ADP3430 datasheet(17/26 Pages) ONSEMI

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ADP3430

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Designing an Inductor

Once the inductance and DCR are known, the next step is

to either design an inductor or to find a standard inductor that

comes as close as possible to meeting the overall design

goals. It is also important to have the inductance and DCR

tolerance specified to control the accuracy of the system.

Reasonable tolerances most manufacturers can meet are

15% inductance and 7% DCR at room temperature. The first

decision in designing the inductor is choosing the

core material. Several possibilities for providing low core

loss at high frequencies include the powder cores (from

Micrometals, Inc., for example, or Kool Mu

® from

Magnetics

®) and the gapped soft ferrite cores (for example,

3F3 or 3F4 from Philips

®). Low frequency powdered iron

cores should be avoided due to their high core loss,

especially when the inductor value is relatively low and the

ripple current is high.

The best choice for a core geometry is a closed−loop type

such as a potentiometer core (PQ, U, or E core) or toroid. A

good compromise between price and performance is a core

with a toroidal shape.

Many useful magnetics design references are available for

quickly designing a power inductor, such as:

• Intusoft Magnetic Designer Software

• Designing Magnetic Components for High Frequency

DC to DC Converters, by William T. McLyman,

Kg Magnetics, Inc., ISBN 1883107008

Selecting a Standard Inductor

The following power inductor manufacturers can provide

design consultation and deliver power inductors optimized

for high power applications upon request.

• Coilcraft®

• Coiltronics®

• Sumida Corporation®

Current Sense Amplifier

The output current is measured by summing the voltage

across each inductor and passing the signal through a

low−pass filter. This summer filter is the CS amplifier

configured with resistors RPH(X) (summers), and RCS and

CCS (filter). The impedance gain of the regulator is set by the

following equations, where RL is the DCR of the output

inductors:

(eq. 6)

CCS +

RL RCS

The user has the flexibility to choose either RCS or RPH(X).

However, it is best to select RCS equal to 110 kW.

In this example, RPH(X) is selected as 61.9 kW, DCR = 0.57

W, use Equation 6 to solve for CCS.

(eq. 7)

CCS +

450 nH

0.57 mW

110 kW +

7.2 nF

It is best to have a dual location for CCS in the layout so

that standard values can be used in parallel to get as close to

the desired value. For best accuracy, CCS should be a 5% or

10% NPO capacitor. This example uses a 5% combination

for CCS of one 3.3 nF capacitor and one 3.9 nF capacitor in

parallel.

Inductor DCR Temperature Correction

When the inductor DCR is used as the sense element and

copper wire is used as the source of the DCR, the user needs

to compensate for temperature changes of the inductor’s

winding. Fortunately, copper has a well known temperature

coefficient (TC) of 0.39%/

°C.

If RCS is designed to have an opposite and equal

percentage change in resistance to that of the wire, it cancels

the temperature variation of the inductor DCR. Due to the

non−linear nature of NTC thermistors, Resistor RCS1 and

Resistor RCS2 are needed. See Figure 8 to linearize the NTC

and produce the desired temperature tracking.

Figure 8. Temperature Compensation Circuit Values

CSREF

CSSUM

CSCOMP

CCS1

CCS2

RCS1 RCS2

RPH1

RTH

RPH2 RPH3

PLACE AS CLOSE AS POSSIBLE

TO NEAREST INDUCTOR

OR LOW−SIDE MOSFET

KEEP THIS PATH

AS SHORT AS

POSSIBLE AND

WELL AWAY FROM

SWITCH NODE LINES

ADP3430

PH4

The following procedure and equations yield values to use

for RCS1, RCS2, and RTH (the thermistor value at 25°C) for

a given RCS value.

1. Select an NTC based on type and value. Because

the value is unknown, use a thermistor with a

value close to RCS. The NTC should also have an

initial tolerance of better than 5%.

2. Based on the type of NTC, find its relative

resistance value at two temperatures. The

temperatures that work well are 50

°C and 90°C.

These resistance values are called

A (RTH(50°C))/RTH(25°C)) and

B (RTH(90°C))/RTH(25°C)). The relative value of the

NTC is always 1 at 25

°C.

3. Find the relative value of RCS required for each of

these temperatures. This is based on the percentage

change needed, which in this example is initially

0.39%/

°C. These temperatures are called

r1 (1/(1 + TC × (T1 − 25°C)))

and

r2 (1/(1 + TC × (T2 − 25°C))), where TC = 0.0039

for copper, T1 = 50°C, and T2 = 90°C. From this,

r1 = 0.9112 and r2 = 0.7978.

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