What is the duty cycle of a chopper ?

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Ton/T Explanation: The duty cycle (often denoted by $\alpha$ or $D$) of a chopper represents the ratio of the time during which the chopper is in the 'ON' state to the total switching period. It is a dimensionless quantity that dictates the average output voltage of the DC-DC converter. Working Principle: In a chopper circuit, a semiconductor switch operates periodically with a total period $T = T_{on} + T_{off}$. By controlling the ratio of $T_{on}$ to $T$, the average output voltage $V_o = D \cdot V_s$ is varied, enabling efficient DC voltage regulation without significant power loss. Key Formulas: $D = \frac{T_{on}}{T_{on} + T_{off}} = \frac{T_{on}}{T}$ — Duty cycle formula; $V_{avg} = D \cdot V_s$ — Average output voltage Key Points: Duty cycle range is typically $0 \le D \le 1$.; It is independent of the switching frequency $f = 1/T$ for a fixed ratio.; Higher duty cycle results in a higher average output voltage for a step-down (Buck) chopper. Advantages: High efficiency due to switching operation; Precise control of output voltage Disadvantages: Generation of electromagnetic interference (EMI); Requires filter components to reduce output ripple Applications: DC motor speed control; Switched-mode power supplies (SMPS) Additional Info: The total time period $T$ is the inverse of the switching frequency ($T = 1/f$).. Option A is incorrect because it uses the ratio of time-on to time-off, which is not the definition of duty cycle. Answer: B is correct — The duty cycle $D$ is defined as the ratio of the ON-time ($T_{on}$) to the total period $T$ of one cycle. Exam Tip: Always remember that $T = T_{on} + T_{off}$. If a question gives you frequency $f$, substitute $T$ with $1/f$ in the duty cycle expression.

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