A centre tap rectifier is connected with transformer whose turn ratio is 5:1connected with 230v/50hz main AC supply.find the average DC voltage ,PIV of diode and ripple frequency
Answers
During negative half cycle of the input signal sin ( ) i m v V t = ω
The diode D is reverse biased and acts as open.
Hence practically no current flows through the load RL
and no voltage across the load.
• The output of HWR is not a perfect DC, but at least unidirectional.
• For HWR, the typical parameters are:
m
dc
V
V
π
=
2
m
rms
V
V =
Ripple factor=
2
,
1 1.21 r rms rms
dc dc
v v
r
V V
= = − =
The output of HWR has high ripple 121%. The ripple content exceeds the DC content.
This is undesirable. Hence there is need for FWR
Rectification efficiency:
Rectification efficiency dc
ac
Power delivered to the load P
Total input AC power P
η = =
The rectification efficiency of a HWR = 40.6 %
Under best conditions, only 40.6 % of the input power is converted into DC power. The rest
59.4% % remains as AC power in the load. This is highly undesired
PIV: The peak inverse voltage is the maximum reverse voltage the diode should withstand
without breakdown.
In HWR, when the diode D is reverse biased, the maximum voltage across it
would be Vm
o In HWR, the PIV of the diode should be Vm
Note: If asked as long answer question, you need to derive values of Vdc , rms v ripple factor