Amateurfunkvorlesung Klasse A -- Lektion 5: Strom und Spannungsversorgung

Added: 2026-05-26

[00:00:00]Right, I think we finished chapters 3 and 4 last time, and now we're supposed to start chapter 5 this week. Hopefully, we'll manage it.

[00:00:10]Maybe we'll even get a bit of chapter 6 done, because I realized today with a [snorting] shock that we're not making as much progress as I'd like. Um, so we'll need a few additional appointments. I wanted to ask beforehand if it's okay if we schedule additional appointments on Tuesdays at the same time every now and then, not always, but I think we need about three. Would Tuesday 1615 work? Okay, good, because you tend to say more than you should, but that's actually a good thing, then you'll understand it better. Exactly. Now we have electricity and voltage supply, and we'll start with internal resistances. Um, we also have internal resistances in measuring instruments and so on. But now we're going to look at it from the perspective of electricity and voltage sources.

[00:01:18]And, um, let's start with a power source.

[00:01:24]Um, what you see here is the equivalent circuit diagram of the power source.

[00:01:29]That means an ideal power source would initially look like this. Yes.

[00:01:34]Um, that doesn't happen in reality. In reality, we will always have some resistance here, uh, very high resistance here.

[00:01:48]Ideally, it would be infinitely high, meaning it wouldn't be there. Yes, then we have the ideal source. But there is no ideal. Um, in practice it's a very high resistance and, uh, the rule here is that this Ri is significantly larger than RL or RL significantly smaller than R, the other way around, and that's also called current matching.

[00:02:18]Um, exactly. Theoretically, the resistance is infinitely large, but in reality, unfortunately, it is not infinitely large.

[00:02:27]Yes, and that also leads to the fact that, um, the current that should actually come out here and flow, branches out, and a slightly smaller current still flows, um, which then doesn't come out here. Yes, that's basically how you can imagine it. Exactly. Yes, in what relationship must the internal resistance of the current source and a directly connected load resistance be related, so that current matching occurs and then the load resistance must be significantly smaller than the internal resistance.

[00:03:02]Yes, when we talk about power sources, we can also talk about the current limiting of such a laboratory power supply.

[00:03:09]Um, we usually have a built-in protection device where you can say, okay, maximum 0.5 amps or maximum 2.5 amps.

[00:03:20]Um, and um, yeah, that thing pretty much always strikes. That means, um, the current can't exceed 2.5 amps, for example. This simply helps to prevent you from damaging something, such as what you connect to it. Well, simply by limiting the current, you have a certain level of safety, especially with a power supply like this. Yes, so even if I somehow create a short circuit here, i.e., directly connect positive and negative, then a maximum of about 0.5 amps will flow through the cable here. Yes, exactly. Then we come to the source of the voltage. With the voltage source, it's exactly the opposite.

[00:04:07]Um, the ideal voltage source basically has a solid line here.

[00:04:14]And here there is also, um, resistance.

[00:04:18]This is the RI. However, it is now connected in series, not in parallel like the power source.

[00:04:24]And ideally, it has 0 oh, meaning it's not there either. In reality, there is a low internal resistance, and if it is a good voltage source, this should be as low as possible. And there too, uh, yes, RI should ideally be smaller than RL, or RL should be significantly larger than R.

[00:04:47]This is then called voltage matching.

[00:04:50]Yes, I have a question about that too. We can quickly go over it. Exactly. Then, combining both into one question: what properties should current and voltage sources have, and current sources should have the highest possible internal resistance, voltage sources the lowest possible internal resistance. Exactly. Um, yes, I can't just measure the internal resistance with a multimeter. However, I can calculate it mathematically, and that's also in the formula collection. Rielta U is divided by Delta I.

[00:05:25]And here we are actually looking at two cases.

[00:05:29]First of all, uh, idle without load.

[00:05:32]So, that means nothing is connected here. If we were to measure, then in principle the UL that we are measuring here is also the UQ. Yes, no load there. Um, that means the entire tension drops off there.

[00:05:50]So, that means we can measure that.

[00:05:52]Then we basically know what the UQ is.

[00:05:56]And now we apply a load and then, in principle, measure the current flowing here.

[00:06:04]And then we'll measure the voltage again. And since this resistance is not zero, we actually have a voltage divider here. That means a certain voltage will drop across it here and a certain voltage will drop across it there. Since RL will be significantly larger than RI, most of the voltage will drop here. But such a small delta is missing, yes, compared to the ideal source. Yes, that means this RI is kind of stealing the excitement from us. Yes, and that is exactly what is meant here.

[00:06:34]We need to start with the actual source voltage, which is what the device has that we can measure under no-load conditions. Then we need to measure the actual voltage across the RL, the UL. This voltage collapses a bit, and down here, in the case where we're not loading it, no current flows, logically, and therefore it's zero. Then you can simply calculate UQ - UL and then I can calculate the internal resistance of the voltage source. Are there any questions? And, uh, you see, it 's nothing other than Occupation's law.

[00:07:13]Yes, you can definitely manage that. Um, here's 13.5 volts. When the voltage source is then loaded and delivers a current of 2 amps, the terminal voltage drops to 13.5. That means our Delta U is 0.5 and then 0.5 is great due to the 2 amps.

[00:07:34]Then we arrive at this figure of 0.25, which is correct. And there are many questions about that, right? um, 13.8 volts, 13.6 and so on.

[00:07:49]Simply think about it briefly, plug it in, and then calculate it. Yes, nothing complicated.

[00:08:00]Okay, I'll skip over all the tasks now, because nothing significant changes there, but you can always see that, um, the internal resistance is also always very small. Yes.

[00:08:13]The only question is about sources of tension.

[00:08:15]Exactly, this is only about voltage sources. Exactly. [sighs] Here the question is asked the other way around.

[00:08:25]Um, no, it's in just as much demand. Okay, nothing changes, it's just a two-stage process, but I trust you can manage that too. So, um, now we've talked about current matching, voltage matching, and current matching. Now we need to talk briefly about performance adjustments. Um, this refers to voltage sources for now, but it can also be applied to the transmitter, and that's why we'll briefly go into some detail.

[00:08:58]The goal now is to ensure that as much performance as possible reaches this RL. Yes, so in this case, power is delivered by, uh, sort of as heat. Yes, well, that thing will get warm somehow.

[00:09:14]Um, but the same applies just as well as an antenna at the transmitter at the end of the day. Yes, and what exactly is the power output of the resistor?

