ADP1073

–6–

REV. 0

COMPONENT SELECTION

General Notes on Inductor Selection

When the ADP1073 internal power switch turns on, current

begins to flow in the inductor. Energy is stored in the inductor

core while the switch is on, and this stored energy is then trans-

ferred to the load when the switch turns off. Both the collector

and the emitter of the switch transistor are accessible on the

ADP1073, so the output voltage can be higher, lower or of

opposite polarity than the input voltage.

To specify an inductor for the ADP1073, the proper values of

inductance, saturation current and dc resistance must be deter-

mined. This process is not difficult, and specific equations for

each circuit configuration are provided in this data sheet.

In general terms, however, the inductance value must be low

enough to store the required amount of energy (when both

input voltage and switch ON time are at a minimum) but high

switch ON time are at their maximum values. The inductor

must also store enough energy to supply the load without satu-

rating. Finally, the dc resistance of the inductor should be low

so that excessive power will not be wasted by heating the

windings. For most ADP1073 applications, an 82

µH to

1000

µH inductor with a saturation current rating of 300 mA to

1 A is suitable. Ferrite core inductors that meet these specifica-

tions are available in small, surface-mount packages.

To minimize Electro-Magnetic Interference (EMI), a toroid or

pot core type inductor is recommended. Rod core inductors are

a lower cost alternative if EMI is not a problem.

Calculating the Inductor Value

Selecting the proper inductor value is a simple three-step process:

1. Define the operating parameters: minimum input voltage,

maximum input voltage, output voltage and output current.

2. Select the appropriate conversion topology (step-up, step-

down or inverting).

3. Calculate the inductor value, using the equations in the

following sections.

Inductor Selection—Step-Up Converter

In a step-up, or boost, converter (Figure 15), the inductor must

store enough power to make up the difference between the

input voltage and the output voltage. The power that must be

stored is calculated from the equation:

OUT

()

(1)

Schottky). Energy is only stored in the inductor while the

ADP1073 switch is ON, so the energy stored in the inductor on

each switching cycle must be must be equal to or greater than:

(2)

in order for the ADP1073 to regulate the output voltage.

When the internal power switch turns ON, current flow in the

inductor increases at the rate of:

R

′

1– e

–R

′t

L









(3)

where L is in henrys and R' is the sum of the switch equivalent

resistance (typically 0.8

Ω at +25°C) and the dc resistance of

the inductor. If the voltage drop across the switch is small com-

L

t

(4)

Replacing t in the above equation with the ON time of the ADP1073

(38

µs, typical) will define the peak current for a given inductor

value and input voltage. At this point, the inductor energy can

be calculated as follows:

1

2

L

(5)

ADP1073 can deliver the necessary power to the load. For best

efficiency, peak current should be limited to 1 A or less. Higher

switch currents will reduce efficiency because of increased satu-

ration voltage in the switch. High peak current also increases

output ripple. As a general rule, keep peak current as low as

possible to minimize losses in the switch, inductor and diode.

In practice, the inductor value is easily selected using the equa-

tions above. For example, consider a supply that will generate

5 V at 25 mA from two alkaline batteries with a 2 V end-of-life

voltage. The inductor power required is, from Equation 1:

On each switching cycle, the inductor must supply:

=

87.5 mW

19 kHz

Since the inductor power is low, the peak current can also be

low. Assuming a peak current of 100 mA as a starting point,

Equation 4 can be rearranged to recommend an inductor value:

L

=

t

=

2V

100 mA

38

µs = 760 µH

Substituting a standard inductor value of 470

µH, with 1.2 Ω dc

resistance, will produce a peak switch current of:

2 V

2.0

Ω

1– e

–2.0

Ω× 38 µs

470

µH











 =149 mA

Once the peak current is known, the inductor energy can be

calculated from Equation 5:

1

2

(470

The inductor energy of 5.2

µJ is greater than the P

quirement of 4.6

µJ, so the 470 µH inductor will work in this

application. The optimum inductor value can be determined

by substituting other inductor values into the same equations.

When selecting an inductor, the peak current must not exceed

the maximum switch current of 1.5 A.

The peak current must be evaluated for both minimum and

maximum values of input voltage. If the switch current is high