Now I am reading the wiki article about Bucker converter.
Buck converter - Wikipedia
A buck converter is a step-down DC to DC converter. Its design is similar to the step-up boost converter, and like the boost converter it is a switched-mode power supply that uses two switches (a transistor and a diode), an inductor and a capacitor.
The simplest way to reduce the voltage of a DC supply is to use a linear regulator (such as a 7805), but linear regulators waste energy as they operate by dissipating excess power as heat. Buck converters, on the other hand, can be remarkably efficient (95% or higher for integrated circuits), making them useful for tasks such as converting the main voltage in a computer (12 V in a desktop, 12-24 V in a laptop) down to the 0.8-1.8 volts needed by the processor.
Theory of operation
Fig. 1: Buck converter circuit diagram.
Fig. 2: The two circuit configurations of a buck converter: On-state, when the switch is closed, and Off-state, when the switch is open (Arrows indicate current as the conventional flow model).
Fig. 3: Naming conventions of the components, voltages and current of the buck converter.
Fig. 4: Evolution of the voltages and currents with time in an ideal buck converter operating in continuous mode.
The operation of the buck converter is fairly simple, with an inductor and two switches (usually a transistor and a diode) that control the inductor converter. In the idealised converter, all the components are considered to be perfect. Specifically, the switch and the diode have zero voltage drop when on and zero current flow when off and the inductor has zero series resistance. Further, it is assumed that the input and output voltages do not change over the course of a cycle (this would imply the output capacitance being infinitely large).
Conceptual Overview
The conceptual model of the buck converter is best understood in terms of an inductor's "reluctance" to allow a change in current. Beginning with the switch open (in the "off" position), the current in the circuit is 0. When the switch is first closed, the current will begin to increase, but the inductor doesn't want it to change from 0, so it will attempt to fight the increase by dropping a voltage. This voltage drop counteracts the voltage of the source and therefore reduces the net voltage across the load. Over time, the inductor will allow the current to increase slowly by decreasing the voltage it drops and therefore increasing the net voltage seen by the load. During this time, the inductor is storing energy in the form of a magnetic field.
If the switch is opened before the inductor has fully charged (i.e., before it has allowed all of the current to pass through by reducing its own voltage drop to 0), then there will always be a voltage drop across it, so the net voltage seen by the load will always be less than the input voltage source.
When the switch is opened again, the voltage source will be removed from the circuit, so the current will try to drop. Again, the inductor will try to fight against it changing, which it does by reversing the direction of its voltage and acting like a voltage source. Put another way, there is a certain current flowing through the load due to the input voltage source: in order to maintain this current when the input source is removed, the inductor will have to take the place of the voltage source and provide the same net voltage to the load. Over time, the inductor will allow the current to decrease gradually, which it does by decreasing the voltage across itself. During this time, the inductor is discharging its stored energy into the rest of the circuit.
If the switch is closed again before the inductor fully discharges, the load will always see a non-zero voltage. The capacitor placed in parallel with the load helps to smooth out voltage waveform as the inductor charges and discharges in each cycle.
Continuous mode
A buck converter operates in continuous mode if the current through the inductor (IL) never falls to zero during the commutation cycle. In this mode, the operating principle is described by the plots in figure 4:
When the switch pictured above is closed (On-state, top of figure 2), the voltage across the inductor is V_L = V_i - V_o. The current through the inductor rises linearly. As the diode is reverse-biased by the voltage source V, no current flows through it;
When the switch is opened (off state, bottom of figure 2), the diode is forward biased. The voltage across the inductor is V_L = -V_o (neglecting diode drop). Current IL decreases.
The energy stored in inductor L is
E= ...
Therefore, it can be seen that the energy stored in L increases during On-time (as IL increases) and then decreases during the Off-state. L is used to transfer energy from the input to the output of the converter.
The rate of change of IL can be calculated from:
V_L= ...
With VL equal to V_i-V_o during the On-state and to -V_o during the Off-state. Therefore, the increase in current during the On-state is given by:
...
Identically, the decrease in current during the Off-state is given by:
...
If we assume that the converter operates in steady state, the energy stored in each component at the end of a commutation cycle T is equal to that at the beginning of the cycle. That means that the current IL is the same at t=0 and at t=T (see figure 4).
So we can write from the above equations:
...
It is worth noting that the above integrations can be done graphically: In figure 4, \Delta I_{L_{on}} is proportional to the area of the yellow surface, and \Delta I_{L_{off}} to the area of the orange surface, as these surfaces are defined by the inductor voltage (red) curve. As these surfaces are simple rectangles, their areas can be found easily: \left( V_i-V_o\right) t_{on} for the yellow rectangle and -V_o t_{off} for the orange one. For steady state operation, these areas must be equal.
As can be seen on figure 4, \scriptstyle t_{on} \,=\, DT and \scriptstyle t_{off} \,=\, (1-D)T. D is a scalar called the duty cycle with a value between 0 and 1. This yields:
...
From this equation, it can be seen that the output voltage of the converter varies linearly with the duty cycle for a given input voltage. As the duty cycle D is equal to the ratio between tOn and the period T, it cannot be more than 1. Therefore, V_o \leq V_i. This is why this converter is referred to as step-down converter.
So, for example, stepping 12 V down to 3 V (output voltage equal to a fourth of the input voltage) would require a duty cycle of 25%, in our theoretically ideal circuit.
Discontinuous mode
...
In some cases, the amount of energy required by the load is small enough to be transferred in a time lower than the whole commutation period. In this case, the current through the inductor falls to zero during part of the period. The only difference in the principle described above is that the inductor is completely discharged at the end of the commutation cycle (see figure 5). This has, however, some effect on the previous equations.
...
The previous study was conducted with the following assumptions:
The output capacitor has enough capacitance to supply power to the load (a simple resistance) without any noticeable variation in its voltage.
The voltage drop across the diode when forward biased is zero
No commutation losses in the switch nor in the diode
These assumptions can be fairly far from reality, and the imperfections of the real components can have a detrimental effect on the operation of the converter.
Output voltage ripple
Output voltage ripple is the name given to the phenomenon where the output voltage rises during the On-state and falls during the Off-state. Several factors contribute to this including, but not limited to, switching frequency, output capacitance, inductor, load and any current limiting features of the control circuitry. At the most basic level the output voltage will rise and fall as a result of the output capacitor charging and discharging:
...
Qualitatively, as the output capacitor or switching frequency increase, the magnitude of the ripple decreases. Output voltage ripple is typically a design specification for the power supply and is selected based on several factors. Capacitor selection is normally determined based on cost, physical size and non-idealities of various capacitor types. Switching frequency selection is typically determined based on efficiency requirements, which tends to decrease at higher operating frequencies, as described below in Effects of non-ideality on the efficiency. Higher switching frequency can also reduce efficiency and possibly raise EMI concerns.
Output voltage ripple is one of the disadvantages of a switching power supply, and can also be a measure of its quality.
Effects of non-ideality on the efficiency
A simplified analysis of the buck converter, as described above, does not account for non-idealities of the circuit components nor does it account for the required control circuitry. Power losses due to the control circuitry is usually insignificant when compared with the losses in the power devices (switches, diodes, inductors, etc.) The non-idealities of the power devices account for the bulk of the power losses in the converter.
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