A boost converter (step-up converter) is a power 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 two semiconductor switches and at least one energy storage element. Filters made of inductor and capacitor combinations are often added to a converter’s output to improve performance.

Overview

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An electrical outlet’s voltage cannot directly power the devices such as computers, clocks, and telephones. The outlet supplies ac voltage and the devices and loads require dc voltage. Power conversion enables DC devices to utilize power from ac voltage sources. A process called ac to dc conversion 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. In addition, the output current is lowered from the source current.

History

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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 as vacuum tubes and electromechanical relays. The major DC to DC converters were developed in the early-1960s when semiconductor switches had become available. The aerospace industry’s need for small, lightweight, and efficient power converters led 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 in 1977 published the models for DC to DC converters used today. Middlebook 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

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Battery powered systems often stack batteries in series to achieve higher voltage. However, stacking batteries 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 batteries to power the motor. However, a real Prius uses only 168 batteries 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 4V 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 operate cold cathode fluorescent tubes (CCFL) in devices such as LCD backlights and some flashlights.

Circuit analysis

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Operating principle

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 the diode D, the capacitor C and the load R. This result in transferring the energy accumulated during the On-state into the capacitor.

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 follow, 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, causing the current in the inductor to increase at a rate given by:

$V\_i=L\backslash frac\{dI\_L\}\{dt\}$

At the end of the On-state, the increase of I_L is therefore:

$\backslash Delta\; I\_\{L\_\{On\}\}=\backslash int\_0^\{D\backslash cdot\; T\}dI\_L=\backslash int\_0^\{D\backslash cdot\; T\}\backslash frac\{V\_i\backslash cdot\; dt\}\{L\}=\backslash frac\{V\_i\; \backslash cdot\; D\backslash cdot\; T\}\{L\}$

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:

$V\_i-V\_o=L\backslash frac\{dI\_L\}\{dt\}$

Therefore, the variation of I_L during the Off-period is:

$\backslash Delta\; I\_\{L\_\{Off\}\}=\backslash int\_0^\{\backslash left(1-D\backslash right)\; T\}dI\_L=\backslash int\_0^\{\backslash left(1-D\backslash right)\; T\}\backslash frac\{\backslash left(V\_i-V\_o\backslash right)\; dt\}\{L\}=\backslash frac\{\backslash left(V\_i-V\_o\backslash right)\; \backslash left(1-D\backslash right)\; T\}\{L\}$

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:

$E=\backslash frac\{1\}\{2\}L\backslash cdot\; I\_L^2$

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

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

Substituting $\backslash Delta\; I\_\{L\_\{On\}\}$ and $\backslash Delta\; I\_\{L\_\{Off\}\}=0$ by their expressions yields:

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

This can be written as:

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

From the above expression it can be seen that the output voltage is always more than the input voltage (as the duty cycle goes from 0 to 1), and that it increases with D, theoritically up to the infinite 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

$I\_\{L\_\{Max\}\}=\backslash frac\{V\_i\backslash cdot\; D\backslash cdot\; T\}\{L\}$

During the off-period, I_L falls to zero after δ.T:

$I\_\{L\_\{Max\}\}+\backslash frac\{\backslash left(V\_i-V\_o\backslash right)\backslash cdot\; \backslash delta\backslash cdot\; T\}\{L\}=0$

Using the two previous equations, δ is:

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

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:

$I\_o=\backslash bar\{I\_D\}=\backslash frac\{I\_\{L\_\{max\}\}\}\{2\}\backslash delta$

Replacing I_Lmax and δ by their respective expressions yields:

$I\_o=\backslash frac\{V\_i\backslash cdot\; D\backslash cdot\; T\}\{2L\}\backslash frac\{V\_i\backslash cdot\; D\}\{V\_o-V\_i\}=\backslash frac\{V\_i^2\backslash cdot\; D^2\backslash cdot\; T\}\{2L\backslash left(V\_o-V\_i\backslash right)\}$

Therefore, the output voltage gain can be written as:

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

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 not only depends on the duty cycle, but also on the inductor value, the input voltage, and the output current.

