Closing the switch causes current to build up through the inductor, as shown in the simplified inductive-boost dc-dc converter circuit (Fig. 1). Opening the switch forces current through the diode to the output capacitor. Multiple switching cycles build the output capacitor voltage due to charge it stores from the inductor current. This results in an output voltage that’s higher than the input.
What determines the output voltage of the inductive-boost dc-dc converter?
In the real circuit of Figure 2, an IC with an integrated power MOSFET replaces the mechanical switch, and pulse-width modulation (PWM) control turns the MOSFET on and off. PWM duty cycle always determines the output voltage, which is twice the input for a 50% duty cycle. Stepping up the voltage by a factor of two causes the input current to be twice the output current. In a real circuit with losses, the input current is slightly higher.
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