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LE3_3c Diode - Schaltverhalten, Doppelpulsschaltung

May 09, 2020
Now I would like to take a quick look at a test circuit, how we can measure such switching losses with the reverse recovery effect written, also other switching losses, to see what effects the reverse recovery effect has at this point. of a voltage source Diode This is the object we want to test, that is, this is the

diode

whose switching behavior we want to examine. In parallel, there is an inductive load and a switch at the bottom which can now switch

diode

1. This is a typical test circuit, reasonably application oriented. The so-called load drops of the power electronics, hard switching with an inductive clarification let, means that here we can maintain something similar to what we have with inverters. or DC-DC converters, which means that switching losses can be generated comparatively similar to those that would occur in practical circuits.
le3 3c diode   schaltverhalten doppelpulsschaltung
Here we use the so-called double pulse process. I would like to explain it briefly. With dual process tools, this switch down here is turned on twice in a row via a USC control signal. The first pulse is used to set the inductance to magnetize a target current and the second pulse we can measure the switching losses of the switch and the ode. in switching processes. Now we want to look at the voltage across the switch, the voltage across the diode, and also the currents through the diode and through the switch to see what caused this reverse recovery effect example.
le3 3c diode   schaltverhalten doppelpulsschaltung

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le3 3c diode schaltverhalten doppelpulsschaltung...

