sort of sloshes back and forth back and forth back and forth forever now let's also look at the current we can look at the current say at the beginning of the transmission line so in the beginning there is no current anywhere on the transmission line so let's draw that out this is i beginning and this is at zero nanoseconds so we have no current however you turn on the voltage source and immediately we start getting an electric field and this part of the this part of the transmission line is at zero volts this part of the transmission line is at one volt that's a potential difference of one volt that means that we have 20 milliamps flowing this way so immediately we're going to start seeing 20 milliamps at the beginning of the line now very quickly of course we realize that two nanoseconds later this field hits the end of the transmission line and we get a reaction at the end of the transmission line in order to cancel out that current that reaction takes two nanoseconds to get back to the beginning of the transmission line but immediately we get a reaction like this so we have one volt over here and two volts over here which means that we have 20 milliamps flowing this way and that would be four nanoseconds later four nanoseconds passes through zero and goes down to negative 20 milliamps and the pattern will just repeat so at eight nanoseconds which is here we do this and at 12 nanoseconds we get this so in other words the current goes this way then it goes this way then it goes this way then it goes this way forever so that's what happens with a transmission line that has the ends open circuited now we can sort of generalize so first of all we can see that when this wave hits the end of an open circuit we get because because of the requirement that there is no current we get an equal voltage traveling back so in other words open circuit equals the reflection is whatever voltage comes in now this is basically a short circuit so and the reason that it's a short circuit is that we require the voltage to be something if this were just a wire we would require the voltage to be zero in this case we require the voltage to be one volt so when the wave hits that we get a reflection that's equal but opposite so closed circuit equals the reflection would be negative all right now what happens if the termination of the transmission line were something in between an open circuit and a closed circuit in other words what happens if you had some load resistance which we'll call zl i'm using z instead of r because it doesn't necessarily have to be a purely resistive component it's just an impedance well there's an equation that we can write down this equation is called the reflection coefficient and it's symbolized by gamma and that's equal to the load minus whatever the incoming impedance is divided by the load plus the incoming impedance and i'll just sort of draw that out here's your transmission line and here is some discontinuity and then here is the rest of the transmission line and this is your load and this is your input so this could represent for example two transmission lines each with a different dielectric or this could represent a transmission line over here and just you know a terminating load over here or this could be a source over here and this could be a transmission line the point is that there's a discontinuity in the impedances so what happens is you have an incoming wave and when it hits the discontinuity this gamma value tells you how much of that wave travels back like this and of course what the sign is now if we do the math we see that let's suppose that z in is just our characteristic impedance z zero and then we can change zl to be something so let's look at the open circuit so zl in that case is is infinite so gamma is just going to be infinite infinity plus z zero divided by infinity oops infinity minus z zero infinity plus z zero well that's just going to be basically infinity minus a small amount divided by infinity plus a small amount which is going to just be one so what we see is that whatever comes in is exactly whatever goes gets reflected and that's what we saw back here where we had this incoming wave going over here and then we had the reflection being exactly equal there's the reflection right there okay now let's look at the other case which is where the load is zero so that's a short circuit and that's what happens at the beginning of this transmission line or it could be if we shorted this that's what would happen at the end of this transmission line so in that case gamma is just going to be equal to zero minus z zero divided by zero plus z zero which is minus z zero divided by z zero which is negative one so in other words it's equal but opposite reflection and that's what we saw at the beginning of the transmission line over here we saw that this wave traveled all the way back so we had basically this one volt adding to the existing one volt field on the transmission line flowing this way and the moment it hit the short circuit over here which requires the voltage to be exactly one volt well we get an equal but opposite reflection so that's negative one volt going this way and of course when we hit the short circuit over here we have a reflection of one an equal reflection so that's negative one volt flowing this way and then of course when the negative one volt hits the end this the short circuit over here we get an equal but opposite reaction that's this one volt flowing back now there is another special case where the load resistance is exactly equal to the characteristic impedance of the transmission line now there are two sort of reasons that you would have that the first reason is that you have a transmission line and another transmission line and the dielectrics are exactly equal so in other words what you've done is you've basically soldered two transmission lines together now intuition would tell you that there would be no reflection right because there's no discontinuity in impedance and if we do the math we see of course that that is exactly correct because the load is z zero and we subtract from that z zero the z zero plus z zero and that's equal to zero so there is no reflection the whole thing gets transmitted across that boundary so if you have a transmission line where zl is exactly equal to the characteristic impedance of the transmission line you