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Boost converter from parts

Power, transformers and three-phase · Advanced · about 20 minutes · part of Electronics Engineering

Close the switch and the inductor stores energy from 5 V; open it and the inductor’s collapsing field adds its voltage on top of the input, pushing current through the diode into the output capacitor. The output rises to V_in ÷ (1 − D).

Boost converter from parts: the circuit on the simulator's bench

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How it works

Ideal: V_out = V_in ÷ (1 − D) = 5 ÷ (1 − 0.50) = 10.00 V.

Less the diode drop (0.38 V) and the coil’s resistance carrying 397 mA: 9.33 V.

Coil ripple V_in·D ÷ (L·f) = 500 mA p-p around 397 mA: continuous mode.

Try it on the bench

  1. Raise the duty to 60 %: about 12 V.
  2. Remove the load (variant): with nothing to take the energy, every cycle pumps the capacitor higher.

Parts on the bench

In the course

Electronics Engineering, module “Diodes and supplies”.

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