[00:09:28]And we have various formulas in the formula collection, but we can say that the power at the resistor PL is, i.e., load resistance RL times I², i.e., the current that flows through the resistor through the entire system. Yes.

[00:09:48]Um, yes, how can you calculate the current?

[00:09:53]Well, we can calculate the total resistance of RE + RL and then, using the law above, we can easily calculate the total current. Yes, so the total current is simply U divided by RE + RL.

[00:10:11]Yes, I think that's clear enough.

[00:10:15]Now I can use it all here as well. Yes, and then, um, I can simply calculate the power using the resistances and the voltage, so this UQ, I'll just abbreviate it as U for now.

[00:10:27]The whole thing would then be the same in PL.

[00:10:32]Okay, and now I'll insert this here.

[00:10:37]Yes, then we have RL times U² divided by Ri + RL squared.

[00:10:50]Yes, you can simply square both the top and bottom values. So, what do we actually want to know now? We actually want to know, um, how do I choose RL if RI is given, for example, so that the power is maximized? That is actually our question. Yes, so there will be some kind of tradeoff. There is a specific value where this P is maximized. So. And how do you solve something like that? Typically, you all learned this at school.

[00:11:28]Yes, I've already heard. Derivation.

[00:11:29]Yes, that means we calculate P as a function of RL.

[00:11:38]So. And you just believe me now, [snorting], that U² times ri-L and then accordingly RE + RL to the power of 3. Yes. Okay. Now how do I find the maximum or minimum? I have to set the derivative equal to 0, right?

[00:12:03]Okay. Now you just look at this term or equation.

[00:12:09]When will the total become zero? Yes, well, if I make the voltage zero, that's trivial, but the whole thing becomes zero if I ensure that EI = RL. That is a very important finding.

[00:12:23]Um, let's write that down too.

[00:12:27]Yes, we have the maximum of PL if and only if Ri = RL.

[00:12:37]Yes, is that clear so far?

[00:12:42]That's important to understand.

[00:12:43]So, we achieve this power adjustment so that as much power as possible is converted into heat by making the internal resistance and the load resistance the same, because that's what the derivative says, yes, so the curve, I ca n't quite get it right, but it looks something like this, this is what it will look like and there is our maximum, where the slope is zero.

[00:13:07]That's exactly what we do with the derivative. Yes, we'll see the curve even more beautifully in a moment.

[00:13:12]Um, now we can of course calculate the maximum power. That is precisely the case. So, if we take the formula here and then, um, yes, actually, I think it's better expressed the other way around. Actually, one should say that if RL = RI, then it's expressed more nicely. Yes, that's the only one I'm choosing now. Yes. [sighs] Uh, and we can now insert that into the formula here. Let's insert this condition now and then we'll figure it out.

[00:13:53]P max is Ri times U².

[00:14:00]And now we substitute the RI for the RL. Then we have RI + RI squared.

[00:14:09]And that simplifies to, um, so I can cancel that out now, right? So, I have one here and then two underneath, so try to make it look a little nicer. So, we have now divided ri times U² by, um, 2* Ri and then squared the whole thing. And what's left then, well, I can just cancel out this RI with one down there.

[00:14:50]One more thing remains, then we have U² divided by 4 Ri.

[00:15:02]That's interesting too. That 's the maximum performance I can get. This also shows that the efficiency is not that good at all. So, the efficiency is only 50%.

[00:15:11]We'll see about that in a moment.

[00:15:14]Um, if I want to visualize this whole thing, um, where do I want to normalize the performance to this maximum performance?

[00:15:25]This is the typical graphic that you can also find on Wikipedia, for example.

[00:15:30]So I take the formula from PL and divide it by P max, and then I get a relatively large fraction here.

[00:15:41]RL times U² divided by RE + RL squared divided by U² divided by 4* RI and then I can divide by multiplying by the reciprocal.

[00:16:04]Then RL times U² is calculated and now I simply copy that back. ri + squared times and now the reciprocal 4 x Ri divided by U² and the whole thing is then 4* Ri* RL divided by Ri + RL squared.

[00:16:32]And now I can express the whole thing in a more relative way, by saying that I am now creating an X, which is RL through Ri. Yes, and if Ri and RL are equal, then this ratio will be 1.

[00:16:52]Yes, well, that says nothing other than that uh RL = X* Ri. And I can now incorporate that into all of this to get rid of the risk.

[00:17:04]That means PL divided by P max is 4* Ri* X. Yes, I'll plug that in there. X times RI. So, previously it said Ri x RL. And now I substitute Ri for RL X times. and then down here riadrat.

[00:17:31]And here too, the risks are essentially eliminated completely. So I can completely exclude and cancel them out then. And what's left is 4x awesome divided by 1 + x squared.

[00:17:47]Yes, and that's exactly what you typically see as a graphic. Let's take a look at this.

[00:18:06][sighs] Um, typically this function would look like this. Yes, so this is basically the maximum.

[00:18:17]The slope is zero there. That's where we have our maximum power (P max) and then the whole thing drops off again.

[00:18:24]However, it is often plotted logarithmically.

[00:18:26]So this ratio of RL to Ri is then represented logarithmically, and there you can see nicely, okay, um, when RL = RI, we are at the maximum here, right? And that is normalized to the maximum as well. And then you can see, okay, if we make it smaller or larger, depending on the situation.

[00:18:50]Um, if we make RL bigger, for example, then we move in that direction and less and less power is transmitted. If we make the RL smaller, then when the other direction is also less power is transferred to the RL.

[00:19:03]And the graph on the right does not show the whole thing logarithmically, and the internal resistance is fixed at 50 ohms.

[00:19:11]Yes, that means we have, quite specifically, an internal resistance of 50 ohms. The reason the whole thing now has such a leftward slant is simply that we now have a linear axis, not a logarithmic one. And that shows that the optimum performance is therefore at 50 ohms. That means, um, if our load also has 50 ohms and the internal resistance is also 50 ohms, then we have a perfect match and most of the power is transferred accordingly.

[00:19:44]Yes, but if you look at it again here in the circuit, okay, if there are 50 at the top here and 50 at the top here, then yes, we have half the power here and half the power here.

[00:19:56]So, one must also be aware that the efficiency is only 50%.

[00:20:02]Yes, one could of course increase the efficiency by, for example, moving further to the right or making the RL larger, but that doesn't help.