get no reflection another way of saying that is that when the wave hits that load it's completely absorbed so there's no reaction at this end and there's no reflected wave going back that way so there's no sloshing back and forth of energy it's just the energy flows from one end of the transmission line to the other and then it just stops and the current remains the same that's the ideal case now what happens if there's an impedance mismatch so what's going to happen is let's suppose that zl the load is just a little bit bigger than the characteristic impedance of the transmission line well then the reflection is going to be between zero and one so it's going to be the reflection is going to be a little bit bigger than zero so there's going to be a little bit of reflection going back so what's that going to look like in terms of voltage well if we have one volt flowing this way then the reflection is going to be slightly bigger than zero volts going this way so that would look maybe something like this so we would get zero volts at the end of the transmission line and then because we have one volt plus a little more you know let's suppose it's 1.1 volt so the end of the transmission line would see something like that and then it would go back it would get reflected with say negative 0.1 that would flow this way so the negative 0.1 would travel this way canceling out the positive 0.1 and then that would be reflected just a little bit over here so in other words this would be going just slightly below one volt and then just slightly above one volt and then just slightly below one volt and it eventually dies out and settles down at one volt if we if we say that this voltage source doesn't instantaneously go up to one volt but it actually has some sort of a rise time then what you're actually going to see is not this sort of step thing but you're going to see something like this and maybe you've seen that on oscilloscopes when you measure a digital signal and that's really the reason that you see this so-called ringing or overshoot or undershoot is because whatever you're measuring doesn't have the same impedance as your oscilloscope probe or you know maybe the signal that you're measuring is traveling down a transmission line where you are perfectly matched to your oscilloscope but whatever is at the end of the transmission line is not perfectly matched that's why you see this ringing it's actually the electrical field that's bouncing back and forth so that's what happens if the load is slightly mismatched in the positive direction let's suppose the load is slightly mismatched and it's a little bit less well then we're going to get less than zero and something between zero and negative one so in that case what's going to happen here is if this load is slightly less than z zero then we're going to get a reaction where we get a very small negative reaction so what that looks like is something like this we see zero zero zero zero zero and then we get one volt minus a little bit so let's suppose that this is I don't know point nine so it goes up to point nine so this negative point one say goes all the way back to the beginning of the transmission line and it gets reflected in an equal and opposite direction so it goes plus one so now we have one minus point one plus point one so we've still got one volt traveling down here and then that positive point one gets reflected negative back a little bit so maybe it's point oh one negative so it's one minus point oh one so it's going to be say point ninety nine so in other words it's going to slowly creep up on one volt and if we consider that the rise time of the initial wave is not instantaneous you're going to see something like this it's kind of like a charging curve and it's kind of charging up the transmission line almost but in steps almost so that's basically my intuitive understanding of the way that a transmission line works and how it reacts and all the weird things that can happen on transmission lines when you don't terminate them properly so let's take a look at a transmission line like this so this is an apple 2e and these are slots and we've got say address lines running this way or data lines running this way now there isn't any termination on here and if we look at this old board over here we can see that there are traces which are by definition transmission lines and we don't see any termination over here either now why is that so let's suppose that you know we have one of these chips here and it's basically a driver and we've got a wire that goes over to some other chip over here so it's a receiver and we have of course at some point a ground and there's the input and there's the output and you know maybe this is like a clock input right so maybe this is a latch or a counter or something like that and this is this is your clock signal so clock signal and this is your say let's make it a counter because there's a point that i want to make so let's suppose we launch a signal from the clock and it goes from zero up to well this is probably a CMOS or TTL compatible CMOS so this probably goes up to five volts somewhere so we've got in the beginning from zero to five volts except it doesn't look like that no no no it looks like this zero five volts now this has a certain rise time and likewise on the opposite end it has a certain fall time the rate at which the voltage climbs is called the slew rate and it's measured in volts over time and you'll typically see this on a data sheet as volts per nanosecond let me see if i have an example here is a 74 LVT 16374 and if you look at its slew rate it's measured at a maximum well this is an input transition rate unfortunately they don't have the output rise and fall rate what they're basically saying is that the input transition rate can be no slower than 10 nanoseconds per volt in other words if it's too slow on the input then the input you know may actually register double clocks but we're talking about the output um and i got this backwards i guess nanoseconds per volt anyway here's the rise time and here's the fall time and the rise time and fall time is typically taken between 10% and 90% of the full range now for for TTL that's something like five nanoseconds so between 10% and 90% is five nanoseconds now think about what that means in terms of the length of a transmission line on a printed