User:Mclausma/Sandbox

A SEPIC (single ended primary inductor converter) is a DC-DC converter which allows the output voltage to be greater than, less than, or equal to the input voltage. The output voltage of the SEPIC is controlled by the duty cycle of the control transistor. The largest advantage of a SEPIC over the buck-boost converter is a non-inverted output (positive voltage).

Continuous Mode
The basic schematic for a SEPIC is shown in Figure 1. A SEPIC is said to be in continuous mode if the current through the inductor L1 never falls to zero. In a SEPIC, during steady-state operation, the average of VC1 is Vin. Since C1 blocks DC current, the average of IC1 is zero. Since IC1 is zero, the only source of the average load current is IL2. Therefore, the average current through L2 is the same as the load current and is independent of the input voltage.

Looking at average voltages, the following can be written: $$ V_{IN} = V_{L1} + V_{C1} + V_{L2}$$

and since the average voltage of VC1 is equal to VIN, VL1 = -VL2. For this reason, the two inductors can be wound on the same core. Since the voltages are the same in magnitude, there effects of the mutual inductance will be zero, assuming the polarity of the windings is correct. Also, since the voltages are the same in magnitude, the average currents from the two inductors will be equal in magnitude.

The average currents can be summed as follows:

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

and since IL1 = -IL2, ID1 = 2IL1 = -2IL2

When switch S1 is turned on, current IL1 increases and the current IL2 decreases(becomes more negative). The energy to increase the current IL1 comes from the input source. Since S1 is a short while closed, and the instantaneous voltage VC1 is approximately VIN, the voltage VL2 is approximately -VIN. Therefore, the capacitor C1 supplies the energy to decrease(more negative) the current IL2.

When switch S2 is turned off, the current IL2 becomes the same as the current IC1. Also, since inductors do not allow instantaneous changes in current, the current IL2 will continue in the negative direction. From Kirchoff's Current Law, it can be shown that ID1 = IC1 - IL2. It can then be concluded, that while S2 is off, power is delivered to the load from L2 and L1. C1, however is being charged by L1 during this off cycle, and will in turn recharge L1 during the on cycle.

The capacitor CIN is required to reduce the effects of the parasitic inductance in internal resistance of the power supply as well. The inductance of the leads of the input supply are not accounted for in this circuit and would add additional inductance that was not accounted for. 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 generate a voltage (VS1) that is higher than VIN, whose magnitude is determined by the duty cycle of the switch S1. Since the average voltage across C1 is VIN, the output voltage (VO) is VS1 - VIN. If VS1 is less than double VIN, then the output voltage will be less than the input voltage. If VS1 is greater than double VIN, then the output voltage will be greater than the input voltage.

Non-Ideal Circuit
The voltage drop and switching time of the diode(D1) is extremely critical. The switching time needs to be extremely fast in order to not generate extremely high voltages across the inductors which could cause damage to components. Since not damaging components is more important than efficiency, a high speed silicon diode is normally used instead of Schottky diodes. The resistances in the inductors and the capacitors can also have large effects on the converter efficiency and ripple. Lower series resistance in the inductors allows for less energy dissipated as heat, which results in a large portion of the energy being transferred to the load. Low ESR capacitors should also be used for C1 and C2 to minimize ripple and prevent heat build up, especially in C1 where the current is changing direction frequently.