Typical Applications (Continued)
Step 1: Calculate the maximum DC current through the
inductor, I
L(max)
. The necessary equations are indicated at
the top of the chart and show that I
L(max)
=I
o(max)
for the
buck configuration. Thus, I
L(max)
= 350 mA.
Step 2: Calculate the inductor Volts-sec product, E-T
op
,
according to the equations given from the chart. For the
Buck:
E-T
op
=(V
in
−V
o
)(V
o
/V
in
) (1000/f
osc
)
=(15 − 5) (5/15) (1000/50)
= 66V-µs.
with the oscillator frequency, f
osc
, expressed in kHz.
Step 3: Using the graph with axis labeled “Discontinuous At
%I
OUT
” and “I
L(max, DC)
” find the point where the desired
maximum inductor current, I
L(max, DC)
intercepts the desired
discontinuity percentage.
In this example, the point of interest is where the 0.35A line
intersects with the 20% line. This is nearly the midpoint of the
horizontal axis.
Step 4: This last step is merely the translation of the point
found in Step 3 to the graph directly below it. This is accom-
plished by moving straight down the page to the point which
intercepts the desired E-T
op
. For this example, E-T
op
is
66V-µs and the desired inductor value is 470 µH. Since this
example was for 20% discontinuity, the bottom chart could
have been used directly, as noted in step 3 of the chart
instructions.
For a full line of standard inductor values, contact Pulse
Engineering (San Diego, Calif.) regarding their PE526XX
series, 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
current 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 LM2578A output transis-
tor.
V is the current limit sense voltage.
B
f
is the forced current gain of transistor Q1 (B
f
= 30 for
Figure 17 ).
I
R4
=V
BE1
/R4
I
p
=I
L(max, DC)
+ 0.5∆I
L
LM2578A/LM3578A
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