DC-DC converters or choppers are power electronic circuits which convert a fixed DC input to a variable DC at the output. The Buck converter is a form of a DC-DC converter which steps down the level of supply voltage and provides it at the output. To understand how a Buck converter works, lets understand a simple step-down converter first. A step-down converter circuit is shown below:
Whenever the switch is on, vo = VS and whenever the switch is off, vo = 0. In every time period T, the switch is on for a certain duration of time DT (where D < 1) and off for the rest of the time remaining (1-D)T. In this case, D is the duty cycle ratio of the switch.
Averaging Vo over a period of T gives,
As D is less than 1, Vo is less than VS, this implies a step-down action. Also, as the average output voltage depends on D the duty cycle ratio of the switch, by adjusting D - the duration for which we keep the switch on in each time period - we can vary the average voltage available at the output. As the stepping down action is achieved wholly by switching on and off a switch, choppers in general are highly efficient.
Step-down converters can be used in applications such as the speed control of DC machines. However, some applications may require a steady DC voltage at the output. For such applications, a low pass filter is connected at the output to provide a constant DC voltage. This circuit is called a Buck Converter. Let’s now better understand the buck converter.
The circuit of a buck converter is shown below. The IGBT is used as a switch here but can be replaced with a MOSFET as well. An inductor and capacitor are connected at the output as shown to form a low pass filter. The capacitance of the capacitor is kept high to ensure that Vo is ripple free and constant. A diode is provided across the low pass filter on the supply side and its purpose will be evident when we study the operation of the circuit.
Operation of the Buck Converter
When the switch is closed (for 0 < t < DT), the diode is reversed biased as VS appears across it. The inductor starts conducting the supply current from 0 and the equivalent circuit is as shown below:
Applying KVL to find vL,
Notice that diL/dt is a positive constant (as VS > Vo) hence the inductor current increases linearly. Some part of iL charges the capacitor to Vo while the rest flows through the load as the load current Io. Note that Vo is the same as what we saw in the step down converter and is equal to DVS.
When the switch is open (for DT < t < T), the inductor tries to maintain current in the circuit in the same direction as before, forward biasing the diode in the process. The diode now provides a freewheeling path for the inductor current.
Applying KVL to find vL, we get
Notice that diL/dt is a negative constant hence the inductor current decreases linearly. The inductor discharges the energy stored in it from when the switch was closed and the current decreases linearly in this duration with time. The capacitor maintains the output at Vo in this duration as it was charged to Vo from when the switch was closed.
iL starts from 0 initially when the switch is closed for the first time. The amount of energy discharged by the inductor when the switch is open may not be exactly equal to the amount of energy it had stored hence some residual energy may be carried on to the next cycle. However, at steady state the inductor current oscillates between iL(min) and iL(max). The behavior of vL , iL and vo can be seen from the timing diagram below.
The inductor ensures the continuity of load current when the switch is off, and the high value of capacitance ensures that the output is held constant at Vo throughout.
The buck converter obtained by implementing a low pass filter at the output of a step-down chopper as we saw, exhibits better performance characteristics while giving the same average voltage at the output as that of a step down converter. Such characteristics make the buck converters well suited for applications like powering up small electronic modules and in battery chargers.