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Battery power systems often stack cells in series to achieve higher voltage. However, sufficient stacking of cells is not possible in many high voltage applications due to lack of space.
Boost converters can increase the voltage and reduce the number of cells. Two battery-powered applications that use boost converters are used in hybrid electric vehicles HEV and lighting systems.
Without a boost converter, the Prius would need nearly cells to power the motor. However, a Prius actually uses only cells [ citation needed ] and boosts the battery voltage from V to V.
Boost converters also power devices at smaller scale applications, such as portable lighting systems. A white LED typically requires 3. This energy would otherwise be wasted since the low voltage of a nearly depleted battery makes it unusable for a normal load.
This energy would otherwise remain untapped because many applications do not allow enough current to flow through a load when voltage decreases.
This voltage decrease occurs as batteries become depleted, and is a characteristic of the ubiquitous alkaline battery. The key principle that drives the boost converter is the tendency of an inductor to resist changes in current by creating and destroying a magnetic field.
In a boost converter, the output voltage is always higher than the input voltage. A schematic of a boost power stage is shown in Figure 1.
Polarity of the left side of the inductor is positive. The magnetic field previously created will be destroyed to maintain the current towards the load.
Thus the polarity will be reversed means left side of inductor will be negative now. As a result, two sources will be in series causing a higher voltage to charge the capacitor through the diode D.
If the switch is cycled fast enough, the inductor will not discharge fully in between charging stages, and the load will always see a voltage greater than that of the input source alone when the switch is opened.
Also while the switch is opened, the capacitor in parallel with the load is charged to this combined voltage. When the switch is then closed and the right hand side is shorted out from the left hand side, the capacitor is therefore able to provide the voltage and energy to the load.
During this time, the blocking diode prevents the capacitor from discharging through the switch. The switch must of course be opened again fast enough to prevent the capacitor from discharging too much.
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.
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 I L is:.
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:. 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:.