[00:20:12]Um, because, um, you can also consider that with the performance formula, the performance will then decrease accordingly. So the optimum is indeed when the two are the same size.

[00:20:26]Good. Um, and the same basically applies to the broadcasters as well. Yes, and that's why I included the 50 ohms here, because 50 ohms is what we typically have. And then the question often arises, yes, why 50 ohms? And I think we had already hinted at that a little bit in class.

[00:20:46]50 ohms is a compromise between high power transmission and low line attenuation.

[00:20:54]That means you can see a curve here for the power output, and actually 30 would be much, much better, but our cables, our coaxial cables for example, have a relatively high attenuation, right? That means we would actually transfer a lot of energy, but then simply lose it due to the attenuation of our line, for example. Yes, on the other side we have something like 77 ohms here. That 's the best for low damping and um yeah, so somewhere between 30 and 77 ohms we find ourselves again, because there's the 50 ohm and that's the compromise. Yes, little damping, but still a lot of power is transferred. That's why we have 50 ohms. Yes, and also why the website is called 50 Oh. Um, the goal of the website is to ensure that your performance in the exam is as good as possible, so that it is dampened as little as possible. Yes, okay.

[00:21:59]And the questions about it are now incredibly simple. Um, what load impedance is required for power matching if the signal source has an output impedance of 50 ohms? Of course, 50 is at the top. Yes, so RI = L, I just explained to you why that has to be the case. Exactly. To ask it another way: what must apply to performance adjustments? Yes, good. RL = RI. A is correct.

[00:22:29]Okay, here's a summary. I think I can go over it now.

[00:22:33]Then take another look at it.

[00:22:35]Okay, but we're actually talking about power supply.

[00:22:39]So, a little bit more, um, yeah, now something about batteries, and you probably know more about that than I do. Um, no, I know that some of you also build your own flying machines and stuff like that. There are batteries in there somehow, and um, the most important or key characteristic is always the number that's written here, and that's usually 4200, in this case, 4200 campers.

[00:23:09]Yes, that means, uh, yes, in principle that's the capacity, it 's again the integral of the current over time.

[00:23:20]And then there are all these different configurations. So, for example, you can see four cells 1P here, which means four cells in series, and yes, one in parallel simply means in series. If this were 2P, then that would essentially be two parallel strands.

[00:23:43]And there's a lot of other data that you might also find there. Um, but these are the most important ones that are currently standing here. Yes, 30C, I think that also has a meaning in terms of, uh, how much, what is the maximum current that I can draw from it for a short time. Um, I do n't really want to go into too much detail right now. There you can see it again.

[00:24:08]Uh, 3.2 volts per cell, then four connected in series. Adding up the voltages, I get 13.2 volts, um, which is also marked here. Exactly. Exactly. Yes. What does this mean now? So 4200 means 4.2 amp-hours. That means, um, I can use the battery for one hour if I load it with 4.2 amps, or for two hours if I load it with 2.1 amps. Yes, you can roughly calculate that.

[00:24:48]And the formula at the end is basically what I said. So Q is the integral of, uh, a current over time. Yes, it simplifies to the equation. Exactly. And that's how I can calculate it. Yes, so if I have 4.2 amp-hours now, uh, and I think that's wrong, right? It should say 4.2 amps here. Yes, then it would be an hour. Yes, otherwise with one amp it would be 4 hours, I think that's clear. Well, that needs some correction, [snorts] the foil. Exactly. What else do we have? The stored energy. This is often also written on the battery. They often also list watt-hours on them.

[00:25:37]And energy is always something that can be converted into money. Exactly, that 's simply charge times voltage.

[00:25:44]So, we could calculate that in this example as well.

[00:25:47]If the device has a capacity of 4.2 ampere-hours times 13.2 volts, then we get an energy content of 55.44 watt-hours. Yes, I think that's pretty clear. Exactly. If I connect batteries in series, then all the values ​​add up.

[00:26:09]So, we also saw that this cell on its own had 3.3 volts and 4200 milliampere-hours. Now I cascade the four of them in series, then I get 13.2 volts here. Yes, stresses add up, and you should be very careful if you're building something like this yourself. So, you have to know what you're doing. Um, very important: only connect cells with the same data, because otherwise unwanted effects may occur. Exactly. In the parallel circuit, which would be the variant where we have two strings in parallel, i.e. 4S uh 2P, the capacitances are added together. That means the tensions, of course, add up. Then I have 13.2 volts, but because of the parallel circuit I get even more current and therefore I have 8400 ampere-hours. Yes, so I can transmit for twice as long with the battery. Yes. Exactly. And there are a few questions about that now. Um, don't let that confuse you either. The question is simply, uh, is there a battery with 7 volts and 2200 mAh and 16.28 watt-hours?

[00:27:24]What term is meant by the abbreviation "uh" here? And that is the nominal capacity. Exactly. Here again, uh, the batteries are connected in series.

[00:27:41]Each cell has 2 volts. We have 1 2 3 4 5 6 pieces, so it must be 12 volts somehow, and each cell has 10 amp-hours, and since we're not connecting anything in parallel, it stays that way. Therefore, A must be correct. Yes indeed.

[00:28:01]Good. Yes, there is another solution. I think we can skip that now. Exactly. And then there are questions, and as you can see, this is a bit more tricky because there's a safety margin built in here.

[00:28:15]Um, exactly. So, how long could you operate an amateur radio receiver with a fully charged battery that has a capacity of 60 amp-hours, if it draws a current of 0.8 amps, but now we need a 10% safety margin because we don't want to completely discharge the thing?

[00:28:37]And yes, then you can see that you can do business there. Exactly.

[00:28:44]So, there too, um, you first have to consider, okay, what is given? 60 amp-hours, but I want to leave 10%. What is 10% of 60? Yes, that 's 6 amp-hours, and I have to subtract that from the 60 and then divide by 0.8, and then I get 76.5 hours. Now you also need to know that 0.5 hours is 30 minutes.

[00:29:09]Yes, and then I can answer the question.

[00:29:14]Okay, here too, energy is a question.

[00:29:20]I'll just ignore it.

[00:29:21]You know that, you'll learn how to calculate 5 x 12 at the end.