Limit between continuous and discontinuous modes

As told at the beginning of this section, the converter operates in discontinuous mode when low current is drawn by the load, and in continuous mode at higher load current levels. The limit between discontinuous and continuous modes is reached when the inductor current falls to zero exactly at the end of the commutation cycle. with the notations of figure 4, this corresponds to :

$D\backslash cdot\; T\; +\; \backslash delta\; \backslash cdot\; T=T$

$D\; +\; \backslash delta\; =\; 1$

In this case, the output current I_olim (output current at the limit between continuous and discontinuous modes) is given by:

$I\_\{o\_\{lim\}\}=\backslash bar\{I\_D\}=\backslash frac\{I\_\{L\_\{max\}\}\}\{2\}\backslash left(1-D\backslash right)$

Replacing I_Lmax by the expression given in the discontinuous mode section yields:

$I\_\{o\_\{lim\}\}=\backslash frac\{V\_i\backslash cdot\; D\backslash cdot\; T\}\{2L\}\backslash left(1-D\backslash right)$

As I_olim is the current at the limit between continuous and discontinuous modes of operations, it satisfies the expressions of both modes. Therefore, using the expression of the output voltage in continuous mode, the previous expression can be written as:

$I\_\{o\_\{lim\}\}=\backslash frac\{V\_i\backslash cdot\; T\}\{2L\}\backslash frac\{V\_i\}\{V\_o\}\backslash left(1-\backslash frac\{V\_i\}\{V\_o\}\backslash right)$

Let's now introduce two more notations:

• the normalized voltage, defined by $\backslash left|V\_o\backslash right|=\backslash frac\{V\_o\}\{V\_i\}$. It corresponds to the gain in voltage of the converter;
• the normalized current, defined by $\backslash left|I\_o\backslash right|=\backslash frac\{L\}\{T\backslash cdot\; V\_i\}I\_o$. The term $\backslash frac\{T\backslash cdot\; V\_i\}\{L\}$ is equal to the maximum increase of the inductor current during a cycle, i.e the increase of the inductor current with a duty cycle D=1. So, in steady state operation of the converter, this means that $\backslash left|I\_o\backslash right|$ equals 0 for no output current, and 1 for the maximum current the converter can deliver.

Using these notations, we have:

• in continuous mode, $\backslash left|V\_o\backslash right|=\backslash frac\{1\}\{1-D\}$;
• in discontinuous mode, $\backslash left|V\_o\backslash right|=1+\backslash frac\{V\_i\backslash cdot\; D^2\; \backslash cdot\; T\}\{2L\backslash cdot\; I\_o\}=1+\; \backslash frac\{D^2\}\{2\backslash left|I\_o\backslash right|\}$;
• the current at the limit between continuous and discontinuous mode is $I\_\{o\_\{lim\}\}=\backslash frac\{V\_i\backslash cdot\; T\}\{2L\}D\backslash left(1-D\backslash right)=\backslash frac\{I\_\{o\_\{lim\}\}\}\{2\backslash left|I\_o\backslash right|\}D\backslash left(1-D\backslash right)$. Therefore, the locus of the limit between continuous and discontinuous mode is given by: $\backslash frac\{1\}\{2\backslash left|I\_o\backslash right|\}D\backslash left(1-D\backslash right)=1$

These expression have been plotted in figure 5. The difference in behaviour between the continuous and discontinuous modes can be seen clearly.

State Space Averaging Analysis

The average model analysis is a method to calculate the average over time of the waveforms in a switching circuit. It consists in writing the corresponding equations in each working state of the converter (here, there are two states: on and off, as shown in figure 2), and then multiply them by the time the converter spends in each state.

