LE3_3c Diode - Schaltverhalten, DoppelpulsschaltungMay 09, 2020
now i would like to take a quick look at a test circuit again, how can we measure such switching losses with the reverse recovery effect described, other switching losses as well, to see what effects the reverse recovery effect has at this point, the circuit here consists of a voltage source a
diodethis is the object we want to test i.e. it is the
diodewhose switching behavior we want to examine parallel to this an inductive load and below it a switch which can now turn off diode 1 or this It's a typical test circuit that is reasonably close to the application, so-called power electronics load drops, hard switching with an inductively clarified let which means we can keep it here similar to what we have with inverters or with dc converters. dc and with that you can then generate switching losses comparatively similar to those that would occur in practical circuits we don't use what is known as the double pulse method which i would like to briefly explain in a moment with the double tools method this switch down here is turned on twice in succession via a usc control signal the first pulse is used to magnetize the inductance and set a specific current and during the second pulse we can measure the switching losses of the switch and the odo in the switching processes now we want to see the voltage across the switch below, the voltage across the diode and also the currents to through the diode and through the switch to see what caused it for example this reverse recovery effect first of all briefly about this double pulse procedure we want to explain here how the circuit will work we will have the control voltage here now we switch twice 1 2 pulses in succession and at this point in case the transistor is on or the switch is actually called this sc halter the d Iodine will be biased in the spa direction with positive voltage ub, that is, when the switch is on, the input voltage will be present across the inductance as a voltage, which means that where we have a switch state, we will be getting a positive voltage. in the inductance this means that in this range for the voltage op applied to the inductance the current increases linearly, suppose we had 0 current in the inductance at this point at the beginning, then the current here during this first phase will increase with a slope which according to the idea calculates itself as ub by adopting the basic equation of inductance the switch opens at this point here the inductance will act as a current source and then momentarily force to the current here to continue to run freely then get a constant current in the activities and the second time you power on, the current v it will come back with the same and actually now I have these points in time here for example here because the only point in time when we have the tm1 here in the second point in time tm2 and at this point in time now we can measure the switching operations which means at this moment, the switch turns it off and the diode turns on which means you could measure the switch turn off at this specified current up here and the diode turn on losses and at this point tm2 I can observe the switch on or the diode off and therefore calculate the losses That means the idea behind this is a practical test circuit by setting the voltage ub I can then set the desired voltage to determine the losses switching at this point and I can determine how the switch is turned on for the duration of this duty cycle with you ultimately the current will be large for you in the one i will present these measurements to understand the case more precisely we were looking at all three switching states which means this circuit basically has three slightly higher states why the trip at first glance can only see two the most obvious are we're here the diode has the switch it each has a state where the switch is closed and a state where the diode is conducting that means it was 1 1 which was off during this phase and we would have the state here where an ace is actually off during this phase here in the transition a third person agrees that the first two 1's means I have here first of all turn off a jew if we paint our circuit there if of course we still have the voltage source here the diode times me as an open switch is off the switch here is on and then we see ok here across the closed switch we have the vol switch to the inductors i.e. the voltage ul is the same and the current is taken as shown above here at the point where all the current through the inductance flows here through the switch then we have the second state now lets say that the switch is also the diode that is on that means we have the freewheel phase here actually when the switch opens the inductance forces the current to continue flowing in the same direction as here because the idle energies it can't jump and current is the power carrying variable so this diode we have up here we have poles in the direction of flow so it turns on when the switch is off and we have the following arrangement then we have the diode closed here when the on turns the switch open and current through the inductance now runs freely here which means here b-elf current flows through l diode the idea diode current is equal across inductance drops now the ideal is n no drop in voltage also puts out a 0 volt voltage and so current doesn't increase any further here that means which explains why we're here in this constant interval you actually have more current at the end and again voltage divided by inductance values shown below but because voltage and current won't change but now it's really like that if we can using real switches that don't turn on and off as fast as we want that means we set for example a fault and the gbt will only have a finite current rate of change it will actually be the case if we still turn the inductances somewhere of a circle so a parasitic inductance will also prevent this from changing the current in this mesh as fast as you want i.e. the current through this switch doesn't actually you can ramp up as fast as you want semiconductor switches have the property that they are a like a power supply and then we have the state so to speak where one of the switches is on but also the diode of one at the same time an amplifier is the mode we don't want in a semiconductor switch in power electronics ie who really has this arrangement here we have the diode still on we have the inductance forcing the current but here we have a current through the switch which is now determined via the control voltage for mosfets and the ig bce and consequently we will get current in the odes and in this state if we assume the ellis will be constant at this point the current through the diode will be the difference here, the idea is the same l - which means in the same way that the current in the switch, the current in the diode will change, the only thing the d does The difference here so to speak is the constant current of the inductance so let's take a look at this procedure first using a simulation and I'd like to derive the results that comes out because it's interesting actually this third state which means I've set this up here on the lgs i have 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 mosfet each now that is swapped with the other i just taped it but that doesnt matter because the components are in series, so to speak.
