Self Powered Buck-Boost Converter for Low-Voltage Energy Applications

DOI : 10.17577/IJERTV3IS040028

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Self Powered Buck-Boost Converter for Low-Voltage Energy Applications

Alina C R

M.Tech Research Scholar Power Electronics & Control

Government Engineering College Idukki,India

Anilkumar V M

Asst. Professor Dept. Of EEE Government Engineering College Idukki,India

AbstractWith electronic circuits now capable of operating at microwatt levels, it is feasible for them to be powered using non- traditional sources. This has led to energy harvesting, which provides the power to charge, supplement, or replace batteries in systems where battery use is inconvenient, impractical, expensive or dangerous. It can also be used for data transmission as well Energy harvesting can power smart wireless sensor nodes, to monitor and optimize complex industrial processes, remote field installations. Which uses its input as the wasted energy from combustion engines, industrial processes. Key to energy harvesting is a power converter that can operate with ultra-low-voltage inputs. The basic converter topology used here is the Buck-Boost Converter

Index TermsEnergy Harvesting, Buck-Boost Converters


Conventional AC-DC converters for energy harvesting consist of two-stages, full bridge diode rectifier and DC-DC converter. However the diode-bridge would incur considerable voltage drop, making the low-voltage rectification infeasible [1]. Much primitive structures uses transformers in the booster stage to overcome diode voltage drops but that also creates noises. Latest topologies uses bidirectional switches and split capacitors, but due to the low- operating frequency range the value of the capacitors have to be large enough to suppress the voltage ripple under the desired level. Conduction losses and losses due to parasitic capacitances contribute 31%- 37% of total losses [2]. Why we are going for Buck-Boost topology rather than Boost topology alone is because this converter has the ability to set up the input voltage with a reverse polarity, hence it is an appropriate candidate to condition the negative voltage cycle[3]. Buck and Boost also shares the common capacitor and inductor hence reduction in no. of components can be achieved. Buck- Boost structures using linear regulators are in use but these were changed when switching converters came in to play. Switching converters are particularly useful in submilliwatt applications because the generated energy will be in the milliwatt operating range [4]. Also linear voltage regulators cant synthesize optimum load impedances. The input is given from a electromagnetic microgenerator. Microgenerators are of three types, 1) Piezoelectric 2)Electrostatic 3) Electromagnetic in which electromagnetic source has the potential to drive an electronic load rather than

any other source. Rather than solar, wind, hydro, thermal this paper mainly focuses on vibrational energy as input, reason because with all the other sources independent modules have been developed. In the physical realization of the circuit conventional diodes cannot be used since it causes considerable voltage drop therefore Schottky diodes have to be used which drops only some millivolts. In this paper a Buck- Boost topology have been proposed which performs the unique action of boost converter in the positive cycle and buck- boost converter in the reverse cycle thus obtaining the required result.

Fig.1.Conventional topology using bidirectional switch and split capacitor.

Fig.2.Classical Buck-Boost Converter

The operation of the basic configuration [3] is as follows, when the input is positive,M1 is turned on and the circuit operates in boost mode with diode S1 in reverse biased condition, when the input supply become negative, M2 conducts reverse biasing S2, the circuit now operates in Buck- Boost mode.


    Fig.3.The proposed Topology

    In this topology an additional inductor and a capacitor is provided L2 &C2 .This filter will stabilize the output voltage without using a output voltage control OPAMP or any linear regulators.

    1. Modes Of Operation

      Fig.4. Mode 1 operation

      Fig.5.Mode 2 Operation

      Fig.6. Mode 3 Operation

      Fig.7. Mode -4 Operation

      Fig.8. Mode-5 Operation

      Fig.9. Mode-6 Operation

      Mode I (Fig.4) – During the positive half cycle,M1 is turned ON, S2 is forward biased. The circuit operates as Boost circuit. The output is supplied from the capacitor C1 to the load thus non-interrupting the flow of output power

      Mode II (Fig.5) -The operation is same but the only difference is that the inductor L1 which was charging previously during the mode-1 operation discharges through

      the diode S2 thus increasing the power output. Also at this point the capacitor C1 charges to its peak value.

      Mode III (Fig. 6) – As the energy stored in the inductor has

      D = D Vi


      2 1 (Vout V


      been completely dissipated, diode S2 gets reverse biased, M2 is also turned OFF but the capacitor C1 is providing the output continuously from the stored charge. Therfore the powerflow remains unidirectional

      In Boost operation the instantaneous power dissipated in the entire circuit is

      V2 V2 sin 2 t

      Mode IV (Fig.7) -Here the input reverses two MOSFETS are

      Pd = in = m


      turned ON and the two diodes OFF as it is reverse biased but R R

      the current circulates in the reverse direction at the input but at

      The instantaneous current flowing through the load resistor is

      the output the capacitor is keeping the current flow unidirectional. Thus it is observing Buck-Boost operation.At this mode the function of L2 and C2 is expressive, L2 will

      I = P =

      (Vout Vi )

      V2 sin 2 t


      R [Vout Vm sin t]


      operate together with C1 as a tank circuit thus parallel resonance operation can be observed.