[00:29:27]Yes, okay. Then it will just happen, I see more and more often now also for Field Days and so on, that people try it out now or also for repeater stations. Yes, I have one more question. Exactly. So nothing complicated here. Um, normally, I think at 50 on top you can read a bit more about the battery management system and stuff like that, and so on. I just skipped over that because it's not relevant to the exam. Yes, exactly. Um, but what I'm seeing more often now are, uh, Autage relays or Autage LoRa nodes and things like that with, uh, photovoltaics, i.e., solar cells.

[00:30:14]Um, that's becoming more and more common now, and also Field Day and so on, um, you can also try that. It's actually quite interesting.

[00:30:24]Um, that goes back to the discovery by Albert Einstein, who actually discovered the photoelectric effect and I think he received the Nobel Prize for it.

[00:30:35]Not for relativity theory at all, but for that.

[00:30:40]Yes, exactly. And that's actually really exciting, but that's just a minor detail. So sunlight hits what is actually a PN junction, actually a diode. [snorting] Yes.

[00:30:53]And um, radiant energy is then converted into electrical energy, and uh, yes, when I connect consumers, a current flows and I also have a certain voltage. Yes, 0.5 to 0.6 volts. Exactly.

[00:31:15]Yes, as we've already said, the primary function of a solar cell is to convert radiant energy into electrical energy.

[00:31:23]And now it's important, okay, what is the open-circuit voltage? Yes, I can measure that in full sunlight. That's the tension that's present. Then I have the short-circuit current. That's basically the maximum current, um, that could theoretically flow if I short-circuited the thing.

[00:31:43]Yes, and one cell is not enough for us. We connect a lot of cells together for a solar panel, for example, 30 in series and then in parallel, and each cell generates 0.6 volts. So, times 30 gives us the total voltage, and one such string provides 1 amp. If we have four such strings, for example, we then have 4 amps, right?

[00:32:14]And that's also what the exam question asks here. um, four parallel rows of 30 solar cells with 0.6 volts open-circuit voltage, 1 ampere short-circuit current. What open-circuit voltage and short-circuit current does the entire module deliver? And we've already said that, actually.

[00:32:35]Um, 0.6 x 30 equals 18 volts, and then a short-circuit current of 4 amps. Yes, I think you can imagine it quite well like that.

[00:32:46]I'll just skip over that solution as well.

[00:32:51]Yes, what else do we have?

[00:32:51]Voltage converter.

[00:32:54]Yes, uh, we have quite a lot of voltage regulators now, don't we? Or rather, first of all, voltage converters. There are now ready-made modules available that you can simply buy, allowing you to adjust the voltage up or down as you wish. Well, mostly due to switching operations, you always have to be a little careful that it also happens to be on the amateur radio band and that you then somehow interfere with yourself through pulse width modulation.

[00:33:23]Exactly, it's a kind of downward-upward converter, so a bug or boost converter, it can go in both directions.

[00:33:31]For example, if I have an amateur radio power supply, which usually has 13.8 volts, and I can somehow generate 5 volts here, then the voltage is reduced accordingly, step down.

[00:33:46]Or it can also work the other way around: if I apply 12 volts here, for example, I can also generate 19 volts here.

[00:33:52]Yes, that works too.

[00:33:55]And that works via a PWM control system with a regulator.

[00:34:03]Yes, such devices usually have some kind of efficiency rating.

[00:34:07]That means, um, yes, power goes in on the spring side and something comes out again, and in between we lose energy, so to speak, or power.

[00:34:19]Yes, and with that we can calculate the efficiency, usually in percent, and uh, yes, the higher the better.

[00:34:30]We usually never reach 100%.

[00:34:32]Yes, exactly. And there are some very simple questions about that. So, we have a voltage converter that converts 12 volts to 5 volts.

[00:34:40]It takes in 2 amps and outputs 3 amps. So, you ca n't see that directly at first glance, because, well, basically voltage is being converted to amperage, you can imagine it like that, right? Um, so I reduce the 12 volts to 5 volts, and for the 2 amps I put in, I get 3 amps on the other side. But what is the actual efficiency?

[00:35:05]And if you calculate that, simply plug it into our power formula P = U times I and then we have 5 x 3, which is 15, 12 * 2 = 24 and then we see, oops, the efficiency is only 62.5%.

[00:35:22]Yes, that means a little bit is lost in this conversion. Exactly.

[00:35:29]And that raises many questions.

[00:35:32]Same principle.

[00:35:34]I'll skip over it due to time constraints.

[00:35:38]So, in class E we had already learned about the concept of the rectifier.

[00:35:44]Yes, the simplest rectifier, um, is the one that only generates a positive half-wave. Yes, that means we connect an alternating current here at the front, I don't know, our 230 volts, and then somehow, I do n't know, 10 volts or something like that comes out at the end, and then the whole thing is straightened here. That means only the positive half-wave passes through the diode, and we should see that here as well. Yes, so I apply something like that and that's what comes out, and of course it's strongly pulsed, and with that we want a DC voltage that is as flat as possible.

[00:36:23]Uh, that's why we have a capacitor here that holds the whole thing together, which then charges up and smooths everything out a bit. Yes, and um, you have to be careful there too. That is also the reason. For example, with a power supply like this, there's usually some kind of capacitor inside, and when I switch it off, the LED in the power supply still glows a little because it's still being supplied with residual current from the capacitor. Yes. Yes. And this is perhaps a bit exaggerated here to make it a little easier to see, but um, there's always a certain amount of residual ripple, right? So, this capacitor will never quite manage to, uh, achieve complete equality.

[00:37:12]Um, you can of course incorporate other tricks, such as voltage stabilization and so on.

[00:37:18]We'll learn about that in a moment, but I always have a certain residual sense of urgency here. Yes, exactly. The larger the capacity, the more the whole thing climbs.

[00:37:30]And in order to correctly dimension the capacitor, I need to calculate this peak voltage.

[00:37:38]And to properly dimension the diode, I even need the peak voltage, because it has to be able to handle it in the other direction as well. Yes, exactly. Yes, there are now some easy questions too. 230 volts are transformed here to 15 volts.

[00:37:56]And now I'm supposed to calculate the operating voltage at the terminals, and in principle, the peak voltage then sets itself up, and so the capacitor will accumulate at some point in the steady state, let's say, that's when it sets itself up. It just keeps accumulating charges until it's sufficiently charged to reach peak voltage. Exactly. Uh, how do we calculate that?

[00:38:23]Um, what's written here, 230 volts, that 's always an RMS value. That means that even if I transform it, I still have an effective value. That means I first have to calculate the peak voltage from the RMS value, and you can find that in the formula collection as well.