I decide if now in this series I decide the parallel connection of diode and inductance exchanged with the switch and that in one case one is up in the other case the other doesn't matter now and we want to compare that now ss means first of all we look at how are the current curves on these inductors are they really as described and then we want to look at the current curves and the voltage curves on the diodes ie now i have shown the two currents here first we see the current overhead lines here so by which the scheidter first turns it on at the moment 5 microseconds then 100 volts is applied to the 100 microamps which means we get a current surge of 10 amps in ten microseconds right here after the current stays constant and then increases on the second pulse closes again and now we measure the second current here we see that they are congruent that means that the commutation is actually ok here and so this double pulse method works two pulses of current are magnetized 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 i have now here are the two curves shown in the upper diagram we have the curves for the circuit with the real jude in the lower diagram with the ideal diode these simulations give a warning here it is important if in many cases they are not quite the truth I have chosen your model , which works roughly, but you have to be careful about free reverse recovery behavior as it's often not modeled correctly, now was the time to take a look and right now, 25 microseconds, that's what we're really interested in . the runs look like at this point I'll look there we see up and down when you can already see the difference relatively clearly the first thing we notice here is now we have this reverse recovery effect here at the actual diode point which it means the current is not actually here at this point when the 0 junction is there make sure the diode stops conducting right away we're going to get this negative current here it is d So the zero line is this negative current on the recovery graphs reverse and this maximum value of negative current here: 24 rp although we had previously turned on a little less than 10 amps and we also see that the voltage here at this point later jumps to the value of the blocking voltage here in front is the given or very low voltage values turn on and only then here in the area of \u200b\u200bthe current peak changes completely differently with the ideal diode with the ideal diode the current also changes It also decreases linearly, so to speak, this is because these mosfets here are not random, the voltage of their games can change rapidly, and therefore they are a certain part of the active area and this changes slowly, but in fact, the diode stopped conducting when the current reached zero at this point and at this point the diode also starts taking voltage that is the first thing we can see now we can go to the top diagram for example reproduce a drawing of the ideal current when you can see it very clearly that means here we briefly see the ideal current curve the color is different and then we see this ideal current curve at zero in the first area it works the same as with the real one current course but here at this point the diversion this stored charge that has to be cleared that is first observable here these rivas effects recovery effect that we have seen now but the question is what makes foreign countries eros be interesting now when we look at the switch which means i have to change the current and voltage on both switches and compare them well i have both now that they have been set up and now i have chosen the same scale on both diagrams with intent i.e. both for voltage as for those in fluxes we can actually see the voltage with the ideal switch with the mosfet breaking there earlier means it just switches earlier because the switching process is not delayed as with the former we can see the current start to increase slowly and then ends at this maximum value.
What does it look like now with the actual change when we have the reverse payback? effect of the diode then we see in this area that so far it has been the same but if we reach the final value it is water current if the current through the diode becomes 0 the current in the switch continues to increase up to this maximum value plus the value of the current that we should experience, that means we get a peak value of the current of 34 amps while at the same time a very high voltage is applied to the switch, the switching process is delayed and that means we have a longer time . period of time, so to speak, current and voltage on the switch at the same time and therefore higher switching losses that occur from the bad switching behavior of the quoted diode and interest ssante now these switching losses do not appear on the diode but actually the switch is on which ensures that the diode turns off so to speak you can see that very well when you compare the performances which means we have to show the performance in a third diagram here both transistors that it meanshere first of all the power in the transistor that turns off the actual o and then the course of the momentary power in the transistor that turns off the ideal diode and then we see here when you compare that the area under these curves is actually a measurement of power is converted to heat during this switching process and we can see very clearly at the ideal switch point that the switching losses are significantly less here the area and the risk is significantly less here where we turn off the true o so Therefore, it can be said that this reverse recovery effect will generate large losses, not in the diode but in the component that is turned on.
Now we want to briefly summarize the findings of this simulation, that is, I would like to set this up here briefly at the point that we want to just observe the progression in the switching process, that is, when the diode is turned off in the first place, Let's say we'll take the ideal diode and then on the one hand here the current and voltage across diode 1 and on the other hand we'll look at the current and voltage across the switch which means we're going to now let's do what we did in the simulation first of all the ideal diode was the current curve turning off from this maximum value the current goes down linearly either because the transistor switches slowly or because of these parasitic inductances that don't allow it so here is the idea current curve that we have there from this point of the section then we have seen that actually the voltage across the diode decreases as soon as we are now at the point and we get its current 0 the voltage here ideally e will reach the value of the blocking voltage which in our case is the value of the applied voltage and b jump in our test circuit, this is then the blocking voltage here on the diode when we have that these were the curves with this ideal diode, in actually a momentary loss occurs we never have a current and voltage overlap in the ideal diode why is it now off with the switch we had with this ideal diode we have considered? the switch in this ideal diode will then be as follows first the current is 0 to the same extent that its current decreases the current in the switch increases attention this is now this phase this state 3 where both the odo and the switches are on applies the diode current is equally bright, or the switch current is, idea which means that if the idea here decreases from the maximum value, we have to bring the fault current to this maximum value which is what happened with this process of switching for the current in the switch, what happens now to the voltage?
