In the process, the electrical energy that was stored is now being released. When the current flowing through the circuit decreases, the magnetic field across the inductor will collapse.
When we OPEN the switch again, the current flowing through the circuit will no longer be a large current due to the high impedance. This means that the inductor is storing energy, and during this process, the potential across the inductor will be positive on the left and negative on the right. Since we now have a larger value of current flowing through the circuit, the magnetic field of the inductor will expand. When we CLOSE the switch, as shown in the “Boost switch ON State” diagram above, a larger current will flow from the battery, through the inductor and through the switch because we now have a path of least resistance. If we OPEN the switch, as shown in the “Boost switch OFF State” diagram above, a smaller current flows from the battery, through the inductor, through the diode and charging the capacitor. We can use this property to control the output voltage of our circuit. Recall that an inductor is a passive electronic component that can store electrical energy in the form of magnetic energy. Here is a general schematic of a boost converter. This configuration steps up the DC voltage to a level determined by the choice of components in your circuit. They can also be used to supply components with different operating voltages from the same battery. Typically, you find DC-DC boost converters in battery chargers or solar panels. For example a boost converter could take a 5V power source and boost it up to 25V. Boost Convertersīoost converters increase the voltage of a power source. This means in reality there will be some power dissipated, but the conversion efficiencies are high despite these dissipations. In a real-world scenario, however, it may be challenging to find an ideal switch. When the switch is at position OFF, no current flows through the switch, so in this case, the dissipated power is 0, too. Since we have an ideal switch, the dissipated power is P_diss = RI 2 = 0 * 9 2 = 0W. When the switch is at position ON, the current flowing through the circuit is 6A. If we assume the switch to be ideal (an ideal switch is a switch that does not consume or dissipate power from the source), we can calculate the efficiency of this circuit to be 100%. The average output voltage of this circuit is 5V, but we can improve the output waveform by using RC filter circuits to get rid of harmonics.
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