IOR(max) =

80 PH x 154 kHz x 60V

5V x (60V - 5V)

= 372 mAp-p

IOR(max) =

L1min x FS(min) x VIN(max)

VOUT x (VIN(max) - VOUT)

L1 =

0.40A x 154 kHz x 60V

5V x (60V - 5V)

= 74.4 PH

L1 =

IOR x FS(min) x VIN(max)

VOUT x (VIN(max) - VOUT)

0 mA

IOR

1/Fs

IPK+

IO

IPK-

16

LM5010A, LM5010A-Q1

SNVS376F – OCTOBER 2005 – REVISED MAY 2016

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Figure 12. Inductor Current

To keep the circuit in continuous conduction mode, the maximum allowed ripple current is twice the minimum

load current, or 400 mAp-p. Using this value of ripple current, the inductor (L1) is calculated using Equation 12

and Equation 13.

where

•

FS(min) is the minimum frequency of 154 kHz (205 kHz - 25%) at VIN(max)

(12)

(13)

Equation 13 provides the minimum value for inductor L1. When selecting an inductor, use a higher standard

value (100 uH). To prevent saturation, and possible destructive current levels, L1 must be rated for the peak

current which occurs if the current limit and maximum ripple current are reached simultaneously (IPK in Figure 9).

The maximum ripple amplitude is calculated by rearranging Equation 12 using VIN(max), FS(min), and the minimum

inductor value, based on the manufacturer’s tolerance. Assume for Equation 14, Equation 15, and Equation 16

that the inductor’s tolerance is ±20%.

(14)

(15)

IPK = ILIM + IOR(max) = 1.5 A + 0.372 A = 1.872 A

where

•

ILIM is the maximum current limit threshold

(16)

At the nominal maximum load current of 1 A, the peak inductor current is 1.186 A.

8.2.2.1.4

RCL

Since it is obvious that the lower peak of the inductor current waveform does not exceed 1 A at maximum load

current (see Figure 12), it is not necessary to increase the current limit threshold. Therefore RCL is not needed for

this exercise. For applications where the lower peak exceeds 1 A, see Increasing The Current Limit Threshold.