## Introduction to Buck converter

A Buck converter is a switch mode DC to DC electronic converter in which the output voltage will be transformed to level less than the input voltage. It is also called as step down converter. The name step down converter comes from the fact that analogous to step down transformer the input voltage is stepped down to a level 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 (P_{in}) = output power (P_{out})

Since V_{in} > V_{out} in a buck converter, it follows then that the output current will be greater than the input current. Therefore in buck converter

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

## Principle of operation of Buck converter

The main working principle of buck converter is that the inductor in the input circuit resists sudden variations in input current. When switch is ON 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 Buck converter

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

## Modes of operation of Buck converter

The buck 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 t_{on}, the diode will be open circuited since it is in reverse biased condition(the n side of diode is at higher voltage(V_{in}) compared to p side which is shorted to ground through the switch). Hence the buck 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 ton seconds which is given by 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 on state is given as

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

Hence ΔI_{L,on} = (1/L)*(V_{in}-V_{out})*D*T_{s}.

**case 2**: When switch is off

When switch in OFF the diode will be forward biased (short circuited to ground) and the buck 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 at the front end of equation 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_{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} also I’_{L, off} =I’’_{L, off}

Using the equations 1 and 2 we get

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

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

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

V_{out}/V_{in} = D(duty cycle)

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.

**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 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 figure

V_{out}*δ*T_{s} = (V_{in}-V_{o})*D*T_{s}

V_{out}/V_{in} =D/ (D+ δ)

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

## Applications of Boost converter

- Buck converters are used in self-regulating power supplies.
- They are used as low-loss current sources to drive LED arrays(solid state lighting applications).
- Used as interface between battery and components in Notebooks.
- Buck converters are used as Point-of-load (POL) converters in servers
- Buck converters are used in advanced telecom and datacom systems