In the case of the boost converter, in the On-state, the rate of change in the inductor current is given by:

$L\backslash frac\{dI\_L\}\{dt\}=V\_i$

In the Off state, the voltage across the switch is equal to the output voltage (we assume zero voltage drop in the forward-biased diode):

$L\backslash frac\{dI\_L\}\{dt\}=V\_i-V\_o$

Therefore, the averaged rate of change in the inductor current is obtained by multiplying the two previous equations by the time spent in the corresponding states (D.T in the on-state and (1-D)T in the off state, assuming the converter operates in continuous mode) and dividing by the switching period:

$L\backslash bar\{\backslash frac\{dI\_L\}\{dt\}\}=\backslash left(D\backslash cdot\; T\backslash cdot\; V\_i\; +(1-D)T\backslash cdot(V\_i-V\_o)\backslash right)\backslash frac\{1\}\{T\}=D\backslash cdot\; V\_i\; +(1-D)\backslash cdot(V\_i-V\_o)$

It is important to note that $\backslash bar\{\backslash frac\{dI\_L\}\{dt\}\}$ represents the changes in inductor current at a timescale slower than the switching frequency. For a converter operating in steady-state mode, $\backslash bar\{\backslash frac\{dI\_L\}\{dt\}\}=0$. Therefore the previous equation becomes:

$D\backslash cdot\; V\_i\; +(1-D)\backslash cdot(V\_i-V\_o)=0$

Which can be rewritten as

$\backslash frac\{V\_o\}\{V\_i\}=\backslash frac\{1\}\{1-D\}$ (same equation as above)

The interest of this method is that it masks the switching behaviour of the converter, allowing it to be analysed with the classical AC or DC techniques.

Non-ideal circuit

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Effect of parasitic resistances

In the analysis above, no dissipative elements (resistors) have been considered. That means that the power is transmitted without losses from the input voltage source to the load. However, parasitic resistances exist in all circuits, due to the resistivity of the materials they are made from. Therefore, a fraction of the power managed by the converter is dissipated by these parasitic resistances.

For the sake of simplicity, we consider here that the inductor is the only non-ideal component, and that it is equivalent to an inductor and a resistor in series. This assumption is acceptable because as an inductor is made of one long winded piece of wire, it is likely to exhibit a non-negligible parasitic resistance (R_L). Furthermore, current flows through the inductor both in the on and the off states.

Using the state-space averaging method, we can write:

$V\_i=\backslash bar\; V\_L\; +\; \backslash bar\; V\_S$

where $\backslash bar\; V\_L$ and $\backslash bar\; V\_S$ are respectively the average voltage across the inductor and the switch over the commutation cycle. If we consider that the converter operates in steady-state, the average current through the inductor is constant. The average voltage across the inductor is:

$\backslash bar\; V\_L=L\backslash frac\{\backslash bar\{dI\_L\}\}\{dt\}+R\_L\backslash bar\; I\_L=R\_L\backslash bar\; I\_L$

When the switch is in the on-state, V_S=0. When it is off, the diode is forward biased (we consider the continuous mode operation), therefore V_S=V_o. Therefore, the average voltage across the switch is:

$\backslash bar\; V\_S=D\backslash cdot\; 0\; +\; (1-D)V\_o=(1-D)V\_o$

The output current is equal to the inductor current during the off-state. the average inductor current is therefore:

$\backslash bar\; I\_L=\backslash frac\{I\_o\}\{1-D\}$

Assuming the output current and voltage heve negligible ripple, the load of the converter can be considered as purely resistive. If R is the resistance of the load, the above expression becomes:

$\backslash bar\; I\_L=\backslash frac\{V\_o\}\{(1-D)R\}$

Using the previous equations, the input voltage becomes:

$V\_i=\; R\_L\backslash frac\{V\_o\}\{(1-D)R\}+\; (1-D)V\_o$

This can be written as:

$\backslash frac\{V\_o\}\{V\_i\}=\backslash frac\{1\}\{\backslash frac\{R\_L\}\{R(1-D)\}+1-D\}$

If the inductor resistance is zero, the equation above becomes equal to the one of the ideal case. But as R_L increases, the voltage gain of the converter decreases compared to the ideal case. Furthermore, the influence of R_L increases with the duty cycle. This is summarized in figure 7.

References

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External links

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• Boost PFC Mathematical Calculations
• Daycounter Switching Converter Power Supplies

Power supplies | Power electronics

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