First of all, briefly about In this double pulse process, we want to explain how the circuit will work. We'll have the control voltage here. Now we will change 1 2 pulses twice one after another and at this point in case the transistor is on or the switch actually has the name of this switch, the diode will be biased in the spa direction when the voltage ub is positive, which which means when the switch is turned on the input voltage will be present in the inductance as a voltage which means here where If we have the switch state we will get a positive voltage across the inductor i.e. in this range for the voltage op in the inductor, the current increases linearly, suppose that at the beginning we had a current of 0 in the inductor at this point, then now it will be here during this first phase of the current to take with a slope that according to the idea that calculated as ub by adopting the basic equation of inductance, the switch opens at this point at For a while here the inductance will work as a current source and then briefly force the current to continue running freely here and then you will get a constant current at the activities and when I turn it on for the second time the current will increase again with the same gradient and actually now I have these time points here for example here because at one time point because we have the tm1 here at the second time point tm2 and at this time we can now measure the switching processes, which is to say, at this time the switch turns it off and the diode turns it on, that means you could measure the switch off at this given current up here and the voltage losses. diode turn on and at this point tm2 can I then observe the switch turning on or the diode turning off and thus calculate the losses i.e. the idea behind this is a practical test circuit by adjusting the voltage ub?
le3 3c diode   schaltverhalten doppelpulsschaltung
I can then use this point to determine the desired voltage Set the switching losses and for the duration of this duty cycle with you the switch is on. I can determine how large the current will ultimately be at which I want to take these measurements to understand the case in more detail. , we were here to see the three switching states. In principle, this circuit has three states. They were a little taller. Why travel at first glance, you can only see two, the obvious ones are that here we have the diode that has the Each of us has a state in which the switch is closed and a state in which the diode is conducting, which which means that it was 1 1 that was off during this phase and we would have the state of an ace off during this phase. , in the transition a third person agrees with the fact that the first two are 1, that is, first of all, I have a switch on here if we paint our circuit there if of course we still have the voltage source here , the diode open switch, which is off, the switch here is on and then we see well, here we have that through the closed switch voltage is applied to the inductor, that means that the voltage ul is the same and the current is taken here in the point As shown above, all the current through the inductor flows through the switch here, so we have the second state, let's say now the switch is also the diode, which means that here we have the freewheeling phase.
le3 3c diode   schaltverhalten doppelpulsschaltung
When the switch is opened, the inductance forces the current to continue flowing in the same direction as here because energies cannot jump during inactivity and current is the quantity that carries energy. So this diode becomes Up here we have poles in the direction of flow so that it turns on when the switch is off and we have the following arrangement. Then we have the diode closed here while the ignition turns the switch open and the current through the inductor will now run freely here, which means that the current right here flows through the diode. of the diodes the current idea is equal to the inductance drops now the ideals are not yet a voltage drop also turns off a voltage of 0 volts and therefore the current here does not increase more that means it explains why we are here at this constant interval, there is actually more current at the end and again voltage divided by the inductance values ​​shown below, but because the voltage and current will not change now, it is actually like this if we use real switches that cannot be turned on at no speed and change. off that means we set for example a fault and the gbt will only allow a finite current rate of change.
In fact, this will be the case even if we rotate the inductors in a circle somewhere, so a stray inductance will also prevent this. The current in this mesh can change at any speed, which means that the current through this switch it can't actually increase at any speed. Switches, semiconductor switches have the property of functioning as a power source and then we have the situation. , so to speak, where one of the switches is on but also at the same time diode one. Heads up, I'm going to put this in quotes because the switch doesn't actually work in the sense of a switch at this point. , but it actually works in the active operating area as an amplifier, which is the mode we don't want to have with a semiconductor switch in power electronics, that is, in this arrangement here we have the diode that is still on, we have the inductance which forces the current and we have a current through the switch which is now determined through the control voltage for mosfets and the ig bce and consequently we enter the odes to obtain current and in this state if we assume that the ellis is constant at At this point the current through the diode will be the difference idea is equal to l - which means it will change in the same way as the current in the switch.
The current in the diode changes the only thing that shows the difference here, so to speak, is the constant current from the inductance, so we first want to take a look at this procedure using a simulation and I would like to derive the results that emerge because this third state is really interesting. That means I have prepared this here at the LGS. I've set up a circuit with a real diode here on the left. On the right I have a circuit on the ideal diode. I use a switch and a mosfet. Now that is changed for the other one that I just recorded there But it doesn't matter because the components are, so to speak, in series.
Decision if I now decide in this series that the parallel connection of the diode and the inductance is exchanged with the switch and that in one case one is on top, in the other case that other now does not play any role and we want to compare it now, that it means that first we look at what are the current curves in these inductors, are they really as described and then we want to look first at the current leads and the voltage leads on the diodes, that is, now I have shown the two currents here, first we see the current overhead lines here, so the separator first turns it on at 5 microseconds and then 100 volts are applied to the diodes. 100 micro amps, which means we get a 10 amp current boost right here.
