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.