circuit board well we know that there is 67 picoseconds per centimeter so how long is five nanoseconds well it's 5000 picoseconds times and then let's do the unit cancellation so picoseconds goes on the bottom 67 picoseconds and let me pull out my calculator so it's 5000 divided by 67 is 74.6 let's call it 74 centimeters that's kind of huge so that means that this signal over here takes um if we had a 74 centimeter line it would take five nanoseconds to get to the end so there's a rule of thumb that basically says that if your length is greater than or equal to your rise time then you have to start worrying about reflections and the reason for that is that it takes too long for the signal to travel from the source to the destination and then get reflected back again and by the time it gets reflected back the signal has risen a significant amount so significant that the reflection will start changing the voltage significantly if on the other hand the length of your transmission line is less than the rise time then you don't have to worry about reflections and the reason for that is that by the time the signal reflects back from the end of your transmission line the voltage difference that it causes is not so much and your signal has not risen very far either so you don't really have to worry about reflections so in this case for TTL because we know that the rise time of TTL is five nanoseconds and that corresponds to 74 centimeters as long as the length of your trace is under 74 centimeters which is you know three quarters of a meter then you don't have to worry about reflections and that's why on boards like this you don't see termination because there are no traces here that are longer than 74 centimeters now there's another rule of thumb that has to do with crosstalk and with crosstalk basically you have um you know some other signal that's next to this signal and what you don't want is for one signal over here to couple into this line and cause a voltage change and the rule of thumb there is that your length has to be less than half your rise time and then you don't have to worry about crosstalk either okay so half of 74 is 37 centimeters so as long as your length is less than 37 centimeters for TTL you don't have to worry about crosstalk either in other words you don't have to worry about your traces being too close to each other and again if we look at well this is an apple if we look at this funny board um we can actually see that um some of the traces go from here up to here oh okay um how about this trace over here it goes up and over and down and if you measure that that's still less than 37 centimeters so no termination whatsoever was needed on this in order to account for reflections or crosstalk now when you're talking about something like an s100 bus if you've seen that video which i'll link down below um there is basically a chassis and a whole bunch of slots like this and this is fairly long i don't exactly recall how long it is but it's getting quite close to 37 centimeters um it's it's on the order of that much and what they did over here was they did put terminating resistors on the lines that go across all of the slots so one of the logic families that i'm working with is the LVT logic family and i'm looking at the data sheet uh to see if there's any rise or fall time measurements and there aren't any um and i guess the reason for that is they want you to go to the uh maybe the characteristics guide or you know the general logic family guide so here's one from i think texas instruments this is the LVT guide um and they do have uh rise time and fall time but they're always for specific circumstances so you need uh the you need the power supply voltage that you're using and they're basically saying for all output switching presumably for just one output switching it goes even faster uh the other thing that i don't quite understand about this is that um they usually also specify the load that you're putting on each output so anyway we can just use this as maybe a minimum value so if we look at the minimum value for the rise time we can see that at room temperature for 3.3 volts it looks like it's about 1.35 nanoseconds and if we look at the fall time for 3.3 volts it's slightly higher it looks like you know maybe 1.7 nanoseconds so and this this is actually maybe 1.3 so let's go with the 1.3 nanosecond figure so what is 1.3 nanoseconds in terms of length well 1300 picoseconds at 67 picoseconds per centimeter is so 1300 divided by 67 is 19 centimeters and half that is about nine centimeters so that's the crosstalk so my lengths have to be less than nine centimeters that's not a whole lot there's nine centimeters right there so if i have a bus that's you know say that big that's certainly going to be bigger than nine centimeters so i'm going to have to worry about crosstalk and reflection and in order to solve that i need to use termination now i suppose it's important to talk about exactly why we want to avoid reflections so let's suppose that we had a source and a destination and again that was going to be a counter and a clock and this is a transmission line now if this counter has say a MOSFET input and it has essentially infinite input impedance let's say and this clock is going to have a signal and it's going to go from say zero to 3.3 volts and it's going to have a rise time of well you know let's call it i don't know two nanoseconds say so what's the critical length of two nanoseconds well 2000 picoseconds divided by 67 is 29.8 centimeters so let's say the length has to be less than 29.8 centimeters okay now let's suppose my length were much larger than 29.8 centimeters or even you know let's suppose it approaches 29.8 centimeters well um if that's the case then this rise time doesn't really matter all that much the point is that the rise time is fast relative to the time it takes to propagate down to the counter so what exactly would the counter see well let's actually remove the counter from over here and put the counter somewhere in the middle like let's suppose this is a bus and you know let's suppose there is some sort of an enable signal on a whole bunch of these just as an example so we enable one and then we send a clock pulse down the line well what's going to happen at the end of the line is that that clock pulse gets reflected back and it's going to look something