- Boost converter
A

**boost converter**(**step-up converter**) is apower converter with an output DC voltage greater than its input DC voltage. It is a class of switching-mode power supply (SMPS) containing at least twosemiconductor switches (adiode and atransistor ) and at least oneenergy storage element. Filters made ofcapacitor s (sometimes in combination withinductor s) are normally added to the output of the converter to reduce output voltage ripple.**Overview**An AC mains voltage cannot directly power devices such as computers, digital clocks, and telephones. The outlet supplies AC and the devices and loads require DC. Power conversion enables DC devices to utilize power from ac voltage sources. A process called ac to dc conversion (rectification) is used to convert an AC voltage to power a DC load.

Power can also come from DC sources such as batteries, solar panels, rectifiers, and DC generators. A process that changes one DC voltage to a different DC voltage is called dc to dc conversion. A boost converter is a

DC to DC converter with an output voltage greater than the source voltage. A boost converter is sometimes called a step-up converter since it “steps up” the source voltage. Since power (V*I) must be conserved, the output current is lower than the source current.**History**For high efficiency, the SMPS switch must turn on and off quickly and have low losses. The advent of a commercial

semiconductor switch in the 1950’s represented a major milestone that made SMPSs such as the boost converter possible. Semiconductor switches turned on and off more quickly and lasted longer than other switches such asvacuum tube s and electromechanical relays. The majorDC to DC converter s were developed in the early-1960 s when semiconductor switches had become available. Theaerospace industry’s need for small, lightweight, and efficient power converters led to the converter’s rapid development.Switched systems such as SMPS are a challenge to design since its model depends on whether a switch is opened or closed. R.D. Middlebrook from

Caltech in1977 published the models for DC to DC converters used today. Middlebrook averaged the circuit configurations for each switch state in a technique called state-space averaging. This simplification reduced two systems into one. The new model led to insightful design equations which helped SMPS growth.**Applications**Battery powered systems often stack cells in series to achieve higher voltage. However, sufficient stacking of cells is not possible in many high voltage applications due to lack of space. Boost converters can increase the voltage and reduce the number of cells. Two battery-powered applications that use boost converters are hybrid electric vehicles (HEV) and lighting systems.

The

Toyota Prius HEV contains a motor which utilizes voltages of approximately 500 V. Without a boost converter, the Prius would need nearly 417 cells to power the motor. However, a Prius actually uses only 168 cells and boosts the battery voltage from 202 V to 500 V. Boost converters also power devices at smaller scale applications, such as portable lighting systems. A white LED typically requires 3.3V to emit light, and a boost converter can step up the voltage from a single 1.5 V alkaline cell to power the lamp. Boost converters can also produce higher voltages to operatecold cathode fluorescent tubes (CCFL) in devices such as LCD backlights and some flashlights.**Circuit analysis****Operating principle**The key principle that drives the boost converter is the tendency of an inductor to resist changes in current. When being charged it acts as a load and absorbs energy (somewhat like a resistor), when being discharged, it acts as an energy source (somewhat like a battery). The voltage it produces during thedischarge phase is related to the rate of change of current, and not to the original chargingvoltage, thus allowing different input and output voltages.

The basic principle of a Boost converter consists in 2 distinct states (see figure 2):

* in the On-state, the switch S (see figure 1) is closed, resulting in an increase in the inductor current;

* in the Off-state, the switch is open and the only path offered to inductor current is through theflyback diode D, the capacitor C and the load R. This results in transferring the energy accumulated during the On-state into the capacitor.

* The input current is discontinous, stepping between a very high inductor current and 0. The large ripple usually requires a large input bypass capacitor to reduce the source impedance.**Continuous mode**When a boost converter operates in continuous mode, the current through the inductor (I

_{L}) never falls to zero. Figure 3 shows the typical waveforms of currents and voltages in a converter operating in this mode.The output voltage can be calculated as follows, in the case of an ideal converter (i.e using components with an ideal behaviour) operating in steady conditions:During the On-state, the switch S is closed, which makes the input voltage (V

_{i}) appear across the inductor, which causes a change in current (I_{L}) flowing through the inductor during a time period (t) by the formula:$frac\{Delta\; I\_L\}\{Delta\; t\}=frac\{V\_i\}\{L\}$

At the end of the On-state, the increase of I

_{L}is therefore:$Delta\; I\_\{L\_\{On=int\_0^\{Dcdot\; T\}frac\{V\_i\}\{L\}d\; t=frac\{V\_i\; cdot\; Dcdot\; T\}\{L\}$

D is the duty cycle. It represents the fraction of the commutation period T during which the switch is On. Therefore D ranges between 0 (S is never on) and 1 (S is always on).

