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 as the input. This is a switched-mode power supply with a similar circuit topology to the boost converterand 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) as 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 as the input, and can be lower or higher than the input. Such a non-inverting buck-boost converter may use a single inductor that is used as both the buck inductor and the boost inductor.

Principle of operation

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  (for an ideal converter). The output voltage ranges for a buck and a boost converter are respectively 0 to  and  to  

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 forward 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 open before it has a chance to recharge again while the switch is closed.

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.