like this well first of all one of the problems is and this is a pretty important problem is that at the end of the line it's going to look like this it's going to go above 3.3 volts and it's going to go up to 6.6 volts and then it's going to go below and then it's going to go above and then it's going to go below and then above and so on like this so the first problem right away is we can see that there's this voltage spike that's being presented to the input of these chips and that could fry the chips so the second problem though is that even if the chip survives that voltage pulse well the threshold of the chip is probably going to be right around here 2.0 volts and we just sent one clock two clocks whoops one clock two clocks three clocks yeah i got that backwards well anyway three clocks because of these reflections that go back and forth so by sending a single clock we've just clocked this thing a bunch of times so that's obviously no good and that's why we want termination first in order to prevent these spikes from happening and second in order to prevent multiple clocks from happening now that's what happens if there's an open circuit over here let's suppose there is um oh i don't know let's let's put another um let's put another clock on the end and let's suppose that for whatever reason the input impedance is actually quite low in fact maybe it's lower than z zero well in that case the problem isn't so great it's just that you get um something that looks like this rather than you know something that looks like that so in other words your signals get delayed so you can't run your system as fast um but this is really a bad situation to be in this this you can deal with as long as you're willing to deal with a slow situation now here's another logic family here's uh lvc and this guide does talk about termination uh so basically it says depending on the trace length special circumstances special consideration may need to be given to the termination of the outputs and they say as a general rule if the trace length is less than four inches no additional components are necessary now i assume that they're talking about reflection yeah they say if the trace length is greater than four inches reflections begin to appear on the line so now this is for the lvc family so what does that actually mean well what they want is they want the length to be less than four inches and then let's translate that to centimeters so four times 2.54 is 10.16 centimeters and 10.16 centimeters times 67 pico seconds per centimeter oops that was divided by okay obviously i'm having technical difficulties 10.16 centimeters times 67 pico seconds is 680 pico seconds or 0.6 nanoseconds so with lvc they're basically saying that they've got a rise time of 0.6 nanoseconds 0.6 nanoseconds which corresponds to 10.16 centimeters so again what they're saying is that if you have anything bigger than this you have to start worrying about reflections and if we take our crosstalk issue you have to go down to here so let's look at some terminations so here they publish a bunch of terminations so this is the typical termination where you've got your you've got your transmission line which is your characteristic impedance of z0 and then you've got a single resistor at the end and here's an alternate version where you've got two resistors one going to plus and one going to minus now the problem with that is that when you have a driver and you have your transmission line and then you have a resistor to ground and there's your input the problem is that if this stays at say 3.3 volts and say you've got a 50 ohm termination over here well if this thing is at you know just no signal dc 3.3 volts then you've got a current flowing here constantly and that current is going to be let's see what it is so 3.3 volts divided by 50 ohms is 66 milliamps 66 milliamps is quite a lot so you need a pretty strong driver over there and if you have the double termination so going up and going down and let's say instead of well let's say it's 100 ohms over here and 100 ohms over here because in parallel they're going to form 50 ohms and that's what really matters for the transmission line because to the transmission line this is going to look like ac not dc so this is going to be properly terminated for reflections so it's reflection proof but again here's 3.3 volts here's 3.3 volts and if this remains at 3.3 volts well then you need 33 milliamps this way and if this is at zero volts then you need 33 milliamps this way so again this still has to be quite a strong driver so in this case you're wasting power on the positive side if this is zero you're not having any current at all and with this double termination you're using less current on each side but you know still overall 33 milliamps one way and 33 milliamps on the other way so it again it averages out to 33 milliamps so you're wasting power on both sides so the next type of termination that they talk about is this rc termination now with an rc termination you have your driver and you have your input over here and you have a resistor and a capacitor to ground and the resistor is going to be 50 ohms and the capacitor is going to be whatever it is now to ac this capacitor is not going to even be there so the transmission line looks like it's properly terminated for dc so if this is steady at 3.3 volts or if it's at zero volts then well the capacitor looks like an open circuit so there is actually no current flowing so that's pretty good the problem is that this forms an rc low pass filter which means that in terms of your frequency this is what the characteristic is going to going to look like so in other words you're going to be frequency limited so what this actually means is that your edges are going to be spread out because you know an edge will have a certain frequency spectrum and you're cutting off the high end of that spectrum so your edges are going to sort of smear out and also you can't go as fast so that's a bit of a problem here we have another type of termination this is a zener diode at the end and the idea behind that I'm sorry that's a shatky diode no that's a zener diode is it a zener diode I think it is well it's some sort of a clamping diode anyway the point is that kind of looks like a shatky diode to me so the point is that what this does is it doesn't actually terminate the line but