Ten microseconds later the current remains constant and then increases again with the second pulse and if we measure the second current here we can see that they are congruent which actually means "That the switching is fine and this double pulse process works. They become magnetized two current pulses and now I can measure the switching losses here at these points in time and now let's see what the voltage and current look like across the diodes. Now I have shown the two curves here. In the diagram above we have the curves for the circuit with the real Jew at the bottom.
With the ideal diode it is important here to give a warning, these simulations if in many cases are not entirely true. Again, I chose your model, which works approximately, but you have to be careful with the behavior free reverse recovery, as it is often not modeled correctly. Now it was time to look at it right now, 25 microseconds, that's what we're really interested in. What the current flow looks like at this point. I'm looking there. We can see the top and bottom. When you can see the difference relatively clearly, the first thing we notice here is that we now have here, at this point, on the actual diode, this reverse. recovery effect which actually means that here the current will not be at this point when the 0 crossing ensures that the diode stops conducting directly we will get this negative current here is the zero line, so this negative current, the reverse recovery boxes and this maximum negative current value Here - 24 RP, although before we had only turned on just under 10 amps and here we also see that the voltage at this point jumps to the value of the blocking voltage here on the front, or on has values ​​of very low voltage. ​and only then here, in the region of the peak current, it switches completely differently in the ideal diode, in the ideal diode the current also decreases linearly, so to speak, because these mosfets cannot change their voltage as fast as they can. you want and So a certain part is in the active area and that changes slowly, but the diode actually stopped conducting when the current reached zero at this point and at this point the diode also starts absorbing the voltage.
That's the first thing we do now. You can see that we can draw the ideal current in the diagram above, for example, if you can see it very clearly, that is, here we now see the ideal current profile, we briefly change the color and then we see that we have this ideal current. profile at zero In the first zone it works the same as with the real current curve, but here at this point the deviation is this stored charge that has to be removed. So first of all, this effect can be observed here: the Rivas Recovery Effect we have seen, but now the question is what does it do?
Overseas it becomes interesting now when we look at the switch, which means I now have to check the current and voltage on both switches and compare them well. now I have set both and have chosen the same scale on both diagrams. Intent, that means for both the voltage and the currents we can now see the voltage across the switch ideal for the Mosfet to break down sooner, which means it just switches sooner. because the switching process is not delayed as in the previous case, we also see here In fact, the current starts to increase slowly and then ends at this maximum value.
What does this look like with the actual switch when we have reverse recovery? effect of the diode, so we see in this area that so far it remains the same, but if we do that, the final value is enough water current so that when the current through the diode reaches 0, the current in the switch continues to increase until this maximum value plus the value of the current that must be experienced, that is, we obtain a peak value of the current of 34 amperes while at the same time a very high voltage is applied to the switch, the switching process is delayed and this means that we have, so to speak, a longer period of time with current and voltage in the switch at the same time and therefore higher losses.
Switching losses that occur due to poor switching behavior of the diode in quotes and the interesting thing is that these switching losses do not appear in the switch The diode is off but the switch is actually loaded which ensures that the diode is turned off . You can see this veryclearly when you compare the performance, which means we have to show the Performance on both transistors in a third diagram. That means here, first of all, the power on the transistor that is actually turned off or and then the course of the instantaneous power on the transistor that turns off the ideal diode and then we see here if we compare the actual area under these curves is a measurement. for the energy that will be converted to heat in this switching process and we can clearly see at the point of the ideal food change, the switching losses are significantly lower here, the area and the risk are significantly lower in this where we turn off the one real more or less and then it can be said that this reverse recovery effect will generate large losses, not in the diode but in the component that is on.
Now we want to briefly summarize the findings of this simulation, that is, I would now like to briefly state here what we have just observed, let's look at the curves in the switching process when the diode is off. First we say that we use the ideal diode, then on one side we look at the current and voltage across diode 1. and on the other side we look at the current and voltage across the switch. That means we will now implement what we saw in the simulation here. First of all, the ideal diode was the current curve that was turned off.
From this maximum value of the current, it decreases linearly either because the transistor changes slowly or because of these parasitic inductances that do not allow this to be the course of the current idea that we have from this section onwards, so we have actually seen the voltage across the diode. decreases as soon as we reach the point and its current is 0, the voltage here will ideally be the value of In our case, the blocking voltage is the value of the applied voltage and B in our test circuit. voltage that is applied to the diode here if we have that.
Those were the curves for this ideal diode. In fact, momentary losses occur. We never have a superposition of current and voltage in the ideal diode. Why is the switch we had on? This ideal diode is now off, that is, we have the switch on this ideal diode, so it will be as follows, first of all there is the current 0 to the same extent as its current decreases, the current in the switch increases. , this is now this phase, this state 3 where both the ode and the switch are on, the diode current is equal to bright - or the switch current is - idea That means that if the idea here decreases from the maximum value that we have , the current at the fault must be brought to this maximum value that is what happened during this switching process for the current in the switch what happens now with the voltage actually the mesh rule applies here in our circuit, that is that is, whenever the diode is tired and we will be, we get a voltage drop of 0 volts across the diode, that is, as long as we are in this position at this point because the current changes and before that, where the switch is off, the voltage drops the switch here, which means that now I can say at this point until this moment you have to draw it here until this moment here the voltage drops either on the switch and only from this point it changes completely That also means attention here , momentary losses will occur at this point, that means even with an ideal switch we get a momentary power loss.
