When the secondary voltage gets its peak positive value and the terminal A is positive, and B is negative as shown in above. So at this instant, the diode D1 and D3 are forward biased and D2 and D4 are in reverse bias they won’t conduct, but only the diodes D1 &D3 will conduct current through them. Therefore, in between the terminal M-L or A’-B’ gets the same voltage as that the terminals A-B.
Therefore the PIV of bridge rectifiers is
PIV of diode D1 and D3 = Vm
Similarly PIV of diode D2 and D4 = Vm
PIV (peak inverse voltage) of Center Tapped Full Wave Transformer
During the first half cycle of the AC power supply, i.e. when the top of the transformer secondary winding is positive, the diode D1 conducts and offers almost zero resistance. So the whole the voltage Vm max of the upper half winding is developed across the load (RL). Now the voltage across the non-conducting diode D2 is the sum of the voltage across the lower half of the transformer secondary and the voltage across the load (RL).
Thus, PIV of diode, D2 = Vm + Vm
PIV of diode, D2 = 2 Vm
Similarly, PIV of diode D1 = 2 Vm
Transformer utilization Factor (TUF)
TUF is defined as the ratio of DC power delivered to the load and the input AC rating of the transformer secondary.
TUF= Poutput.dc/Pinput.ac
Transformer Utilization Factor (TUF) of the Center Tapped Full-wave Rectifier
Pdc = VL(dc)*IL(dc) => VLM/Ï€ *VLM/RL
=> VLM2 /Ï€RL
=> Vsm2 /Ï€RL (if drop over R0 is neglected)
Now, the rated voltage of the transformer secondary is given by Vsm/√2 but the actual current flowing through the secondary is IL = ILM/2 (not ILM /√2) since it is a Half-Wave rectifier current.
Pac.rated => Vsm/√2 *ILM/2
=> Vsm/√2* VLM/2RL
=> Vsm/2√2RL
Its value is found by considering the primary and secondary winding of the transformer separately. Its value is 0.693.
Transformer Utilization Factor of Bridge Rectifier
Pdc => VL(dc).IL(dc)
=> VLM/Ï€ * VLM/RL => VLM2 /Ï€RL
=> Vsm2 /Ï€RL (if drop over R0 is neglected)
Now, the rated voltage of the transformer secondary is Vsm/√2 but the actual current flowing through the secondary is IL = ILM/2 (not ILM /√2) since it is a Half-Wave rectifier current.
Pac = Vsm/√2 *ILM/2
=> Vsm/√2 *VLM/2RL
=> Vsm/2√2RL
Its value is found by considering the primary and secondary winding of the transformer separately. Its value is 0.812
No comments:
Post a Comment