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37DC-DC Converter Design and Magnetics
synonymous. Therefore, in general, we just refer to all of them as simply the peak curr
ent
I
PK
where
III
PK DC AC

(3-4)
The peak current is in fact the most critical current component of all, because it is not just
a source of long-term heat buildup and consequent temperature rise, but a potential cause
of immediate destruction of the switch. We will show later that the inductor current is
instantaneously proportional to the magnetic fi eld inside the core. So at the exact moment
when the current reaches its peak value, so does this fi eld. We also know that real-world
inductors can saturate (start losing their inductance) if the fi eld inside them exceeds a
certain safe level, that value being dependent on the actual material used for the core
(not on the geometry, or number of turns or even the air-gap, for example). Once saturation
occurs, we may get an almost uncontrolled surge of current passing through the switch
because the ability to limit current (which is one of the reasons the inductor is used in
switching power supplies in the fi rst place), depends on the inductor behaving like one.
Therefore, losing inductance is certainly not going to help! In fact, we usually cannot
afford to allow the inductor to saturate even momentarily . And for this reason, we need to
monitor the peak current closely (usually on a cycle-by-cycle basis). As indicated, the peak
is the likeliest point of the inductor current waveform where saturation can start to occur.
Note: A slight amount of core saturation may turn out to be acceptable on
occasion, especially if it occurs only under temporary conditions, like power-up, for
example. This will be discussed in more detail later.
3.4 Understanding the AC, DC and Peak Currents
We have seen that the AC component ( I
AC
= I /2) is derivable from the voltseconds law.
From the basic inductor equation V L dI/dt , we get
2II
AC
voltseconds
inductance
(3-5)
So the current swing I
PP
I can be intuitively visualized as voltseconds per unit
inductance . If the applied voltseconds doubles, so does the current swing (and AC
component). And if the inductance doubles, the swing (and AC component) is halved.
Let us now consider the DC level again. Note that any capacitor has zero average (DC)
current through it in steady-state , so all capacitors can be considered to be missing
altogether when calculating DC current distributions. Therefore, for a buck, since energy
38
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Chapter 3
ows into the output during both the on-time and off-time, and via the inductor, the
average inductor current must always be equal to the load current. So
II
LO
(buck) (3-6)
On the other hand, in both the boost and the buck-boost, energy fl ows into the output only
during the off-time, and via the diode. Therefore, in this case, the averag e diode current
must be equal to the load current. Note that the diode current has an average value equal
to I
L
when it is conducting (see the dashed line passing through the center of the down-
ramp in the upper half of Figure 3.3 ). If we calculate the average of this diode current
over the entire switching cycle, we need to weight it by its duty cycle, that is, 1 D .
Therefore, calling I
D
the average diode current, we get
II DI
DL O
()1
(3-7)
solving
I
I
D
L
O
1
(3-8)
I
O
I
O
Average diode current
is I
L
(1 D) I
O
Average inductor
current is I
L
I
O
I
L
I
O
/(1 D)
I
L
I
O
At V
INMAX
, the AC component
decreases, DC component
decreases even more, and so
peak current decreases
V
INMAX
corresponds to D
MIN
V
INMIN
corresponds to D
MAX
V
INMAX
corresponds to D
MIN
V
INMIN
corresponds to D
MAX
At V
INMAX
, the AC component
increases, DC component remains
the same, and so peak current
increases
Buck-Boost
Buck
V
INMIN
V
INMAX
Current
Current
Time
Time
Figure 3.3 : Visualizing the AC and DC components of the
inductor current as input voltage varies

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