Buck–boost converter

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 allow 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 own resistance. In an ideal circuit the voltage drop across the inductor would remain constant. When the inherent resistance of wires and the switch is taken into account then the voltage drop across the inductor will also decrease as the current increases. Also during this time, the inductor will store energy in the form of a magnetic field.

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 ${\displaystyle t=0}$ to ${\displaystyle t=DT}$, 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

${\displaystyle {\frac {\operatorname {d} I_{L}}{\operatorname {d} t}}={\frac {V_{i}}{L}}}$

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

${\displaystyle \Delta I_{{\text{L}}_{\text{On}}}=\int _{0}^{D\,T}\operatorname {d} I_{\text{L}}=\int _{0}^{D\,T}{\frac {V_{i}}{L}}\,\operatorname {d} t={\frac {V_{i}\,D\,T}{L}}}$

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 I_L is:

${\displaystyle {\frac {\operatorname {d} I_{\text{L}}}{\operatorname {d} t}}={\frac {V_{o}}{L}}}$

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

${\displaystyle \Delta I_{{\text{L}}_{\text{Off}}}=\int _{0}^{\left(1-D\right)T}\operatorname {d} I_{\text{L}}=\int _{0}^{\left(1-D\right)T}{\frac {V_{o}\,\operatorname {d} t}{L}}={\frac {V_{o}\left(1-D\right)T}{L}}}$

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:

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

it is obvious that the value of I_L at the end of the Off state must be the same with 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:

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

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

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

This can be written as:

${\displaystyle {\frac {V_{o}}{V_{i}}}=-{\frac {D}{1-D}}}$

This in return yields that:

${\displaystyle D={\frac {V_{o}}{V_{o}-V_{i}}}}$

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.

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:

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

${\displaystyle I_{L_{\text{max}}}={\frac {V_{i}\,D\,T}{L}}}$

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

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

Using the two previous equations, δ is:

${\displaystyle \delta =-{\frac {V_{i}\,D}{V_{o}}}}$

The load current ${\displaystyle I_{o}}$ is equal to the average diode current (${\displaystyle I_{D}}$). 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:

${\displaystyle I_{o}={\bar {I_{D}}}={\frac {I_{L_{\text{max}}}}{2}}\delta }$

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

${\displaystyle I_{o}=-{\frac {V_{i}\,D\,T}{2L}}{\frac {V_{i}\,D}{V_{o}}}=-{\frac {V_{i}^{2}\,D^{2}\,T}{2L\,V_{o}}}}$

Therefore, the output voltage gain can be written as:

${\displaystyle {\frac {V_{o}}{V_{i}}}=-{\frac {V_{i}\,D^{2}\,T}{2L\,I_{o}}}}$

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.

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 :

${\displaystyle D\,T+\delta \,T=T}$

${\displaystyle D+\delta =1\,}$

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

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

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

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

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

${\displaystyle I_{o_{\text{lim}}}={\frac {V_{i}\,D\,T}{2L}}{\frac {V_{i}}{V_{o}}}\left(-D\right)}$

Let's now introduce two more notations:

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

Using these notations, we have:

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

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

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 (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:

${\displaystyle V_{i}={\bar {V}}_{\text{L}}+{\bar {V}}_{S}}$

where ${\displaystyle \scriptstyle {\bar {V}}_{\text{L}}}$ and ${\displaystyle \scriptstyle {\bar {V}}_{S}}$ 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:

${\displaystyle {\bar {V}}_{\text{L}}=L{\frac {\bar {dI_{\text{L}}}}{dt}}+R_{\text{L}}{\bar {I}}_{\text{L}}=R_{\text{L}}{\bar {I}}_{\text{L}}}$

When the switch is in the on-state, ${\displaystyle \scriptstyle V_{S}=0}$. When it is off, the diode is forward biased (we consider the continuous mode operation), therefore ${\displaystyle \scriptstyle V_{S}=V_{i}-V_{o}}$. Therefore, the average voltage across the switch is:

${\displaystyle {\bar {V}}_{S}=D\,0+(1-D)(V_{i}-V_{o})=(1-D)(V_{i}-V_{o})}$

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

${\displaystyle {\bar {I}}_{\text{L}}={\frac {-I_{o}}{1-D}}}$

Fig 6: Evolution of the output voltage of a buck–boost converter with the duty cycle when the parasitic resistance of the inductor increases.

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:

${\displaystyle {\bar {I}}_{\text{L}}={\frac {-V_{o}}{(1-D)R}}}$

Using the previous equations, the input voltage becomes:

${\displaystyle V_{i}=R_{\text{L}}{\frac {-V_{o}}{(1-D)R}}+(1-D)(V_{i}-V_{o})}$

This can be written as:

${\displaystyle {\frac {V_{o}}{V_{i}}}={\frac {-D}{{\frac {R_{\text{L}}}{R(1-D)}}+1-D}}}$

If the inductor resistance is zero, the equation above becomes equal to the one of the ideal case. But when 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 6.