 # LE3_3c Diode - Schaltverhalten, Doppelpulsschaltung

May 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

#### diode

this is the object we want to test i.e. it is the

#### diode

whose 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.   