ADP1173

–6–

REV. 0

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. In most applications, where the voltage drop across

used:

L

t

(4)

Replacing “t” in the above equation with the ON time of the

ADP1173 (23

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

(5)

ADP1173 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

saturation voltage in the switch. High peak current also increases

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

sible 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

9 V at 50 mA from a 3 V source. The inductor power required

is, from Equation 1:

On each switching cycle, the inductor must supply:

=

325 mW

24 kHz

=13.5µJ

The required inductor power is fairly low in this example, so the

peak current can also be low. Assuming a peak current of 500 mA

as a starting point, Equation 4 can be rearranged to recommend

an inductor value:

L

=

t

=

3V

500 mA

23

µs =138 µH

Substituting a standard inductor value of 100

µH, with 0.2 Ω dc

resistance, will produce a peak switch current of:

3V

1.0

Ω

1– e

–1.0

Ω× 23 µs

100

µH











 = 616 mA

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

calculated from Equation 5:

1

2

(100

The inductor energy of 19

µJ is greater than the P

quirement of 13.5

µJ, so the 100 µH inductor will work in this

application. By substituting other inductor values into the same

equations, the optimum inductor value can be selected.

When selecting an inductor, the peak current must not exceed

the maximum switch current of 1.5 A. If the equations shown

above result in peak currents > 1.5 A, the ADP1073 should be

considered. This device has a 72% duty cycle, so more energy is

stored in the inductor on each cycle. This results in greater

output power.

The peak current must be evaluated for both minimum and

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

current limit feature can be used to limit switch current. Simply

select a resistor (using Figure 4) that will limit the maximum

section of this data sheet for more information.

Note that the switch current limit feature does not protect the

circuit if the output is shorted to ground. In this case, current is

only limited by the dc resistance of the inductor and the forward

voltage of the diode.

Inductor Selection—Step-Down Converter

The step-down mode of operation is shown in Figure 15. Unlike

the step-up mode, the ADP1173’s power switch does not

saturate when operating in the step-down mode. Therefore,

switch current should be limited to 650 mA in this mode. If the

used to limit the maximum switch current. If higher output

current is required, the ADP1111 should be considered.

The first step in selecting the step-down inductor is to calculate

the peak switch current as follows:

DC









(6)

where DC = duty cycle (0.55 for the ADP1173)

As previously mentioned, the switch voltage is higher in step-

The inductor value can now be calculated:

L

=

× t

ON

(7)

If the input voltage will vary (such as an application that must

rent constant as the input voltage rises. Note that there are separate