Buck-boost converter

From Wikipedia, the free encyclopedia

The buck-boost converter is a type of DC-DC converter that has an output voltage magnitude that is either greater than or less than the input voltage magnitude. It is a switch mode power supply with a similar circuit topology to the boost converter and the buck converter. The output voltage is adjustable based on the duty cycle of the switching transistor. One drawback of this converter is that the switch does not have a terminal at ground; this complicates the driving circuitry. Also, the polarity of the output voltage is opposite the input voltage.

The basic schematic of a buck-boost converter.

1 Principle of operation
2 Non-ideal circuit
- 2.1 Effect of parasitic resistances
3 See also
4 References

[edit] Principle of operation

Fig 1: Schematic of a Buck-Boost converter.

Fig 2: The two operating states of a buck-boost converter: When the switch is turned-on, the input voltage source supplies current to the inductor and the capacitor supplies current to the resistor (output load). When the switch is opened (providing energy is stored into the inductor), the inductor supplies current to the load via the diode D.

The basic principle of the buck-boost converter is fairly simple (see figure 2):

while in the On-state, the input voltage source is directly connected to the inductor (L). This result in accumulating energy in L. In this stage, the capacitor supplies energy to the output load;
while in the Off-state, the inductor is connected to the output load and capacitor, so energy is transferred from L to C and R.

Compared to the buck and boost converters, the characteristics of the buck-boost converter are mainly:

polarity of the output voltage is opposite to that of the input;
the output voltage can vary continuously from 0 to $-\infty$ (for an ideal converter). The output voltage ranges for a buck and a boost converter are respectively 0 to V_i and V_i to $\infty$ .

[edit] Continuous Mode

Fig 3: Waveforms of current and voltage in a Buck-Boost converter operating in continuous mode.

If the current through the inductor L nevers falls to zero during a commutation cycle, the converter is said to operate in continuous mode. The current and voltage waveforms in an ideal converter can be seen in figure 3.

From t=0 to D.T, the converter is in On-State, so the switch S is closed. The rate of change in the inductor current (I_L) is therefore given by:

$\frac{dI_L}{dt}=\frac{V_i}{L}$

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

$\Delta I_{L_{On}}=\int_0^{D\cdot T}dI_L=\int_0^{D\cdot T}\frac{V_i\cdot dt}{L}=\frac{V_i \cdot D\cdot T}{L}$

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 assume zero voltage drop in the diode (we consider an ideal diode), and a capacitor large enough for its voltage to remain constant, the evolution of I_L is:

$\frac{dI_L}{dt}=\frac{V_o}{L}$

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

$\Delta I_{L_{Off}}=\int_0^{\left(1-D\right) T}dI_L=\int_0^{\left(1-D\right) T}\frac{V_o\cdot dt}{L}=\frac{V_o \left(1-D\right) T}{L}$

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. As the energy in an inductor is given by:

$E=\frac{1}{2}L\cdot I_L^2$

it is obvious that the value of I_L at the end of the Off state must be the same as the value of I_L at the beginning of the On-state, i.e the sum of the variations of I_L during the on and the off states must be zero:

$\Delta I_{L_{On}} + \Delta I_{L_{Off}}=0$

Substituting $\Delta I_{L_{On}}$ and $\Delta I_{L_{Off}}=0$ by their expressions yields:

$\Delta I_{L_{On}} + \Delta I_{L_{Off}}=\frac{V_i \cdot D\cdot T}{L}+\frac{V_o\left(1-D\right)T}{L}=0$

This can be written as:

$\frac{V_o}{V_i}=-\frac{D}{1-D}$

From the above expression it can be seen that the polarity of the output voltage is always negative (as the duty cycle goes from 0 to 1), and that its absolute value increases with D, theoritically up to minus infinite as D approaches 1. Apart from the polarity, this converter is either step-up (as a boost converter) or step-down (as a buck converter). This is why it is referred to as a buck-boost converter.

[edit] Discontinuous Mode

Fig 4: Waveforms of current and voltage in a Buck-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=D.T) is

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

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

$I_{L_{Max}}+\frac{-V_o\cdot \delta\cdot T}{L}=0$

Using the two previous equations, δ is:

$\delta=\frac{V_i\cdot D}{V_o}$

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:

$I_o=\bar{I_D}=\frac{I_{L_{max}}}{2}\delta$

Replacing I_Lmax and δ by their respective expressions yields:

$I_o=\frac{V_i\cdot D\cdot T}{2L}\frac{V_i\cdot D}{V_o}=\frac{V_i^2\cdot D^2\cdot T}{2L\cdot V_o}$

Therefore, the output voltage gain can be written as:

$\frac{V_o}{V_i}=\frac{V_i\cdot D^2 \cdot T}{2L\cdot I_o}$

Compared to the expression of the output voltage gain 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.

[edit] Limit between continuous and discontinuous modes

Fig 5: Evolution of the normalized output voltage with the normalized output current in a buck-boost converter.

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 :

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

$D + δ = 1$

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

$I_{o_{lim}}=\bar{I_D}=\frac{I_{L_{max}}}{2}\left(1-D\right)$

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

$I_{o_{lim}}=\frac{V_i\cdot D\cdot T}{2L}\left(1-D\right)$

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:

$I_{o_{lim}}=\frac{V_i\cdot D\cdot T}{2L}\frac{V_i}{V_o}\left(-D\right)$

Let's now introduce two more notations:

the normalized voltage, defined by $\left|V_o\right|=\frac{V_o}{V_i}$ . It corresponds to the gain in voltage of the converter;
the normalized current, defined by $\left|I_o\right|=\frac{L}{T\cdot V_i}I_o$ . The term $\frac{T\cdot V_i}{L}$ 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 $\left|I_o\right|$ equals 0 for no output current, and 1 for the maximum current the converter can deliver.

Using these notations, we have:

in continuous mode, $\left|V_o\right|=-\frac{D}{1-D}$ ;
in discontinuous mode, $\left|V_o\right|=-\frac{D^2}{2\left|I_o\right|}$ ;
the current at the limit between continuous and discontinuous mode is $I_{o_{lim}}=\frac{V_i\cdot T}{2L}D\left(1-D\right)=\frac{I_{o_{lim}}}{2\left|I_o\right|}D\left(1-D\right)$ . Therefore the locus of the limit between continuous and discontinuous modes is given by $\frac{1}{2\left|I_o\right|}D\left(1-D\right)=1$ .

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

[edit] Non-ideal circuit

[edit] Effect of parasitic resistances

Evolution of the output voltage of a buck-boost converter with the duty cycle when the parasitic resistance of the inductor increases.

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:

$V_i=\bar V_L + \bar V_S$

where $\bar V_L$ and $\bar V_S$ 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:

$\bar V_L=L\frac{\bar{dI_L}}{dt}+R_L\bar I_L=R_L\bar I_L$

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_i-V_o. Therefore, the average voltage across the switch is:

$\bar V_S=D\cdot 0 + (1-D)(V_i-V_o)=(1-D)(V_i-V_o)$

The output current is the opposite of the inductor current during the off-state. the average inductor current is therefore:

$\bar I_L=\frac{-I_o}{1-D}$

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:

$\bar I_L=\frac{-V_o}{(1-D)R}$

Using the previous equations, the input voltage becomes:

$V_i= R_L\frac{-V_o}{(1-D)R}+ (1-D)(V_i-V_o)$

This can be written as:

$\frac{V_o}{V_i}=\frac{-D}{\frac{R_L}{R(1-D)}+1-D}$

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.