Match the two lists and choose the correct answer from the code given below:
List I |
List II |
(a) v_{rms} of half-wave rectifier output |
(i)\(\frac{(2V_m)}{π}\) |
(b) V_{rms }of bridge rectifier output |
(ii)\(\frac{(V_m)}{π}\) |
(c) V_{dc} (half-wave rectifier output) |
(iii) \(\frac{(V)}{\sqrt2}\) |
(d) V_{dc} (full-wave rectifier output) |
(iv)\(\frac{(V)}{2}\) |
half-wave rectifier:
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 the negative part of the input signal is present
full-wave rectifier:
A bridge rectifier is of two types:
1) Bridge Type Full Wave Rectifier
2) Center-Tap Full Wave Rectifier
A Bridge type full wave rectifier contains 4 diodes as shown:
Parameters |
FWR (Center tap) |
FWR (Bridge) |
HWR |
Vrms |
\(\frac{{{V_m}}}{{\sqrt 2 }}\) |
\(\frac{{{V_m}}}{{\sqrt 2 }}\) |
\(\frac{{{V_m}}}{2}\) |
VDC |
\(\frac{{2{V_m}}}{\pi }\) |
\(\frac{{2{V_m}}}{\pi }\) |
\(\frac{{{V_m}}}{\pi }\) |
Ripple factor |
0.48 |
0.48 |
1.21 |
PIV |
2Vm |
Vm |
Vm |
Output frequency |
2f |
2f |
f |
Form factor |
1.11 |
1.11 |
1.57 |
efficiency |
81.2% |
81.2% |
40.5% |
Hence, Option 4 is correct.