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Freewheeling diode; step-up chopper principle

Short answer

A freewheeling diode provides a path for inductive current after switch-off, preventing voltage spikes and improving efficiency. A step-up chopper boosts voltage: during on-time inductor stores energy, during off-time inductor voltage adds to supply, giving Vo=Vs/(1−k) where k is duty cycle.

What is the importance of Freewheeling Diode? Explain the principle of step up operation in Chopper with necessary figures and expressions.

Importance of freewheeling diode

A freewheeling diode (FD) is connected across an inductive load (cathode to the positive end) in a rectifier or chopper.

  • When the switch turns OFF, the load inductance tries to keep current flowing. The FD gives this current a path, so the inductor does not produce a large L·di/dt voltage spike that could damage the switch.
  • Stored energy in the inductor is delivered to the load instead of back to the source, improving efficiency and power factor.
  • Load current becomes smoother and can be continuous.
  • In controlled rectifiers it prevents the output voltage from going negative, raising the average output voltage.

Step-up chopper

A step-up chopper gives an average output voltage greater than the input, Vo > Vs.

       L     iL            D
 +Vs--uuuu--+----------->|---+-----+
            |                |     |
            CH (switch)      C   LOAD  Vo
            |                |     |
 −  --------+----------------+-----+

Principle:

  1. CH ON (0 < t < Ton): The inductor is connected directly across Vs. Current rises linearly and L stores energy. The diode is reverse biased; the capacitor supplies the load.
L·di/dt = Vs   →   ΔI = Vs·Ton / L
  1. CH OFF (Ton < t < T): The inductor current cannot change suddenly, so the inductor voltage reverses and adds to Vs. Current flows through D to the load and capacitor. Load voltage = Vs + L·di/dt > Vs.
L·di/dt = Vs − Vo  (negative, current falls)
ΔI = (Vo − Vs)·Toff / L

Expression for output voltage (steady state, rise = fall):

Vs·Ton/L = (Vo − Vs)·Toff/L
Vs(Ton + Toff) = Vo·Toff
Vo = Vs·T/Toff = Vs/(1 − k),   k = Ton/T

Waveforms:

CH    |‾‾‾‾|__|‾‾‾‾|__|‾‾‾‾|__
iL      /|\  /|\  /|\  (rises when ON,
       / | \/ | \/ |    falls when OFF)
vL  +Vs ‾‾‾‾      ‾‾‾‾
        −(Vo−Vs) __    __

Example: k = 0.5 gives Vo = 2Vs; k = 0.75 gives Vo = 4Vs. As k → 1, Vo theoretically → ∞, but in practice losses limit it. Used in regenerative braking of DC motors and boost converters.

✓ Answer checked by Claude Opus 5.5 (high effort)

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