      The instantaneous power delivered to the load

      Mode V (Fig.8) -In this mode MI MOSFET turned OFF because of the current reversal owing to natural commutation/line commutation. But as the inductor L1

      Po = V0 * I (9)

      V2 sin 2 t

      discharges the power flow is still maintained and it also charges the capacitor.

      Mode VI (Fig.9) – As the inductor discharges completely the entire stored energy, the diode S1 gets reverse biased. But the capacitor discharges and the power flow is maintained as


      = V m

      oR [Vout Vm sin t]

      Therefore the total energy delivered to the load




For the steady-state analysis several assumptions are taken in to account [8].

    1. The output capacitor C1 is assumed to be very high so that it filters out the voltage ripple

    2. The input voltage for simulation is assumed to be sinusoidal

    3. The switching frequency of M1 and M2 is assumed to be much larger than the input supply

      Eo = 2 Po dt (11)


      The total energy consumed by the circuit



      Ed = 2 Pd dt (12)

      According to the energy conservation principle energy supplied must be equal to the energy delivered. As a result equating (11) & (12)

      V2 T V2 D2 T T


      0 = m 1 s i


    4. Internal resistances and diode drops are neglected for simplification of analysis

According to the boost operation, shown in modes 1-3, the average voltage across the switch can be expressed as

2R 8L

The inductance L and switching frequency fs is used to determine the duty ratio D,from (13) the equation can be determined as

V (t) = (1-D1-D2) ViTs + D2TsV0ut (1)

D = 2V0 Lfs


Where Vi=Vmsint (2)

The average switch current during the switching period

1 s in

I1(t) = D 2T V /2L (3)

Vm R


The design was carried out based on the requirement the inputs and outputs required are Vin =400mV,f = 100Hz, fs=

Since the waveform for the inductor current is triangular in shape as seen in the simulation result (fig. 15) The average diode voltage and current can be expressed as :

V2(t) = D1TsV0ut +[ (1-D1-D2) (Vout-Vin)] (4)

50000Hz Iin=0.8A

Applying Buck-Boost Operation Equation,

V0 = D

Vin 1D



D2 Ts (Vout V )

I2(t) = i



From this we will get the value of D (duty ratio). The average inductor current may be obtained from the average source current and the equation is:

Applying Voltage second balance rule ie. Energy across the inductor is equal to zero. We get

I = DIo 1D



Supposing Discontinuous Conduction Mode

V0 = D

Vd 1



The simulation was carried out for designed values of inductance and capacitance for an output voltage of 3.3V from

From this obtain the value of discontinuous conduction interval 1 then the value of inductance can be found out as

s 1

L = Vd D T (D+ ) (18)


From the obtained value of L findout the commercially available inductor then recalculate the average currents inorder to obtain the actual values of the same. Again calculate the value of load resistor or the limit of load resistance,

400mV,100Hz input sinusoidal supply. The values of parameters are tabulated as below:





Input Supply



Output Voltage







Il Il

= (1D)2RTs



The charging and discharging currents of the capacitor decides the voltage ripple. Consider that the entire AC part of the inductor current flows in to the capacitor. This relation gives the value of capacitance.



= DTs



Choose the value of the capacitance appropriately . The validity of this assumptions can be checked by using the two given equalities.

Ts << RC

<< 1 (21)

Since the required power and voltage limit is specified the inductance value have to be designed from the given parameters .If the switching frequency increases, the size of the inductor decreases. But we cannot increase the switching frequency above a specified limit because the switching loss will increase considerably. Therefore the actual size of the inductor should compensate the loss incurred. In the efficiency calculation the forward voltage drop and ON state drop of MOSFET plays an important role. The duty ratio for the Boost and Buck- Boost operation is taken as D itself.


    The SinePWM control strategy can be used for the pulse generation in the physical implementation as well as simulation because it produces perfect square pulses for triggering that will limit the output waveform distortion. To make output voltage constant we can use a comparator or P/PI/PID compensator.

    Fig.10. Open loop Simulation Setup of the Proposed topology


    Fig.11. Input Voltage

    Fig.12. Gate Current and Voltage of M1

    Fig.13.Gate Current and Voltage of M2

    Fig.14. Input Current

    Fig.15. Inductor current waveform during Boost operation

    Fig.16. Inductor current during Buck-Boost Operation

    Fig.17. Output Voltage using LC filter at the output stage


      A unique combination of Buck and Buck- Boost Converter is presented in this paper which can produce a constant output voltage .The design was carried out for an output voltage of 3.3V, from an input supply of 400mV ,this potential can drive an LED or it can be used for battery charging purpose provided current gain is optimum. Further expansion can be brought about by providing an active feedback circuit and using instrumentation amplifier at the booster stage. The instrumentation amplifier have to be provided with an input supply as Vcc for driving it this will incur the use of a backup. But providing an active feedback circuitary with minimal losses makes the circuit completely self reliable and we can operate the given circuit without any backup supply that is why the paper is titled as Self Powered Buck-Boost Converter. Since Infrared Data Transmission is also using the same converters at the receiver stage by widening the output range, the industrial relevance of this converter can also be widened.


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