To provide another example of modeling switching loss, let's drive the equivalent circuit model of a boost
converter including the diode reverse recovery. So here's a boost converter.
And we're going to, assume that all of the components are
ideal, except for the diode and it's reverse
recovery. And we'll, derive the averaged, or DC
equivalent circuit model with terms to, incorporate the
affects of reverse recovery and its loss. So here's the blues converter and the
transistor and diode waveforms, they look
qualitatively the same as the waveforms we drew in the last lecture for the butt converter with
reverse recovery. The only real difference is that the
values are a little different so the voltages
across the, the semi conductors, when they're off, is
equal to the output voltage V rather than the input
voltage. Just as in the buck converter, we need to correctly define the duty cycle to
obtain the effective duty cycle that the power
converter operates with, and we define this using the voltage wave
form. So the voltage at this node, the switching
node, or the voltage across the MOSFET is this blue line shown here, and the
effective duty cycle is judged from the MOSFET
voltage is the time when the MOSFET voltage is low. So we will define this, this is the
interval DTs. And, D prime Ts is the remainder of the
interval, including the diode switching time.
So, we have these wave forms with the reverse recovery peak, peak in current and
the and the reverse recovery time T sub r. So we the first step, then, given these
transistor and diode waveforms, is to relate the inductor voltage in capacitor
current waveforms to these waveforms. So, I'll define the inductor voltage here
and you can see from writing the equation in this loop the inductor voltage, V sub L
is equal to the input voltage Vg minus the Transistor Voltage V sub t. likewise for capacitor current, we can
write the node equation here. And the capacitor current is equal to, for
this converter, the diode current, i sub d, minus the load current, which is
V over R, okay. So, we can find the average values of these quantities and apply volt second
balance and charge balance, accounting for the effect of the
reverse recovery of the the diode on these wave
forms. one other small point we in general need to write the equation of the input current
Ig. But for the boost converter IG is the same
as IL. So, if we find the average IL or DC
component of inductor current, that's the same as Ig and we don't have to
do any further work. Okay, so let's do in, inductor volt's
second balance first. So, VL is Vg minus Vt, and Vt is the blue
waveform, right here. So, on the next slide, that waveform is
drawn again. And in fact, here the waveform looks the
same as in the ideal case. the reverse recovery time doesn't change
this waveform, assuming we correctly define the
duty cycle, as I previously mentioned.
So, we can apply volts second balance. The average inductor voltage is the duty
cycle times Vg for the first interval plus D prime times this
Vg minus V. And we get the same equation as usual for
the inductor volts second balance. And we can even construct an equivalent
circuit model from this. what would we get? Vg would be the input voltage source. It has a current capital Ig, and this is,
this voltage is equal to minus D prime V, so we have a dependent source that depends on the
output voltage V, multiplied by D prime. Okay, so here's our loop equation
corresponding to the inductor equation. And again, Ig is the same as the inductor current capital IL, which is the current
in this loop. Okay, for the capacitor we apply charge
balance to the capacitor and basic, basically the
easy way to do this is to write the node equation at the node where the capacitor
is connected Which says that the diode current is equal to the capacitor current, plus
the load current. So one way to apply charge balance is to
average these currents. So, we can say that the DC component of
diode current equals the DC component of capacitor current plus
the DC component of load current. Now, DC component of capacitor current is
zero in steady state by charge balance, which says that the DC
component of the diode current is equal to the DC load current. So, we've drawn the diode current waveform
already for the diode. We simply find its average and equate that
to the load current, and that effectively gives us the
equation from capacitor charge balance. So, here is that equation repeated. And to find the average diode current we
need to average this current. So the average diode current is 1 over Ts times the integral of Id of t over the
period, so we get 1 over Ts times the area under
the curve. Okay, so what is the area under the curve?
Well first we need the area of this rectangle and this rectangle has a height
of IL. And it has a length or base that is this
tongue, this time, and that time is in fact what we normally call D prime,
but then minus the reverse recovery time. So this length here is D prime Ts minus
the reverse recovery time t sub r. And then to that we also have to add the, recovered charge QR, and in this case it's
a negative quantity. We usually have data sheets express QR as a positive number so we should subtract
QR. Okay, we can now divide through by the TS
and what we get is D prime IL minus ILtr over TS minus QR over TS. Which is the value the expression shown
here. Okay, let's construct an equivalent
circuit, then, to go with that. So, this is the node equation at the
output node of the capacitor, and it says that the load current, which
is the current coming out of the node. And is the well we have a voltage, V
across the resistor R, that is, that is our load
resistor. That is equal to the sum of these three
terms. the first term is D prime IL.
which is the usual dependent source. That will become part of a DC transformer.
The second term, is IL times tr over TS and it has a minus sign.
So it's drawn out of the node, and I'll draw it as a independent source.
It is a loss element. A value IL times tr over Ts.
And then the last term is minus Qr over Ts which is also a loss term. So we for the boost converter we get two extra loss terms from the diode
reverse recovery. And in the case of this converter. These currents are actually at the output
node where the capacitor is, in fact the reverse recovery
causes current to be drawn out of the output capacitor as if
it's loading the output, so switching loss effectively is
another load on the output, but it's a loss element. If we take the, the two models together,
for the inductor volt second balance and the
capacitor charge balance, and combine them, including
combining the dependent sources into the usual transformer, we get this
model now. So we have the ideal boost transformer
that is D prime to one and then we have our two independent current sources or
sinks actually that model the loss caused by the reverse recovery.
And these are actual power losses. The voltage across these sources is the
output voltage V. And they cause a power loss equal to that
voltage V times the sum of the two currents like
this. here's a sample prediction of this model
this sample here actually includes an inductor
resistance and so inductor resistance, if we included it, would
become an additional resistor in the loop part of the inductor
is. And so, with that resistor included,
here's a, a solution V over Vg for this boost converter, with
and without the reverse recovery. And so, the effect of the diode reverse
recovery is to actually appear to load down the
output. which reduces the output voltage
especially at high duty cycle. Okay so to summarize we can model or
incorporate switching loss into our averages model, what we have to
do is draw the actual transistor and diode voltage and current wave forms.
And then relate those waveforms to the inductor voltage and capacitor current
waveforms, and perhaps also to Ig. We then apply volt second balance and
charge balance as usual, and proceed to compute the equations of the DC conditions
in the circuit including the effects of this
diode reverse recovery. And we can then reconstruct the equivalent
circuit by a simple extension of, of how we did it in chapter three to get equivalent circuit models that
incorporate switching loss [BLANK_AUDIO]