Final answer:
In a half-wave rectifier circuit, the average current through the load is 9.6mA and the DC output voltage is 689V. The RMS value of the load current is 13.6mA. The efficiency of rectification is 37.3% and the ripple factor is 0.471.
Step-by-step explanation:
A half-wave rectifier circuit is supplied from a 230V, 50Hz supply with a step-down ratio of 3:1 to a resistive load of 10kΩ. The diode forward resistance is 75Ω while the transformer secondary resistance is 10Ω. Here are the calculations:
- First, calculate the peak voltage of the input signal: Peak Voltage = RMS Voltage × √2 = 230V × √2 = 325V
- Next, calculate the peak voltage of the output signal: Peak Voltage = (Step-down Ratio) × (Peak Voltage of Input Signal) = (3:1) × 325V = 975V
- Now, calculate the average current through the load: Average Current = (Peak Voltage of Output Signal) / (Total Resistance) = 975V / (10kΩ + 75Ω + 10Ω) = 9.6mA
- Calculate the DC output voltage: DC Output Voltage = (Peak Voltage of Output Signal) / √2 = 975V / √2 = 689V
- Calculate the RMS value of the load current: RMS Current = (Average Current) × √2 = 9.6mA × √2 = 13.6mA
- Calculate the efficiency of rectification: Efficiency of Rectification = (DC Output Power) / (Input Power) = ((DC Output Voltage) × (Average Current)) / ((RMS Voltage) × (RMS Current)) = ((689V) × (9.6mA)) / ((325V) × (13.6mA)) = 37.3%
- Finally, calculate the ripple factor: Ripple Factor = (RMS Value of AC Component) / (DC Value) = (RMS Voltage) / (DC Output Voltage) = (325V) / (689V) = 0.471