Buck Converter Design


Posted on Feb 5, 2014

The top MOSFET switches on creating a short circuit between the input voltage (IN) and the left hand side of the inductor, L1. The inductor current ramps up according to the equation where V is the voltage across the inductor, L is the inductance value and di/dt is the change in current with time through the inductor. Thus with a fixed input volta


Buck Converter Design
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ge and a fixed output voltage, there is a fixed voltage across the inductor thus the change in current with time is constant (i. e. a ramp waveform). The output voltage on startup is 0V, so the initial voltage across the inductor is equal to the input voltage. However, as the output voltage changes (and then reaches regulation) the above equation becomes The peak inductor current is sensed by a small series resistor, R4, and when the voltage across this resistor equals a certain value (see the specific converter`s datasheet) the IC switches off the top MOSFET. Now, inductors do not like having their current interrupted, so when the top MOSFET switches off, the inductor behaves like a battery to try to maintain the current flow. Referring to FIG 1, output side of the inductor tries to fly positive (to push current out of the right hand side of the inductor) and its switched side (the left hand side) flies negative (to try and sink current into its left hand side) in an effort to maintain the left to right current flow. Since the output side of the inductor is clamped by a capacitor, the left hand side flies negative. At this point the IC switches on the bottom MOSFET, Q2, to clamp the left hand side of the inductor to ground and enable the inductor to maintain its current flow. When MOSFET Q2 switches on, it also provides a short circuit to 0V at the bottom of capacitor C6. Since the top of C6 is connected to the LTC3891`s...




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