JAJSA67I April 2004 – February 2019 LM2743
PRODUCTION DATA.
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:
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 mΩ.
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:
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.