Buck Boost converter-principle of operation-applications

Introduction to Buck Boost converter

A Buck converter is a switch mode DC to DC converter in which the output voltage can be transformed to a level less than or greater than the input voltage. The magnitude of output voltage depends on the duty cycle of the switch.It is also called as step up/step down converter. The name step up/step down converter comes from the fact that analogous to step up/step down transformer the input voltage can be stepped up/down to a level greater than/less than the input voltage. By law of conservation of energy the input power has to be equal to output power (assuming no losses in the circuit).

                                             Input power (Pin) = output power (Pout)

In step up mode Vin < Vout in a Buck Boost converter, it follows then that the output current will be less than the input current. Therefore for a Buck Boost converter in step up mode

                                                     Vin < Vout and Iin >Iout

In step down mode Vin > Vout in a Buck Boost converter, it follows then that the output current will be greater than the input current. Therefore for a Buck Boost converter in step down mode

                                                     Vin >Vout and Iin <Iout

Principle of operation of Buck converter

The main working principle of Buck Boost converter is that the inductor in the input circuit resists sudden variations in input current. When switch is ON the inductor stores energy from the input in the form of magnetic energy and discharges it when switch is closed. The capacitor in the output circuit is assumed large enough that the time constant of RC circuit in the output stage is high. The large time constant compared to switching period ensures that in steady state a constant output voltage Vo(t) = Vo(constant) exists across load terminals.

Circuit diagram of Buck Boost converter

The circuit diagram of Buck Boost converter is shown in the figure below

 

Modes of operation of Buck Boost converter

The Buck Boost converter can be operated in two modes

a) Continuous conduction mode in which the current through inductor never goes to zero i.e. inductor partially discharges before the start of the switching cycle.

b) Discontinuous conduction mode in which the current through inductor goes to zero i.e. inductor is completely discharged at the end of switching cycle.

Circuit analysis of Buck converter

Assume in the entire analysis that the current swing (maximum to minimum value) through inductor and voltage swing through capacitor is very less so that they vary in a linear fashion. This is to ease the analysis and the results we will get through this analysis are quite accurate compared to real values.

Continuous conduction mode

case-1: When switch S is ON

When switch in ON for a time ton, the diode will be open circuited since it does not allow currents in reverse direction from input to output. Hence the Buck Boost converter can be redrawn as follows

During this state the inductor charges and the inductor current increases. The current through the inductor is given as

                                             \dpi{120} \fn_cm I_{L} = (1/L)*\int V*dt

Assume that prior to the opening of switch the inductor current is I’L, off. Since the input voltage is constant

                                           I_{L, on} = (1/L)*\int V_{in}*dt + I^{'}_{L,on}

Assume the switch is open for ton seconds which is given by D*Ts where D is duty cycle and Ts is switching time period. The current through the inductor at the end of switch on state is given as

                     IL, on = (1/L) *Vin*D*Ts + I’L, on (equation 1)

Hence ΔIL,on = (1/L)*Vin*D*Ts.

case 2: When switch is off

When switch in OFF the diode will be forward biased as it allows current from output to input (p to n terminal) and the Buck Boost converter circuit can be redrawn as follows

The inductor now discharges through the diode and RC combination. Assume that prior to the closing of switch the inductor current is I’’L, off. The current through the inductor is given as     

                                \dpi{120} \fn_cm I^{'''}_{L,off} = -(1/L)*\int V_{out}*dt + I^{''}_{L, off}

Note the negative sign at the front end of equation signifies that the inductor is discharging. Assume the switch is open for toff seconds which is given by (1-D)*Ts where D is duty cycle and Ts is switching time period. The current through the inductor at the end of switch off state is given as

                                I’’’L, off = -(1/L) *Vout*(1-D)*Ts + I’’L, off (equation 2)

In steady state condition as the current through the inductor does not change abruptly, the current at the end of switch on state and the current at the end of switch off state should be equal. Also the currents at the start of switch off state should be equal to current at the end of switch on state. Hence

                                             I’’’L, off =IL, on also I’L, off =I’’L, off

Using the equations 1 and 2 we get

(1/L) *Vin*D*Ts =  (1/L) *Vout*(1-D)*Ts

Vin*D =Vout*(1-D) 

Vout/Vin = D/(1-D)

Since D < 1, Vout can be greater than or less than Vin.For D>0.5 the Buck boost converter acts as boost converter with Vout >Vin.For D<0.5 the Buck boost converter acts as buck converter with Vout >Vin.

Assuming no losses in the circuit and applying the law of conservation of energy

                                                            Vout*Iout = Vin*Iin

This implies Iout/Iin = (1-D)/D, Thus Iout > Iin for D<0.5 and Iout < Iin for D<0.5 . As the duty cycle increases the output voltage increases and output current decreases. 

Discontinuous conduction mode

As mentioned before the converter when operated in discontinuous mode the inductor drains its stored energy completely before completion of switching cycle. The current and voltage wave forms of Buck Boost converter in discontinuous mode is shown in the figure below

The inductor in discontinuous mode drains all the current which it piled up in charging interval of same switching cycle. The current through the inductor is given as

           \dpi{120} \fn_cm I_{L} = (1/L) \int V_{L}*dt = (1/L)*area under the curve of voltage v/s time.  Hence from the wave forms shown in the figure

Vout*δ*Ts = Vin*D*Ts

                                                       Vout/Vin =D/δ

and the ratio of output to input current from law of conservation of energy is Iout/Iin = δ/D.

Applications of Buck Boost converter

  • Buck Boost converters are used in self-regulating power supplies.

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