[00:38:37]That simply means that 15 volts times 1.1 = 1.41 equals 21.21 volts, and that is what we were looking for.

[00:38:49]So, not complicated.

[00:38:52]Same question. Um, but now it does n't explicitly say the voltage, but rather a transformation ratio of 20:1.

[00:39:02]That means this needs to be taken into account. So, 230 divided by 20 equals 11.5 vol, and then multiplying the whole thing by the square root of 2 gives me 16.26. Exactly. Ah, there was also a 50% safety margin included here. So, as you can always tell with Class A, it's actually not really more difficult than Class E, you just have to be incredibly careful not to overlook things. Yes, so the 50% is added on top, and then you end up at about 25. That 's 16 too, right? Oh, that's awful. Yes, that's why you should always read the question carefully. Yes, and better to look twice.

[00:39:55]So. Ah, here's another one with a 20% safety margin. I'll go over that now. Yes, everyone in the room simply has to be able to do that. [snorting] So, that was the simple rectifier. But that's not so good, because we're already throwing away half of the momentum of our oscillation, and that's why there's the so-called bridge rectifier, and that's a relatively common rectifier circuit, you see it quite often, but you must n't confuse it with the ring mixer. Yes, it looks similar, but it's a little different.

[00:40:29]Um, we're now using both half-waves.

[00:40:32]Yes, and that's an important takeaway. We'll take a closer look at that in a moment. The pulsating DC voltage at the output is, uh, with the double frequency response voltage, because what the bridge rectifier basically does is, in principle, it calculates the magnitude of our sine wave and flips the negative half-wave upwards.

[00:40:58]And an important mnemonic to correctly understand the structure of this object is to remember that the cathodes of the diodes are located at the positive terminal, and with that you can actually solve the tasks. That makes sense. Uh, we'll see in a moment. Yes, so the cathode is always connected to positive. Um, then I know that 's the right circuit.

[00:41:24]So, what happens now? Um, I already said that. What the circuit will do is we have our sine wave and it will now flip everything that is negative upwards. So, it forms the amount. This is essentially a component that generates a certain amount of charge, the circuit here.

[00:41:40]So, now we actually need to look at two cases. Um, the first case is that we're somehow at the positive half-wave, and the other case is that we're at the negative half-wave. And let's assume ideal diodes for now. Um, and let's ignore the fact that you need at least 0.7 volts for the diode to become conductive.

[00:42:03]Um, and now let's just look at these two cases. Um, yes, well, there you can see it, those are the two cases.

[00:42:11]Firstly, we are now in the positive half-wave.

[00:42:16]And with the diode, we already said P to + = plus. This means that if the anode is positive, then the diode conducts. Yes.

[00:42:26]Um, and that's exactly the case now.

[00:42:28]So now we are at the positive half-wave, which arrives here and can then pass through the diode accordingly. It continues here, down there, and is positive again, then goes back to the negative pole here. Yes, so regarding the journey, if we now come to the negative half-wave, then we are negative at this point and then in principle the current can no longer flow through here, but it can through. Yes, or to put it another way, things are positive down there now. Okay, let's just take a look.

[00:43:10]Down here it's positive, so from the perspective up there it's negative. It's negative here, but it 's positive down there. And then the diode switches on, and we essentially go back along this path, ensuring that the whole thing is always positive up here and always negative here.

[00:43:32]That's how it all works.

[00:43:37]So, regarding the questions, you simply have to recognize it, and we said that this is always the case when the cathodes are at the positive terminal.

[00:43:47]Yes, that's the right one. Just remember that and you can think about it. Yes.

[00:43:54]Um, but I think the mnemonic device is okay.

[00:43:59]So, there too, you want as little pulsation as possible on the voltage. That's why, in principle, all of this is filtered out again here. Um, yes, these capacitors here are attached to the burrs and they charge up to the secondary peak voltage in the same way. But the efficiency here is naturally higher because we essentially take both half-waves with us. Exactly. Another question: what is the voltage across the filter capacitor CS when open circuit, with a mains AC voltage of 230 volts and a turns ratio of 8:1?

[00:44:41]And now you see, you have a complete chain. So, first you have to divide the 230 volts by 8, and then basically calculate the peak voltage again, or it's actually no different than the other task, right?

[00:44:56]So 230 divided by 8 is 28, and then again the square root of 2, which gives us approximately 40, and that's what we wanted. Yes, so it's really not rocket science.

[00:45:10]So, then we have another alternative concept for a rectifier, um, that only needs two diodes. This is the full-wave rectifier.

[00:45:21]We'll take another look at that one too. This only works, however, if I have a transformer with such a center tap.

[00:45:30]Yes, well, that's a tradeoff.

[00:45:32]Either I have such a transformer or I don't. And then, if I don't have it, I'll just take four diodes for the ring mixer, and oh, not a ring mixer for the, uh, four, what was it called again? No way, now I'm confused myself.

[00:45:48]Bridge rectifier. Thank you very much. So, uh, not for the ring wiper.

[00:45:52]Please don't install the ring wiper there, okay?

[00:45:56]So. Um, exactly.

[00:46:01]Positive half-wave. Um.

[00:46:05]On the upper winding opposite the center tap, then D1 conducts, otherwise D2 blocks, and then, when it reverses, it's the other way around. Um, can you see it again in this example? So, we have UE, the input voltage, and UA as the output voltage, and UA is actually nothing other than U1 here.

[00:46:30]Um, and here you can basically see that U1 and U2 are behaving in opposite directions. And down there, the total amount is essentially a combination of both. So. Uh, with the positive half-wave, when we have that one up front, then it's positive here for now. Then the whole thing flows back around here to the, um, potential or our reference potential, which we place on mass. Yes, when the negative half-wave arrives, it's exactly the opposite. Yes, then the current flows in that direction and we ensure that the amount is always calculated here. Uh, yeah, down there. Exactly. And now you have to recognize the right thing, and I think there's the same hint there again. The positive terminal must be where the cardboard of the diode is. Yes, because only if there is a positive signal here will a positive signal get through. Okay, that makes sense. Exactly. Yes, what else do we have? Uh, this is that special case. You have to be careful. Now take another look at the picture.

[00:48:14]Here at this point, we have now said that where the cathode is, that's positive.

[00:48:20]And that's what makes this exam question a bit tricky.

[00:48:24]Um, the diodes are twisted.

