Single-ended primary-inductor converter

A SEPIC is said to be in continuous-conduction mode ("continuous mode") if the currents through inductors L1 and L2 never fall to zero during an operating cycle. During a SEPIC's steady-state operation, the average voltage across capacitor C1 (V_C1) is equal to the input voltage (V_in). Because capacitor C1 blocks direct current (DC), the average current through it (I_C1) is zero, making inductor L2 the only source of DC load current. Therefore, the average current through inductor L2 (I_L2) is the same as the average load current and hence independent of the input voltage.

Looking at average voltages, the following can be written:

${\displaystyle V_{IN}=V_{L1}+V_{C1}+V_{L2}}$

Because the average voltage of V_C1 is equal to V_IN, V_L1 = −V_L2. For this reason, the two inductors can be wound on the same core, which begins to resemble a flyback converter, the most basic of the transformer-isolated switched-mode power supply topologies. Since the voltages are the same in magnitude, their effects on the mutual inductance will be zero, assuming the polarity of the windings is correct. Also, since the voltages are the same in magnitude, the ripple currents from the two inductors will be equal in magnitude.

The average currents can be summed as follows (average capacitor currents must be zero):

${\displaystyle I_{D1}=I_{L1}-I_{L2}}$

When switch S1 is turned on, current I_L1 increases and the current I_L2 goes more negative. (Mathematically, it decreases due to arrow direction.) The energy to increase the current I_L1 comes from the input source. Since S1 is a short while closed, and the instantaneous voltage V_L1 is approximately V_IN, the voltage V_L2 is approximately −V_C1. Therefore, D1 is opened and the capacitor C1 supplies the energy to increase the magnitude of the current in I_L2 and thus increase the energy stored in L2. I_L is supplied by C2. The easiest way to visualize this is to consider the bias voltages of the circuit in a d.c. state, then close S1.

Figure 2: With S1 closed current increases through L1 (green) and C1 discharges increasing current in L2 (red)

When switch S1 is turned off, the current I_C1 becomes the same as the current I_L1, since inductors do not allow instantaneous changes in current. The current I_L2 will continue in the negative direction, in fact it never reverses direction. It can be seen from the diagram that a negative I_L2 will add to the current I_L1 to increase the current delivered to the load. Using Kirchhoff's Current Law, it can be shown that I_D1 = I_C1 - I_L2. It can then be concluded, that while S1 is off, power is delivered to the load from both L2 and L1. C1, however is being charged by L1 during this off cycle (as C2 by L1 and L2), and will in turn recharge L2 during the following on cycle.

Figure 3: With S1 open current through L1 (green) and current through L2 (red) produce current through the load

Because the potential (voltage) across capacitor C1 may reverse direction every cycle, a non-polarized capacitor should be used. However, a polarized tantalum or electrolytic capacitor may be used in some cases,[2] because the potential (voltage) across capacitor C1 will not change unless the switch is closed long enough for a half cycle of resonance with inductor L2, and by this time the current in inductor L1 could be quite large.

The capacitor C_IN has no effect on the ideal circuit's analysis, but is required in actual regulator circuits to reduce the effects of parasitic inductance and internal resistance of the power supply.

The boost/buck capabilities of the SEPIC are possible because of capacitor C1 and inductor L2. Inductor L1 and switch S1 create a standard boost converter, which generates a voltage (V_S1) that is higher than V_IN, whose magnitude is determined by the duty cycle of the switch S1. Since the average voltage across C1 is V_IN, the output voltage (V_O) is V_S1 - V_IN. If V_S1 is less than double V_IN, then the output voltage will be less than the input voltage. If V_S1 is greater than double V_IN, then the output voltage will be greater than the input voltage.

A SEPIC is said to be in discontinuous-conduction mode or discontinuous mode if the current through either of inductors L1 or L2 is allowed to fall to zero during an operating cycle.