Boost Converter

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 (a diode and a transistor) and at least one energy storage element. Filters made of capacitors (sometimes in combination with inductors) are normally added to the output of the converter to reduce output voltage ripple.

Operating principle:

  • in the On-state, the switch S 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 flyback 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 the same as the inductor current as can be seen in figure 2. So it is not discontinuous as in the buck converter and the requirements on the input filter are relaxed compared to a buck converter.

File:Boost conventions.svg

Fig. 1 Boost converter circuit

Fig. 2 Two configurations when the switch S is turned on or off

Continuous mode

File:Boost chronogram.svg

Fig. 3 Waveforms of current and voltage in a boost converter operating in continuous mode.

When a boost converter operates in continuous mode, the current through the inductor (IL) 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 (Vi) appear across the inductor, which causes a change in current (IL) 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 IL is therefore:

\Delta I_{L_{On}}=\frac{1}{L}\int_0^{D T}V_i d t=\frac{D T}{L} V_i

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 IL is:

V_i-V_o=L\frac{dI_L}{dt}

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

\Delta I_{L_{Off}}=\int_0^{\left(1-D\right) T}\frac{\left(V_i-V_o\right) dt}{L}=\frac{\left(V_i-V_o\right) \left(1-D\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=\frac{1}{2} L I_L^2

So, the inductor current has to be the same at the start and end of the commutation cycle. This means the overall change in the current (the sum of the changes) is zero:
\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 D T}{L}+\frac{\left(V_i-V_o\right)\left(1-D\right)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

File:Boost chronogram discontinuous.png

Fig. 4 Waveforms of current and voltage in a boost converter operating in 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 = DT) is

I_{L_{Max}}=\frac{V_i D T}{L}

During the off-period, IL falls to zero after δT:

I_{L_{Max}}+\frac{\left(V_i-V_o\right) \delta T}{L}=0

Using the two previous equations, δ is:

\delta=\frac{V_i D}{V_o-V_i}

The load current Io is equal to the average diode current (ID). 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=\bar{I_D}=\frac{I_{L_{max}}}{2}\delta

Replacing ILmax and δ by their respective expressions yields:

I_o=\frac{V_i D T}{2L}\cdot\frac{V_i D}{V_o-V_i}=\frac{V_i^2 D^2 T}{2L\left(V_o-V_i\right)}

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

\frac{V_o}{V_i}=1+\frac{V_i D^2 T}{2L 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.

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