[00:48:29]Yes, that means we ca n't say that so simply anymore. It's actually a matter of definition, but since 0 volts are defined here, we get a negative voltage at the output. Therefore, B is correct.

[00:48:44]Yes, admittedly a bit tricky, but once you understand it and remember this rule of thumb – it's positive where the cathode points – then you know, okay, this is actually positive, that 's actually negative, and that's why I turn my reference potential here, our ground, because it clearly says 0 volts here, and therefore the output must be a negative voltage.

[00:49:13]Yes, once you understand it, it's super easy. Exactly. Yes, now we have the problem that if we operate with a bridge rectifier like the one we just saw, our mains frequency is 50 Hz, right?

[00:49:37]Um, if I calculate the magnitude again, if I look at this, yes, this oscillation here has 50 Hz. Yes, and if I calculate the magnitude now, then I get something that oscillates twice as fast. Yes, so one period up here was this whole piece, and now one period here will only be this piece. Yes, it oscillates much faster, exactly twice as fast. And that is exactly what we are concerned with here regarding this question. If the mains frequency is 50 Hz and we perform rectification, then the resulting signal is 100 Hz.

[00:50:23]Yes, and that's also the issue that now comes into play with the so-called ripple; that's the next issue that comes into play here.

[00:50:32]No circuit works perfectly. Um, and a rectifier, although we filter it well, will still have some residual decay. And you can see that on an oscilloscope.

[00:50:43]We've already seen this graphic.

[00:50:46]And now this question essentially asks, yes, what is the remaining due date?

[00:50:53]Yes, what is the frequency now and what is the residual maturity?

[00:50:59]And this is basically the direct current voltage we can reliably obtain. And there's something else dancing around up there. That means one box represents 3 volts. This means that this residual ripple has a value of 3 vol and the frequency, yes, you also have to read that off.

[00:51:19]The period here, one box is 5 milliseconds, the period is 10 milliseconds and 10 milliseconds 1 dur F, um, we then arrive at 100 Hz, so 3 volts and 100 Hz is the solution here. Yes.

[00:51:38]Good. Um, maybe there's something else here so you can get a feel for it.

[00:51:52]I hope that's over now.

[00:51:55]This is what 50 Hz sounds like and this is what 100 Hertz sounds like. Yes. M, that means, um, if you ever hear something like that somewhere, then it's usually in the circuit or you hear it somewhere in the audio branch, then you have a kind of humming sound in there. That means, somehow mains voltage is transferred to the system [snorting]. Yes, and then you have to figure out how to get rid of it.

[00:52:29]Good. Yes, the power supply continues.

[00:52:34]Switching power supply. We've seen that before in class E. What's the idea behind that?

[00:52:41]Uh, first of all, we have 230 volts here and we want to end up with, I don't know, 13.8 volts or something like that. Yes, what we'll do here is, well, we'll fix this right away and try to smooth things over here as well.

[00:53:06]And then we somehow have a direct current voltage that is relatively hollow. And now there's also something here, formerly called Zahacker. There's a switch here that turns the thing on and off.

[00:53:20]And that too, since it 's also an alternating voltage that is switched on and off abruptly, um, I can transform that as well. Yes, so I can also transform that down somehow, from, uh, 230 volts times, uh, the square root of 2.

[00:53:38]And if I switch quickly or slowly here, I can basically also adjust what arrives over there at the end of the day by changing the switching duration. This will also be fixed here, for example, and then smoothed out again. Yes, that means here we make the transition from mains voltage, alternating voltage, to direct voltage, and here the whole thing is specifically clocked and switched so that we then achieve a desired voltage. And all of this up here, this switch E is a so-called pulse width modulator, and here at the back is the capacitor that simply integrates it away again. Yes, it 's simply another low-pass filter that removes these high frequency components. Exactly. Yes, what is it? The switch, that's the pulse width modulator, and unfortunately, this isn't always perfect, um, a switching power supply, because these switching processes can cause interference signals that might not be filtered out very well. Yes, sometimes a manufacturer of an electronic device might cut corners a bit and not include capacitors and coils for filtering in order to save money, and then you have a typical power supply that causes such interference in the spectrum. Yes, I believe this is a disturbance that was recorded directly on such a power supply using an antenna. If you move 1 meter away from it, it's already gone again. Yes.

[00:55:25]But that's why you still have to be careful not to connect such a power supply directly to, uh, the transistor input, or anywhere near it. Yes, there are some interesting approaches, for example in the construction of radio equipment. For example, QRP Labs offers these kits by Hans Summers, and he solved the problem very cleverly. It also has switching power supplies in it.

[00:55:55]Um, but since the vaccination radio is integrated, it always knows the frequency that is set in the radio. And then the switching frequency of the switching power supply is adjusted so that it does n't collide with the frequency of the receiver. It's actually a pretty clever solution. Yes. Yes. Um, you should really try to filter out these interference signals from the power supply as much as possible, because if it goes back into the power grid, the power grid is a very large antenna and then you're radiating that stuff back out. Um, that's definitely something you want to avoid. Exactly. Yes, what is the main disadvantage of such a switching power supply? Yes, um, it creates an undesirable signal spectrum because it causes the chopping up of high-frequency components, and these have to be filtered out accordingly.

[00:56:52]Exactly, even in an amateur radio receiver, unwanted signals are detected every 120 kHz. This is probably due to unwanted radiation from a switched-mode power supply, which then, um, as you can see here, um, you can see the harmonics of this switching process at regular intervals. Yes, well, I mean, a rectangle is generated, and you know, with a rectangle I always get the harmonics of the signal included.

[00:57:27]Good.

[00:57:30]Yes, to prevent such things from getting back into the network, there are so-called network filters. Um, there you can see how it's structured. At the end of the day, that's nothing more than a low pass. Yes, you already know all of this, right? So, I think we already discussed second-order low-pass filters in class E, right? So, in principle, the first order was that.

[00:57:58]And then we said we could make the whole thing even better by basically installing a coil here. And that's exactly what you'll find there again now, right? That's exactly what this structure is, it 's symmetrically built. It's basically a coil with a capacitor. Here you can usually see that it's the same in the devices. Here you have your IEC connector, and then the mains filter is directly attached to the back.

[00:58:38]Yes, that also comes up in exam questions somehow. Um, you'll just have to recognize that accordingly. This is a bit tricky, but there are things down there that you should never do, like connecting PE and mass here. Um, yes, so we can rule that out.