what it does is it prevents this sort of overshoot from happening so you know the wave flows this way and then the reflection flows back that way at double the voltage and what this is supposed to do is let's suppose this is a 3.3 volt zener then what this is supposed to do is as the voltage rises and then it goes above 3.3 volts because of the reflection all of a sudden this line gets short circuited down to 3.3 volts so the reflection cannot go above 3.3 volts but the other problem is that this zener diode has to react very very quickly and that's kind of expensive and hard to find so there's this final type of termination called series termination now what series termination looks like is this we have a driver which has some sort of source resistance already let's call it rs and then we've got a series resistor let's just call it r and then you've got your line which is z0 and then you've got your receiver over here now if the resistance over here is equal to z0 then you have a properly terminated line there is just a bit of an issue here so because this is actually on the source what it's going to look like is this so you have z0 over here and then you have your transmission line over here and then let's suppose this is just an open circuit because maybe there's just a mosfet input over here with no input impedance with infinite input impedance say for example so what's going to happen is that when this thing goes up to 3.3 volts then over here this is basically looking like a voltage divider so you're only going to see 1.55 volts over here so you're going to get 1.55 volts traveling towards the end of the line and then that gets reflected back to form the 3.3 volts that you wanted in the first place 3.3 volts so the thing is that if the logic threshold of this is 2 volts 2.0 volts then when the wave travels this way no receiver that's connected over here is going to get triggered until the wave bounces back off against the end of the line then the receivers get triggered and the funny thing is is that they get triggered in the reverse order from where they are in distance from the source so they get triggered back to front and again this relies on the reflection to go back and the advantage of this is that no power is wasted if this voltage source is simply stuck at 3.3 or at zero because well there's no termination over here so there's no resistor so that's kind of a nice way to handle the case where you've got basically a bus full of receivers there can be as many receivers as you want and not have to worry about terminating the end in fact what you're relying on is the reflection and then the question is you know now that you've got this wave traveling back and it adds up to 3.3 volts well what happens at the source well for all intents and purposes this is a properly terminated line because it matches the impedance so there is no not going to be any further reflection going back here which reduces the voltage and then you know raises the voltage when the reflection comes back in to cause this sort of multiple clocking behavior so this is pretty good there is a slight delay because you are adding resistance to the line and of course the more devices that you add the more capacitance that you add to the line which means that you're not going to be able to run your system as fast but overall you're not wasting any power so this is pretty good now what I'd love to do is get a printed circuit board that has you know a long distance and let's suppose that we were using what and let's suppose that we were going to use one of the LVT chips so you know we have this rise time of 1.3 nanoseconds so we need to make sure that the length of the line is under I think I said 9 centimeters so it'd be nice to have something like a you know I don't know 12 centimeter or 15 centimeter printed circuit board printed up with an LVT chip on one end and another counter chip on the other and just start sending pulses without any resistor and see what happens well actually I know what's going to happen it's going to fry the it's going to fry the the the component but you know assuming that I don't fry the component see just see what happens to it you know and maybe actually observe this double clocking event and then I can replace that zero ohm resistor that I'm going to put in series with a proper termination resistor now with LVT you're not going to have 50 ohms in the driver so this source is probably going to be something like 20 or 30 ohms so all I'm going to need is about 20 or 30 ohms over here and it doesn't have to be exact because if it's not exact I am going to get a little bit of reflection but it's not enough to exceed the threshold that would be required for a double clocking so it would be kind of nice to have such a printed circuit board and to just you know test it and just start sending clock pulses and see and make sure that I get one clock pulse every single time that I send a single clock pulse but that'll take some time and I have run out of time for this video boy this video sure took a long time to make this is the third time that I've made this video and each time I think I got better and better at explaining the the various issues and you know using these pieces of paper to you know show what happens with the field so it took three times as long to do this you know and we're talking about like six or eight hours of time to do this and it's it's kind of it takes a lot of energy to do that so if you'd like you can go to my patreon and you can look at the link down below and what that does is it keeps me motivated to make more of these videos and I kind of like making these videos which you know sort of educate you and the reason that I make this particular video is that I am planning on using a particular family of logic on a bus and so I ran into this transmission issue this transmission line issue so I did want to make the video and I suppose what I could have done is just work through the problem myself and then just you know do the solution but I thought it was interesting enough to share so please consider becoming a patron I I don't you know necessarily want you to feel obligated to become a patron but it it does motivate me to make more of these videos so anyway thanks very much and I hope to see you on the next video where maybe I'll have a printed circuit board to show see ya you