Question

An a. c. supply of 220V is applied to a half wave rectifier through a transformer of turn ratio 20:1. In this circuit the peak inverse voltage is:

• A. 11 V
• B. 311.08 V
• C. 1.5554 V
• D. 15.554 V

Hint:

Always be careful whether the given AC voltage is an RMS value or a peak value. In standard practice (like mains supply), the value given (e.g., 220V, 110V) is always the RMS value unless specified otherwise. Remember PIV values for standard rectifiers:
Half-wave: PIV = \(V_{s,peak}\)
Full-wave (center-tapped): PIV = \(2V_{s,peak}\) (where \(V_{s,peak}\) is across half the secondary)
Full-wave (bridge): PIV = \(V_{s,peak}\)

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Peak Inverse Voltage (PIV) is the maximum reverse voltage that a diode in a rectifier circuit must be able to withstand without breaking down. For a simple half-wave rectifier, during the negative half-cycle of the input AC voltage, the diode is reverse-biased. The maximum voltage it experiences is the peak voltage of the AC waveform at the secondary of the transformer.
Step 2: Key Formula or Approach:
1. Calculate the RMS voltage at the secondary of the transformer (\(V_{s,rms}\)). 2. Calculate the peak voltage at the secondary (\(V_{s,peak}\)). 3. For a half-wave rectifier, PIV = \(V_{s,peak}\).
The relevant formulas are:
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \implies V_{s,rms} = V_{p,rms} \times \frac{N_s}{N_p} \] \[ V_{peak} = V_{rms} \times \sqrt{2} \] Step 3: Detailed Explanation:
1. Find the secondary RMS voltage (\(V_{s,rms}\)):
The primary voltage is given as \(V_{p,rms} = 220\) V. The turns ratio is given as 20:1, which means \( \frac{N_p}{N_s} = \frac{20}{1} \). \[ V_{s,rms} = V_{p,rms} \times \frac{N_s}{N_p} = 220 \, V \times \frac{1}{20} = 11 \, V \] 2. Find the secondary peak voltage (\(V_{s,peak}\)):
The relationship between peak voltage and RMS voltage for a sinusoidal AC supply is:
\[ V_{s,peak} = V_{s,rms} \times \sqrt{2} \] \[ V_{s,peak} = 11 \, V \times \sqrt{2} \approx 11 \times 1.414 \] \[ V_{s,peak} \approx 15.554 \, V \] 3. Determine the Peak Inverse Voltage (PIV):
In a half-wave rectifier, during the negative cycle, the diode is open and the full secondary peak voltage appears across it.
\[ \text{PIV} = V_{s,peak} \approx 15.554 \, V \] Step 4: Final Answer:
The peak inverse voltage for the diode in this circuit is 15.554 V. This corresponds to option (D).