Boost Converter Design

Posted on Feb 5, 2014

Ignore components C1, C2, R3. The MOSFET, Q1, switches on creating a short circuit between the right hand side of the inductor, L1, and 0V. Thus a fixed voltage of 3. 3V is applied across the inductor, so its current will ramp up according to When the MOSFET switches off, the inductor tries to maintain its current flow. It does this by generating a

Boost Converter Design
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voltage across its terminals very similar to a battery, where the current flows from the negative terminal, through the battery, to the positive terminal. In the circuit of FIG 1, we can see that to maintain current flow, the right hand side of the inductor has to increase in voltage with respect to the left hand side. The left hand side is connected to the input voltage (so cannot change), thus the right hand side voltage increases above the input voltage and continues to do so until something conducts. Theoretically, this voltage will rise to an infinite value, making the inductor very good at generating high voltages from low voltages. In FIG 1, the inductor voltage increases until diode D1 conducts after which the energy in the inductor flows into the output capacitor C3, causing the voltage across C3 to increase slightly. It is worth noting that even before the MOSFET has started to switch, there is a dc path flowing from the input, through L1 and diode D1 into C3, so at startup C3 will have a voltage across it (equal to Vin Vdiode). When the inductor has discharged, the MOSFET switches on and the process starts again. Repeating this process produces pulses of energy from the inductor into the output capacitor making the voltage across the output capacitor rise. In FIG1, resistors R1 and R2 monitor the output voltage and when the voltage at the FB pin reaches a certain point, the chip terminates the drive to the...

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