Homework assignment two is concerned with simulation of a buck converter that steps 12 V down to 1 V at 20 A to supply a microprocessor. You will add a synchronous rectifier to a buck converter to improve its efficiency. In this lecture, I'm going to illustrate how to use the schematic capture capabilities Of LTspice. And I'm going to work an example in which I will add a synchronous rectifier and the associated circuitry in a boost converter example. Here's the circuit that we'll start with. This is a boost power stage having input voltage VG and inductor L1, with a MOSFET and diode and then the output R and C. This converter is intended to step 12 volts up to 24 volts and it operates at a duty cycle of 0.5. Here we have a voltage source to set the duty cycle to 0.5. This voltage goes into a pulse-width modulator that then generates the switched waveform going to the driver, which then turns the MOSFET on and off. To insert a synchronous rectifier in the circuit, what we need to do is to take the diode and replace it with a MOSFET. We need to drive it with a gate driver circuit. The gate driver needs a power supply to supply the gate driver and we're going to derive that power supply from the 12 volts at the input. We need to generate complementary drive signals for the two transistors. So we're going to add what's called a dead time generator that takes the pulse-width modulator logic output signal and generates a second complementary signal to drive the synchronous rectifier FET. The switching converter library that accompanies these LTspice files for this course includes what's called a dead time generator block, which is shown right here. This dead time generator takes the control input signal coming from the pulse-width modulator c, which is shown here in green. And it generates two drive signals, c1 and c2. So c1 is intended to drive the main power FET of the converter. And c2 is intended to drive the synchronous rectifier FET. So these are complementary signals. When one is high, the other is low and vice versa, so that they switch at opposite times. This dead time generator also generates what's called a dead time. And here you can see the output. This is the c1 output in red and the c2 output in purple. And there are dead times here, produced by the dead time generator block, that can be set as parameter for the block. In this case it's set to 100ns. So the object is here is to turn off one FET, wait for the dead time and then turn on the other FET. So this is called break before make operation. We want to make sure that both of FETS are not on at the same time. And the dead time here, set to 100ns, is supposed to be large enough to make sure that the switching transition of the first device being turned off can complete before the second device is told to turn on. Here is the LTspice boost converter. To remove the diode, we use the scissors tool. So I click on that tool and then I can just cut away the diode. On the Macintosh, it doesn't have the toolbar like this. Instead you have to right-click with the mouse to bring up menus that include all of these functions. Okay, we'll insert a device here. So I click on that one. And let's find the MOSFETs. Okay, so I will insert an NMOS transistor right here. Okay, and then to stop this, we will hit escape. I want to change this transistor into a specific device. So I right-click on it, and I can pick a new MOSFET. So here's a library of parts that are built into the LTSpice libraries. These transistors, their models are optimized so that the LTSpice simulation will run quickly. And I strongly suggest you choose one from this list. Okay, so I'm going to choose this IRFZ44N. That is a 55.0 volt rated transistor with an on resistance of 14 milliohms. Next thing that I'm going to do is add the dead time driver from our library. So we go here to the component list. And I need to go to the directory that has the downloaded files from the course website. And here is the block called dead time that we can insert. So I'll put it right there. And then we get the little pencil to draw some wires to connect this up, so. Connect its input up to c. And take the c1 output and we'll connect that to the driver input of the main transistor. Now we need the driver for the synchronous rectifier. I can either insert a driver from the library or I can copy the one we already have. Let's take the duplicate or Copy function. Copy this driver and I'll just move or place one up here. Lastly, we need to add the bootstrap power supply for the high side driver. So we derive this from the 12 volt input. What we'll do is add a diode. And I think I'll just use another one of these Shockley diodes for that purpose. So I'll copy that, And place it over here, okay? We need to add a capacitor for this bootstrap power supply. And I'll place right about there. And we need to connect these components up. Okay, let's see, we need to set the value of the capacitor. I'm going to make it the 10 microfarads. All right, I think we're ready to go. Let's press the run button and see how it works. Here I've zoomed in on the switching transition. So this upper waveform is the switch node voltage, right at the switch node between the two FETS. It's high, so the synchronous rectifier is on here. The gate driver turns the synchronous rectifier off right there. And you can see the voltage increase a little bit. What's happening is the body diode of the synchronous rectifier is being forward biased by the inductor current, and it's turning on. So we get a higher forward voltage drop during this short time during the dead time. Here you can see the gate drive or the gate voltage of the synchronous FET is going low right there, which coincides with the jump in the switch node voltage. The synchronous FET is fully turned off here. And the green waveform is the gate voltage of the main FET, this M1 FET in green. So, the gate driver starts turning it on right here. And there's some rise time on the gate voltage. At this point here, the switch node voltage falls. And we have a low voltage after that. Here's a magnified view of the current in the output capacitor at the time when we're turning off the synchronous rectifier, so that same switch transition. And what happens, as we just saw, was that the body diode of the synchronous rectifier is turned on during the dead time. And then when we turn on the lower MOSFET, the body diode of the upper FET goes through a reverse recovery. And that reverse recovery comes out of the capacitor and goes through the two MOSFETs. So it goes around this loop. The green waveform is that capacitor current. So it's It's near zero here. And it goes down to a very large negative value. What is that, about minus 55 amps. The area of this current spike, Is the charge that is pulled out of the capacitor and flows through the two MOSFETs. So this current is caused primarily by the reverse recovery of the body diode of the synchronous FET. Also, some of this charge represents the change in charge on the output capacitances of the two MOSFETs. Okay, so if we define this area as Q, you can estimate what Q is. It's the area of this triangle, so if we find the switching time and this amplitude, we can work out what Q is. We could also, if we're clever, figure out how to get LTspice to tell us the area of this triangle. But in any event, its area we can define as some charge Q. And we get this charge drawn out of the capacitor once per switching period. So it represents an average current, That is the charge divided by the switching period. We get that much average current flowing out of the capacitor as a result of this. And this then represents a power loss, or at least a power flowing out of the capacitor of this current multiplied by the output voltage. So, Q over Ts times the output voltage, and this average power essentially is switching loss. It's the switching loss induced by the reverse recovery of the body diode, primarily, as well as some power loss induced by charging or discharging the output capacitances of the MOSFETs. A little bit of that capacitive power actually is not lost. But this primarily is from the reverse recovery. So on the homework you will work out or estimate this power loss predicted by LTspice due to the diode reverse recovery. And compare that to the total loss in your circuit. So the homework assignment is to start with a basic buck converter. So this circuit is given on the course website. It has a main FET and a diode with the output filter and the output load. Again, in this microprocessor we want to supply 1 V at 20 A. So we'll have to adjust the duty cycle so that we get 1 V with high accuracy. And you can measure the efficiency of this circuit, including both the input power, the gate driver power, and the output power in your calculation. And then as we just did in this boost example, what we'll want to do is in the buck, replace this diode with a MOSFET as a synchronous rectifier like this, along with its gate driver. And we'll want to add a bootstrap power supply to the upper driver. [COUGH] And then work out the efficiency in this case. And look at things like the current spike from the reverse recovery. There's also some discussion questions at the end of the assignment which are optional. But if you wish, you may want to further investigate how to improve the efficiency of this circuit. If you look here, we have two different MOSFETs. In the power electronics field, we figured out that synchronous rectifier power supplies like this can be better optimized if the MOSFETs are different. So you might consider how this was optimized. And in fact, look at whether you can find other MOSFETs in the LTspice library that would further optimize it. We can also optimize the dead time and the gate driver resistors to get some improvements and efficiency as well. And it's worth looking at that and trying it in LTspice if you have the time.