Question

If \(\alpha\) is ratio \(\frac{\text{T}_{\text{ON}}}{\text{T}}\) of chopper, then the ratio of input voltage to output voltage of step down and step-up chopper respectively are

• A. \( \frac{1}{\alpha} \text{ and } (1-\alpha) \)
• B. \( \frac{1}{\alpha} \text{ and } \frac{1}{(1-\alpha)} \)
• C. \( \alpha \text{ and } (1-\alpha) \)
• D. \( \alpha \text{ and } \frac{1}{(1-\alpha)} \)

Hint:

Always read the question carefully during exams! Most problems ask for the voltage gain ratio (\(\frac{V_0}{V_s}\)), which would be \(\alpha\) and \(\frac{1}{1-\alpha}\) (Option 4). However, this question explicitly asks for the inverse ratio (\(\frac{V_s}{V_0}\)), which turns the answers upside down into \(\frac{1}{\alpha}\) and \((1-\alpha)\). Don't lose points to a reading trap!

Answer 1

The Correct Option is A

Concept: A DC chopper (or DC-to-DC buck/boost converter) acts like a transformer for DC voltages. Its conversion ratio is controlled by adjusting the duty cycle ($\alpha$), which is defined as the ratio of the switch on-time ($T_{\text{ON}}$) to the total operating period ($T$): \[ \alpha = \frac{T_{\text{ON}}}{T} \] Let us look at the input-to-output relationships for both configurations:
• Step-Down Chopper (Buck): Lowers the voltage. Output voltage is given by: \[ V_0 = \alpha \cdot V_s \]
• Step-Up Chopper (Boost): Raises the voltage. Output voltage is given by: \[ V_0 = \frac{Vs}{1 - \alpha} \] Critical Warning: Pay close attention to the wording in the question prompt. It asks for the ratio of input voltage to output voltage ($\frac{V_s}{V_0}$), which is the inverse of the standard transfer gain formulas.

Step 1: Calculating the ratio for the Step-Down Chopper.
The output voltage formula for a step-down chopper is: \[ V_0 = \alpha \cdot V_s \] Rearranging this to find the ratio of input voltage ($V_s$) to output voltage ($V_0$): \[ \frac{V_s}{V_0} = \frac{1}{\alpha} \quad \cdots (1) \]

Step 2: Calculating the ratio for the Step-Up Chopper.
The output voltage formula for a step-up chopper is: \[ V_0 = \frac{V_s}{1 - \alpha} \] Rearranging this to find the ratio of input voltage ($V_s$) to output voltage ($V_0$): \[ \frac{V_s}{V_0} = 1 - \alpha \quad \cdots (2) \]

Step 3: Combining the two ratio terms.
The required ratios are $\frac{1}{\alpha}$ for the step-down chopper and $(1 - \alpha)$ for the step-up chopper, which matches option (1). Hence, the correct choice is option (1).