Step 2: Calculate the inductor Volts-sec product, E-Top, ac-
cording to the equations given from the chart. For the Buck:
Typical Applications (Continued)
=
E-Top (Vin − Vo) (Vo/Vin) (1000/fosc
)
=
=
(15 − 5) (5/15) (1000/50)
66V-µs.
with the oscillator frequency, fosc, expressed in kHz.
DS008711-5
FIGURE 14. Basic Buck Regulator
Figure 15 is a 15V to 5V buck regulator with an output cur-
rent, Io, of 350 mA. The circuit becomes discontinuous at
20% of Io(max), has 10 mV of output voltage ripple, an effi-
ciency of 75%, a load regulation of 30 mV (70 mA to 350 mA)
and a line regulation of 10 mV (12 ≤ Vin ≤ 18V).
Component values are selected as follows:
DS008711-6
=
=
R1 (Vo − 1) x R2 where R2 10 kΩ
=
=
=
R3 0.15Ω
R3 V/Isw(max)
Vin 15V
=
=
=
C1 1820 pF
Vo 5V
R3 0.15Ω
=
=
C2 220 µF
Vripple 10 mV
where:
=
=
C3 20 pF
Io 350 mA
V is the current limit sense voltage, 0.11V
=
=
L1 470 µH
fosc 50 kHz
Isw(max) is the maximum allowable current thru the output
transistor.
=
=
D1 1N5818
R1 40 kΩ
=
R2 10 kΩ
L1 is the inductor and may be found from the inductance cal-
culation chart (Figure 16) as follows:
FIGURE 15. Buck or Step-Down Regulator
=
Given Vin 15V
Step 3: Using the graph with axis labeled “Discontinuous At
% IOUT” and “IL(max, DC)” find the point where the desired
maximum inductor current, IL(max, DC) intercepts the desired
discontinuity percentage.
=
Vo 5V
=
Io(max) 350 mA
=
fOSC 50 kHz
Discontinuous at 20% of Io(max)
.
In this example, the point of interest is where the 0.35A line
Note that since the circuit will become discontinuous at 20%
of Io(max), the load current must not be allowed to fall below
70 mA.
%
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-Top. For this example, E-Top 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.
Step 1: Calculate the maximum DC current through the in-
ductor, IL(max). The necessary equations are indicated at the
=
top of the chart and show that IL(max) Io(max) for the buck
=
configuration. Thus, IL(max) 350 mA.
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