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Chopper classification; step-up converter

Classify chopper on the basis of power flow. Explain the principle of step-up converter with resistive load.

Classification of chopper on the basis of power flow (quadrant of operation)

ClassQuadrant(s)VoIoPower flow
AI++Source → load (motoring)
BII+−Load → source (regenerating)
CI and II+±Both directions
DI and IV±+Both directions
EAll four±±Both directions

So by power flow, choppers are unidirectional (Class A, B) or bidirectional (Class C, D, E). By output level they are also classed as step-down (Vo < Vs) and step-up (Vo > Vs).

Principle of step-up converter with resistive load

       L     iL           D
 +Vs--uuuu--+----------->|---+-----+
            |                |     |
            S (switch)       C     R   Vo
            |                |     |
 −  --------+----------------+-----+
  1. Switch ON (0 < t < Ton): The inductor is placed directly across Vs; current rises linearly and energy is stored in L. The diode is reverse biased, and the capacitor supplies the resistive load.
vL = Vs  →  ΔI = Vs·Ton/L
  1. Switch OFF (Ton < t < T): Inductor current cannot change suddenly; its voltage reverses and adds to Vs. Current flows through D into C and R. Load voltage is now higher than Vs.
vL = Vs − Vo (negative)  →  ΔI = (Vo − Vs)·Toff/L
  1. Steady state: current rise during ON equals fall during OFF:
Vs·Ton = (Vo − Vs)·Toff
Vo = Vs·T/Toff = Vs/(1 − k),  k = Ton/T

Load current Io = Vo/R and, ignoring losses, input current Is = Io/(1 − k).

Waveforms:

S    |‾‾‾‾‾‾|___|‾‾‾‾‾‾|___
iL   |  /‾‾\   /‾‾\
     | /    \_/    \_   rises ON, falls OFF
vo   |‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾  ≈ Vs/(1−k), small ripple

Example: Vs = 12 V, k = 0.6 → Vo = 12/0.4 = 30 V. Since 0 < k < 1, Vo is always greater than Vs.

✓ Answer checked by Claude Opus 5.5 (high effort)

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