[00:58:59]That can also be ruled out because there is no transmitter. Yes, there's a filter here, so it has to be the other way around.

[00:59:06]Therefore, the correct solution must already be stated here. And uh yes, if you look back again, it's also very important that PE is on the network side and not with us over there in the device. Yes, this should definitely be grounded.

[00:59:28]Perhaps this is the best way to remember it. Yes.

[00:59:31]So, through the rotation of the transformer here and through the position of PE.

[00:59:42]So, now we have our voltage after the rectifier and we already have capacitors in there to smooth it out as much as possible, but we want it to be really stable. Yes, we really want to have a very clear DC voltage, and that's why there's also the issue of voltage stabilization.

[01:00:01]And there are actually three options that we will look at. Firstly, there's the Zehner diode, then the linear regulator, and then, as is typically done nowadays, an integrated circuit. Exactly. Yes. Initial state, input voltage can vary, e.g. due to battery, but also due to rectification. And what we need is voltage stabilization.

[01:00:28]And the first case is now stabilization with the Zehner diode. It's a very simple circuit.

[01:00:37]However, it is no longer used for power supply in circuit technology nowadays, but it is still sometimes used. So, it's often used to, I don't know, precisely determine a bias voltage from the transistor amplifier or something like that.

[01:00:54]Um, yes, we use the characteristic curve of the Zehner diode here, and it is also installed in reverse bias. So, we use them in the blocking case, because then it's actually the case that, um, if you look at the characteristic curve in this, um, blocking case, well, normally we look at this characteristic curve here and here at the back it's like this, that it drops off so drastically at the Zehner diode, right? And then, in principle, the current is underneath it.

[01:01:31]And here's the tension, and you can see that I can never get past a certain level of tension.

[01:01:38]As much electricity can flow as it wants.

[01:01:40]I can't get over this tension. And that is essentially the effect that is being used here, so that a very constant voltage is obtained at this point with this voltage divider. Yes, exactly. Um, and there are a few questions surrounding that, right? What is the efficiency?

[01:02:04]And that brings us back to the question: okay, how much power is ultimately converted into resistance here at the end of the day, and how much power went in at the front?

[01:02:14]And then it's specified here, okay, RL is 470 ohms, the current here should be 10 milliamps, and then I can basically calculate what the power will be back here. Yes, I can manage this with I and R.

[01:02:30]And then I also know that a certain current flows through the Zehner diode, and with that I can basically calculate the total current here, or I can also calculate the voltage that must drop across the Zehner diode. Um, that would also be possible.

[01:02:52]Um, yes, here's the calculation method: first, the power of the load.

[01:02:58]Resistance 74 MW.

[01:03:01]Um, exactly. And here is the input power. I need to know that the two currents add up and what the input voltage is. And so, in principle, I get the result that, okay, I send 345 m in and only get 74 m out the back, and then I have the efficiency again, and that's relatively bad. Yes, you can actually use that to set a kind of basic bias or something like that, but in a circuit where a lot of power is supposed to pass through it, that might be rather counterproductive.

[01:03:41]Yes, a better option in that location would be, for example, the annual voltage regulator.

[01:03:49]Um, here we use a power transistor, which we operate as a variable resistor, and it receives this reference via a Zener diode, so 5.6 volts at the base.

[01:04:04]Um, and basically we have a voltage divider here across the transistor and the load here, or yes, load resistor.

[01:04:18]Uh, the efficiency is n't that good here either. Um, so, unfortunately the scaling on the foil got a bit messed up.

[01:04:27]Um, but here you should also recognize, okay, if I apply this voltage here, so somehow we are at 2 volts, we have here uh per division, that means we have here 1 2 3 4, so somehow 8 volts at that point.

[01:04:46]That has to be extracted first.

[01:04:48]Then we have the 10-diode with 5.6 volts, and that is basically the target value we want. However, it should be noted that another 0.6 volts are lost at the transistor, and therefore what I will set here is 5 volts. Isn't there a solution for that?

[01:05:16]Yes, well, I don't know if you've seen it yet. Um, for almost every question we now try to outline a solution that may be a little different from what is written in the text. Exactly. But it's written there again. Um, between the base and emitter there's another 0.6 volt drop, and then we have the 5.6 volts that are specified here, and then the 0.6 that drops here, and then what's left is the plate 5 volts. Yes, exactly. Exactly.

[01:05:57]We haven't defined power loss at all.

[01:06:00]Sometimes the question is about efficiency, and sometimes about power loss. Power loss is basically what we lose along the way, right? So, power loss is essentially P in - P out, right? And that allows me to calculate it accordingly here.

[01:06:17]Um, okay. I can also use tension. That's also possible.

[01:06:21]I don't find that particularly elegant. Um, basically I would calculate one output and the other output and then subtract them. But the result is the same.

[01:06:31]Yes, in that case, from a logical and didactic point of view, I think it's better to consider performance. Exactly. Here, the question of efficiency is raised again. This is all just a lot of smoke and mirrors. At the end of the day, it's always just the same efficiency formula combined with the power formula that we already know. Yes, don't let that unsettle you. Okay.

[01:06:57]And the third case is a fixed voltage regulator. That's what's typically, um, installed these days. What do I know, uh, what's there? 8705 or LM XYZ, I have no idea what they're all called.

[01:07:12]Um, and this is basically an integrated circuit that also contains a linear voltage regulator.

[01:07:20]And um, it's often the case that they also get very hot because the efficiency isn't perfect. That's why they often have heat sinks. Yes, you often see that in power supplies when you open one of those up. Exactly. So, for example, here's a 12-volt fixed voltage regulator. That means the goal here is basically 12 volts. What's important is that this input voltage must always be a little higher than what's supposed to come out the other end. Only then does the thing work really well. Uh, that's another important takeaway. Um, exactly. How large is the voltage fluctuation at the output? This is kind of the trick question here, because the thing is a voltage stabilizer. Of course, even if the voltage fluctuates between 15 and 18 volts here at the front, 15 volts is still higher than 12 volts. Therefore, nothing will sway back there. The fluctuation should be almost 0 volts. Yes. Exactly. Um, this question will be asked again here.

[01:08:42]So just remember, the input voltage needs to be a little higher than what I would like. It can also be significantly higher. Uh, you always have to look in the datasheet of this component to see what it actually wants. So what is the minimum requirement? Yes, the input voltage must be higher than the desired output voltage. That 's generally the case when it's warm.

