11.2.1.2.3.4 Inductor Selection

The inductor in a boost converter serves as an energy storage element. The energy stored equals ½ L × I². Therefore, the inductor must not be saturated as the inductance will drop and the energy stored will be reduced causing bad efficiency. The converter operates with typically 15-µH to 22-µH inductors. A design example for an application powering 6LEDs in one string given below:

Vin = 2.8 V — minimum input voltage for the boost converter
Vo = 6 × 3.2 V = 19.2 V — assuming a forward voltage of 3.2 V per LED
Vf = 0.5 V — forward voltage of the Schottky diode
Io = 25 mA maximum LED current
Fsw = 1.125 MHz — switching frequency — T=890 ns
Rds(on) = 0.6R — drain-source resistance of the internal NMOS switch
Vsw — voltage drop at the internal NMOS switch
I_AVG — average current in NMOS when turned on

The duty cycle for a boost converter is:

With:

A different approach to calculate the duty cycle is based on the efficiency of the converter. The typical number can be found in the graphs, or as a first approach, we can assume to get an efficiency of about 80% as a typical value.

With the values given above

ton = T × D = 890 ns × 0.89 = 792 ns

toff = 890 ns – 792 ns = 98ns

When the NMOS switch is turned on, the input voltage is forcing a current into the inductor. The current slope can be calculated with:

The minimum and maximum inductor current can be found by adding half of the inductor current ripple (di) to the average value, which gives:

Given the values above, an inductor with a current rating greater than 290 mA is needed. Plenty of margin should be kept to the rating in the inductor vendors data sheets as the maximum current is typically specified at a inductance drop of 20% or even 30%. A list of tested inductors is given in Table 13 with the following test conditions.

Test conditions:

• Vin = 2.8 V
• Vf = 3.2 V (per LED)
• Vf = 0.5 V (Schottky diode)
• Iout = 25 mA per string; no dimming

Other inductors, with lower or higher inductance values can be used. A higher inductance will cause a lower inductor current ripple and therefore will provide higher efficiency. The boost converter will also stay in continuous conduction mode over a wider load current range. The energy stored in an inductor is given by E=1/2L × I² where I is the peak inductor current. The maximum current in the inductor is limited by the internal current limit of the device, so the maximum power is given by the minimum peak current limit (see electrical specifications) times the inductance value. For highest output power, a large inductance value is needed. The minimum inductor value possible is limited by the energy needed to supply the load. The limit for the minimum inductor value is given during the on-time of the switch such that the current limit is not reached.

Example for the minimum inductor value:

Vin = 4.2 V, Vout = 19.7 V, Iout = 5 mA, Vsw =0.1 V
→ D = 79%
→ ton = 703 ns

During the on-time, the inductor current should not reach the current limit of 1.4 A.

With V… voltage across the inductor (V = Vin–Vsw)

→ L = V × dt/di = (4.2 V–0.1 V) × 703 ns/1.4 A = 2µH