During the Off-state, the switch S is open, so the inductor current flows through the load. If we consider zero voltage drop in the diode, and a capacitor large enough for its voltage to remain constant, the evolution of I

_{L}is:$V\_i-V\_o=Lfrac\{dI\_L\}\{dt\}$

Therefore, the variation of I

_{L}during the Off-period is:$Delta\; I\_\{L\_\{Off=int\_0^\{left(1-D\; ight)\; T\}frac\{left(V\_i-V\_o\; ight)\; dt\}\{L\}=frac\{left(V\_i-V\_o\; ight)\; left(1-D\; ight)\; T\}\{L\}$

As we consider that the converter operates in steady-state conditions, the amount of energy stored in each of its components has to be the same at the beginning and at the end of a commutation cycle. In particular, the energy stored in the inductor is given by:

$E=frac\{1\}\{2\}Lcdot\; I\_L^2$

Therefore, the inductor current has to be the same at the beginning and the end of the commutation cycle. This can be written as

$Delta\; I\_\{L\_\{On\; +\; Delta\; I\_\{L\_\{Off=0$

Substituting $Delta\; I\_\{L\_\{On$ and $Delta\; I\_\{L\_\{Off$ by their expressions yields:

$Delta\; I\_\{L\_\{On\; +\; Delta\; I\_\{L\_\{Off=frac\{V\_i\; cdot\; Dcdot\; T\}\{L\}+frac\{left(V\_i-V\_o\; ight)left(1-D\; ight)T\}\{L\}=0$

This can be written as:

$frac\{V\_o\}\{V\_i\}=frac\{1\}\{1-D\}$

Which in turns reveals the duty cycle to be:

$D=\{1-frac\{V\_i\}\{V\_o$

From the above expression it can be seen that the output voltage is always higher than the input voltage (as the duty cycle goes from 0 to 1), and that it increases with D, theoretically to infinity as D approaches 1. This is why this converter is sometimes referred to as a "step-up" converter.

**Discontinuous mode**In some cases, the amount of energy required by the load is small enough to be transferred in a time smaller 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 waveforms in figure 4). Although slight, the difference has a strong effect on the output voltage equation. It can be calculated as follows:

As the inductor current at the beginning of the cycle is zero, its maximum value $I\_\{L\_\{Max$ (at t=D.T) is

$I\_\{L\_\{Max=frac\{V\_icdot\; Dcdot\; T\}\{L\}$

During the off-period, I

_{L}falls to zero after δ.T:$I\_\{L\_\{Max+frac\{left(V\_i-V\_o\; ight)cdot\; deltacdot\; T\}\{L\}=0$

Using the two previous equations, δ is:

$delta=frac\{V\_icdot\; D\}\{V\_o-V\_i\}$

The load current I

_{o}is equal to the average diode current (I_{D}). As can be seen on figure 4, the diode current is equal to the inductor current during the off-state. Therefore the output current can be written as:$I\_o=ar\{I\_D\}=frac\{I\_\{L\_\{max\}\{2\}delta$

Replacing I

_{Lmax}and δ by their respective expressions yields:$I\_o=frac\{V\_icdot\; Dcdot\; T\}\{2L\}frac\{V\_icdot\; D\}\{V\_o-V\_i\}=frac\{V\_i^2cdot\; D^2cdot\; T\}\{2Lleft(V\_o-V\_i\; ight)\}$

Therefore, the output voltage gain can be written as flow:

$frac\{V\_o\}\{V\_i\}=1+frac\{V\_icdot\; D^2\; cdot\; T\}\{2Lcdot\; I\_o\}$

Compared to the expression of the output voltage for the continuous mode, this expression is much more complicated. Furthermore, in discontinuous operation, the output voltage gain not only depends on the duty cycle, but also on the inductor value, the input voltage, the switching frequency, and the output current.

**See also****References****External links*** [

*http://www.smps.com/Knowledge/Boost_PFC/Boost_PFC.shtml Boost PFC Mathematical Calculations*]

* [*http://www.daycounter.com/Calculators/Switching-Converter-Calculator.phtml Daycounter Switching Converter Power Supplies*]

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