SNVSBZ3 June 2021 LM5168-Q1
For this fly-buck application, a coupled inductor (sometimes called a transformer) is required. The first step is to decide upon the turns ratio. In a fly-buck, the secondary output voltage is slightly less than the reflected primary output voltage scaled by the turns ratio. Equation 17 can be used to calculate the turns ratio for a given VOUT1 and VOUT2. The nearest integer ratio should be selected. VOUT2 will be slightly less than calculated due to the secondary diode drop and other parasitic voltage drops in the secondary. Also, keep in mind that the secondary voltage is not fed back to the controller, and is, therefore, not well regulated. For this example, it is required that VOUT2 is equal to VOUT1, therefore, use a 1:1 turns ratio.
Next, the primary inductance must be calculated. This is the same as calculating the inductance for an ordinary buck regulator, and is based on the desired primary ripple current. Typically, a ripple current of between 20% and 40% of the primary current is used. Equation 18 gives the primary current in a fly-buck and Equation 19 gives the required primary inductance. Using an input voltage of 24 V and the other parameters in Table 9-2, the user arrives at a value of 38 μH. A standard value of 33 μH for this example is selected. Although the inductance can be selected based on the maximum input voltage and lower values of K, a somewhat smaller value of inductance is used in this example to save space on the PCB.
Finally, the maximum currents in the transformer must be checked. A transformer with a saturation current equal to or greater than the device current limit must be selected. Also, the maximum primary current, and, therefore, the output current, is limited by the current limit if the device. Equation 20 can be used to calculate the maximum output current for a given inductance and application parameters.