We can observed from the circuit diagram that when the diode is
reversed biased then full transformer secondary voltage gets impressed
across it. The drop across conducting diode is assumed zero. Thus the
peak value of the inverse voltage to which diode gets subjected is
voltage across both the parts of the transformer secondary. This is
shown in the Fig. 1.
 |
| Fig. 1 |
When D2 is reverse biased, point A is at - Esm with respect to ground while point B is at + Esm with respect to ground, neglecting diode drop. Thus total peak voltage across D2 is Esm.
If the diode drop is considered to be 0.7 V then the PIV of reverse biased diode is,
1.12 Transformer Utilization Factor (T.U.F.)
In full wave rectifier, the secondary current flows through each half
separately in every half cycle. While the primary of transformer carries
current continuously. Hence T.U.F. is calculated for primary and
secondary windings separately and then the average T.U.F. is determined.
The primary of the transformer is feeding two half-wave rectifiers
separately. These two half-wave rectifiers work independently of each
other but feed a common load. We have already derived the T.U.F. for
half wave circuit to be equal to 0.287. Hence
T.U.F. for primary winding = 2 x T.U.F. of half wave circuit = 2 x 0.287 = 0.574.The average T.U.F. for fullwave circuit will be
Average T.U.F. for full wave rectifier circuit = (T.U.F. of primary + T.U.F. of secondary)/2
= (0.574 + 0.812)/2 = 0.693
... Average T.U.F. for full-wave rectifier = 0.693
Note : Thus in full-wave circuit, transformer gets utilized more than the half wave rectifier circuit.
1.13 Voltage Regulation
For a full wave circuit,
where Rf = Forward resistance of the diode.
The regulation characteristics is drooping, as we see earlier in case
of half wave rectifier as output voltage decreases as load increases
from no load to full load.
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