## Introduction to Boost converter

A Boost converter is a switch mode DC to DC converter in which the output voltage is greater than the input voltage. It is also called as step up converter. The name step up converter comes from the fact that analogous to step up transformer the input voltage is stepped up to a level greater 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 (P_{in}) = output power (P_{out})

SinceV_{in} < V_{out} in a boost converter, it follows then that the output current is less than the input current. Therefore in boost converter

V_{in} < V_{out }and I_{in} >I_{out}

## Principle of operation of Boost converter

The main working principle of boost converter is that the inductor in the input circuit resists sudden variations in input current. When switch is OFF the inductor stores energy 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 a constant output voltage V_{o}(t) = V_{o}(constant)

## Circuit diagram of Boost converter

The circuit diagram of boost converter is shown in the figure below

## Modes of operation of Boost converter

The 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 Boost 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 the diode will be open circuited since the n side of diode is at higher voltage compared to p side which is shorted to ground through the switch. Hence the 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

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

Assume the switch is open for t_{on} seconds which is given by D*T_{s }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

(equation 1)

Hence ΔI_{L} = (1/L)*V_{in}*D*T_{s}.

**case 2**: When switch is off

When switch in OFF the diode will be short circuited and the 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

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

I’’’_{L, off} = – (1/L) *(V_{in}-V_{out)}*(1-D)*T_{s} + 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 }=I_{L, on}, alsoI’’_{L, off }=I’’_{L, off}

Using the equations 1 and 2 we get

(1/L) *V_{in}*D*T_{s }= – _{(}1/L) *(V_{in}-V_{out)}*(1-D)*T_{s}

V_{in}*D=- _{(}V_{in}-V_{out)}*(1-D)

V_{in}* (D-1+D) = V_{out}*(1-D)

V_{out}/V_{in} = 1/ (1-D)

Since D < 1 V_{out} > V_{in}. Assuming no losses in the circuit and applying the law of conservation of energy

V_{out}*I_{out} = V_{in}*I_{in}

This implies I_{out}/I_{in} = (1-D), Thus I_{out} < I_{in}. As the duty cycle increases the output voltage increases and output current decreases. But due to parasitic elements in the lumped elements resistor, inductor, capacitor the step up ratio Vout/Vin decreases at higher duty cycles and approaches zero at unit duty cycle.

**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 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

= (1/L)*area under the curve of voltage v/s time. Hence from the wave forms shown in the

V_{in}*D*T_{s} = -(V_{in}-V_{o})*δ*T_{s}(negative sign signifies that the inductor is discharging)

V_{out}/V_{in }=(D+ δ)/ δ

and the ratio of output to input current from law of conservation of energy is I_{out}/I_{in} =δ/ (D+ δ).

## Applications of Boost converter

- They are used in regulated DC power supplies.
- They are used in regenerative braking of DC motors
- Low power boost converters are used in portable device applications
- As switching regulator circuit in highly efficient white LED drives
- Boost converters are used in battery powered applications where there is space constraint to stack more number of batteries in series to achieve higher voltages.