LM5010ASD datasheet(12/25 Pages) TI

12 / 25 page

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+

IPK-

RON =

5V x (8V - 1.4V)

8V x 175 kHz x 1.18 x 10

-10 - 1.4 k: = 198 k:

LM5010A

LM5010A-Q1

SNVS376E – OCTOBER 2005 – REVISED FEBRUARY 2013

www.ti.com

APPLICATIONS INFORMATION

EXTERNAL COMPONENTS

The procedure for calculating the external components is illustrated with a design example. Referring to the

Block Diagram, the circuit is to be configured for the following specifications:

•

VOUT = 5V

•

VIN = 6V to 60V

•

FS = 175 kHz

•

Minimum load current = 200 mA

•

Maximum load current = 1.0A

•

Softstart time = 5 ms

R1 and R2: These resistors set the output voltage, and their ratio is calculated from:

R1/R2 = (VOUT/2.5V) - 1

(10)

R1/R2 calculates to 1.0. The resistors should be chosen from standard value resistors in the range of 1.0 k

Ω - 10

Ω. A value of 1.0 kΩ will be used for R1 and for R2.

RON, FS: RON can be chosen using Equation 7 to set the nominal frequency, or from Equation 6 if the on-time at

a particular VIN is important. A higher frequency generally means a smaller inductor and capacitors (value, size

and cost), but higher switching losses. A lower frequency means a higher efficiency, but with larger components.

Generally, if PC board space is tight, a higher frequency is better. The resulting on-time and frequency have a

±25% tolerance. Using Equation 7 at a nominal VIN of 8V,

(11)

A value of 200 k

Ω will be used for RON, yielding a nominal frequency of 161 kHz at VIN = 6V, and 205 kHz at VIN

= 60V.

L1: The guideline for choosing the inductor value in this example is that it must keep the circuit’s operation in

continuous conduction mode at minimum load current. This is not a strict requirement since the LM5010A

regulates correctly when in discontinuous conduction mode, although at a lower frequency. However, to provide

an initial value for L1 the above guideline will be used.

Figure 11. 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 the following:

(12)

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

(13)

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