[01:09:04]How many people actually look at the datasheet? Exactly. Here too, the question of power loss is raised.

[01:09:11]Well, that's obvious, 13.8 volts somehow go in here. Um, 5 volts come out here. The load resistance is given. That means I can already calculate the power output here.

[01:09:25]Um, it's also important to know that there isn't any significant electricity branching off here or anything like that. Yes, so direct current is already applied there. Then it charges up, since no current flows through it. It becomes charged, and then no current flows through it anymore. And nothing gets lost here either. And that's why I can now use this information to calculate it.

[01:09:46]Yes, that means, uh, first calculating the current that is actually flowing here.

[01:09:51]And this current flows both here and back there. Yes, that means 5 volts divided by 10 ohms equals 500 amperes.

[01:10:03]And now I'm calculating that it's also good here that he's going over the tensions again. The difference is 8.8 volts, and then, um, 8.8 volts times the um, 500 amperes, I also get the 4.4 watts that are being sought, which I have now calculated differently. So, basically, total power, and then, um, the power at the back is subtracted. Um, but that works too. Yes, I do n't even know what the model solution we have here does, whether it does it the other way around. Let's take a look.

[01:10:48]Which one was that? Was that her? How much power is lost? Exactly.

[01:10:56]No, do it that way too.

[01:11:02]Okay, now we've talked a lot about power supply. There are a few special cases, and we already encountered one of them last week at the MMIC.

[01:11:12]This is the so-called remote power supply or bias T.

[01:11:17]When do I need this?

[01:11:20]This happens quite often.

[01:11:22]For example, I want to install a camera somewhere, or something like that, an analog camera, and I don't have a power supply there. Then you can do something like using the camera's coaxial cable, for example, to power the camera itself.

[01:11:39]We see the same thing here, for example at a Qo 100 station. That's how it's solved there too. That would be stupid if I had to go to the antenna now, since there's an LNB there, which is like an amplifier, yes, on our satellite dish here.

[01:11:54]Um, if I had to bring my own power supply there again, that would be kind of annoying. Yes, and then they thought, well, we just take the cable and a DC voltage is very low frequency, so 0 Hz, yes, and then we somehow have our 10 GHz or 2.4 GHz, depending on what we are doing. In this case, 10 GHz does n't interfere with each other. I can have a DC voltage there, and then I basically modulate the high frequency onto it. um, it works. That's the idea behind such a remote power supply switch.

[01:12:31]That means we have our Korx cable here, and now we're using the Korx cable to essentially provide the power supply for this LNB. And that's what we're doing with this circuit. And that's basically it. Um, you can basically see everything in it. The coil blocks high frequencies.

[01:12:55]This means that the high frequency does not enter the power supply, but the coil allows direct current to pass through and the capacitor blocks direct voltage. So the track voltage ca n't get any further, but the high frequency can pass through the capacitor. That's basically all it does at ST. Um, of course, the circuit diagram looks a bit more complicated here, but at the end of the day, it's really only this trick that interests us. Yes, so the DC voltage is somehow arriving here.

[01:13:28]She won't get any further here for now. Yes, it also charges up. Then no direct current will pass through here either.

[01:13:37]That means it turns here and then comes directly or directly supplies the LNA or LNB.

[01:13:44]And this is no further way forward. So the coil for the DC voltage is simply a piece of wire. But for the high frequency, the high frequency is essentially suffocated by the coil; it can no longer proceed. And if there's still a little bit of high frequency present, then the high frequency simply flows to ground via the very low resistance of this capacitor, right?

[01:14:10]And that's exactly where the high frequency simply comes through, and it works great, right? I also use this with my Q100 station.

[01:14:21]Um, every Astra sentence bowl uses that, right? So even if you have a satellite dish at your house, your receiver generates a high DC voltage of 12 volts or 14 volts, depending on the model, and thus supplies the preamplifier directly at the antenna. Exactly. Yes, so what is the purpose of all this? For DC power supply and RF signal transmission over a common line.

[01:14:53]Clever.

[01:14:55]Then they ask what it is.

[01:14:57]Hopefully you won't forget this now. Bias T because of this T-shape, in order to generate a bias or a supply voltage for the LNA. Exactly. It explains how it works again. But I've already done that.

[01:15:16]Um, yes, now there are a few more questions: what does each component do? C1. Yes, C1 is there to allow the high frequency to pass through to the receiver, but to prevent the DC voltage from damaging the receiver input, and therefore to separate the DC voltage from the receiver input.

[01:15:42]Yes, what needs to be considered when dimensioning the coil? Well, okay, the LNA might use a bit too much power. This means the coil must be able to handle the corresponding current load.

[01:15:55]Yes, good. And with that, we're already at the end of the chapter on power supply, and I would actually like to be much further along, because I would have even said that we should go into chapter 6 now, because that is also a relatively long chapter.

[01:16:15]Um, I'll just open it briefly, sort of as a teaser for next time.

[01:16:22]Basic formwork. This section will revisit all the basic circuitry concepts that we did n't cover in Class E. And that's why it's a relatively extensive topic.

[01:16:32]Let's take another look at oscillation circles. Um, but we've already derived that, e.g., how to arrive at the formula. Um, we still need to calculate cutoff frequencies and so on, but it's all a bit... um, do n't be intimidated by such a circuit. That's always the problem. You think, wow, that's a really complex circuit. But at the end of the day, all I have to do is calculate the low pass we have up front. Yes.

[01:17:01]Um, and the rest is just a smokescreen. Yes, so most of the more complicated circuit problems here in class A can always be reduced to something you already know from class E.

[01:17:13]You just need to put it in the right context for yourself. Yes. Um, exactly. And then we'll just repeat a little bit more.

[01:17:22]We already know this; we've discussed it all here in the lecture. For class E, there's actually nothing new.

[01:17:30]Um, that's why we can maybe go through it a little faster.

[01:17:35]There are just a few more details that need to be added. Yes, we recently learned about the varactor diode, so we can see one of its applications. Exactly. And to filter out a few more details, so to speak. But as you can see, there are 223 slides. We really need to step on the gas to get through this.

[01:17:57]Uh, we'll actually do that next week. I don't think it makes sense to start doing that now in the last 3 minutes. Yes, okay. Okay, that concludes our day. Okay, thanks for your attention, and see you next week.

[01:18:17]Thank you very much. Yes.