ESRMAX =

'VOUT

'IOUT

VIN(MAX) - VO

'IOUT =

FSW x LACTUAL

x D

L =

3.3V - 1.2V

0.4 x 4A x 300 kHz

x

1.2V

3.3V

L =

VIN - VOUT

'IOUT x fSW

x D

21

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Typical Applications (continued)

where n is the number of capacitors, and ESR is the equivalent series resistance of each capacitor. The equation

above indicates that power loss in each capacitor decreases rapidly as the number of input capacitors increases.

The worst-case ripple for a Buck converter occurs during full load and when the duty cycle (D) is 0.5. For this

3.3V to 1.2V design the duty cycle is 0.364. For a 4A maximum load the ripple current is 1.92A.

8.2.1.2.4

Output Inductor

The output inductor forms the first half of the power stage in a Buck converter. It is responsible for smoothing the

square wave created by the switching action and for controlling the output current ripple (ΔIOUT). The inductance

is chosen by selecting between tradeoffs in efficiency and response time. The smaller the output inductor, the

more quickly the converter can respond to transients in the load current. However, as shown in the efficiency

calculations, a smaller inductor requires a higher switching frequency to maintain the same level of output current

ripple. An increase in frequency can mean increasing loss in the MOSFETs due to the charging and discharging

of the gates. Generally the switching frequency is chosen so that conduction loss outweighs switching loss. The

equation for output inductor selection is:

(21)

(22)

L = 1.6 µH

(23)

Here we have plugged in the values for output current ripple, input voltage, output voltage, switching frequency,

and assumed a 40% peak-to-peak output current ripple. This yields an inductance of 1.6 µH. The output inductor

must be rated to handle the peak current (also equal to the peak switch current), which is (IOUT + (0.5 x ΔIOUT)) =

4.8 A, for a 4 A design. The Coilcraft DO3316P-222P is 2.2 µH, is rated to 7.4-A peak, and has a direct current

resistance (DCR) of 12 m

Ω.

After selecting an output inductor, inductor current ripple should be re-calculated with the new inductance value,

as this information is needed to select the output capacitor. Re-arranging the equation used to select inductance

yields the following:

(24)

VIN(MAX) is assumed to be 10% above the steady state input voltage, or 3.6V. The actual current ripple will then

be 1.2A. Peak inductor/switch current will be 4.6A.

8.2.1.2.5

Output Capacitor

The output capacitor forms the second half of the power stage of a Buck switching converter. It is used to control

the output voltage ripple (ΔVOUT) and to supply load current during fast load transients.

In this example the output current is 4 A and the expected type of capacitor is an aluminum electrolytic, as with

the input capacitors. Other possibilities include ceramic, tantalum, and solid electrolyte capacitors, however the

ceramic type often do not have the large capacitance needed to supply current for load transients, and tantalums

tend to be more expensive than aluminum electrolytic. Aluminum capacitors tend to have very high capacitance

and fairly low ESR, meaning that the ESR zero, which affects system stability, will be much lower than the

switching frequency. The large capacitance means that at the switching frequency, the ESR is dominant, hence

the type and number of output capacitors is selected on the basis of ESR. One simple formula to find the

maximum ESR based on the desired output voltage ripple, ΔVOUT and the designed output current ripple, ΔIOUT,

is:

(25)