What does an overlay mean here? If I had product images around Molly, I would have them in the course. of tea and the area under This curve is a measure of the energy that is converted into heat during the switching processes, which means that it is very important that even with an ideal switch we have switching losses, especially when the current in this switch ideal cannot change as quickly as desired and in fact because we would always have parasitic inductances here we would also have losses arising at this point even if we had an ideal diode now you can see the second state of what we had there, this real, the eau de vie sa there we color the preliminary heats here again 1 that means I'm doing it here now again in the testicles strom.de has here the switch current that will do it.
We will see the voltage ud and the switch voltage and then we will also see again the current power loss, that is, first of all the current curve, the current first of all we have seen in the simulation with a slight gradient up to step 0 current if the parasitic inductances or controls are at this point and are equal so that this current change is the same then we will get this peak reverse current this decrease. Now I have to do this a little bit directly this point to this value or - at this point and we had I've seen that the switching process is delayed, in fact the diode now just turns off a little bit later, what you have here at this point this jump maybe only here where the reverse current peak is at this value.
Now losses are also occurring at this time. "I could draw them due to bad switching behavior. We have a power loss here at this point and also a power loss here at this point, which ensures that the diode here will acquire additional losses compared to the suffering and now I want to see what what this does to the switch and at first everything will be the same and it will turn off again until this moment, then the current increases until now everything remains the same and now the attention is applied here again the rule of note in these nodes happen, it will continue to apply here the idea is equal to yes or resolved is the idea l, which means that it actually now adds to this current value that we have here, this negative current value in the inductance in the diode means that I actually have it in idea of negative value which means that in terms of quantity this value will be greater than the current, the negative value will make a value which is greater than half of the current, that means that here the current will continue to increase and only then it will decrease, that means that only here we will reach it, so the value that we had here before actually the voltage that we had seen now will change here later just this way the voltage will jump back to 0 10 at this point before We still have the voltage and b applied here and we see clearly, now we have a longer interval here where these losses occur at the same time and the current and the voltage occur at the same time, which means the losses will be larger.
I'll do it again like this. The current power loss is this energy, which here is again the integral pdt, which would now be a power loss here, which is significantly higher at this point than the power we have up here because there was no loss with the diode and That is now the real problem. This reverse recovery effect means that we have higher losses here and, especially in the switch, there are no directly on the diodes. There are also course losses but most of the losses will arise at the switch and that is why I already talked about this in the last part.
Diode selection is carried out After the reverse recovery behavior and now we want to get into it in more detail, there are basically three ways in which you can classify it in power electronics, that is, we say that there are standard diodes that have very long recovery times, about 0 to 5 microseconds and two microseconds or more, that's a lot, they are not suitable for high switching frequencies only for mains frequencies that are optimized for low forward losses, but typically only 50 hertz up to, say, 400 hertz, so they are suitable and have relatively poor switching behavior. There are fast diodes or also known as fast recovery.
We will use them in power switching. supplies, for example, or in the inverter and register very quickly. They have recovery times in this range, roughly maybe single digits, other seconds vary up to 100 nanoseconds, ten to a hundred seconds longer. So in terms of size, it's small, they're fast diodes and they're for fast sound applications and now there's a third type of diode at this point that doesn't have any reverse recovery behavior and these are Schottky diodes which means that in There are actually Schottky diodes too. Power diodes, perhaps As a reminder, Schottky diodes consist of a metal semiconductor junction, which means we have a metal somewhere here, then here we have a doped semiconductor layer when they are in the power period where the positive layer and these now do not have a pronounced reverse.
Recovery Behavior They basically only have capacitive charging current and no reverse recovery behavior, so they are now significantly faster. What is the disadvantage? There are some restrictions at this point when we look at these diodes, we usually only have values. ​up to a breakdown voltage of 200 volts, which means we have breakdown voltages of less than 200 volts, but then we have burnout voltages that are lower than with silicon pin diodes, but there are special Schottky diodes, for For example, the silicon carbide material, where we can use up to a breakdown voltage of several hundred volts, I will write it down in the range that occurs to me, even 600 volts to two kilo volts in this order of magnitude.
Of course they have the disadvantage of being a little more expensive, having a slightly higher flow voltage, also higher than that of the Indy Oden, but it has the advantage that there are practically no losses, which means that in power electronics we select the diodes depending on the different recovery times depending on the application. low frequency mains frequencies where light losses need to be optimized, we can use diodes that have long blocking delay times and large charging delays, but lower losses at the high frequency synchronized with the DC, were from inverters, for so we use a fast recovery that looks fast. diodes and for small voltages we can use silicon schottky diodes or for special applications where switching losses are really crucial we can use faster silicon carbide schottky diodes which do not have a reverse recovery effect

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