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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

2II

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

ﬂ

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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