Actually, the rule of thumb applies here at this point in our circuit, which means as long as I gave the diode yes, I'm sorry, and we're going to get a 0 volt drop across the diode here, which means as long as we're at this state at this point because the current changes and before that where the switch is probably off it drops above the switch here the voltage drops that means now I can tell at this point up to this point in time it has to draw it here up to this point in time here the voltage drops either across the switch and only from this point on does it fully turn on that means actually be careful here too losses will occur at this point right now that means even with an ideal switch we have a momentary power loss pvt here which means here such an overlap if i had pictures of products around molly i have in the course of T and area under this curve which has a measure of the energy that is converted into heat during the switching processes which is very important even with an ideal switch we have switching losses and between ar then if the current in this ideal switch, so to speak, does not can change at will and in reality because we would always have parasitic inductances we would also have losses here that arise at the point even if we had an ideal dioe now we can assume the second state of what we there these real ones had the schnapps so there we paint again here the preliminaries 1 that means i'm doing it again here at strom.de testicles you have the current switch here which will make it look at the ud voltage and the switch voltage and then we'll also see the current power loss again it's say, first of all, the current course, the current we have initially seen in the simulation with a course slightly up to the step current 0 if the parasitic inductances or the control and the same is the same here the current change is the same so we'll get this peak reverse current this decay I have to do that now mal a little bit angular at this point at this value or - on this point and we had seen the switching process lag the diode actually just turns off a bit later you have here at the point of this jump possibly just here where this reverse current spike at this value now we'll also experience losses at the moment it could draw them due to bad switching behavior we're having a power loss here right now and also here a power loss at this point which ensures that the diode is additionally acquisition will lose compared to suffering and now we want to see what it does with the switch with the switch and initially everything will be the same it will turn off again up to this point in time then the current to will increase so far everything stays the same and now attention applies here again note rule on these nodes what happens will continue to apply here idea equals yes or idea l has been resolved meaning this value of current that we have here now actually adds this negative current value in the inductance in the diode to the fact that it actually has a negative value idea which means this value is greater than the current becomes the negative value and has a value that is greater than half of the current that means the current will keep increasing here and only then will it go down that means only here at the back will we reach the value that we actually had before here the voltage that we had seen just it will change later here with just this route the voltage will jump back to 0 10 at this point before we still have the voltage yb here and we see it very clearly we actually now have an interv to the longer here where these losses occur at the same time where the current and voltage occur at the same time that means the losses get big no I'll do that now it's kind of like what we see here it will be the course of energy loss right now this energy which is the pdt integral here again what now would be the energy loss here which is significantly higher at this point than the energy we have up here because there were no losses you're welcome with the diode and that's now the real problem with this reverse recovery effect which means we have higher losses here and especially in the switch not directly in which the diodes are also price losses but the Most of the losses are in the switches that are created and that is why I already discussed this in the last part.
Diode selection is carried out according to reverse recovery behavior and now we want to go into this in more detail. There are basically three options that can now be divided into power electronics, that is, we say it. There are standard diodes that have very long recovery times, around 0 to 5 micros. clients and two microseconds or more that's a long time not suitable for high switching frequencies just for network frequencies that are optimized for low transmission loss but usually only 50 hertz to maybe say 400 hertz here so which are suitable because they have relatively poor switching behavior then there are fast diodes aka fast recovery we'll use them in switch mode power supplies for example or in the inverter and they work very fast they have recovery times in the range, very roughly, maybe one-digit other seconds up to 100 nanoseconds, ten to a hundred other seconds the order of magnitude that's small these are fast diodes and they're for fast sounding applications and now there's a third type of diodes in this point they don't have reverse recovery behavior at all and these are schottky diodes which means schottky actually doesn't There are no power diodes, maybe as a reminder Again, Schottky diodes consist of a metal-semiconductor transition, which means we have a metal somewhere here, so here we have one in this case: a semiconductor layer doped si you have a period of performance where the layer appears positive and now they have no pronounced reverse recovery behavior essentially just a capacitive charging current and no reverse recovery behavior they are now much faster what is the downside now there is some restrictions at this point if we look at these diodes it seems like the issue normally we only have values up to a breakdown voltage of 200 volts then we have breakdown voltages less than 200 volts but then we have latent voltages which are lower than with diodes silicon pin, but there are special Schottky diodes, for example, silicon carbide material, we can u put it there up to a breakdown voltage of several hundred volts I'll write it down to that in that range that comes to mind even 600 volts to two kilovolts in this order of magnitude which of course have the disadvantage that they're a bit more expensive, they will have a slightly higher flow voltage, also larger than those of the indy ode, but they have the advantage that there is practically no loss, that is, in power electronics we select the diodes according to the different times of recovery depending on the application, i.e. for low mains frequency frequencies where light losses need to be optimized, we can take diodes that have long reverse recovery times and large heavy delay loads but lower losses at high frequency dc sync they were from inverters so we use fast recovery diodes which seem fast and for small voltages we can use silicon or application schottky diodes It is special where switching losses are really crucial we can use faster silicon carbide Schottky diodes that have no reverse recovery effect to point out
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