The SEPIC converter
The principle diagram of the DC/DC SEPIC converter is shown in figure : in comparison to the previously seen converters, it is possible to immediately notice the double inductor and the C1 capacitor, that represent the “complication” with respect to the other ones.
A positive note concerning the buck converter is that in this one, if the switch short-circuits, all the input voltage ends on the load, that has many odds of taking damage; on the other hand in the SEPIC – thanks to the C1 capacitor – the generator’s DC component is blocked and, in the case of malfunction, the output voltage is zeroed, thus protecting the powered devices.
However, the functioning is decisively much more complex and deserves a detailed analysis; let’s start by examining the functioning in continuous mode (CCM) that is obtained when the current in the L1 inductor is never zeroed; even here, the discussion converning the functioning in the discontinuous mode will be omitted.
In conditions of stability, the average voltage at the C1 (VC1) capacitor’s ends is equal to the input voltage (VIN). Since the C1 capacitor blocks the continuous component, the average current flowing through it (IC1) is null, thus the L2 inductor turns out to be the only source of direct current for the load.
Therefore, the average current that flows through the L2 (IL2) inductor is the same of the average current on the load and therefore it is independent from the input voltage.
By analyzing the average voltages on the circuit, we may write as follows:
VIN = VC1 + VL1 + VL2
and, since the average voltage, VC1 is equal to VIN:
VL1 = –VL2
This makes it possible to wind up the inductors on a single nucleus, given that the previous equation says that the influence of the mutual inductance between the two is null.
It won’t be our case – for problems of availability of the components – but this represents a great advantage at an industrial level.
Even the peak currents in the two inductors will be equal as an absolute value.
The average currents may therefore be expressed by:
ID1 = IL1 – IL2
with the average current in C1 being null.
When the switch is closed we have the condition depicted in previous figure
The input voltage charges the L1 inductor, while the L2 inductor is charged by C1’s voltage that, as we said, is equal to the input voltage, in the beginning.
Please notice that, arithmetically speaking, the two currents flowing into L1 and L2 are opposite, i.e. the current that charges the L2 coil has a “negative sign”: so, actually, the charging current is “discharging” the coil during this phase.
