Inductor selection for SEPIC designs

Basic operation

The single ended primary inductor converter (SEPIC) allows the output voltage to be greater than, less than, or equal to the input voltage in DC-DC conversion. Some typical applications include digital cameras, mobile phones, CD/DVD players, portable devices and GPS systems.

Figure 1 depicts a typical SEPIC circuit. During the switch (SW) ON time the voltage across both inductors is equal to V_in. When the switch is ON capacitor C_p is connected in parallel with L₂. The voltage across L₂ is the same as the capacitor voltage, -V_in. Diode D₁ is reverse bias and the load current is being supplied by capacitor C_out. During this period, energy is being stored in L₁ from the input and in L₂ from C_p.

Application conditions:

    Input voltage (Vin) – 2.8 V – 4.5 V

    Output (Vout & Iout) – 3.3 V, 1 A

    Switching Frequency (Fs) – 250 kHz

    Efficiency - 90%

Step 1. Calculate the duty cycle

    D = Vout/(Vout + Vin)

The worst case condition for inductor ripple current is at maximum input voltage D = 3.3/(3.3 + 4.5) = 0.423.

The output inductor is sized to ensure that the inductor current is continuous at minimum load and that the output voltage ripple does not affect the circuit that the converter is powering. In this case we will assume a 20% minimum load thus allowing a 40% peak-to-peak ripple current in the output inductor L2.

Step 2. Calculate the value of L2

    V = L di/dt

• V is the voltage applied to the inductor
• L is the inductance
• di is the inductor peak to peak ripple current
• dt is the duration for which the voltage is applied

    L = V.dt/di

• dt = 1/Fs x D
• dt = 1/(250 x 103) x 0.423 = 1.69 μ-sec
• V = Vin during the switch ON time so;
• L2 = 4.5 x (1.69 x 10-6/0.4)
• L2 = 19 μH

Result: Using the nearest preferred value would lead to the selection of a 22 μH inductor. It is common practice to select the same value for both input and output inductors in SEPIC designs although when two separate parts are being used it is not essential.

Step 3. Calculate RMS and peak current ratings for both inductors

Input inductor L1

• Irms = (Vout x Iout)/(Vin (min) * efficiency)
• Irms = (3.3 x 1)/(2.8 x 0.9) = 1.31 A
• Ipeak = Irms + (0.5 x Iripple)
• Iripple = (V.dt)/L
• Iripple = (2.8 x 2.2 x 10-6)/22 x 10-6 = 0.28 A
• Ipeak = 1.31 + 0.14 =1.45 A

Although worst case ripple current is at maximum input voltage the peak current is normally highest at the minimum input voltage.

Result: 22 μH, 1.31 Arms and 1.45 Apk rated inductor is required. For example the Eaton DR73-220-R which has 1.62 Arms and 1.67 Apk current ratings.

Output inductor L2

Irms = Iout = 1 A

Iripple = (4.5 x 1.69 x 10-6)/22 x 10-6 = 0.346 A

Ipeak = 1 + 0.173 = 1.173 A

Result: A 2 2μH, 1 Arms and 1.173 Apk rated inductor is required, which for simplicity could be the same DR73-220-R

Step 1. Perform Step 1 and the Irms Portion of Step 3 from the two separate inductor selection.

The application information listed for the two inductor selection will be used.

Step 2. Calculate the inductance value

L = V.dt/di

From our earlier example the output ripple current needs to be 0.4 Apk-pk, so now we calculate for 0.8 A as the ripple current is split between the two windings

L = 4.5 x (1.69 x 10-6/0.8) = 9.5 μH

• A coupled inductor has the current flowing in one inductor and if the two windings are closely coupled the ripple current will be split equally between them.
• Using a coupled inductor reduces the required inductance by half.
• Since the two winding are on the same core they must be the same

Step 3. Calculate the peak current

Continuing with the example using an inductance value of 10 μH we now need to calculate the worst case peak current requirement. The RMS current in each winding is already known.

• Input inductor RMS current = 1.31 A
• Output inductor RMS current = 1 A
• Ipeak = Iin + Iout + (0.5 x Iripple)
• Iripple = (2.8 x 2.2 x 10-6)/10 x 10-6 = 0.62 A
• Ipeak = 1.31 + 1 + 0.31 = 2.62 A @ minimum input voltage

Result: A 10 μH coupled inductor with 2.31 Arms and 2.62 Apk current ratings is required, for example the Eaton DRQ74-100-R.

Using a coupled inductor takes up less space on the PCB and tends to be lower cost than two separate inductors. It also offers the option to have most of the inductor ripple current flow in either the input or the output. By doing this the need for input filtering can be minimized or the output ripple voltage can be reduced to very low levels when supplying sensitive circuits.