Figure 15 is a 15V to 5V buck regulator with an output cur-

rent, I

20% of I

ciency of 75%, a load regulation of 30 mV (70 mA to 350 mA)

and a line regulation of 10 mV (12

≤ V

Component values are selected as follows:

R1 = (V

R3 = V/I

sw(max)

R3 = 0.15

Ω

where:

V is the current limit sense voltage, 0.11V

I

transistor.

L1 is the inductor and may be found from the inductance cal-

culation chart (

Figure 16) as follows:

Given

V

V

I

f

Discontinuous at 20% of I

Note that since the circuit will become discontinuous at 20%

of I

70 mA.

Step 1: Calculate the maximum DC current through the in-

ductor, I

top of the chart and show that I

configuration. Thus, I

Step 2: Calculate the inductor Volts-sec product, E-T

cording to the equations given from the chart. For the Buck:

E-T

=(15 − 5) (5/15) (1000/50)

= 66V-µs.

with the oscillator frequency, f

Step 3: Using the graph with axis labeled “Discontinuous At

% I

maximum inductor current, I

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

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.

DS008711-5

FIGURE 14. Basic Buck Regulator

DS008711-6

R3 = 0.15

Ω

C1 = 1820 pF

C2 = 220 µF

C3 = 20 pF

L1 = 470 µH

R1 = 40 k

Ω

D1 = 1N5818

R2 = 10 k

Ω

FIGURE 15. Buck or Step-Down Regulator

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