Typical Applications
(Continued)
For a full line of standard inductor values, contact Pulse En-
gineering (San Diego, Calif.) regarding their PE526XX se-
ries, or A. I. E. Magnetics (Nashville, Tenn.).
A more precise inductance value may be calculated for the
Buck, Boost and Inverting Regulators as follows:
BUCK
L = V
o
(V
in
− V
o
)/(∆I
L
V
in
f
osc
)
BOOST
L = V
in
(V
o
− V
in
)/(∆I
L
f
osc
V
o
)
INVERT
L = V
in
|V
o
|/[∆I
L
(V
in
+ |V
o
|)f
osc
]
where
∆I
L
is the current ripple through the inductor.
∆I
L
is
usually chosen based on the minimum load current expected
of the circuit. For the buck regulator, since the inductor cur-
rent I
L
equals the load current I
O
,
∆I
L
= 2
•
I
O(min)
∆I
L
= 140 mA for this circuit.
∆I
L
can also be interpreted as
∆I
L
= 2
•
(Discontinuity Factor)
•
I
L
where the Discontinuity Factor is the ratio of the minimum
load current to the maximum load current. For this example,
the Discontinuity Factor is 0.2.
The remainder of the components of
Figure 15
are chosen
as follows:
C1 is the timing capacitor found in
Figure 1.
C2
≥
V
o
(V
in
− V
o
)/(8f
osc 2
V
in
V
ripple
L1)
where V
ripple
is the peak-to-peak output voltage ripple.
C3 is necessary for continuous operation and is generally in
the 10 pF to 30 pF range.
D1 should be a Schottky type diode, such as the 1N5818 or
1N5819.
BUCK WITH BOOSTED OUTPUT CURRENT
For applications requiring a large output current, an external
transistor may be used as shown in
Figure 17.
This circuit
steps a 15V supply down to 5V with 1.5A of output current.
The output ripple is 50 mV, with an efficiency of 80%, a load
regulation of 40 mV (150 mA to 1.5A), and a line regulation
of 20 mV (12V
≤
V
in
≤
18V).
Component values are selected as outlined for the buck
regulator with a discontinuity factor of 10%, with the addition
of R4 and R5:
R4 = 10V
BE1
B
f
/I
p
R5 = (V
in
− V − V
BE1
− V
sat
) B
f
/(I
L(max, DC)
+ I
R4
)
where:
V
BE1
is the V
BE
of transistor Q1.
V
sat
is the saturation voltage of the LM1578A output transis-
tor.
V is the current limit sense voltage.
B
f
is the forced current gain of transistor Q1 (B
f
= 30 for
Fig-
ure 17
).
I
R4
= V
BE1
/R4
I
p
= I
L(max, DC)
+ 0.5∆I
L
DS008711-8
V
in
= 15V
V
o
= 5V
V
ripple
= 50 mV
I
o
= 1.5A
f
osc
= 50 kHz
R1 = 40 kΩ
R2 = 10 kΩ
R3 = 0.05Ω
R4 = 200Ω
R5 = 330Ω
C1 = 1820 pF
C2 = 330 µF
C3 = 20 pF
L1 = 220 µH
D1 = 1N5819
Q1 = D45
FIGURE 17. Buck Converter with Boosted Output Current
BOOST REGULATOR
The boost regulator converts a low input voltage into a
higher output voltage. The basic configuration is shown in
Figure 18.
Energy is stored in the inductor while the transis-
tor is on and then transferred with the input voltage to the
output capacitor for filtering when the transistor is off. Thus,
V
o
= V
in
+ V
in
(t
on
/t
off
).
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