Question
Download Solution PDFA half-wave rectifier circuit using ideal diode has an input voltage of 20 sin ωt Volt. Then average and rms values of output voltage respectively, are
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
For the half-wave rectifier, the output is present is only for one half of the input signal and clipped for the other half.
Positive half-wave rectifier clips the negative half of the input signal and only a positive part of the input signal is present.
Negative half-wave rectifier clips the positive half of the input signal and only negative part of the input signal is present
In a half-wave rectifier circuit, the average value of output voltage is:
\({V_{avg}} = \frac{{{V_m}}}{\pi }\)
The RMS value of the output voltage is:
\({V_{rms}} = \frac{{{V_m}}}{2}\)
Where Vm is the maximum value of supply voltage
Calculation:
Input voltage = 20 sin ωt Volt
Vm = 20 V
The average value of output voltage:
\({V_{avg}} = \frac{{20}}{\pi }\;V\)
The RMS value of output voltage:
\({V_{rms}} = \frac{{20}}{2} = 10\;V\)
Parameters |
Half wave |
Center Tap FWR |
Bridge FWR |
Type of Transformer |
Step Down |
Center tapped |
Step Down |
VDC |
\(\frac{V_m}{\pi}\) |
\(\frac{2V_m}{\pi}\) |
\(\frac{2V_m}{\pi}\) |
IDC |
\(\frac{I_m}{\pi}\) |
\(\frac{2I_m}{\pi}\) |
\(\frac{2I_m}{\pi}\) |
Vrms |
\(\frac{V_m}{2}\) |
\( \frac{{{{\rm{V}}_{\rm{m}}}}}{{\sqrt 2 }}\) |
\( \frac{{{{\rm{V}}_{\rm{m}}}}}{{\sqrt 2 }}\) |
Ripple factor |
1.21 |
0.48 |
0.48 |
PIV |
Vm |
2Vm |
Vm |
O/P Frequency |
F |
2f |
2f |
Number of Diodes |
1 |
2 |
4 |
Form factor |
1.57 |
1.11 |
1.11 |
Crest factor |
2 |
1.41 |
1.41 |
Last updated on Jul 2, 2025
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