Inductor selection for SEPIC designs A few simple calculations can remove the mystery of a coupled-inductor approach BY CHRIS LIKELY Cooper Electronic Technologies Boynton Beach, FL http://www.cooperet.com The single-ended primary inductance converter (SEPIC) is an increasingly popular topology, particularly in battery-powered applications, as the input voltage can be higher or lower than the output voltage. While this offers obvious design advantages, the circuit operation and component selection is a mystery for many engineers. Even for those who understand the basics, the addition of a coupled inductor adds complexity. This article looks at the operation of the SEPIC and compares the design procedure for two single winding inductors with a coupled inductor approach. Coupled inductors can simplify your SEPIC circuit design. Basic operation In a simple SEPIC circuit (see Fig. 1) during switch-on (SW-ON), the voltage across both inductors is equal to Vin. This is obvious for L1, however it is not so clear for L2. Page 1 of 4Inductor selection for SEPIC designs - Electronic Products4/14/2011http://www2.electronicproducts.com/PrintArticle.aspx?ArticleURL=cooper.jun2005.htmlFig. 1. In a simple SEPIC circuit, both inductors will always see the same applied voltage. To understand this, we first need to look at the voltage across Cp. Neglecting ripple voltage, this voltage is constantly at the value of Vin. The simplest way to see this is when the circuit is at equilibrium. Under these conditions, there is no Vdc across L1 or L2, so one side of the capacitor is at Vin and the other at 0 V. When the switch is on, capacitor Cp is connected in parallel with L2, hence the voltage across L2 is the same as the capacitor voltageVin. This, in turn, means that diode D1 is reverse-bias and the load current is being supplied by capacitor Cout. During this period energy is being stored in L1 from the input and in L2 from Cp. When the switch turns off, the current in L1 continues to flow through Cp, D1, and into Cout, and the load recharges Cp to prepare for the next cycle. The current in L2 also flows into Cout and the load, ensuring that Cout is recharged and ready for the next cycle. During this period, the voltage across both L1 and L2 is equal to Voutthis is fairly clear for L2 but no so for L1. However, we already know that the voltage across Cp is equal to Vin and that the voltage on L2 is equal to Vout. For this to be true, the voltage at the node of Cp and L1 must be Vin + Vout. This in turn means that the voltage across L1 is (Vin + Vout) – Vin = Vout. The output inductor value and ratings are calculated using the same basic equations. Inductor selection First let us look at the selection of two separate inductors for L1 and L2, where input voltage is 2.8 to 4.5 V, and output is 3.3 V at 1 A. Switching frequency is 250 kHz, and efficiency is 90%. Calculating the duty cycle (D=Vout/(Vout+Vin) is the first step. The worst-case condition for inductor ripple current is at maximum input voltage, meaning D = 3.3/(3.3 + 4.5) = 0.423. Normally, output inductors are sized to ensure that inductor current is continuous at minimum load and output voltage ripple does not affect the circuit that the converter is Page 2 of 4Inductor selection for SEPIC designs - Electronic Products4/14/2011http://www2.electronicproducts.com/PrintArticle.aspx?ArticleURL=cooper.jun2005.htmlpowering. In this case we will assume a 20% minimum load, thus allowing a 40% peak-to-peak ripple current in the output inductor L2. To calculate the value of L2, we use V = L di/dt, where V is the voltage applied to the inductor, L is the inductance, di is the inductor's peak-to-peak ripple current, and dt is the duration the voltage is applied. Hence, L = V dt/di. dt = 1/Fs x D dt = 1/(250 x 103) x 0.423 = 1.69 µs Since V = Vin during SW ON time, L2 = 4.5 x (1.69 x 10–6/0.4) = 19 µH. Typically, we can use the nearest preferred value, which 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. Having selected the inductance value we now need to calculate the required RMS and peak current ratings for both inductors. For input inductor L1, Irms = (Vout x Iout)/Vin (min) * efficiency, so Irms = (3.3 x 1)/(2.8 x 0.9) = 1.31 A Ipeak = Irms + (0.5 x Iripple) Although worst-case ripple current is at maximum input voltage, the peak current is normally highest at the minimum input voltage. So a 22-µH 1.31-Arms/1.45-Apk inductor is required. Coupled inductor selection When calculating the value for a coupled inductor you need to bear in mind that all the current is effectively flowing in one inductor. If the two windings are closely coupled, the ripple current will be split equally between them. So we use L = V dt/di to calculate the inductance value. From our earlier example, the output ripple current needs to be 0.4 A peak-to-peak, 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 From this equation, it is evident using a coupled inductor halves the required inductance. It is also important to note that because the two winding are on the same core they must be the same value. If they are not, the voltage across each winding will not be equal and Cp will act as a short circuit to the difference. Continuing with the example using an inductance value of 10 µH, we now need to calculate the worst-case peak-current requirement. We already know the input-inductor rms current is Page 3 of 4Inductor selection for SEPIC designs - Electronic Products4/14/2011http://www2.electronicproducts.com/PrintArticle.aspx?ArticleURL=cooper.jun2005.html1.31 A and the Output-inductor rms current is 1 A, meaning Ipeak = Iin + Iout + (0.5 x Iripple) Iripple = (V.dt)/L 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 at minimum input voltage Under these circumstances, a 10-µH coupled inductor with 2.31-Arms and 2.62-Apk current ratings is required. Using a coupled inductor typically takes up less space on the pc board and tends to be