For the PDF version of this article, click here.
In all switching regulators, the inductor is used as an energy storage device. When the semiconductor switch is on, the current in the inductor ramps up and energy is stored. When the switch turns off, this energy releases into the load. The amount of energy stored is given by Energy = ½L.I2 (Joules)
Where L is the inductance in Henrys and I is the peak value of inductor current.
The amount by which the current in the inductor changes during a switching cycle is known as the ripple current and is defined by the following equation:
Vl = L.di/dt
Where Vl is the voltage across the inductor, di is the ripple current, and dt is the duration for which the voltage is applied. From this, we can see that the value of ripple current is dependent on the value of the inductance.
Buck Converter Considerations
For a buck converter, choosing the correct value of inductance is important to obtain acceptable inductor and output capacitor sizes and sufficiently low output voltage ripple.
As you can see from Fig. 1, the inductor current is made up of ac and dc components. Because the ac component is high frequency, it will flow through the output capacitor, which offers a low HF impedance. This will produce a ripple voltage due to the capacitor equivalent series resistance (ESR), which appears at the output of the buck converter. This ripple voltage needs to be sufficiently low — so not to affect the operation of the circuit the regulator is supplying. It is normally in the order of 10-500mVpk-pk.
Selecting the correct ripple current also affects the size of the inductor and the output capacitor. This capacitor must have a sufficiently high ripple current rating, or it overheats and dries out. To get a good compromise between inductor and capacitor size, you should choose a ripple current value of 10% to 30% of maximum load current. This also implies that the current in the inductor will be continuous for output currents greater that 5% to 15% of full load.
You can operate buck converter inductors in continuous or discontinuous mode. This means that the inductor current can flow continuously or can fall to zero during the switching cycle (discontinuous). However, operating in discontinuous mode is not recommended as it makes for a more complex converter design. Selecting an inductor ripple current less than two times the minimum load ensures continuous mode operation.
When selecting an inductor for a buck converter, as with all switching regulators, you'll need to define or calculate the following parameters:
- Maximum input voltage
- Output voltage
- Switching frequency
- Maximum ripple current
- Duty cycle
For the buck converter shown in Fig. 2, for example, let's assume a switching frequency of 200 kHz, input voltage range of 3.3V±0.3V and output of 1.8V at 1.5A with a minimum load of 300mA.
For an input voltage of 3.6V, the duty cycle will be:
D = Vo/Vi = 3.6/1.8 = 0.5
Where Vo is the output voltage and Vi is the input voltage.
Voltage across the inductance:
Vl = Vi - Vo = 1.8V when the switch is on;
Vl = - Vo = -1.8V when the switch is off.
Selecting a 600mA ripple current, the required inductance is: L = Vl.dt/di = (1.8 × 0.5/200 × 103)/0.6
L = 7.5µH
To allow some margin, you should select a value of 10µH. This gives a nominal peak-to-peak ripple current of 450mA. In the finished design, this can be seen as an output ripple voltage of 0.45 × output capacitor ESR.