Many applications require positive and negative supply voltages, with only one voltage requiring tight regulation. This Design Idea describes a dual-output, *hybrid* **SEPIC–Ćuk** converter whose positive output voltage can be lesser or greater than the input voltage. The unregulated negative output is a mirrored replica of the positive output.

To find out the output voltages we apply the principle of inductor volt-second balance and capacitor charge balance. To simplify the calculus, we neglect the voltage drop over the **MOSFET** and diode and we consider only continuous conduction mode.

When the **MOSFET** is on we have this equivalent network:

**Figure 2 **Equivalent circuit in switch on-state

The inductor voltages and capacitor currents for this interval, using small-ripple approximation, are:

*V1 = V _{IN}*

*V3 = V2+V5*

*I3 = -I2*

*I4 = -V4/R*_{L1}

*I5 = -I2-V5/R*_{L2}When the MOSFET is off we have this equivalent network:

**Figure 3**Equivalent circuit in switch off-state

and the inductor voltages and capacitor currents for this interval are:

*V1 = V*_{IN}-V2-V4

*V3 = V5-V4*

*I3 = I1*

*I4 = I1+I2-V4/R*_{L1}

*I5 = -I2-V5/R*_{L2}Equating the average inductor voltages and capacitor currents over one switching period to zero, we get:

*v1_avg = D × V*_{IN }+ (1-D) × (V_{IN}-V2-V4) = 0

*v3_avg = D × (V2+V5) + (1-D) × ( V5-V4) = 0*

*i3_avg = D × (-I2) + (1-D) × I1 = 0*

*i4_avg = D × (-V4/R*_{L1}) + (1-D) × ( I1+I2-V4/R_{L1}) = 0

*i5_avg = D × (-I2-V5/R*_{L2}) + (1-D) × ( -I2-V5/R_{L2}) = 0where D is the duty cycle.

Solving for V4 and V5:

*V4 = Vin × R*_{L1}/(R_{L1}+R_{L2}) × D/(1-D)

*V5 = -Vin × R*_{L2}/(R_{L1}+R_{L2}) × D/(1-D)In practice, due to the feedback, the positive output voltage, V4, is fixed.

Extracting the duty cycle from the V4 equation and inserting it into

**V5**results in:

*V5 = -R*_{L2}/R_{L1}× V4Therefore, this topology is most suitable when the output currents do not differ much.

When the two loads are equal, then:

*V4 = V*_{IN}/2 × D/(1-D)*and*

*V5 = -V*_{IN}/2 × D/(1-D)A variation of the topology can supply a “floating” load.

The op-amp converts the differential output voltage to single-ended for regulation. The differential output voltage is:

V_{OUT}= V_{IN}× D/(1-D)

**For** **Figure 4** :

V_{OUT}= V_{FB}× (1+16.7/3.9) × 2.2/0.82 = 17.86V

where V_{FB} of the** LM3488** is **1.26V.**

I measured **V _{OUT} = 18V** for

**R**and

_{LOAD}= 47Ω,**18.02V**with a

**94Ω**load. That gives a load regulation of about

**0.1 V/A,**or

**0.55%**of

**V**

_{OUT}.**Read More: SEPIC/Ćuk converter sprouts second output**