# Buck–boost converter

The basic schematic of an inverting buck–boost converter.

The buck–boost converter is a type of DC-to-DC converter that has an output voltage magnitude that is either greater than or less than the input voltage magnitude.

Two different topologies are called buck–boost converter. Both of them can produce a range of output voltages, from an output voltage much larger (in absolute magnitude) than the input voltage, down to almost zero.

The inverting topology
The output voltage is of the opposite polarity than the input. This is a switched-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 possible drawback of this converter is that the switch does not have a terminal at ground; this complicates the driving circuitry. Neither drawback is of any consequence if the power supply is isolated from the load circuit (if, for example, the supply is a battery) because the supply and diode polarity can simply be reversed. The switch can be on either the ground side or the supply side.
A buck (step-down) converter followed by a boost (step-up) converter
The output voltage is of the same polarity of the input, and can be lower or higher than the input. Such a non-inverting buck-boost converter may use a single inductor which is used for both the buck inductor and the boost inductor.[1][2][3]

## 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, 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 results 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 $\scriptstyle -\infty$ (for an ideal converter). The output voltage ranges for a buck and a boost converter are respectively 0 to $\scriptstyle V_i$ and $\scriptstyle V_i$ to $\scriptstyle \infty$.

### Conceptual overview

Like the buck and boost converters, the operation of the buck-boost is best understood in terms of the inductor's "reluctance" to allow rapid change in current. From the initial state in which nothing is charged and the switch is open, the current through the inductor is zero. When the switch is first closed, the blocking diode prevents current from flowing into the right hand side of the circuit, so it must all flow through the inductor. However, since the inductor doesn't like rapid current change, it will initially keep the current low by dropping most of the voltage provided by the source. Over time, the inductor will allow the current to slowly increase by decreasing its voltage drop. Also during this time, the inductor will store energy in the form of a magnetic field.

When the switch is then opened, the inductor will be cut off from the input voltage supply, so the current will tend to drop to zero. Again, the inductor will fight such an abrupt change in current. To do so, it must now act like a voltage source to the rest of the circuit, which it can do using the energy it stored while charging. Since current was previously flowing "down" the inductor, it will want to maintain this direction, and so the voltage that it provides will be inverted relative to input supply. During this time, the inductor will discharge through the load and the rest of the circuit, which will cause its voltage to decrease over time. Also during this time, the capacitor in parallel with the load will charge up to the voltage presented by the inductor.

When the switch is once again closed, the diode is reverse biased by the input supply, cutting the load off from the left hand side of the circuit. During this time, the capacitor will discharge into the load, providing energy and voltage to it. By cycling the switch fast enough, the inductor can be allowed to charge and discharge only slightly in each cycle, maintaining a relatively steady voltage to the load. Similarly, the capacitor will only need to discharge slightly while the switch is closed before it has a chance to recharge again while the switch is open.

The voltage presented by the inductor to the load depends on how long the switch is opened and closed. When the switch is closed and the inductor is charging, the current through the inductor is ramping up linearly. The longer the switch is closed, the higher the current will get. When the switch is then opened, it is the end current that the inductor will try to maintain by acting like a voltage source. The higher this current is, the more voltage the inductor will need to provide in order to produce it. Thus, the longer the switch is closed during the on stage, the higher the output voltage will be.

### 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 never 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 $\scriptstyle t=0$ to $\scriptstyle t=D\,T$, the converter is in On-State, so the switch S is closed. The rate of change in the inductor current (IL) is therefore given by

$\frac{\operatorname{d}I_{\text{L}}}{\operatorname{d}t}=\frac{V_i}{L}$

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

$\Delta I_{\text{L}_{\text{On}}}=\int_0^{D\,T}\operatorname{d}I_{\text{L}}=\int_0^{D\, T}\frac{V_i}{L} \, \operatorname{d}t=\frac{V_i\,D\,T}{L}$

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, and a capacitor large enough for its voltage to remain constant, the evolution of IL is:

$\frac{\operatorname{d}I_{\text{L}}}{\operatorname{d}t}=\frac{V_o}{L}$

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

$\Delta I_{\text{L}_{\text{Off}}}=\int_0^{\left(1-D\right) T}\operatorname{d}I_{\text{L}}=\int_0^{\left(1-D\right) T}\frac{V_o\, \operatorname{d}t}{L}=\frac{V_o \left(1-D\right) T}{L}$

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:

$E=\frac{1}{2}L\, I_{\text{L}}^2$

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

$\Delta I_{\text{L}_{\text{On}}} + \Delta I_{\text{L}_{\text{Off}}}=0$

Substituting $\Delta I_{\text{L}_{\text{On}}}$ and $\Delta I_{\text{L}_{\text{Off}}}$ by their expressions yields:

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

This can be written as:

$\frac{V_o}{V_i}=\left(\frac{D}{1-D}\right)$

This in return yields that:

$D=\frac{V_o}{V_o+V_i}$

From the above expression it can be seen that the polarity of the output voltage is always negative (because the duty cycle goes from 0 to 1), and that its absolute value increases with D, theoretically up to minus infinity when D approaches 1. Apart from the polarity, this converter is either step-up (a boost converter) or step-down (a buck converter). Thus it is named a buck–boost converter.

### 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 like follows:

Because the inductor current at the beginning of the cycle is zero, its maximum value $I_{L_{\text{max}}}$ (at $t=D\,T$) is

$I_{L_{\text{max}}}=\frac{V_i\, D\, T}{L}$

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

$I_{L_{\text{max}}}+\frac{V_o\, \delta\, T}{L}=0$

Using the two previous equations, δ is:

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

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:

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

Replacing $I_{L_{\text{max}}}$ and δ by their respective expressions yields:

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

Therefore, the output voltage gain can be written as:

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

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.

### 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 :

$D\, T + \delta \, T=T$
$D + \delta = 1\,$

In this case, the output current $\scriptstyle I_{o_{\text{lim}}}$ (output current at the limit between continuous and discontinuous modes) is given by:

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

Replacing $\scriptstyle I_{L_{\text{max}}}$ by the expression given in the discontinuous mode section yields:

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

As $\scriptstyle I_{o_{\text{lim}}}$ 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:

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

Let's now introduce two more notations:

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

Using these notations, we have:

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

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

## Non-ideal circuit

### Effect of parasitic resistances

Fig 6: 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 an inductor is made of one long wound piece of wire, so it is likely to exhibit a non-negligible parasitic resistance (RL). Furthermore, current flows through the inductor both in the on and the off states.

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

$V_i=\bar V_{\text{L}} + \bar V_S$

where $\scriptstyle \bar V_{\text{L}}$ and $\scriptstyle \bar V_S$ 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:

$\bar V_{\text{L}}=L\frac{\bar{dI_{\text{L}}}}{dt}+R_{\text{L}}\bar I_{\text{L}}=R_{\text{L}}\bar I_{\text{L}}$

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

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

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

$\bar I_{\text{L}}=\frac{-I_o}{1-D}$

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

$\bar I_{\text{L}}=\frac{-V_o}{(1-D)R}$

Using the previous equations, the input voltage becomes:

$V_i= R_{\text{L}}\frac{-V_o}{(1-D)R}+ (1-D)(V_i-V_o)$

This can be written as:

$\frac{V_o}{V_i}=\frac{-D}{\frac{R_{\text{L}}}{R(1-D)}+1-D}$

If the inductor resistance is zero, the equation above becomes equal to the one of the ideal case. But when RL increases, the voltage gain of the converter decreases compared to the ideal case. Furthermore, the influence of RL increases with the duty cycle. This is summarized in figure 6.